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## 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). \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}, \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}, \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. \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. 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 $E\text{.}$ 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 $E\text{.}$