##### Example21.29

Let \(p(x) = x^4 + 2x^2 - 8\) be in \({\mathbb Q}[x]\). Then \(p(x)\) has irreducible factors \(x^2 -2\) and \(x^2 + 4\). Therefore, the field \({\mathbb Q}( \sqrt{2}, i )\) is a splitting field for \(p(x)\).

Let \(F\) be a field and \(p(x)\) be a nonconstant polynomial in \(F[x]\). We already know that we can find a field extension of \(F\) that contains a root of \(p(x)\). However, we would like to know whether an extension \(E\) of \(F\) containing all of the roots of \(p(x)\) exists. In other words, can we find a field extension of \(F\) such that \(p(x)\) factors into a product of linear polynomials? What is the “smallest” extension containing all the roots of \(p(x)\)?

Let \(F\) be a field and \(p(x) = a_0 + a_1 x + \cdots + a_n x^n\) be a nonconstant polynomial in \(F[x]\). An extension field \(E\) of \(F\) is a *splitting field* of \(p(x)\) if there exist elements \(\alpha_1, \ldots, \alpha_n\) in \(E\) such that \(E = F( \alpha_1, \ldots, \alpha_n )\) and
\begin{equation*}p(x) = ( x - \alpha_1 )(x - \alpha_2) \cdots (x - \alpha_n).\end{equation*}
A polynomial \(p(x) \in F[x]\) *splits* in \(E\) if it is the product of linear factors in \(E[x]\).

Let \(p(x) = x^4 + 2x^2 - 8\) be in \({\mathbb Q}[x]\). Then \(p(x)\) has irreducible factors \(x^2 -2\) and \(x^2 + 4\). Therefore, the field \({\mathbb Q}( \sqrt{2}, i )\) is a splitting field for \(p(x)\).

Let \(p(x) = x^3 -3\) be in \({\mathbb Q}[x]\). Then \(p(x)\) has a root in the field \({\mathbb Q}( \sqrt[3]{3}\, )\). However, this field is not a splitting field for \(p(x)\) since the complex cube roots of 3, \begin{equation*}\frac{ -\sqrt[3]{3} \pm (\sqrt[6]{3}\, )^5 i }{2},\end{equation*} are not in \({\mathbb Q}( \sqrt[3]{3}\, )\).

Let \(p(x) \in F[x]\) be a nonconstant polynomial. Then there exists a splitting field \(E\) for \(p(x)\).

The question of uniqueness now arises for splitting fields. This question is answered in the affirmative. Given two splitting fields \(K\) and \(L\) of a polynomial \(p(x) \in F[x]\), there exists a field isomorphism \(\phi : K \rightarrow L\) that preserves \(F\). In order to prove this result, we must first prove a lemma.

Let \(\phi : E \rightarrow F\) be an isomorphism of fields. Let \(K\) be an extension field of \(E\) and \(\alpha \in K\) be algebraic over \(E\) with minimal polynomial \(p(x)\). Suppose that \(L\) is an extension field of \(F\) such that \(\beta\) is root of the polynomial in \(F[x]\) obtained from \(p(x)\) under the image of \(\phi\). Then \(\phi\) extends to a unique isomorphism \(\overline{\phi} : E( \alpha ) \rightarrow F( \beta )\) such that \(\overline{\phi}( \alpha ) = \beta\) and \(\overline{\phi}\) agrees with \(\phi\) on \(E\).

Let \(\phi : E \rightarrow F\) be an isomorphism of fields and let \(p(x)\) be a nonconstant polynomial in \(E[x]\) and \(q(x)\) the corresponding polynomial in \(F[x]\) under the isomorphism. If \(K\) is a splitting field of \(p(x)\) and \(L\) is a splitting field of \(q(x)\), then \(\phi\) extends to an isomorphism \(\psi : K \rightarrow L\).

Let \(p(x)\) be a polynomial in \(F[x]\). Then there exists a splitting field \(K\) of \(p(x)\) that is unique up to isomorphism.