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 $K_S/k(t)$.