(Theorem) The number of extensions of a field homomorphism is at most the degree of the field extension

Overview

The isomorphism extension theorem says that if $F,F’$ are fields with finite extensions $E,K$, respectively, then any ring homomorphism $\sigma :F\to F’$ has at most $[E:F]$ extensions to homomorphisms $E\to K$, with equality if $F$ is of characteristic zero, finite, or perfect.

The proof of the theorem for simple extensions is a direct corollary of (Theorem) Universal property of simple field extensions, and the proof for general finite extensions proceeds by induction.

The theorem is especially interesting when $F’=F$ and $\sigma ={\text{id}}_F$, in which case it counts the automorphisms of $K=E$ that fix $F$:

$$\#(\text{Gal}(E/F))\le [E:F].$$

In particular, this is an equality when $E$ is a Galois extension of $F$. This is an important preliminary result for (Theorem) Fundamental theorem of Galois theory, which links intermediate fields to subgroups. wip

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Preliminaries: Isomorphism extension theorem for simple extensions

Isomorphism extension theorem for simple extensions

Let $E$ be a simple extension of a field $F$, meaning $E=F(\alpha )$ for some $\alpha \in E$ where $\alpha$ is algebraic over $F$, and let $\sigma :F\to F’$ be a homomorphism. Then we have the following:

• (i) For every extension $K$ of $F’$, there exist at most $[E:F]$ homomorphisms $\phi :E\to K$ extending $\sigma$, i.e., such that $\phi (\alpha )=\sigma (\alpha )$ for all $\alpha \in F$.
• (ii) There exists an extension field $K$ of $F’$ and a homomorphism $\phi :E\to K$ extending $\sigma$.
• (iii) If $F$ has characteristic $0$ (or $F$ is finite or perfect) and $E’$ is an extension field of $F’$, then there exists an extension field $K$ of $E’$ such that there are exactly $[E:F]$ homomorphisms $\phi :E\to K$ extending $\sigma$.

Proof from Modern Algebra II.

• (i) From (Theorem) Universal property of simple field extensions, we know that the extensions of $\sigma$ to a homomorphism $F(\alpha )\to K$ are in one-to-one correspondence with the $\beta \in K$ where $\beta$ is a root of $\sigma (f)$, with $f=\text{irr}(\alpha ,F)$. Since $\sigma (f)$ has at most $\mathrm{deg}(f)=\mathrm{deg}(\text{irr}(\alpha ,F))=[E:F]$ roots in any extension field $K$ of $F’$, this implies that there are at most $[E:F]$ such extensions.
• (ii) Since (Theorem) Every nonconstant polynomial over a field has a root in some extension field, we can choose an extension field $K$ of $F’$ such that $\sigma (f)$ has a root $\beta$ in $K$. Thus (Theorem) Universal property of simple field extensions implies that there will be at least one homomorphism $\phi :F(\alpha )\to K$ extending $\sigma$.
• (iii) From a corollary of (Theorem) Every nonconstant polynomial over a field has a root in some extension field, we can choose an extension field $K$ of $F’$ such that $\sigma (f)$ factors into a product of linear factors in $K[x]$. If $F$ is perfect (i.e., $F$ is finite, has characteristic 0, or has characteristic $p>0$ and the Frobenius homomorphism is surjective), then from a result of Formal derivatives on polynomial rings we know that the irreducible polynomial $f$ has no multiple roots in any extension field. Then there are $\mathrm{deg}(f)$ distinct roots of $f$ in $J$, and hence (Theorem) Universal property of simple field extensions implies that there are $\mathrm{deg}(f)$ different extensions of $\sigma$ to a homomorphism $F(\alpha )\to K$. $\square$

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Statement and proof of the theorem

Isomorphism extension theorem

Let $E$ be a finite extension of a field $F$, and suppose $\sigma :F\to F’$ is a homomorphism to another field $F’$. Then we have the following:

• (i) For every extension field $K$ of $F’$, there exist at most $[E:F]$ homomorphisms $\phi :E\to K$ extending $\sigma$, i.e., such that $\phi (\alpha )=\sigma (\alpha )$ for all $\alpha \in F$.
• (ii) There exists an extension field $K$ of $F’$ and a homomorphism $\phi :E\to K$ extending $\sigma$.
• (iii) If $F$ has characteristic $0$ (or $F$ is finite or perfect) and $E’$ is an extension field of $F’$, then there exists an extension field $K$ of $E’$ such that there are exactly $[E:F]$ homomorphisms $\phi :E\to K$ extending $\sigma$.

Proof from Modern Algebra II. Let

$$E=F({\alpha }_1,\dots ,{\alpha }_n),{\alpha }_i\in E$$

be a finite extension. We proceed by induction on $n$, with the base case $n=1$ already proven for simple extensions. Let $F_1=F({\alpha }_1,\dots ,{\alpha }_{n-1})$ and $\alpha ={\alpha }_n$, so we view $E=F_1(\alpha )$ as a simple extension, and we have a sequence of extensions $F\le F_1\le E$. We first observe that that any extension $\phi :F_1\to K$ of $\sigma$ can itself be extended to a homomorphism $\psi :E\to K$, where $\psi$ is clearly also an extension of $\sigma$; conversely, if $\psi :E\to K$ is an extension of $\sigma$, then we can find an extension of $\sigma$ to a homomorphism $\psi :F_1\to K$ by simply taking the restriction $\psi {|}_{F_1}=\phi$.

• (i) By the inductive hypothesis, any homomorphism $\sigma :F\to F^{\prime }$ has at most $[F_1:F]$ extensions to homomorphisms $F_1\to K$; let $\{{\phi }_1,\dots ,{\phi }_m\},m\le [F_1:F]$ be the set of all such extensions. For each ${\phi }_i:F_1\to K$, the $n=1$ case implies there can be at most $[E:F_1]=[F_1(\alpha ):F_1]$ extensions of ${\phi }_i$ to homomorphisms $E=F_1(\alpha )\to K$. Then summing this for all $m$ extensions and applying the tower law for finite extension degrees, we see that there are $[F_1:F][F(\alpha ):F_1]=[E:F_1][F_1:F]=[E:F]$ extensions of $\sigma$ to homomorphisms $E\to K$, as claimed.
• (ii) The inductive hypothesis implies that for $\sigma :F\to F’$, there is an extension ${\phi }_1:F_1\to K_1$, where $K_1$ is some field containing $F’$. Let $f_1=\text{irr}(\alpha ,F_1)$. By (Theorem) Every nonconstant polynomial over a field has a root in some extension field, we can find an extension $K$ of $K_1$ such that ${\sigma }_1(f_1)$ has a root in $K$ (e.g., by formally adjoining a root $\beta$ to define $K=K_1(\beta )\cong K_1[x]/({\sigma }_1(f_1))$, if there is not a root in $K_1$ already). Then it follows from the $n=1$ case that there is an extension of ${\phi }_1$ to a homomorphism $\phi :E=F(\alpha )\to K$, hence extending $\sigma$ as well.

wip

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Corollary: An upper bound for the size of the Galois group

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Code snippets

\text{ev}_\beta \circ \sigma