Abstract Algebra: Theory and Applications, Remixed!

Proof.

We will use mathematical induction on the degree of \(p(x)\text{.}\) If \(\deg p(x) = 1\text{,}\) then \(p(x)\) is a linear polynomial and \(E = F\text{.}\) Assume that the theorem is true for all polynomials of degree \(k\) with \(1 \leq k \lt n\) and let \(\deg p(x) = n\text{.}\) We can assume that \(p(x)\) is irreducible; otherwise, by our induction hypothesis, we are done. By Theorem 8.10, 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 \(\deg q(x) = n -1\text{,}\) there exists a splitting field \(E \supset K\) of \(q(x)\) that contains the zeros \(\alpha_2, \ldots, \alpha_n\) of \(p(x)\) by our induction hypothesis. Consequently,

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

Proof.

If \(p(x)\) has degree \(n\text{,}\) then by Theorem 8.18 we can write any element in \(E( \alpha )\) as a linear combination of \(1, \alpha, \ldots, \alpha^{n - 1}\text{.}\) Therefore, the isomorphism that we are seeking must be

where

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

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

This extension agrees with the original isomorphism \(\phi : E \rightarrow F\text{,}\) since constant polynomials get mapped to constant polynomials. By assumption, \(\phi(p(x)) = q(x)\text{;}\) hence, \(\phi\) maps \(\langle p(x) \rangle\) onto \(\langle q(x) \rangle\text{.}\) Consequently, we have an isomorphism \(\psi : E[x] / \langle p(x) \rangle \rightarrow F[x]/\langle q(x) \rangle\text{.}\) By Proposition 8.17, we have isomorphisms \(\sigma: E[x]/\langle p(x) \rangle \rightarrow E(\alpha)\) and \(\tau : F[x]/\langle q(x) \rangle \rightarrow F( \beta )\text{,}\) defined by evaluation at \(\alpha\) and \(\beta\text{,}\) respectively. Therefore, \(\overline{\phi} = \tau \psi \sigma^{-1}\) is the required isomorphism.

We leave the proof of uniqueness as a exercise.

Proof.

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 \(\deg p(x) = 1\text{,}\) 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 9.4, there exists an isomorphism \(\overline{\phi} : E(\alpha ) \rightarrow F( \beta)\) such that \(\overline{\phi}( \alpha ) = \beta\) and \(\overline{\phi}\) agrees with \(\phi\) on \(E\text{.}\)

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 \(\psi : K \rightarrow L\) such that \(\psi\) agrees with \(\overline{\phi}\) on \(E( \alpha)\text{.}\) Hence, there exists an isomorphism \(\psi : K \rightarrow L\) such that \(\psi\) agrees with \(\phi\) on \(E\text{.}\)