(Theorem) Universal property of simple field extensions

Overview

The universal property of simple extensions says that every homomorphism $\sigma :F\to F^{\prime }$ can be extended to a homomorphism out of $F(\alpha )$, where $\alpha$ is algebraic over $F$ with minimal polynomial $f=\text{irr}(\alpha ,F)$, by sending $\alpha$ to any root of the polynomial $\sigma (f)$ in some extension field of $F’$.

This is an important preliminary result for proving the isomorphism extension theorem for simple extensions.

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Statement and proof of the theorem

By a lemma in Automorphisms and polynomial roots, we know that when $E=F(\alpha )$ is a simple extension of a field $F$ with $f=\text{irr}(\alpha ,F)$, then given a homomorphism $\sigma :F\to F’$, $F’\le K$ a subfield, and $\phi :E\to K$ an extension of $\sigma$, the element $\phi (\alpha )$ is a root of $\sigma (f)$. The following lemma states the converse.

Universal property of simple extensions

Let $E$ be a simple extension of a field $F$, meaning $E=F(\alpha )$ for some $\alpha \in E$ that is algebraic over $F$. Write $f=\text{irr}(\alpha ,F)$ for the minimal polynomial. Suppose $\sigma :F\to F’$ is a homomorphism and $K$ is an extension of $F’$. Under these conditions, if $\beta \in K$ is a root of $\sigma (f)$, then there is a unique extension of $\sigma$ to a homomorphism $\phi :E\to K$ such that $\phi (\alpha )=\beta$.

In other words, there is a bijection from the set of homomorphisms $\phi :E\to K$ that restrict to $\sigma$ on $F$ (i.e., $\phi (a)=\sigma (a)$ for all $a\in F$) to the set of roots of the polynomial $\sigma (f)$ in $K$, where $\sigma (f)\in F^{\prime }[x]$ is the polynomial obtained by applying the homomorphism $\sigma$ to the coefficients of $K$.

Proof from Modern Algebra II. Key steps are emphasized.

Recall the following basic construction: if $E=F(\alpha )$ is a simple extension, then the kernel of the evaluation homomorphism

$${\text{ev}}_{\alpha }:F[x]\to F(\alpha ),g\mapsto g(\alpha )$$

is precisely the principal ideal generated by the minimal polynomial $f=\text{irr}(\alpha ,F)$:

$$\ker ({\text{ev}}_{\alpha })=\{g\in F[x]:g(\alpha )=0\}=(f).$$

Then (Theorem) First isomorphism theorem implies that the induced map

$$\overline{{\text{ev}}_{\alpha }}:F[x]/(f)\to F(\alpha ),g+(f)\mapsto g(\alpha )$$

is an isomorphism, and there is a well-defined inverse $\rho ={\overline{{\text{ev}}_{\alpha }}}^{-1}$ satisfying

$$\rho (a)=a+(f)\text{ for all }a\in F,\rho (\alpha )=x+(f).$$

Let $\beta \in K$ be a root of $\sigma (f)$, and define the homomorphism

$${\text{ev}}_{\beta }\circ \sigma :F[x]\to F^{\prime }[x]\to K,g\mapsto \sigma (g)\mapsto \sigma (g)(\beta )$$

for all $g\in F[x]$, where $\sigma (g)$ is the polynomial obtained by applying $\sigma$ to the coefficients of $g$ and $\sigma (g)(\beta )$ is the evaluation of $\sigma (g)$ at $\beta$. By construction¹,

$$({\text{ev}}_{\beta }\circ \sigma )(a)=\sigma (a)\text{ for all }a\in F,({\text{ev}}_{\beta }\circ \sigma )(x)=\beta .$$

Moreover, the hypothesis that $\sigma (f)(\beta )=0$ implies $f\in \ker ({\text{ev}}_{\beta }\circ \sigma )$, hence $(f)\subseteq \ker ({\text{ev}}_{\beta }\circ \sigma )$, hence $(f)=\ker ({\text{ev}}_{\beta }\circ \sigma )$ since $f$ irreducible implies $(f)$ is a maximal ideal and ${\text{ev}}_{\beta }\circ \sigma \ne 0$ implies $\ker ({\text{ev}}_{\beta }\circ \sigma )$ is a proper, nontrivial ideal. This ensures that $\ker ({\text{ev}}_{\beta }\circ \sigma )$ factors through $F[x]/(f)$, meaning there is an induced isomorphism $e:F[x]\to K$. Then there exists a homomorphism $\phi =e\circ \rho$ satisfying the desired

$$\phi (a)=\sigma (a)\text{ for all }a\in F,\phi (\alpha )=\beta .$$

For uniqueness, note that every element of $E=F(\alpha )$ can be uniquely written as a polynomial in $\alpha$ with coefficients in $F$, i.e., ${\sum }_{i=0}^N{a}_i{\alpha }^i$ for some $a_i\in F$ where $N=\mathrm{deg}(f)-1$ ². Since $\phi$ is defined by its action on $F$ (via $\sigma$) and on $\alpha$, we see that its action an element of $E$ is fully determined:

$$\phi \left(\sum _{i=0}^N{a}_i{\alpha }^i\right)=\sum _{i=0}^N\phi (a_i)\phi (\alpha {)}^i=\sum _{i=0}^N\sigma (a_i){\beta }^i.$$

To summarize, for every extension $\phi$ of $\sigma$, we know that:

• $\phi (\alpha )\in K$ is a root of $\sigma (f)\in F^{\prime }[x]$, since by definition $\sigma (f)(\sigma (\alpha ))=\phi (f(\alpha ))=\phi (0)=0;$ thus, there is a well-defined function from the set of extensions $\phi$ of $\sigma$ to the set of roots of $\sigma (f)$ in $K$ given by $\phi \mapsto \phi (\alpha )$.
• $\phi$ is uniquely determined by the value $\phi (\alpha )\in K$; thus, the function is injective.
• All possible roots of $\sigma (f)$ in $K$ arise as $\phi (\alpha )$ for some extension $\phi$ of $\sigma$ (from the construction above); the function is surjective.

We conclude that when $E=F(\alpha )$ for some $\alpha \notin F$ with $f=\text{irr}(\alpha ,F)$, for any homomorphism $\sigma :F\to F’\le K$, there is a bijection from the set of extensions $E\to K$ of $\sigma$ to the set of roots of $\sigma (f)$ in $K$. $\square$

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Examples

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Footnotes

1. For $({\text{ev}}_{\beta }\circ \sigma )(a)=\sigma (a)$, note that $a\in F$ is a constant (i.e., degree 0) polynomial in $F[x]$, so $\sigma (a)\in F’$ is a constant polynomial at $F’[x]$, and therefore evaluates to itself on $\beta$. For $({\text{ev}}_{\beta }\circ \sigma )(x)=\beta$, note that the degree 1 polynomial $x$ has coefficients $a_1=1$ and $a_i=0$ for all $i\ne 1$. Since $\sigma$ is a homomorphism, it preserves these identities, so we have $\sigma (0)=0$ and $\sigma (1)=1$ in $F’$ and hence $\sigma (x)=x\in F’[x]$. Evaluating $x$ at $\beta$ is clearly $\beta$. ↩

2. The upper bound $N=\mathrm{deg}(f)-1$ comes from the fact that higher powers of $\alpha$ can be reduced using $f(\alpha )=0$. concept-question WHY? ↩