Abstract Algebra: Theory and Applications

We will use mathematical induction on the degree of $p(x)\text{.}$ If $\mathrm{deg}p(x)=1\text{,}$ then $p(x)$ is a linear polynomial and $E=F\text{.}$ Assume that the theorem is true for all polynomials of degree $k$ with $1\le k<n$ and let $\mathrm{deg}p(x)=n\text{.}$ We can assume that $p(x)$ is irreducible; otherwise, by our induction hypothesis, we are done. By Theorem 21.5, there exists a field $K$ such that $p(x)$ has a zero ${\alpha }_1$ in $K\text{.}$ Hence, $p(x)=(x-{\alpha }_1)q(x)\text{,}$ where $q(x)\in K[x]\text{.}$ Since $\mathrm{deg}q(x)=n-1\text{,}$ there exists a splitting field $E\supset K$ of $q(x)$ that contains the zeros ${\alpha }_2,\dots ,{\alpha }_n$ of $p(x)$ by our induction hypothesis. Consequently,

is a splitting field of $p(x)\text{.}$

If $p(x)$ has degree $n\text{,}$ then by Theorem 21.13 we can write any element in $E(\alpha )$ as a linear combination of $1,\alpha ,\dots ,{\alpha }^{n-1}\text{.}$ Therefore, the isomorphism that we are seeking must be

where

is an element in $E(\alpha )\text{.}$ The fact that $\bar{\varphi }$ is an isomorphism could be checked by direct computation; however, it is easier to observe that $\bar{\varphi }$ is a composition of maps that we already know to be isomorphisms.

We can extend $\varphi$ to be an isomorphism from $E[x]$ to $F[x]\text{,}$ which we will also denote by $\varphi \text{,}$ by letting

This extension agrees with the original isomorphism $\varphi :E\to F\text{,}$ since constant polynomials get mapped to constant polynomials. By assumption, $\varphi (p(x))=q(x)\text{;}$ hence, $\varphi$ maps $⟨p(x)⟩$ onto $⟨q(x)⟩\text{.}$ Consequently, we have an isomorphism $\psi :E[x]/⟨p(x)⟩\to F[x]/⟨q(x)⟩\text{.}$ By Proposition 21.12, we have isomorphisms $\sigma :E[x]/⟨p(x)⟩\to E(\alpha )$ and $\tau :F[x]/⟨q(x)⟩\to F(\beta )\text{,}$ defined by evaluation at $\alpha$ and $\beta \text{,}$ respectively. Therefore, $\bar{\varphi }=\tau \psi {\sigma }^{-1}$ is the required isomorphism (see Figure 21.33).

We leave the proof of uniqueness as a exercise.

We will use mathematical induction on the degree of $p(x)\text{.}$ We can assume that $p(x)$ is irreducible over $E\text{.}$ Therefore, $q(x)$ is also irreducible over $F\text{.}$ If $\mathrm{deg}p(x)=1\text{,}$ then by the definition of a splitting field, $K=E$ and $L=F$ and there is nothing to prove.

Assume that the theorem holds for all polynomials of degree less than $n\text{.}$ Since $K$ is a splitting field of $p(x)\text{,}$ all of the roots of $p(x)$ are in $K\text{.}$ Choose one of these roots, say $\alpha \text{,}$ such that $E\subset E(\alpha )\subset K\text{.}$ Similarly, we can find a root $\beta$ of $q(x)$ in $L$ such that $F\subset F(\beta )\subset L\text{.}$ By Lemma 21.32, there exists an isomorphism $\bar{\varphi }:E(\alpha )\to F(\beta )$ such that $\bar{\varphi }(\alpha )=\beta$ and $\bar{\varphi }$ agrees with $\varphi$ on $E$ (see Figure 21.35).

Now write $p(x)=(x-\alpha )f(x)$ and $q(x)=(x-\beta )g(x)\text{,}$ where the degrees of $f(x)$ and $g(x)$ are less than the degrees of $p(x)$ and $q(x)\text{,}$ respectively. The field extension $K$ is a splitting field for $f(x)$ over $E(\alpha )\text{,}$ and $L$ is a splitting field for $g(x)$ over $F(\beta )\text{.}$ By our induction hypothesis there exists an isomorphism $\psi :K\to L$ such that $\psi$ agrees with $\bar{\varphi }$ on $E(\alpha )\text{.}$ Hence, there exists an isomorphism $\psi :K\to L$ such that $\psi$ agrees with $\varphi$ on $E\text{.}$