Algebraic Inseparable Extension

 

Proposition. Suppose $\alpha$ is algebraic over $k$ and $f$ is its minimal polynomial over $k$. If $\text{char} k = 0$ then $f$ is separable. Otherwise, if $\text{char} k = p$ then there exists an integer $\mu$ such that every root of $f$ has multiplicity $p^{\mu}$. In other words
$$[k(\alpha): k] = p^{\mu}[k(\alpha): k]_s$$
and $\alpha^{p^{\mu}}$ is separable over $k$.

Proof. 

Comment: In particular, we see that if $\text{char} k = p$ then the minimal polynomial  $f$ of  $\alpha$ over $k$ must be a power of some polynomial $f = g^{p^{\mu}}$ where $g = \prod_{i=1}^m (x- \alpha_i).$
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We now assume $\text{char} k = p$.

Example. Consider $k(\alpha)/k$ where the minimal polynomial of $\alpha$ over $k$ is $(x- \alpha)^{p^{\mu}}.$ Then $[k(\alpha):k]_s = 1$ and $[k(\alpha): k]_i = p^{\mu}$ and we say that $k(\alpha)/k$ is purely inseparable.


Definition. An element $\alpha$ algebraic over $k$ is called purely inseparable if there exists $n$ such that $\alpha^{p^n}= c$ lies in $k$ (so the minimal polynomial of $\alpha$ over $k$ must divide $x^{p^n} - c$.)

Proposition/Definition. Let $E/k$ be an algebraic extension. TFAE:

1. $[E:k] = 1$;
2. Every element of $E$ is purely inseparable over $k$;
3. The irreducible polynomial over $k$ of every $\alpha \in E$ is of the form $X^{p^{\mu}}- c$ for some $\mu$ and some $c \in k$.
4. $E = k(\{\alpha_i\}_i)$ for some $\alpha_i$'s purely inseparable over $k$. 

If $E/k$ satisfies any of the above, we call it a purely inseparable extension.

Proposition. Purely inseparable extensions form a distinguished class of extensions.

Proposition. Let $E/k$ be an algebraic extension. Let $E_0$ be the compositum of all separable subextensions $F/k$. Then $E/E_0$ is purely inseparable and $E_0/k$ is separable.

Corollary. If an algebraic extension $E/k$ is both separable and purely inseparable then $E= k$.

Algebraic Separable Extensions

Example.

Consider $k(\alpha)/k$ where $\alpha$ is algebraic over $k$ with minimal polynomial $f$. 

Then there exists an integer $r$ such that $f(x) = \prod_{i=1}^m(x -\alpha_i)^{r}$ where $\alpha_i$ ranges over all distinct roots of $f$. We call

• $r =[k(\alpha): k]_{i}$ the inseparable degree of $k(\alpha)/k$;
• $m = [k(\alpha): k]_s$ the separable degree of $k(\alpha)/k$;
  Notice that as $rm$ is equal to the degree of $f$, we have $[k(\alpha): k]_i [k(\alpha): k]_s = [k(\alpha): k].$

In general, let $E/k$ be an algebraic field extension. Fix an embedding $\sigma: k \to L$ for some algebraically closed field $L$. We define the separable degree $[E: k]_s$ of $E$ over $k$ to be number of extensions of $\sigma$ to $E/k \to L/k$. It is independent of the choice of $L$ and $\sigma$. 

Comment: It turns out that if $\text{char} k = 0$ then $r = 1$, and if $\text{char} k = p$ then $r$ is a power of $p$.

Separability for Finite Extension

Let $E$ be a finite extension of $k$. We define  the inseparable degree $[E:k]_i$ of $E/k$ to be $[E:k]/[E:k]_s$. We say that $E$ is separable over $k$ if $[E:k]_s = [E:k]$. If $\alpha$ is algebraic over $k$, we say $\alpha$ is separable over $k$ if $k(\alpha)/k$ is (i.e. if the irreducible polynomial of $\alpha$ over $k$ has no multiple roots). A polynomial $f(x) \in k[x]$ is called separable if it has no multiple root.

Theorem. Let $F\supset E \supset k$. Then
$$[F:k]_s = [F:E]_s [E:k]_s.$$
If $F/k$ is finite then so is $[F:k]_s$ and $[F:k]_s \leq [F:k].$

Claim. Moreover if $k \subset E \subset F$ and $\alpha \in F$ is separable over $k$ then it is also separable over $E$.

Theorem. A finite extension $E/k$ is separable iff every element of $E$ is separable over $k$.

Proof. Suppose $E/k$ is separable and $\alpha \in E$. Then consider $E \supset k(\alpha) \supset k$. By tower law, $[E:k] = [E: k]_s = [E:k(\alpha)]_s [k(\alpha):k]_s \leq [E:k(\alpha)] [k(\alpha):k] = [E: k]$ so we must have $[k(\alpha): k]_s = [k(\alpha):k].$

Conversely, suppose every element of $E$ is separable over $k$ and $E= k(\alpha_1, \ldots, \alpha_n).$ Consider the tower $k \subset k(\alpha_1) \subset k(\alpha_1, \alpha_2) \subset \cdots \subset k(\alpha_1, \ldots, \alpha_n) = E$.

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Separability for Algebraic Extension

Let $E/k$ be an algebraic extension. We define $E$ to be separable over $k$ if every finitely generated subextension is separable over $k$.

Theorem.  Suppose $E/k$ is an algebraic extension generated by a (not necessarily finite) family $\{\alpha_i\}$. If each $\alpha_i$ is separable ver $k$ then $E/k$ is separable.

Proof.
Every element $x \in E$ will lie in some $E_x= k(\alpha_{i_1}, \ldots, \alpha_{i_n})$ finitely generated by a finite subset of the $\alpha_i$'s. As each $\alpha_i$ is separable over $k$, so must $E_x$. Now let $F/k$ finitely generated subextension of $E/k$ is of finite degree (as $E/k$ is finite). From the above all of its elements are separable over $k$, so $L/k$ must be separable.

Theorem. Separable extensions form a distinguished class of extensions.

In particular, fixed an algebraic closure $k^a$ of $k$. Then the compositum of all separable subextensions $k^a/k$ is separable over $k$. We call it the separable closure of $k$ and denote it by $k^s$.

A field $k$ is called perfect if $k^p = k$.

Proposition. If $k$ is a perfect field then every algebraic extension of $k$ is separable and perfect.

 

Integral = Reduced and Irreducible

A scheme $X$ is integral if for every $U \subset X$ open we have $O_X(U)$ is integral.

Proposition. A scheme $X$ is integral iff it is reduced and irreducible.

Proof.
Suppose $X$ is integral. Then $O_X(U)$ is integral and hence reduced for all $U$ open in $X$.  It suffices to show that every two nonempty open subsets of $X$ intersect.

Indeed, suppose $U$ and $V$ are nonempty open subsets of $X$ that do not intersect. By shrinking $U$ and $V$, we can assume that they are affine $U = \text{Spec} A$ and $V = \text{Spec} B$. Then $O_X(U \cup V) = O_X(U) \times O_X(V) = A \times B$ is not integral.

Conversely, suppose $X$ is reduced and irreducible. Let $U =\text{Spec} A$ be an affine open of $X$. Then $U$ is reduced and irreducible so $A$ is a reduced ring i.e. the nilradical of $A$ is $0$ and $U = \mathbb{V}(\mathfrak{p})$ for some prime ideal $\mathfrak{p}$, i.e. the nilradical of $A$ is prime. Thus $0$ is a prime ideal of $A$ so $A$ is integral domain.

Now let $U$ be an open subset of $X$. Let $V$ be an affine open contained in $U$. Then the restriction map $O_X(U) \to O_X(V)$ is an injection so $O_X(U)$ must be integral.


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Corollary. An affine scheme $X =\text{Spec} A$ is integral iff $A$ is an integral domain.

Proof. Suppose $A$ is an integral domain. As reduced-ness is an affine-local condition,  $X$ is reduced. On the other hand, $\text{Spec} A = \mathbb{V}(0)$ and $0$ is a prime ideal so $X$ is irreducible.
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Note: Integrality is not a stalk-local condition, as the disjoint union of integral scheme is not integral.
($\text{Spec} A \sqcup \text{Spec} B = \text{Spec} (A \times B)$). However it is almost stalk-local.

Underlying set of affine schemes

Examples: (using division algorithm we can find points of the first three schemes)

• $\text{Spec} \mathbb{C}[x]$: closed points and generic pt;
• $\mathbb{A}^1_k$ for algebraically closed $k$: same;
• $\mathbb{A}^1_k$ for any $k$ has infinitely many point (imitate Euclid's proof of infinitude of primes). 
• $\text{Spec}\mathbb{Z}$;
• $\text{Spec} k$;
• $\text{Spec}k[\epsilon]/(\epsilon^2)$ where $k[\epsilon]/(\epsilon^2)$ is called the ring of dual numbers;
• $\mathbb{A}^1_{\mathbb{R}}= \text{Spec} \mathbb{R}[x] = \{(0), (x-a), (x^2 + ax + b) \mid \text{irreducible}\}$ (use the fact that $\mathbb{R}[x]$ is a UFD.)
  Note that the $(x-a)$ are maximal ideals $(x^2 + ax + b)$ as the corresponding quotients are always fields. In particulat, $\mathbb{R}[x]/(x^2 + ax + b) \cong \mathbb{C}$.
  So $\mathbb{A}^1_{\mathbb{R}} is the complex plane folded along real axis, where Galois-conjugate points are glued. 
• $\mathbb{A}^1_{\mathbb{Q}}$
• Closed points of $\mathbb{A}^n_{\mathbb{Q}}$ are just Galois-conjugate glued together
• $\mathbb{A}^1_{\mathbb{F}_p} = \{(0), (f(x))\mid \text{ irreducible }\}$ (use the fact that $\mathbb{F}_p[x]$ is a Euclidean domain)
  Think of $f \leftrightarrow$ roots of $f$ (i.e. set of Galois conjugates in $\overline{\mathbb{F}_p}.$
  $\mathbb{A}^1_{\mathbb{F}_p}$ is bigger than $\mathbb{F}_p$. A polynomial $f(x)$ is not determined by its value at $\mathbb{F}_p$ (e.g. $x^p - x$), but it is uniquely determined by its valued on $\mathbb{A}^1_{\mathbb{F}_p}$ (as it is a reduced scheme). 
• $\mathbb{A}^2_{\mathbb{C}} = \text{Spec} \mathbb{C}[x,y] = \{(0), (x-a, y-b), f(x,y) \mid f \text{ irreducible}\}.$
•  $\mathbb{A}^n_{\mathbb{C}}$

Interpreting Fibers as Fiber Product

Proposition. Let $f: Y \to Z$ be a continuous map of topological spaces.  Let $z$ be a point of $Z$. Then $f^{-1}(\{z\}) = \{z\} \times_Z Y$ as topological spaces.

More generally, let $g: X\to Z$ is a morphism and $x \in X$. Let $\pi: X \times_Z Y \to X$ be the pullback of $f$ along $g$. Then $\pi^{-1}(x) = f^{-1}(g(x))$ as topological spaces.

Explanation:  The scheme structure on $\{z\}$ is $\text{Spec}\kappa(z)$, spectrum of the residue field at $z$.
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Defn. Scheme-theoretic Fiber. Let $f: Y \to Z$ be a continuous map of topological spaces.  Let $z$ be a point of $Z$. Then we could assign $f^{-1}(\{z\})$ the scheme structure of  $\{z\} \times_Z Y$ as topological spaces via the natural identification in the above proposition. We call $\{z\} \times_Z Y$  is scheme-theoretic preimage of $z$, or fiber of $f$ above $z$, and also denote it by $f^{-1}(z).$

If $Z$ is irreducible, the fiber above the generic point of $Z$ is called the generic fiber of $f$.

Note: Finite morphisms have finite fibers.

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Example. Projection of the parabola $y^2 = x$ to the $x$-axis over $\mathbb{Q}$. This corresponds to a map of Spec $f: \text{Spec} \mathbb{Q}[x,y]/(y^2 - x) \to \text{Spec} \mathbb{Q}[x]$ induces by the ring map $\mathbb{Q}[x] \to \mathbb{Q}[x,y]/(y^2 - x)$ given by $x \mapsto x$.

Preimage of $1$ is two points $\pm 1$.  $\kappa(x-1) = \mathbb{Q}[x]/(x-1)$ (since $(x-1)$ is maximal). So
$$f^{-1}(x-1) =  \text{Spec} \mathbb{Q}[x,y]/(y^2 - x) \times_{\text{Spec} \mathbb{Q}[x]$} \text{Spec}\mathbb{Q}[x]/(x-1)$$
i.e. $$\text{Spec }\left( \mathbb{Q}[x,y]/(y^2 - x) \otimes_{ \mathbb{Q}[x]}\mathbb{Q}[x]/(x-1) \right).$$
The latter is equal to
$$\text{Spec}\mathbb{Q}[x,y]/(y^2 -x, x- 1) = \text{Spec}\mathbb{Q}[y]/(y^2-1).$$
This is simply
$$\text{Spec} \mathbb{Q}[y]/(y-1) \sqcup \text{Spec} \mathbb{Q}[y]/(y+1).$$

Preimage of $0$ is a non-reduced point
$$\text{Spec}\mathbb{Q}[x,y]/(y^2 -x, x) = \text{Spec}\mathbb{Q}[y]/(y^2).$$

Preimage of $-1$ is a non-reduced point  of "size 2 over the base field"
$$\text{Spec}\mathbb{Q}[x,y]/(y^2 -x, x+1 ) = \text{Spec}\mathbb{Q}[y]/(y^2+1) \cong \text{Spec}\mathbb{Q}[i] = \text{Spec}\mathbb{Q}(i).$$

Preimage of generic point is a non-reduced point of "size 2 over the residue field"
$$\text{Spec }\left( \mathbb{Q}[x,y]/(y^2 - x) \otimes_{ \mathbb{Q}[x]}\mathbb{Q}(x)\right) \cong \text{Spec }\left( \mathbb{Q}[y] \otimes_{ \mathbb{Q}[y^2]}\mathbb{Q}(y^2)\right) .$$
The latter is simply $\text{Spec} \mathbb{Q}(y)$. Note that $[\mathbb{Q}(y): \mathbb{Q}(x)] = [\mathbb{Q}(y): \mathbb{Q}(y^2)] = 2$.


Notice: In all cases above, the fiber is an affine scheme whose vector space dimension over the residue field is $2$.

Normal Schemes

Normality = "not too far from smooth"

Recall that the punctured plane $U= \mathbb{A}^2- \{(0,0)\}$ viewed as an open subscheme of $X = \mathbb{A}^2$ is NOT affine. In proving this, we computed the global section of the punctured plane and found that $O_X(U) = O_X(X)$. This means in particular that every function on $U$ extends to the whole of $X$. This is an analogue of Hartogs's Lemma in complex geometry: we can extend a holomorphic function defined on the complement of a set of codimension at least two on a complex manifold over the missing set.

In algebraic geometry, there is an analogue. We can extend functions over points (in codim at least 2) not only if they are smooth, but also if they are "mildly singular"Hart, i.e. normal. Locally Noetherian normal schemes satisfy Hartogs's Lemma. Consequently rational functions without poles are defined everywhere.

Defn. A scheme $X$ is called normal if all of its stalks are normal, i.e. integral domains that are integrally closed in its field of fraction.

Claim. Normal schemes are reduced.
Proof. A scheme is reduced iff all of its stalks are reduced.
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Recall that if $A$ is integrally closed then so is its nontrivial localization $S^{-1}A$. From this we can deduce the following.

Claim. If $A$ is integrally closed then $\text{Spec} A$ is normal.

Claim. If $X$ is quasi-compact then $X$ is normal iff its normal at every closed point.
Proof. Here we simply use the fact that if a property $P$ is compatible with localization, then to check that $P$ holds for a quasi-compact scheme, it suffices to check it at closed points.  We replicate the proof from previous post here.

Suppose $X$ is normal at every closed point. Let $x \in X$. Then $\overline{\{x\}}$ contains a closed point $y$. Let $U= \text{Spec} A$ be an affine neighborhood of $y$. Then $U$ must contains $X$ as $y$ is in the closure of $x$. Thus we have $\mathbb{V}(\mathfrak{p}_x) \ni \mathfrak{p}_y$ so $\mathfrak{p}_y \supset \mathfrak{p}_x$ so we have a localization map
$$A_{\mathfrak{p}_y} \to A_{\mathfrak{p}_x}.$$
As $A_{\mathfrak{p}_y}$ is normal and normality is compatible with localization, we have $A_{\mathfrak{p}_x}$ is also normal.
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From commutative algebra, we know that for an integral domain $A$, TFAE:

• $A$ is integrally closed;
• $A_{\mathfrak{p}}$ is integrally closed for every $\mathfrak{p}$ prime in $A$;
• $A_{\mathfrak{m}}$ is integrally closed for every $\mathfrak{m}$ max in $A$.

Thus we have.

Proposition. Let $X$ be a scheme. TFAE

1. $X$ is normal;
2. For every affine open $U$ in $X$, $O_X(U)$ is normal; (in particular, $U$ itself is normal)
3. There is a cover of $X$ by open affine $U_{\alpha}$ such that $O_X(U_{\alpha})$ is normal

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Proposition. Let $X$ be a scheme. TFAE

1. $X$ is normal;
2. There exists an open covering $X = \bigcup X_i$ such that each open subscheme $X_i$ is normal;
3. Every open subscheme of $X$ is normal.

Proof.
(1) $\implies$ (2). Suppose $X$ is normal. Then for every open affine $U$ of $X$, $O_X(U)$ is normal. This implies $U$ itself is normal.
(2) $\implies$ (1). Cover $X$ by affine covers of $X_i$. Each of this affine piece is normal so $X$ is normal.
(1) $\implies$ (3). Every open subscheme of $X$ can be covered by affine opens of $X$. 

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Warning: Normal schemes are NOT necessarily integral.

Example. $X = \text{Spec} k \sqcup \text{Spec} k$ is a normal scheme. (Note: disjoint union of normal schemes is normal since normality is a stalk-local condition). However $X = \text{Spec} (k \times k) = \text{Spec} k[x]/(x(x-1))$ so its global section is not an integral domain.
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Claim: Global sections of an irreducible normal scheme is normal.

Proof.  Note a normal scheme is reduced so an irreducible normal scheme is integral. So we can embed all sections of $O_X$ to its function field $K(X)$ and all restriction maps are inclusion. In particular, we have $O_X(X) = \bigcap_{U \subset X, \text{open}}O_X(U) =\bigcap_{U \subset X, \text{affine}}O_X(U)$.

Suppose $s \in K(X)$ is integral over $O_X(X)$. Then it is integral over $O_X(U)$  for $U$ affine, so $s \in O_X(U)$, as it is normal by Proposition above. Thus $s \in \bigcap_{U \subset X, \text{affine}}O_X(U) = O_X(X).$

Constants are Definable

Let $K/k$ be a geometric function field, i.e. a finitely generated extension of an algebraically closed field $k$. We claim that there exists a one-parameter formula $\Theta$ such that
$$K \models \Theta(a) \iff a \in k.$$

Let $\mathcal{P}^c(k)$ denote the set of finite subsets of $k$ of odd cardinality greater than $c$. Let $\mathcal{P} = \mathcal{P}^0(k)$ denote the set of finite subsets of $k$ of odd cardinality.

Pairing $K \times \mathcal{P}'(k) \to k[t].$

For every $S \in \mathcal{P}(k)$ we can associate a polynomial $P_S(t) = \prod_{a \in S} (t-a) \in k[t]$.

For $(S, x) \in  K \times \mathcal{P}(k)$ we can associate the polynomial $p_{S,x}(T)$ defined as

• $T^2 - P_S(x)$ if $\text{char} k \neq 2$;
• $T^2 - T - P_S(x)$ if $\text{char} k = 2$. 

If $p_{S,x}$ has a root in $K$, we say that $(S, x)$ is a good pair.

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There exists $\mathcal{P}^c$ that capture $k$.

Lemma. Let $K/k$ be geometric function field. Then there exists $c= c_{K/k}$ such that for all $x \in K$,
if $(S, x)$ is a good pair for some $S \in \mathcal{P}^c(k)$ then $x \in k$.

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Application. $k$ is definable.

Let $c= c_{K/k}$ be as in the above lemma. Let $S$ in $\mathcal{P}^c$ be any set of absolute algebraic elements.
L et $\Theta(a)$ be the following formula in the language of fields:
$$\exists T, p_{S,a}(T) = 0.$$

Proposition. $k = \{ x \in K \mid \Theta(x)\}.$

Proof.  Clearly by the above lemma, if $x$ is a root of $\Theta$ then $x \in k$.
Conversely, let $x \in k$. Then?
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Proof of Lemma.

Observation. Let $x \in K$ and $S$ be a finite subset of $k$. Then $P_{S,x}(T)$ has a root iff $K_S$ has a $k(x)$-embedding in $K$ where $K_S$ is an extension if $k(t)$ by the roots of $P_{S,t}(T)$ in $K$. This is equivalent to a dominant rational $k$-map $X \to \dashrightarrow C_S$ where $X\to k$ is a projective normal model of $K/k$ and $C_S \to \mathbb{P}^1_t$ is the normalization of $\mathbb{P}^1_t$ in the Galois extension $K_S/k(t)$.

Difficulty in Proving Elementary Equiv of Geometric Function Fields implies Isomorphism

We proved that elementary equivalence between arithmetic function fields implies isomorphism. We will explain why we cannot apply this proof directly to the case of geometric function fields.

Let $k$ and $l$ be algebraically closed fields. Let $K/k$ and $L/l$ be finitely generated extensions. Suppose $K$ and $L$ are elementarily equivalent. Then

• $k$ and $l$ have the same prime field $F$;
• $\text{td}(K/k) = \text{td}(L/l)$

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Let $(t_1, \ldots, t_d)$ be a separable transcendence basis of $K/k$.

Notice: Unlike in the arithmetic case, we cannot in general write $K = F(t_1, \ldots, t_d)[x]$. (Why?)

Suppose $K = k(t_1, \ldots, t_d)[x]$ and $P(\underline{T}) \in k[\underline{T}]$ is an irreducible polynomial such that $P(t_1, \ldots, t_d, x) = 0.$ (In particular, $K$ is the function field of the irreducible variety $\text{Spec} k[\underline{T}]/(P)$.)

--------------------

Cutting out $x$ by a polynomial with coefficient in $F$.

As $P$ does not have coefficients in $F$, we cannot use $P$ to cut out $x$. Our idea is to try to replace $P$ with a polynomial $Q$ with coefficients in $F$. 

Let $\alpha_i$'s be the coefficients of $P$, then $P$ is defined over $F[\underline{\alpha}] \subset k$, i.e. we can view it as an element of $F[\underline{\alpha}][\underline{T}].$ Now $F[\underline{\alpha}= F[\underline{Z}]/(\underline{f})$ for some finite system $\underline{f}$ of relations $f_j$ of the variables $Z_i$.
So we can view $P$ as the equivalence class of an irreducible (why?) polynomial $Q \in F[\underline{Z}, \underline{T}].$

-----------

Cutting out $\underline{\alpha}, \underline{t}, x$ by a formula.

Let $\Psi(\underline{\zeta},\underline{\xi}, z)$ be the conjunction of

• $\Xi (\underline{\xi})$, i.e. $\underline{\xi}$ is a transcendental basis;
• $\underline{f}(\underline{\zeta})$;
• $P(\underline{\zeta}, z).$

Define $\iota: K \to L$.

As $\Psi$ has roots in $K$ and $K \equiv L$, $\Psi$ must have roots $(\underline{\beta}, \underline{u},y)$ in $L$. Thus we define $\iota: F[\underline{\alpha}, \underline{t}, x] \to L$ by

$\alpha_i \mapsto \beta_i$, $t_i \mapsto u_i$ and $x \mapsto y$.

Then as $\underline{u}$ is a root of $\Xi$, it must be a separable transcendence basis of $L/l$. 

What is the problem?

We want to extend $\iota$ to a function field embedding $K/k \to L/l$ where $K = k[t_0, \ldots, t_d][x]$. To this end, it suffices to extend 

$\iota': R=F[\underline{\alpha}] \to F[\underline{\beta}]$ to a field extension $k \to l$. 

The problem is:

1. Image of $\iota'$ might not lie in $l$
2. $\iota'$ might not be injective

Even when these two problems are resolved, we could only guarantee an embedding of $k_0 = \text{Frac} R$ into $l$. This would extend to a function field embedding $K_0/k_0 \to L/l$ where $K_0 = k_0(\underline{t},x).$

We do not know if (2) can be resolved, but (1) can be resolved by replacing $\Psi$ with a more complicated formula. This is because the constant  are definable. In other words, there exists a formula $\Theta(a)$ such that $\Theta(a)$ holds in $K$ iff $a \in k$.

Take $\Psi'$ to be $\Psi(\underline{\zeta},\underline{\xi}, x) \vee \Theta(\zeta_i).$

Then the roots $\beta_i$ have to all satisfy $\Theta(\beta_i)$ in $L$ and thus must lie in $l$.  

 

Elementary Equivalence of Arithmetic Function Fields of General Type implies Isomorphism

Let $K$ and $L$ be arithmetic function fields, i.e. finitely generated field extensions over their prime subfields. Let $k$ and $l$ be their absolute subfields, resp. Suppose $K$ and $L$ are elementarily equivalent. Then:

• $k \cong l$ and $\text{td}(K/k) = \text{td}(L/l)$
•  they have the same prime subfield $F$.

Theorem. There exists a field embedding $\iota: K \to L$ such that $L$ is finite separable over $\iota(K)$.

Proof. 

Let $(t_1, \ldots, t_d)$ be a separating transcendence basis of $K/k$, i.e. $K$ is algebraic separable over the polynomial ring $k[t_1, \ldots, t_d]$. Since $k = \overline{F} \cap K$ is algebraic separable over $F$, we must have $K$ is algebraic separable over $F[t_1,\ldots, t_d].$ Moreover, as $K$ is finitely generated over  $F$, it must be finitely generated over $F[t_1,\ldots, t_d].$ Thus $K = F(t_1, \ldots, t_d)[x]$ for some $x \in K$.
Question: Is the above the correct description of $x$?


To define a field embedding $\iota: K \to L$, it suffices to define $\iota(t_i)$ and $\iota(x)$. The idea is as follows. We want a formula $\Psi(\underline{\psi}, y)$ that holds for $(\underline{t}, x)$. Thus the sentence $\exists (\underline{\xi}, z), \Psi(\underline{\xi}, z)$ holds in $K$, so as $K \equiv L$, it must also holds in $L$. The "roots" of this sentence will be taken to be images of $t_i$ and of $x$ under $\iota$.

From before, we know that the exists a formula $\Xi(\underline{xi})$ such that  $\Xi(\underline{xi})$ holds in $K$ iff $\xi_1, \ldots, \xi_d$ is a transcendental basis of $K/k$. So this formula cuts out the $t_i$.

On the other hand, let $P(\underline{T}) \in k[\underline{T}]$ be an irreducible polynomial such that $P(t_1, \ldots, t_d, x) = 0$ (so that $K$ is canonically isomorphic to the functional field of the affine irreducible $F$-variety $\text{Spec} F[\underline{T}]/(P).$

Question: Since $x$ is algebraic over $F$, we know that $k= F(x)$. So why not just let $P \in F[T]$ (one-variable) be the minimal polynomial of $x$ over $F$? 

Then we can define the formula $\Psi(\underline{\xi}, z)$ to be the conjunction of:

• $\Xi(\underline{\xi})$
• $P(\underline{\xi}, z) = 0$ (notice as $P$ has coefficient in $F$, it is a first order sentence as $F$ is definable).

As $\Psi$ has roots $(\underline{t}, x)$ in $K$, it must have roots $(\underline{u}, y)$ in $L$. Define $\iota: K \to L$ by $t_i \mapsto u_i$ and $x \mapsto y$.

Then $(u_1, \ldots, u_d)$ is a separable transcendence basis of $L/l$ (being roots of $\Xi$), so $L$ is separable over $l(u_1, \ldots, u_d)$ and in particular over $\iota(K)$. 

Question: Does $j$ map $k$ isomorphically to $l$?

Proposition. If additionally, $K$ is of general type, then $K \cong L$ as fields.

Proof. By symmetry, we also have field embedding $\iota': L \to K$.

Let $j = \iota'\circ \iota: K \to K$.

Recall that if $K/k$ is a function field of general type, then every field $k$-embedding $K \hookrightarrow K$ is an isomorphism. Be careful: even though $j$ is a field embedding, it might not be a function field embedding! (It might not be identity on $k$).  It suffices then to show that some power $j^n$ of $j$ is a function field embedding. This is because then, as $K$ is of general type, $j^n$ must be an isomorphism. But then $j$ and thus $\iota$ itself must be isomorphisms. 

Indeed, notice that $j$ maps $k$ isomorphically to itself. (Why?) However $k$ is either a number field or a finite field, so it has only finitely many automorphisms. Thus for some $n$, $j^n$ is indeed identity on $k$.

Integral Morphisms

Defn. A morphism $\pi: X \to Y$ of schemes is integral if $\pi$ is affine and for every affine open $\text{Spec} B \subset Y$ with $\pi^{-1}\text{Spec} B = \text{Spec} A$, the induced map $B \to A$ is an integral ring homomorphism.

Claim. Finite morphisms are integral.

Note: the converse is not true. e.g. $\text{Spec} \overline{\mathbb{Q}} \to \text{Spec}\mathbb{Q}.$

Claim. Integrality is an affine-local condition

Claim: Integrality is closed under composition.

Fibers. Unlike finite morphisms, fibers of integral morphisms might not be finite. Fibers of integral morphisms have the property that no point is in the closure of any other point.

Proposition. Integral morphisms are closed. (Thus finite morphisms are closed)

Proof.
First we prove the proposition for affine case. Suppose $\pi: \text{Spec} A \to \text{Spec} B$ is an integral morphism induced by $\phi: B \to A$. Let $\mathbb{V}(I) \subset \text{Spec} A$ be a closed subset. It's image is the set $S$ of $\phi^{-1}(\mathfrak{p})$ where $\mathfrak{p} \supseteq I$. We claim that is this just $\mathbb{V}(\phi^{-1}(I)).$

Indeed, every prime ideal in $S$ lies on $\mathbb{V}(\phi^{-1}(I)).$ It thus suffices to show that every prime ideal containing $J = \phi^{-1}(I)$ is the contraction of a prime ideal containing $I$.

Key point: Lying Over (Going-up).
We have $\phi$ induces an embedding $B/J \to A/I$, and $A/I$ is integral over $B/J$.  QED

Now let $\pi: X \to Y$ be an integral morphism between schemes. Let $Z$ be a closed subset of $X$. Let $V_{\alpha}$ be an affine cover of $Y$ and $U_{\alpha} = \pi^{-1}(V_{\alpha})$ be their affine open preimages. Then $\pi(Z) \cap V_{\alpha}= \pi(Z \cap U_{\alpha})$ is closed in $V_{\alpha}$ for each $\alpha$, so $\pi(Z)$ is closed in $Y$.

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Theorem. Lying Over. Suppose $\phi: B \to A$ is an integral extension. Then for any prime ideal $\mathfrak{q}$ of $B$ there is a prime ideal $\mathfrak{p}$ of $A$ such that $\varphi^{-1}(\mathfrak{p}) = \mathfrak{q}.$

In other words, $\text{Spec} A \to \text{Spec} B$ is surjective. 

(Its generalization is Going-up).

Lemma. Let $U_{\alpha}$ be an open cover of $X$. If $Z \cap U_{\alpha}$ is closed in every $U_{\alpha}$ then $Z$ is closed in $X$.
Proof.
Let $V = X - Z$. Then $V \cap U_{\alpha}$ is open in every $U_{\alpha}$ and thus must be open in $X$. So their union, $V$, must be open in $X$.

Finite Morphisms have Finite Fibers

A morphim $\pi: X \to Y$ is finite if for every affine open $\text{Spec} B$ of $Y$, $\pi^{-1}(\text{Spec}B)$ is the spectrum of a $B$-algebra that is a finitely generated $B$-module.

Notice that finite morphisms are automatically affine.

We will show that all finite morphisms have finite fibers. However, the converse is not true.

Example. Finite fibers does NOT imply finite.
Let $\pi: \mathbb{A}^2 - \{(0,0)\} \to \mathbb{A}^2$ be the open embedding. It is not affine as  $\pi^{-1}(\mathbb{A}^2) =  \mathbb{A}^2 - \{(0,0)\}$ is not affine. So it cannot be finite. However it does have finite fibers.

Example. Finite fibers and affine does NOT imply finite.
Let $\pi: \mathbb{A}_{\mathbb{C}}^1 - \{0\} \to \mathbb{A}_{\mathbb{C}}^1$ be the open embedding. It clearly has finite fiber, and it is affine as $\pi^{-1}  \mathbb{A}^1 = \text{Spec} \mathbb{C}[x]_{(x)}.$ However, it is not finite as $\mathbb{C}[x]_{(x)}$ is not a finite $\mathbb{C}[x]$-module.

Proposition. Finite morphisms have finite fibers.

Proof. Let $\pi: X \to Y$ be a finite morphism. Suppose $Y = \text{Spec} B$. Then as $\pi$ is affine, $X$ must equal $\text{Spec} A$ for some $A$ that is finitely generated as a $B$-module. Let $[q]$ be a point of $Y$. We want to show that $\pi^{-1}(q)$ is finite.


Claim. We can assume $B$ is an integral domain and that $q = (0)$ is the generic point of $Y$.

Indeed, notice that $\pi^{\sharp}: B \to A$ induces $B/q \to A/ \pi^{\sharp}(q) A$ and thus a morphism
$\pi_1: \text{Spec} (A/\pi^{\sharp}(q) A) \to \text{Spec}(B/q).$  The fiber over $0$ in $\pi_1$ consists of exactly the prime ideals of $A$ containing $\pi^{\sharp} q$ that contracts to $q$. Thus $\pi_1^{-1}(0) \leftrightarrow \pi^{-1}(q)$.

Now suppose $B$ is an integral domain and $q = (0).$

Notice that $[q]$ lies in $U = \text{Spec} B_{q} \subset \text{Spec} B$ (here we think of $\text{Spec} B_{q}$ as the  set of prime ideals of $B$ contained inside $q$). So $\pi^{-1}(q)$ is the same as fiber over $q$ of the map $\pi: \pi^{-1}(U) \to U$.

Actually, as $B_{q}$ is a field, $U = \{[q]\}.$

Now $\pi^{-1}(U)$ are prime ideals of $A$ whose contractions under $\pi^{\sharp}$ is $q = (0)$. These are prime ideals $p$ such that $p \cap \pi^{\sharp} B = \{0\}$, i.e. prime ideals of $(\pi^{\sharp} B)^{-1} A = A'$.

As $A$ is a finite $B$-module, we must have $A'$ is also a finite $B_{(0)}$-module. Indeed suppose $a_1, \ldots, a_n$ generate $A$ over $B$. Let $a/\pi^{\sharp}b$ be in $A'$. Then $a$ is generated by $a_i$'s over $B$ and $1/\pi^{\sharp}(b) = 1/b \cdot 1$ is generated by $1$ over $B_{(0)}.$

As the map $\text{Spec} A' \to \text{Spec} B_{(0)}$ is finite and $B_{(0)}$ is a field, we must have $\text{Spec} A'$ is a finite discrete space.

Finite Morphism to Spec k

Proposition. Suppose $\pi: X \to \text{Spec} k$ is a finite morphism. Then $X$ is a finite union of point with discrete topology. The residue at each point is a finite extension of $k$.

Proof.  
As $\pi$ is finite, it is in particular affine, so we must have $X = \text{Spec} A$ for some $A$.

Claim: Every point of $X$ is closed.

Indeed, for all point $\mathfrak{p}$ of $X$, we have $A/\mathfrak{p}$ is an integral domain that is a finite $k$-algebra (i.e. finite dimensional $k$-vector space), so it must be a field.

Thus the points of $A$ must be the irreducible components of $A$. But $X= \text{Spec} A$ has only finitely many irreducible components (being a Noetherian space). So $X$ is finite and discrete.

On the other hand, the residue field at every point $\mathfrak{m}$ of $A$ is $A/\mathfrak{m}$ which is  a finite field extension of $k$.

Note: $A$ is Artinian.
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Lemma. Suppose $B$ is an integral domain such that $B$ is a finite $k$-algebra (i.e. finitely generated as a $k$-module). Then $B$ is a field.
Proof. Suppose $b_1, \ldots, b_n$ is a basis of $B$ as a $k$-vector space. Let $b \in B$ be non-zero. Consider the multiplication by $b$-map $B \to B$. As $B$ is an integral domain, this map must be injective. Thus linear transformation must be also surjective. In particular, $1$ must be in the image.
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Example.
$$\text{Spec}\left(\mathbb{F}_8 \times \frac{\mathbb{F}_4[x,y]}{(x^2, y^4)} \times \frac{\mathbb{F}_4[t]}{t^9}\times \mathbb{F}_2 \right) \to \text{Spec} \mathbb{F}_2.$$
LHS is disjoint union of
$\text{Spec} \mathbb{F}_8$ which is a point,
$\text{Spec}\frac{\mathbb{F}_4[x,y]}{(x^2, y^4)}$ which is a fat point $(x,y)$,
$\text{Spec}\frac{\mathbb{F}_4[t]}{t^9}$, which is another fat point $(t)$,
and  $\text{Spec}  \mathbb{F}_2$ which is a point.

Examples of Finite Morphisms

A morphim $\pi: X \to Y$ is finite if for every affine open $\text{Spec} B$ of $Y$, $\pi^{-1}(\text{Spec}B)$ is the spectrum of a $B$-algebra that is a finitely generated $B$-module.

Notice that finite morphisms are in particular affine.

Claim: Finite-ness is affine local on target. Thus it suffices to check that $\pi^{-1}(U_{\alpha})$ satisfies the above for one affine cover $\{U_{\alpha} \}_{\alpha}$ of $Y$.

In particular, a morphism $\pi: \text{Spec} A \to \text{Spec} B$ is a finite iff the induced homomorphism $B \to A$ gives $A$ the structure of a finitely generated $B$-module.

Claim: Finite morphisms are always closed, and they always have finite fibers. 

Example 1. Finite Field Extenion.

 If $L/K$ is a field extension, then $\text{Spec} L \to \text{Spec} K$ is finite iff $L/K$ is finite.

Example 2. Branched Cover.

Let $\pi: \text{Spec} k[t] \to \text{Spec} k[u]$ be the morphism induced by $u \mapsto p[t]$, where $p(t)$ is a polynomial of degree $n$.

Thus the $k[u]$ algebra structure on $k[t]$ is given by $u \cdot f(t) := p(t) f(t)$.
Therefore $k[t]$ is generated as a $k[u]= k[p(t)]$-module by $1, t, t^2, \ldots, t^{n-1}$ (i.e. $[k[t]: k[p(t)]] = n.$) so $\pi$ is finite.

Indeed, let $M$ be the $k[p(t)]$-submodule of $k[t]$ generated by $1, t, t^2, \ldots, t^{n-1}.$. Then clearly $M$ contains all polynomial in $t$ of degree less than or equal to $n$. Suppose $k[t] \backslash M \neq \emptyset$. Then there is a minimal degree among all the elements in $k[t]$ outside of $M$. Suppose $g$ is such a polynomial with minimal degree.  Dividing $g$ by $p(t)$, we get $g = fp + r(t)$ for some $r$ either equal to $0$ or of degree less than $p$. As $r \in M$, we must have $fp \not \in M, in particular$f$must not be in$M$, contradicting minimality of$g$.

Example 3. Closed Embedding.

The morphism $\text{Spec}(A/I) \to \text{Spec} A$ is finite since $A/I$ is a finitely generated $A$ module (generated by $1 \in A/I$.)

In particular, the embedding $\text{Spec} k \to \text{Spec} k[t]$ is finite. (Here we take $I= (t)$).

Example 4. Normalization. 

Let $\pi: \text{Spec} k[t] \to \text{Spec} k[x,y]/(y^2 - x^2 - x^3)$ induced by
$x \mapsto t^2 - 1$ and $y \mapsto t^3 - t^2$.

This is finite as $k[t]$ is a finite $k[t^2-1, t^3- t^2]$-module generated by $1$ and $t$.

This is an isomorphism between $D(t^2 -1)$ and $D(x)$ (i.e isomorphism away from the node of the target).

Affine Morphisms: affine-local on target

A morphism $\pi: X \to Y$ is affine if for every affine open $U \subset Y$, $\pi^{-1}(U)$ is affine.

Clearly, affine morphisms are quasi-compact and quasi-separated (since affine schemes are).

Proposition. $\pi$ is affine iff there is a cover of $Y$ by affine opens $U_{\alpha}$ such that $\pi^{-1}(U_{\alpha})$ is affine.

The proposition has some nonobvious consequence. Recall that if $Z \subset \text{Spec} A$ is globally cut out by an equation (i.e. $Z = \mathbb{V}(V)$) then its complement is affine. It turns out that is is also true if $Z$ is locally cut out by an equation.

Corollary. Let $Z$ be a closed subset of $X = \text{Spec} A$. Suppose $X$ can be covered by affine open sets on each of which $Z$ is cut out by one equation. Then the complement of $Z$ is affine.

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Proof of Proposition. By the Affine Communication Lemma, it suffices to check that the condition $\pi^{-1}(U)$ (variable $U$) is affine-local. This is equivalent to checking the following to criteria.

1. Suppose $\pi^{-1}(\text{Spec} B) = \text{Spec} A$ is affine. Then for all $g \in B$, $\pi^{-1}(\text{Spec} B_g)$ is affine.
   Proof. We have $\pi$ restricts to a map $\text{Spec} A \to \text{Spec} B$ which must be induced from some $\phi: B \to A$. Localization gives $\phi_g: B_g \to A_\phi(g)$  which must induce the restriction of $\pi: \text{Spec} A_\phi(g) \to \text{Spec} B_g$. So $\pi^{-1}(\text{Spec} B_g) = \text{Spec} A_{\phi(g)}.$
2. Suppose $(g_1, \ldots, g_n) = B$ and $\pi^{-1}(\text{Spec} B_{g_i}) = \text{Spec} A_i$ is affine. Then $\pi^{-1}(\text{Spec} B)$ is affine.
   Proof. Let $Z = \pi^{-1}(\text{Spec} B)$. Let $A = O_X(Z)$. We want to show that $Z$ is affine, i.e. $(Z, O_Z)$ is isomorphic  to $(\text{Spec} A, O_{\text{Spec} A})$ via the canonical map $\alpha: Z \to  \text{Spec} A$ which is induced by $A \to O_Z(Z).$ (Think of the case $Z$ is affine. For general case, use gluing).
   
   The factorization $B \to A \to O_Z(Z)$ gives a factorization $$\pi: Z \xrightarrow{\alpha} \text{Spec} A \xrightarrow{\beta} \text{Spec} B.$$ As $\alpha$ and $\beta$ are both surjective, and since $D(g_i)$'s cover $\text{Spec} B$, we must have $\beta^{-1}(D(g_i))$'s cover $\text{Spec} A$.
   
   Thus it suffices to show that $\alpha|_{\pi^{-1}(D(g_i)} : \pi^{-1}(D(g_i)) \to \beta^{-1}(D(g_i))$ is an isomorphism for each $i$. Then by gluing, $\alpha$ is an isomorphism $Z \to \text{Spec} A$ (or by stalk).
   
   Abusing notation, we denote both $\beta^{\sharp}g_i$ (an element of $A$) and $\pi^{\sharp}g_i$ (an element of $O_Z(Z)$) by $f_i$. Then $\beta^{-1}D(g_i) = D(f_i) = \text{Spec} A_{f_i} \subset \text{Spec} A$.
   
   On the other hand, $\pi^{-1}(D(g_i)) = Z_{f_i} = \text{Spec} A_i$. It suffices to show that $\alpha^{\sharp}$ induces an isomorphism $A_{f_i} \to A_i$.
   
   Indeed, notice that $Z$ is quasi-compact and quasi-separated since these notion are affine-local, and we have assume that $\pi$ is affine (hence quasi-compact and quasi-separated) on an open cover of $\text{Spec} B$. Then by the QCQS Lemma, we have the canonical map $A_{f_i} = (O_Z(Z))_{f_i} \to O_Z(Z_{f_i}) = A_i$ is an isomorphism.

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Proof of Corollary.

Suppose $(U_{\alpha})_{\alpha}$ is an affine cover of $X= \text{Spec} A$ such that on each $U_{\alpha}$, $Z$ is cut out by an equation $f_{\alpha}$. Let $V = X - Z$ and $\pi: V \to X$ be the inclusion map. Then $\pi^{-1}(U_{\alpha}) = U_{\alpha} \backslash Z$ is affine so $\pi$ is an affine map.  Thus $\pi^{-1} (X)$ must also be affine.  QED.

Affine Communication Lemma

Definition. A property $P$ enjoyed by some affine open sets of a scheme $X$ is called affine-local if

• If $\text{Spec} A \hookrightarrow X$ has property $P$ then so does $\text{Spec} A_f$. 
• If $A = (f_1, \ldots, f_n)$ and if $\text{Spec} A_{f_i} \hookrightarrow X$ has property $P$ for all $i$ then so does $\text{Spec} A.$

Affine Communication Lemma

Lemma. Suppose $P$ is an affine-local property. Suppose $X = \bigcup_{i} \text{Spec} A_i$ where $\text{Spec} A_i$ all have property $P$. Then every open affine subset of $X$ has property $P$.

Proof. Let $U = \text{Spec} A$ be an affine open of $X$. Cover $U_i = \text{Spec} A_i \cap U$ by simultaneous distinguished open sets of both $U$ and $\text{Spec} A_i$ (possible by Lemma below). By the first property of affine-local $P$, these distinguished open sets all have property $P$. But together they cover $U$, so by the second property of affine-local $P$, $U$ also has $P$. QED.
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Examples of Affine-Local Properties:

• reduced-ness
• Noetherian
• finite type $B$-scheme

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Covering by Simultaneous Distinguished Open Sets

Lemma. Let $\text{Spec} A$ and $\text{Spec} B$ be affine open subschemes of a scheme $X$. Then $\text{Spec} A \cap \text{Spec} B$  can be covered by open sets that are simultaneously distinguished open subschemes of both $\text{Spec} A$ and $\text{Spec} B$.
Proof.
In the proof below we need to use the following observation: Suppose $g \in O_X(U)$ and $V \subset U$. Let $U_g$ denote the non-zeros of $g$, i.e. $x \in U$ such that $g$ is not zero in $\kappa(x) = O_x/\mathfrak{m}_x.$. Then $$U_g \cap V = V_{g|_V}.$$

Let $p \in \text{Spec} A \cap \text{Spec} B$. We claim that there is a simultaneous distinguished open of both $\text{Spec} A$ and $\text{Spec} B$ containing $x$.

Let $\text{Spec} A \supset \text{Spec} A_f \ni p$ and let $p \in \text{Spec} B_g \subset \text{Spec} A_f$. We claim that $\text{Spec} B_g$ is also a distinguished open of $\text{Spec} A$.

Thus the idea is to get rethink of $\text{Spec}B_g$, the non-zeros of a section $g$ of $B$, first as non-zeros of a section of $\text{Spec} A_f$ (restriction of $g$), then as the non-zeros of a section of $A$.  

Indeed $\text{Spec} B_g$ are just the non-zero locus of $g \in B= O_X(V)$.  The problem is that $g$ is not a section on $\text{Spec} A$. However, as $\text{Spec} A_f \subset \text{Spec} B$, we can restrict $g$ to  $\text{Spec} A_f$. Let $g'$ denote this restriction. Then $\text{Spec} B_g$ are exactly the nonzeros of $g'$. Thus it is $\text{Spec} (A_f)_g' = \text{Spec} A_{fh}$ if $g' = h/f^n$ for some $h \in A$. 

Quasi-separated Schemes

Defn. A topological space is quasi-separated if the intersection of any two quasi-compact open sets is quasi-compact.  (In a separated space, the intersection of any two affine open sets is affine).

Claim. Affine schemes are quasi-separated.
Proof.  The quasi-compact open subsets of an affine schemes are all finite unions of distinguished open sets. So their intersection will again be a finite union of distinguished open sets.

(Note: quasicompact scheme $\leftrightarrow$ has a finite cover by affine opens).

Lemma. A scheme is quasi-separated iff the intersection of any two affine open subsets is a finite union of affine open subsets.

Proof. Suppose a scheme $X$ is quasi-separated. Take any two affine open subsets of $X$. Since they are quasi-compact, their intersection is also quasi-compact. As their intersection can be covered by affine open sets (e.g. by distinguished open sets), it can be covered by finitely many distinguished open sets.

Conversely, suppose the intersection of any two affine open subsets of $X$ is a finite union of affine open subsets. Take two quasi-compact open sets of $X$. Then each of this open set, being a quasi-compact scheme, is a finite union of affine opens. Their intersection will be the union of intersection of their affine open sets, which by hypothesis is a finite union of affine opens.
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In practice, we will see the hypothesis "$X$ is quasi-compact and quasi-separated" appear often. Usually, we would transfer such a hypothesis to the following statement, by using the above Lemma.

Proposition. A scheme $X$ is quasi-compact and quasi-separated iff $X$ can be covered by a finite number of affine open subsets, any two of which have intersection also covered by a finite number of affine open subsets.

Proof. $X$ is quasi-compact iff $X$ has a finite cover by affine opens. Then combine with Lemma above.
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Claim. Projective $A$-schemes are quasi-compact and quasi-separated.
Proof.  Let $S$ be a finitely generated graded ring over $A$ (i.e. $S_0 = A$) (by definition of projective $A$-schemes). Suppose $f_1, \ldots, f_n$ are generators of $S$ over $A$. Let $X = \text{Proj} A$. Then $D_{+}(f_i)$ cover $X$.  Notice $D_+{f_i} = \text{Spec} ((S_{f_i})_0)$ where $(S_{f_i})_0$ denotes the $0$-th graded piece of $S_{f_i}$, and $D_+(f_i) \cap D_+(f_j) = D_+(f_i f_j)$.
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Example. (Non-quasi-separated Scheme).

Let $X = \text{Spec}k[x_1, x_2, \ldots]$. Let $\mathfrak{m}$ be the maximal ideal $(x_1, x_2, \ldots).$. Let $U = X - [\mathfrak{m}].$ Let $Y$ be the result of gluing two copies of $X$ along $U$. Then $Y$ is not quasi-separated.
Proof. The two copies of $X$ are both quasi-compact, being affine schemes. However, their intersection is $U$, which is not quasi-compact.

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Claim. Locally Noetherian schemes are quasi-separated.

Open Subset of Affine NOT Quasi-compact

Let $X = \text{Spec} k[x_1, x_2, \ldots].$ Let $\mathfrak{m}$ be the maximal ideal $(x_1, x_2, \ldots).$. Let $U = X - [\mathfrak{m}].$ Then $U = \bigcup_{i} D(x_i)$.

Claim: $U$ is not quasi-compact

Proof.
 Suppose $U$ can be covered by finitely many of the $D(x_i)$'s say $U= \bigcup_{i=1}^n D(x_i)$. Then $(x_1, \ldots, x_n) \in U$ but $(x_1, \ldots, x_n) \not \in D(x_i)$, a contradiction.

Morphisms of Function Fields do NOT induce Rational Maps

We have seen that dominant rational maps between integral schemes induce morphisms of function fields.  The converse however, is not true.  Morphisms of function fields of two integral schemes do not always induce rational maps.

Example. Let $X = \text{Spec} k[x]$ and $Y = \text{Spec} k(x)$. Then their function fields are canonically isomorphic (both being $k(x)$).

Claim. There is no rational map $X \dashrightarrow Y$.

Proof. Suppose we have a rational map $f: U \to Y$ for some open dense subset $U$ of $X$. Shrinking $U$ if necessary (as all nonempty subsets of $X$ are dense), assume that $U = D(f)$. Thus $f$ induces a morphism $\varphi: k(x) \to k[x, 1/f]$.   Note that on stalks, this morphism should be identity $\varphi_{\eta}: k(x) \to k(x)$, so $\varphi$ must be identity on $k[x]$. This is impossible.
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For integral finite type $k$-schemes (in particular, irreducible varieties), morphisms of function fields do indeed determine rational maps between the schemes.

Proposition. Let $X$ be an integral $k$-scheme and $Y$ be an integral finite type $k$-scheme. Let $\varphi: K(Y) \hookrightarrow K(X)$ be a $k$-homomorphism. Then there exists a dominant $k$-rational map $\phi: X \dashrightarrow Y$ inducing $\varphi$.

Proof. The map $\phi$ is equivalent to a map from a non-empty open subset of $X$ to a subset $Y$ sending the generic point $\eta_X$ to $\eta_Y$. Moreover, as function fields of nonempty open subsets of $X$ (resp. $Y$) are canonically isomorphic to $K(X)$ (resp. $K(Y)$), we can assume $Y = \text{Spec} B$ and $X = \text{Spec} A$ are affine. We have a morphism $\varphi: B \hookrightarrow K(Y) \hookrightarrow K(X)$.

Warning: Clearly $\varphi$ induces a morphism $\text{Spec} \varphi(B) \to \text{Spec} B$. However we do NOT have a way to take $U = \text{Spec}\varphi(B)$ as an open subset of $X$.

So instead, we want to find some $A_g \subset K(X)$ such that $\varphi: B \hookrightarrow A_g$. This would induce a map a map $D(g) \to Y$.

Why can we find such an $A_g$? This is because $Y$ is a $k$-scheme of finite type. Thus $B = k[b_1, \ldots, b_n]$ for some $b_i$'s. Let $\varphi(b_i) = a_i/f_i$. Let $g = f_1 \cdots f_n$. Then as $\varphi$ is a $k$-homomorphism, the image of $B$ indeed lies in $A_g$.
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In particular if both $X$ and $Y$ are integral finite type $k$-schemes then a rational map $X \dashrightarrow Y$ that induces isomorphism $K(Y) \cong K(X)$ must be birational.

Function Field of an Integral Scheme

Let $X$ be an integral scheme. Then it is irreducible so it has a generic point $\eta$. Let $\text{Spec} A$ be a nonempty affine open of $X$ (thus must contain $\eta$). Let $K(A)$ denote the fraction field of $A$.

Claim. $(O_X)_{\eta} \cong K$ canonically.

Proof. We have $\text{Spec} A$ is itself irreducible, being a nonempty open subspace of an irreducible space. Thus $\eta$ corresponds to the generic point of $\text{Spec} A$, i.e. the nilradical of $A$ (which must be prime). However, since $A$ is integral, $\sqrt{0_A} = (0)$.
Therefore we have $(O_X)_{\eta} = A_{\sqrt{0_A}} = A_{(0)} = K(A)$.
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Let $V \supset U$ be nonempty open subsets of $X$.

Claim. The restriction map $\text{res}: O_X(U) \to O_X(V)$ is an inclusion.

Proof. Let $s \in O_X(U)$ be such that $s|_V = 0$. Notice that $U$ is irreducible, being a nonempty open subset of an irreducible space, so $V$ is dense in $X$. Thus as $s$ vanishes on $V$, it also vanishes on $U$. (Warning: this is NOT enough to conclude that $s = 0$). However, as $O_X(U)$ is reduced, its sections are uniquely determined by their values at points of $U$, so $s$ must indeed be $0$.
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Claim. The map $O_X(U) \to (O_X)_{\eta}= K(A)$ is an inclusion.

Proof. Suppose $s_{\eta} = 0$ for some $s \in O_X(U)$. Then $s|_V = 0$ for some $\emptyset \neq V \subset U$. Thus by the previous claim, $s = 0$.
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Advantage of Irreducible Varieties. 

Irreducible varieties form an important class of integral schemes. We thus see that for such varieties, all $O_X(U)$'s can be embedded into the same ring $K(X)$. As restriction maps are simply inclusion, sections glue iff they are the same element of $K(X)$.
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Dominant maps of Integral Schemes induce morphism of Function Fields.

Let $f: X\dashrightarrow Y$ be a dominant rational map. Then $f$ sends $\eta_X$ to $\eta_Y$, as dominant rational map between irreducible schemes preserve generic pt.  Thus it induces a map on stalk $(O_Y)_{\eta_Y} = K(Y) \to (O_X)_{\eta_X} = K(X).$

Dominant Maps and Generic Points

Proposition. Let $\varphi: A \to B$ and $f: \text{Spec} B \to \text{Spec} A$ be the induced map. Then $\text{Im} f$ is dense in $\text{Spec} A$ iff $\varphi^{-1}(\sqrt{0_B}) \subset \sqrt{0_A}.$
Equivalently, $\ker \varphi \subset \sqrt{0_A}.$

Proof.  Note that $\text{Im} f = \{ \varphi^{-1}(q) \mid q \in \text{Spec} B \}.$ So
$$\overline{\text{Im} f} = \mathbb{V}\left( \bigcap_{q \in \text{Spec} B} \varphi^{-1} q \right)$$
But $$\bigcap_{q \in \text{Spec} B} \varphi^{-1} q = \varphi^{-1} \left( \bigcap_{q \in \text{Spec} B} q\right) = \varphi^{-1}(\sqrt{0_B}).$$
Thus $$\overline{\text{Im} f} = \text{Spec} A = \mathbb{V}(\sqrt{0_A}) \iff \varphi^{-1}(\sqrt{0_B}) \subset \sqrt{0_A}.$$
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Note that for a set $S \in \text{Spec} B$, $$\overline{S} = \bigcup_{s \in S} \overline{\{s\}} = \bigcup \mathbb{V}(s) = \mathbb{V}\left( \bigcap \mathfrak{p}_s\right).$$

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If $\text{Spec} A$ and $\text{Spec} B$ are both irreducible, then their nilradicals are their generic points, so the above statement is equivalent to saying that $\text{Im} f$ is dense iff $f$ maps generic pt to generic pt. There is a generalization of this.

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Proposition. A rational map $\pi: X \dashrightarrow Y$ of irreducible schemes is dominant iff $\pi$ sends the genetic point $\eta_X$ of $X$ to the generic point $\eta_Y$ of Y. 

Proof. Let $(U, \pi)$ be a representation of $\pi$. Note that every open set of $X$ contains $\eta_X$.

Suppose $\pi(\eta_X) = \eta_Y$. Then $\pi(U) \ni \eta_Y$ so must be dense in $Y$.

Conversely, suppose that $\pi(U)$ is dense in $Y$.  Let $V\cong \text{Spec} B$ be an affine open of $Y$.   Cover $\pi^{-1}(V)$ by (finitely many) affine opens $U_i\text{Spec} A_i$. Then $\pi$ gives a morphism
$$\pi': \bigsqcup U_i \to V.$$
As $\pi$ is dominant, we must have image of $\pi'$ (which is equal to image of $\pi|_{\pi^{-1}(V)}$) is dense in $V$. On the other hand
$$\bigsqcup U_i = \bigsqcup \text{Spec} A_i = \text{Spec} \left(\prod A_i\right).$$ Let $A = \prod A_i$. Suppose $\varphi: B \to A$ is the induced map on global sections.

As the image of $\pi'$ is dense we know that $\varphi^{-1}(\sqrt{0_A}) \supset \sqrt{0_B}$, which is $\eta_Y$. On the other hand $(\sqrt{0_A}) = \bigcap_{p \in A_1, A_2, \ldots, A_n} p$  as $\text{Spec} A =  \bigsqcup \text{Spec} A_i$. Thus for each $A_i$, as $\bigcap_{p \in A_i} p \supset \bigcap_{p \in A_1, A_2, \ldots, A_n} p$ we have $\varphi^{-1}(\bigcap_{p \in A_i}) \supset \sqrt{0_B})$. But the former is just $\pi(\eta_X).$
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Note: Recall that every open subset of an irreducible space is irreducible.

Note: If $\eta$ is the generic point of an irreducible scheme $X$ and $U = \text{Spec} A$ is a affine open of $X$ then $\eta$ is the generic point of $U$ (i.e., corresponding to the nilradical of $A$).
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Quicker method: Suppose $\pi(\eta_X) \neq \eta_Y$. Then $U = Y - \overline{\pi(\eta_X)} \neq \emptyset$. As $\pi(X)$ is dense in $Y$, $U \cap \pi(X) \neq \emptyset$ so $\pi^{-1}(U)$ is not empty and thus must contain $\eta_X$, a contradiction.

Hypersurfaces have no Embedded Points

In this post, we assume the following properties of associated points of an $A$ module $M$ (for some Noetherian ring $A$).

1. Associated points of $M$ are precisely the generic points of irreducible components of support of some elements in $M$ (viewed as global section on $X = \text{Spec} A$).
2. $M$ has finitely many associated points
3. An element of $A$ annihilates some non-zero element of $M$ iff it vanishes at some associated point of $M$. 

Notice that $m_p = 0$ iff $x m = 0$ for some $x \not \in p$, i.e. $\text{ann}(m) \not \subset p$. In other words, $p \in \text{Supp}(m)$ iff $\text{ann}(m) \subset p$. 

So $\text{Supp}(m)$ consists of exactly the prime ideals of $A$ containing $\text{ann}(m)$. Thus its irreducible components should correspond to the minimal such prime ideals. In other words, the associated primes of $A$ lying inside support of $m$ are exactly minimal prime ideals containing $\text{ann}(m)$. Thus from (1) and (2), we already can deduce that if $f \in A$ annihilates some nonzero $m \in M$ then $f$ lies inside some associated point of $M$ (namely those lying inside $\text{Supp}(m)$).

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Lemma. Let $A = k[x_1, \ldots, x_n]/(f)$. Then $A$ has no embedded points.

Proof. Let $p$ be an associated prime. Then $p$ is a minimal prime ideal containing $\text{ann}(g)$ for some $g \in A^*$. We claim that $p$ is a minimal prime ideal containing $f$. Indeed, if $\text{ann}(g) \neq 0$, then $f \mid hg$ for some $h$ but $f \not \mid g$, so by unique factorization, $f$ and $g$ must have some greatest common divisor $1 \neq f_1 \neq f$. Suppose $f = f_1 f_2$. Then $\text{ann}(g) =  (f_2).$ Thus $p$ is a minimal prime ideal containing $f_2$, and must also be a minimal prime ideal containing $f$.  

Locus of Nonreduced points & Associated Points.

Before, we demonstrated that reduced-ness is not an open condition (though for quasi-compact schemes, we can still check reducedness at closed points). However, for Noetherian schemes, reduced-ness is indeed an open condition.

For this post, we only use two properties of associated points of an $A$-module $M$ for $A$ Noetherian.

• The associated points of $M$ are precisely generic points of irreducible components of supports of some global sections of $\widetilde{M}$.
• $M$ has finitely many associated points.
  
  Warning: we only talk about supports of global sections.

Proposition. Let $X = \text{Spec} A$. Show the the set of nonreduced points of $X$ is the closure of the nonreduced associated points. In other words, let $S$ be the set of non-reduced points of $X$. Then
$$S = \overline{S \cap \text{Ass} X}.$$

Proof.  Let $p$ be in the closure of the set of non-reduced associated points.
Then $p \in \overline{\{q\}}$ for some non-reduced associated point $q$, i.e.$p \supset q$. Thus we have  $A_q = (A_p)_q$. If $A_p$ is reduced then so must $A_q$, therefore $A_p$ is nonreduced.

On the other hand, suppose $p$ has nonreduced stalk, i.e. there is some (not necessarily global) section $s_p \neq 0$ such that $s_p^n = 0$ for some $n$. On a sufficiently small affine open $D(f) \ni p$, we could represent $s$ as $a/f$ for some $a \in A$. Thus we have $(af)$ is nilpotent (by proof of Lemma). Now $(af)_p \neq 0$ so $p \in \text{Supp}(af)$ thus it must be in the closure of some associated point $q$. (In particular, $(af)_q \neq 0$). On the other hand, as $D(f) \ni q$, and $(af)^n = 0$ on $D(f)$. Thus  $(af)^n_q = 0$, so $q$ is a non-reduced point.



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Thus we see that in particular, for an affine scheme $X$, the ``reduced locus'' of $X$ is open.

Recall that reduced-ness in in general not an open condition. However, from the above, we see that, for Noetherian schemes $X$, the "reduced locus" of $X$ is an open subset of $X$.
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In the above proof we have used the following Lemma.

Lemma. Localization of reduced rings is reduced.

Proof. Suppose $A$ is a reduced ring. Let $B = S^{-1} A$.
Suppose $a/s$ is a nilpotent in $B$. This means $a/s \neq 0$ in $B$ and $(a/s)^m = 0$ in $B$. In other words, $x a \neq 0$ for all $x \in S$ and $ya^m = 0$ for some $y \in S$. Thus $ya \neq 0$ but $ya^m = 0$, so $ya$ is a nilpotent in $A$, contradiction.

Lemma. If $s_p \neq 0$ then $p$ is in the support of some global section.

Proof. If $s_p \neq 0$, then $s$ is not $0$ on some neighborhood of $p$. WLOG, assume $s \neq 0$ on some $D(f) \ni p$. Suppose $s$ is represented by $a/f$ on $D(f)$. Then $a_p \neq 0$.

Reduced Schemes have no Embedded Points

In this Post, we only talk about associated points of an affine scheme $X = \text{Spec} A$.

We only use the following axiom of associated points: associated points are generic points of irreducible components of the support of sections.
Note that $X$ is itself is always a possible support (of the function $1$, for example), so the generic points of its irreducible components are associated points. The other associated points are called embedded points.

Example: Let  $A= k[x,y]/(xy, y^2)$.

 Possible supports of sections:
$\emptyset$ = $\text{Supp}(0).$
$X = \text{Supp} f$, for $f \not \in (y)$.
$\{(x,y)\} = \text{Supp}(f)$, $f \in (y) \backslash (y^2, xy)$.

Thus the associated points of $X$ are $(y)$ and $(x,y)$. Among these, only $(x,y)$ is an embedded point.

Notice that  $A$  has an embedded point corresponding exactly to the unique non-reduced point

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Reduced Schemes have no Embedded Points

Let $A$ be an integral domain. Then as the only possible support of sections of $A$ is $\text{Spec} A$,  the the generic point is the only associate point of $\text{Spec}A$.
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Lemma. If $A$ is reduced then $X = \text{Spec}A$ has no embedded points.

Proof. First consider the case when $\text{Spec} A$ is irreducible, i.e. $A$ is a domain. Then $(0)$ is the only associated point, but it is also the generic point of $X$, so there is no embedded point.

 Now let $X =\text{Spec}A$ be any reduced scheme. Let $f \in A^*$. Then $\text{Supp} (f) = \overline{D(f)}$.We claim that its irreducible components are also irreducible components of $X$ (thus all associated points are generic pts of irreducible components of $X$ and hence there is no embedded points).

Note that the irreducible components of a closed subspace $Y \subset X$ are always closed and irreducible in $X$, but might not be maximal.

Claim: the irreducible components of $\overline{D(f)}$ are exactly the irreducible components of $X$ that meet $D(f)$.

Clearly all irreducible components of $\overline{D(f)}$ are contained in some irreducible components of $X$ that meet $D(f)$. Thus it suffices to show that if $\mathbb{V}(\mathfrak{p})  \cap D(f) \neq \emptyset$ then $$\overline{\{\mathfrak{p}\}} =\mathbb{V}(\mathfrak{p})  \subset \overline{D(f)}, \text{i.e.}$$
$$\mathfrak{p} \in \overline{D(f)}.$$
Suppose $\mathbb{V}(\mathfrak{p})  \cap D(f) \neq \emptyset$, i.e. for some $\mathfrak{q} \supset \mathfrak{p}, \mathfrak{q} \not \ni f$. Then in particular, $\mathfrak{p} \not \ni f$. (Or we can see that $\mathbb{V}(\mathfrak{p}) \not \subset \mathbb{V}(f)$.)
Thus $\mathfrak{p} \in D(f)$.
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Another way to phrase it is that if $D(f) \cap \mathbb{V}(\mathfrak)(p)$ is not empty then it is dense in $\mathbb{V}(\mathfrak{p})$ so its closure in $\mathbb{V}(\mathfrak)(p)$ must be $\mathbb{V}(\mathfrak)(p)$. Hence $\overline{D(f)} \supset \mathbb{V}(\mathfrak)(p)$
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Support of a section

Defn. Let $\mathcal{F}$ be a sheaf (or a separated presheaf) of abelian groups on a scheme $X$. Let $s$ be a global section of $\mathcal{F}$. Define support of $s$ to be the set of points $p \in X$ where $s_p \neq  0 \in \mathcal{F}_p$.

Lemma: $\text{Supp} s$ is closed

Proof.  $s_p = 0$ iff $s$ is $0$ (as a section) on a neighborhood of $p$. Thus the complement of  $\text{Supp} s$ in $X$ is open.

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In general, for an $A$-module $M$ and $p \in \text{Spec} A$, we see that $p \in \text{Supp}(m)$ iff $m \neq 0$ in $M_p$ $\iff$ $\text{ann}(m) \subset p$.

Claim: If $X = \text{Spec} A$ for a integral domain  $A$ and $f \in A^*$ then $\text{Supp} f = X.$

Proof.  For every $f \in A$, if $f \neq 0$ then $\text{ann}(f) = 0$ so $\text{Supp}(f) = \text{Spec} A$.
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Notice that if $s_p = 0$ iff $s$ vanishes on a neighborhood of $p$, so $s(p) = 0$. In other words, if $s(p) \neq 0$ then $p \in \text{Supp}(s)$, i.e.
$$\overline{D(s)} \subset \text{Supp}(f).$$
 However, the other direction is not true in general. For $X = \text{Spec} k[x,y]/(y^2, xy)$, the global section $s = y$ is $0$ at $p=(x,y)$ but $s_p$ is not zero. (Because there are extra differential operators at $p$).

Claim: If $X = \text{Spec} A$ for a reduced ring $A$ then $\text{Supp} f = \overline{D(f)}.$

Proof. Clearly if $p \in D(f)$ then $f(p) \neq 0$ so $f_p \neq 0$ and thus $\text{Supp} f \supset D(f)$, and thus $\text{Supp} f \supset \overline{D(f)}.$

(Another way to state the same thing: Suppose $p \in \overline{D(f)}$. Then every neighborhood of $p$ contains points of $D(f)$ so $f$ cannot vanish on any neighborhood of $p$.)

Conversely, suppose $p \in \text{Supp} f$. Then $f_p \neq 0$. Thus for all (sufficiently small) neighborhood $W \ni p$, we have $f|_W \neq 0$.

Warning: The common mistake here is to deduce from this that $W \cap D(f) \neq \emptyset$, by arguing that as $f|_W$ is not equal to the zero function, $f$ cannot vanish on all of $W$, i.e, $W \not \supset \mathbb{V}(f)$. However in general schemes, functions are NOT determined by their values at points. That is why we need to reduced hypothesis.  (The difference between two functions that equal pointwise lies in the nilradical, so if the nilradical is $0$, functions are determined by their values at points.)

As $W$ is itself a reduced scheme, and $f|_W$ is uniquely determined by its values at points of $W$, we do have that $W \not \subset \mathbb{V}(f)$ and thus $W \cap D(f) \neq \emptyset$.



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Example. Let $X = \text{Spec} k[x]/(x^2)$ be the fat point.  Let $p = (x)$ be the only point of $X$. Then
$Supp f = \emptyset$ if $f \in (x^2)$ (i.e. if $f \sim 0$ as elements of $k[x]/(x^2)$).
$Supp f = X$, otherwise.

Indeed, if $f$ does not vanish at $p$ to begin with, then $f_p \neq 0$. On the other hand, if $f(p) = 0$ and $f \neq 0$ then $f$ is a constant multiple of $x$. Now $x_p \neq 0$. This can be checked algebraically, but geometrically it is because a neighborhood of $p$ contains the operator: $f \mapsto$ the linear term of $f$. Notice that $x$ does not vanish under this operator.

See What does the Stalk Perceive for more explanation.

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Let $X = \text{Spec} A$. Let $f \in A$. Then $p$ is in $\text{Supp} f$ iff  $f$ is not in the kernel of
$A \to A_p$.

Notice $\text{Supp} (fg) = \text{Supp}(f) \cap \text{Supp} (g)$.


Example. Let $A = \frac{k[x,y]}{(y^2, xy)}$ and $X = \text{Spec} A$. Let $f \in A$. We claim that the support of $f$ is either $\emptyset, (x,y)$ or $X$.

Proof. Note the elements of $\text{Spec} A$ are prime ideals of $k[x,y]$ containing $(y)$. In other words, $(y)$ is the generic point of $X$.

Case 1. $f \not \in (y)$. Then $D(s)\ni (y)$  so $\text{Supp}(s) \supset \overline{D(s)} = X$.

Case 2. $f \in (y)$. Then $f$ vanishes at all points of $X$ since all prime ideals of $A$ contains $(y)$. (Another way to see this is to notice that $D(f)$ is an open subset of $X$ not containing the generic point so must be empty).

For all $p \neq (x,y)$, $p$ is a reduced point of $X$. As $f$ vanishes on a neighborhood of $p$ and $p$ is reduced, $f$ must actually be equal to the zero section on a neighborhood of $p$, i.e. $f_p = 0$. So $p \not \in \text{Supp}(f)$.

Thus we only need to check if $f_{(x, y)} = 0$

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Some other random reasonings that achieve the same thing:
Suppose $p \neq (x, y)$, i.e. $p \in D(x)$.

We have $s_p$ iff $s$ is $0$ on some neighborhood of $p$. Without loss of generality, assume that neighborhood is $D(xf) \subset D(x)$ for some $f$.

We have a morphism $A \to  O_X(D(x)) = A_x = k[x] \to O(D(xf))$ (given by $y \mapsto 0$). As $k[x]$ is an integral domain, the localization map $O_X(D(x)) \to O_X(D(xf))$ is injective,  so $s \in A$ restricts to $0$ in $D(xf)$ iff it restricts to $0$ on $D(x)$, i.e. iff it is in the kernel of
$$A \mapsto A_x$$
i.e.
$$k[x,y]/(y^2, xy) \to k[x]$$ given by $y \mapsto 0$ and $x \mapsto 0$.
The kernel of this map is $(y)$.

So for all $s \in (y)$, $s_p = 0$ at all $p \in D(x)$. We cannot conclude yet for all such $s$, that $\text{Supp} (s) = X - D(x) = (x, y)$. We need to check if $s_{(x,y)} = 0$.
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Let $p =(x,y)$. At the stalk of $X$ at $p$ there are extra differential vectors generated by the $x, y, x^2$. Thus $s_p = 0$ iff the $y$ term, $x$ term and $x^2$ term of $s$ is zero. A way to rephrase it is to think of the ``$y$-operator'', as a point $\text{Spec}k[y]/(y^2) \to X$. As its image is the point $(x, y)$, we can pullback maps on neighborhoods of $(x,y)$ to maps on neighborhood of $(y)$. In other words, we have a local homomorphism of stalks $A_p \to (k[y]/(y^2))_{(y)}$. Thus $s$ is $0$ in $(O_X)_p$ iff its pullback is $0$ in $k[y]/(y^2))_{(y)}$. Among the element $s \in (y)$, only those in $y^2$ pullsback to $0$. So if $s$ is a multiple of $y$, then $s_{(x,y)} \neq 0$, and thus $\text{Supp} (s) = (x,y)$.



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Lemma. Let $M$ be an $A$-module. Then the natural map
$$M \to \prod_{\text{associated }$p$} M_p$$
is an injection.

What does the stalk perceive?

Let $\mathcal{F}$ be a sheaf (or a separated presheaf) of abelian groups on a scheme $X$. Let $s$ be a global section of $\mathcal{F}$.


Notice that even $s(p) = 0$, $s_p$ could still be nonzero.

Example. We give an example to illustrate the fact that two sections become the same element in the stalk at $p$ iff they restrict to the same section on a neighborhood of $p$ (but not that they agree pointwise) (Of course, this is simply by definition of stalk).

 Let $X = \text{Spec} A$. Two global sections $f, g \in A$ satisfies $f_p = g_p$ iff $f_p - g_p$ is $0$ in $A_p$. So let's look at the kernel of $A \to A_p$.

These consist of $s \in A$ such that $h s = 0$ for some $h \not \in p$. Thus $s$ is $0$ on $D(h)$, which is a neighborhood of $p$. Thus if $s_p = 0$ then $s$ vanishes on a neighborhood of $p$.

The converse is not true.

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We see that if $s_p = 0$ then $s$ is $0$ on a neighborhood of $p$ and thus $s(p) = 0$. Is the converse true?

The answer is No. Warning:
 

$$s(p) = 0 \not \implies  s_p = 0.$$

Example. Let $X = \text{Spec} k[x]/(x^2)$. Then $X$ has only one point, but there are extra first derivative vectors at the single point of $X$ also. As in, two global sections of $X$ are identical iff they agree on points of $X$ and also have the same first derivative. Now let $s = x$. At the point $p= (x)$, we have $s(p)$ is $0$ in $A/(x)$ (or equivalent, in $\text{Frac} A/(x) = A_{(x)}/(x)$). However $s_p$ is not $0$ in $A_{(x)} = (k[x])_{(x)}/(x^2)$. This is because $s= x$ is  $0$ in the latter iff $x f = x^2 g$ for some $f \not \in (x)$, which is impossible as $k[x]$ is a UFD. Thus $\text{Supp} s = X$.

Intuitively, what happened is that for $s_p$ to be $0$, $s$ needs to be $0$ on a neighborhood of $p$. In our case, a neighborhood of $p$ would include the extra derivatives at $p$.   So since $s$ only vanish at $p$ with order $1$, its stalk at $p$ is not $0$.

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The Property of being Reduced

click here

A scheme $X$ is said to be reduced if $O_X(U)$ is reduced for all open $U$. 

Key point

Reduced-ness is stalk-local and affine local. However, it is not an open condition . 

Stalk-local.

Reduced-ness is a stalk-local property: $X$ is reduced iff $(O_X)_x$ is reduced for all $x$.

Paraphrasing this, a scheme is not reduced iff its talk at some point is not reduced. 

proof. (Trivial) Suppose $X$ is not reduced. Then there exists a section nonzero $s \in O_X(U)$ such that $s^n = 0$ for some $n$. As $s$ is not $0$ and $O_X$ is a sheaf, $s_x \neq 0$ for some $x \in U$, but $s_x^n = 0$ in $(O_X)_x$.

On the other hand, suppose $[(s, U)]^n = 0$ in $(O_X)_x$ and $[(s, U)] \neq 0$. Then $s|_W \neq 0$ on some $x \in W \subset U$.

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Affine Local

Affine Communication Lemma.

A property $P$ enjoyed by some affine open sets of a scheme $X$ is called affine-local if

• If $\text{Spec} A \hookrightarrow X$ has property $P$ then so does $\text{Spec} A_f$.
• If $A = (f_1, \ldots, f_n)$ and if $\text{Spec} A_{f_i} \hookrightarrow X$ has property $P$ for all $i$ then so does $\text{Spec} A.$

Suppose $P$ is an affine-local property. Suppose $X = \bigcup_{i} \text{Spec} A_i$ where $\text{Spec} A_i$ all have property $P$. Then every open affine subset of $X$ has property $P$.

Reducedness is affine-local

Suppose $A$ is a ring generated by $f_1, \ldots, f_n$. Then $A$ is reduced iff each $A_i$ is.

$X$ is reduced iff X can be covered by affine open sets $\text{Spec} A$ where $A$ is reduced.

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Example: The scheme $X = \text{Spec} k[x,y]/(y^2, xy)$ is non-reduced. 

The ring $k[x,y]/(y^2, xy)$  contains the element $y \neq 0$ such that $y^2 = 0$. At the stalk of every point $p$, we will still have $y_p^2 = 0$. So $(O_X)_p$ is reduced iff $y_p = 0$. This is possible if we invert $x$, i.e. if $p \in D(x) =$ $x$-axis minus origin.

We show the origin is the only point at which $X$ is non-reduced.

Fuzzy Origin

Proof.  This is equivalent to saying that $\text{Spec} \left(k[x,y]/(y^2, xy)\right)_x$ is reduced, as 

the points of $X$ are prime ideals containing $y^2$ and $xy$, i.e. $(y)$- the $x$-axis.  We think of  $X = \text{Spec} k[x,y]/(y^2, xy)$ then as the $x$ axis with some derivatives at $(0,0)$. 

To say that the origin is the only non-reduced point is equivalent to saying that that on $x$-axis minus the origin, $X$ is reduced. In other words $O_X(D(x))$ is reduced. 

We have $A_x := \left(k[x,y]/(y^2, xy)\right)_x = k[x, y, x^{-1} ]/(y^2, xy)$. As $x$ is a unit in this ring and $xy = 0$, we must have $y = 0$ in this ring. So $A_x = k[x, y, x^{-1}]/(y) = k[x, x^{-1}]$ is reduced. 

At the origin, we have $$A_{(x, y)} = \left(k[x,y]/(y^2, xy)\right)_{(x, y)}  = \frac{k[x,y]_{(x,y)}}{(y^2, xy)}.$$

Thus $y$ is zero in $A_{(x,y)}$ iff the image of $y$ in $k[x,y]_{(x,y)}$lies in$(y^2, xy)$. Thus for some$h$outside of$(x, y) we have

$$h y = y^2 f + xy g.$$

Dividing both sides by $y$ gives $h \in (x, y)$ a contradiction. 

More Illuminating Explanation.   Remember that $\text{Spec} k[x,y]/(y^2, xy)$ is just the $x$-axis together with (the span of ) some extra derivative vectors at the origin $\displaystyle \frac{\partial}{\partial_x}, \frac{\partial}{\partial_y}, \frac{\partial^2}{\partial_{x^2}}$ (cf. Visualizing Nilpotents). For $y$ to be $0$ at the stalk of $\mathfrak{p}$ it has to be zero in a neighborhood of $\mathfrak{p}$ (where neighborhood here include the derivative vectors). So for  $y$ to be zero at the stalk of the origin, we not only need $y|_{(0,0)} = 0$ but also that $\frac{\partial}{\partial_x}, \frac{\partial}{\partial_y}$ and $\frac{\partial^2}{\partial_{x^2}}$ applied to $y$ is $0$. The later does not hold as $\frac{\partial}{\partial_y}(y) \neq 0.$  (It gives $y$, the $y$-term)

To say this rigorously, remember that $\frac{\partial}{\partial_y}(y)$ is just the image of $y$ in $k[x, y] \to k[x,y]/(y^2)$ and clearly the image of $y$ is not $0$.

Now how to formalize  "if $y$ vanishes at the stalk of the origin then $\frac{\partial}{\partial_y}(y) =0$"?  To view $\frac{\partial}{\partial_y}$ as an element of $\text{Spec} k[x,y]/(y^2, xy)$ is to have a morphism $k[x,y]/(y^2, xy) \to k[y]/(y^2)$ given by $x \mapsto 0$ and $y \mapsto y$, i.e every $F$ is mapped to its $y$-term.

 (Two polynomials $f, g \in k[x,y]$ are equal in the first ring iff $\displaystyle \frac{\partial}{\partial_x}, \frac{\partial}{\partial_y}, \frac{\partial^2}{\partial_{x^2}}$ applied to them are equal, whereas in the second ring, functions are identified if they agree under $\frac{\partial}{\partial_y}$. As $x$ has no $y$ term, it becomes $0$ in the latter.)

The map $\text{Spec} k[y]/(y^2) \to X$ induces  a pull-back map on stalks, so we have
$$\left(\frac{k[x,y]}{(y^2, xy)}\right)_{(x,y)} \to \left(\frac{k[y]}{(y^2)}\right)_{(y)}$$  

which is a local homomorphism.
If $y$ is $0$ in $\left(\frac{k[x,y]}{(y^2, xy)}\right)_{(x,y)}$ then its image in $\left(\frac{k[y]}{(y^2)}\right)_{(y)}$ must also be $0$. In other words, as the latter is just $\frac{k[y]_{(y)}}{(y^2)}$, we must have the image of  $y$ in $k[y]_{(y)}$ lie in $y^2$ and thus $yh = y^2 f$ for some $h, f \not \in (y)$. This is impossible by unique factorization in $k[y]$. Another way to see this is to see that if $y$ is $0$ in the ring above then $\frac{k[y]_{(y)}}{(y^2)} =  k$, i.e. $y^2$ is kernel of $k \to k[y]_{(y)}$ which is not possible as the latter is injective. 

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Not an open condition:

Reducedness is in general, not an open condition (see example below). However if $X$ is a locally Noetherian scheme then reducedness is open.

Example.
Let $X$ be $\text{Spec} A$ where

$$A = \text{Spec} \frac{\mathbb{C}[x, y_1, \ldots]}{(y_1^2, y_2^2, y_3^2, \ldots, (x-1)y_1, (x-2)y_2,  (x- 3)y_2, \ldots)}$$

Notice that prime ideals of $A$ are exactly prime ideals $\mathfrak{p}$ of $\mathbb{C}[x, y_1, y_2, \ldots]$ containing $(y_1^2, y_2^2, y_3^2, \ldots, (x-1)y_1, (x-2)y_2,  (x- 3)y_2, \ldots)$. These are exactly prime ideals containing $(y_1, y_2, \ldots,)$.

So $\text{Spec} A$ as a set is homeomorphic to the $$\text{Spec} \frac{\mathbb{C}[x, y_1, \ldots]}{(y_1, y_2, y_3, \ldots } = \text{Spec} \mathbb{C}[x].$$

Claim: the nonreduced points in $X$ correspond to the set $\{\text{x - m}\mid m \in \mathbb{Z}\}$ in $\text{Spec} \mathbb{C}[x]$. In particular the set of  reduced points in $\text{Spec} A$ is therefore not open.

Proof:  Indeed similar to the above example, $A_{\mathfrak{p}}$ is not reduced iff one of $y_1, y_2 \ldots$ is not zero in the stalk at $\mathfrak{p}$.

Clearly, at points $\mathfrak{p}$ where $(x-m)$ is invertible in $\mathfrak{p}$, we have $y_m = 0$ in $\mathfrak{p}$, so $y_m$ is not nilpotent. Thus $A$ is reduced at $\mathfrak{p}$ if for all $m$, $x- m$ is invertible in $\mathfrak{p}$. This is true if $x-m$ all lie outside $\mathfrak{p}$, i.e. $\mathfrak{p} \neq x-m$ for any $m$. (As $\mathfrak{p}$ is maximal).

Let $\mathfrak{p}$ correspond to $(x-m)$ for some $m$. Then in $A_{\mathfrak{p}}$, we invert $(x - n)$ for every $n\neq m$, so for all such $n$, $y_n = 0$. Thus
$$A_{\mathfrak{p}} = \text{Spec}\frac{\mathbb{C}[x,y_m]}{(y_m^2, (x-m)y_m)},$$
which reduces the problem to the previous example.

Relationship with quasi-compact schemes: Recall that for every open condition $P$, to check if all points of a  quasi-compact scheme $X$ has $P$, it suffices to check that all closed points do. Though reducedness is not an open condition, it is still possible to check the reducedness of a quasi-compact schemes by checking it at closed points.