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 $h \in A$.