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Section21.2Splitting Fields

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 \(E[x]\).

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 \(p(x)\).

Example21.30

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 \({\mathbb Q}( \sqrt[3]{3}\, )\).

Proof

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.

Proof