## Section21.2Splitting Fields

Let $$F$$ be a field and $$p(x)$$ be a nonconstant polynomial in $$F[x]\text{.}$$ We already know that we can find a field extension of $$F$$ that contains a root of $$p(x)\text{.}$$ 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 $$p(x)\text{?}$$

Let $$F$$ be a field and $$p(x) = a_0 + a_1 x + \cdots + a_n x^n$$ be a nonconstant polynomial in $$F[x]\text{.}$$ 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)\text{.} \end{equation*}

A polynomial $$p(x) \in F[x]$$ splits in $$E$$ if it is the product of linear factors in $$E[x]\text{.}$$

###### Example21.29.

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

###### Example21.30.

Let $$p(x) = x^3 - 3$$ be in $${\mathbb Q}[x]\text{.}$$ Then $$p(x)$$ has a root in the field $${\mathbb Q}( \sqrt{3}\, )\text{.}$$ However, this field is not a splitting field for $$p(x)$$ since the complex cube roots of 3,

\begin{equation*} \frac{ -\sqrt{3} \pm (\sqrt{3}\, )^5 i }{2}\text{,} \end{equation*}

are not in $${\mathbb Q}( \sqrt{3}\, )\text{.}$$

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 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 $$\deg q(x) = n -1\text{,}$$ 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,

\begin{equation*} E = K(\alpha_2, \ldots, \alpha_n) = F(\alpha_1, \ldots, \alpha_n) \end{equation*}

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

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 $$p(x) \in F[x]\text{,}$$ there exists a field isomorphism $$\phi : K \rightarrow L$$ that preserves $$F\text{.}$$ In order to prove this result, we must first prove a lemma.

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 $$1, \alpha, \ldots, \alpha^{n - 1}\text{.}$$ Therefore, the isomorphism that we are seeking must be

\begin{equation*} \overline{\phi}( a_0 + a_1 \alpha + \cdots + a_{n - 1} \alpha^{n - 1}) = \phi(a_0) + \phi(a_1) \beta + \cdots + \phi(a_{n - 1}) \beta^{n - 1}\text{,} \end{equation*}

where

\begin{equation*} a_0 + a_1 \alpha + \cdots + a_{n - 1} \alpha^{n - 1} \end{equation*}

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

\begin{equation*} \phi( a_0 + a_1 x + \cdots + a_n x^n ) = \phi( a_0 ) + \phi(a_1) x + \cdots + \phi(a_n) x^n\text{.} \end{equation*}

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 21.12, 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 (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 $$\deg p(x) = 1\text{,}$$ 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 $$\overline{\phi} : E(\alpha ) \rightarrow F( \beta)$$ such that $$\overline{\phi}( \alpha ) = \beta$$ and $$\overline{\phi}$$ agrees with $$\phi$$ 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 $$\psi : K \rightarrow L$$ 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 $$E\text{.}$$