2.2 Splitting fields

Let $f$ be a polynomial with coefficients in $F$ . A field $E$ containing $F$ is said to split $f$ if $f$ splits in $E[X]$ , i.e.,

f(X)=a\prod\nolimits_{i=1}^{m}(X-\alpha_{i})\text{ with all }\alpha_{i}\in E.

If $E$ splits $f$ and is generated by the roots of $f$ ,

E=F[\alpha_{1},\ldots,\alpha_{m}],

then it is called a splitting or root field for $f$ .

Note that $\prod f_{i}(X)^{m_{i}}$ ( $m_{i}\geq 1$ ) and $\prod f_{i}(X)$ have the same splitting fields. Note also that $f$ splits in $E$ if it has $\deg(f)-1$ roots in $E$ because the sum of the roots of $f$ lies in $F$ (if $f=aX^{m}+a_{1}X^{m-1}+\cdots$ , then $\sum\alpha_{i}=-a_{1}/a$ ).

Example 2.3

(a) Let $f(X)=aX^{2}+bX+c\in\mathbb{Q}{}[X]$ , and let $\alpha=\sqrt{b^{2}-4ac}$ . The subfield $\mathbb{Q}[\alpha]$ of $\mathbb{C}$ is a splitting field for $f$ .

(b) Let $f(X)=X^{3}+aX^{2}+bX+c\in\mathbb{Q}[X]$ be irreducible, and let $\alpha_{1},\alpha_{2},\alpha_{3}$ be its roots in $\mathbb{C}$ . Then $\mathbb{Q}[\alpha_{1},\alpha_{2},\alpha_{3}]=\mathbb{Q}[\alpha_{1},\alpha_{2}]$ is a splitting field for $f(X)$ . Note that $[\mathbb{Q}[\alpha_{1}]\colon\mathbb{Q}]=3$ and that $[\mathbb{Q}[\alpha_{1},\alpha_{2}]\colon\mathbb{Q}[\alpha_{1}]]=1$ or $2$ , and so $[\mathbb{Q}[\alpha_{1},\alpha_{2}]\colon\mathbb{Q}]=3$ or $6$ . We’ll see later (LABEL:cg2) that the degree is $3$ if and only if the discriminant of $f(X)$ is a square in $\mathbb{Q}{}$ . For example, the discriminant of $X^{3}+bX+c$ is $-4b^{3}-27c^{2}$ , and so the splitting field of $X^{3}+10X+1$ (discriminant $-4027)$ has degree $6$ over $\mathbb{Q}$ .

Proposition 2.4

Every polynomial $f\in F[X]$ has a splitting field $E_{f}$ , and

[E_{f}\colon F]\leq(\deg f)!\quad(\text{factorial }\deg f).

Proof.

Let $F_{1}=F[\alpha_{1}]$ be a stem field for some monic irreducible factor of $f$ in $F[X]$ . Then $f(\alpha_{1})=0$ , and we let $F_{2}=F_{1}[\alpha_{2}]$ be a stem field for some monic irreducible factor of $f(X)/(X-\alpha_{1})$ in $F_{1}[X]$ . Continuing in this fashion, we arrive at a splitting field $E_{f}$ . Let $n=\deg f$ . Then $[F_{1}\colon F]=\deg g_{1}\leq n$ , $[F_{2}\colon F_{1}]\leq n-1,...$ , and so $[E_{f}\colon F]\leq n!$ . _□

Aside 2.5

Let $F$ be a field. For a given integer $n$ , there may or may not exist polynomials of degree $n$ in $F[X]$ whose splitting field has degree $n!$ — this depends on $F$ . For example, there do not exist such polynomials for $n>1$ if $F=\mathbb{C}$ (see LABEL:ag5), nor for $n>2$ if $F=\mathbb{R}$ or $F=\mathbb{F}_{p}$ (see LABEL:cg18). However, later (LABEL:cg24) we’ll see how to write down infinitely many polynomials of degree $n$ in $\mathbb{Q}[X]$ with splitting fields of degree $n!$ .

Example 2.6

(a) Let $f(X)=(X^{p}-1)/(X-1)\in\mathbb{Q}{}[X]$ , $p$ prime. If $\zeta$ is one root of $f$ , then the remaining roots are $\zeta^{2},\zeta^{3},\ldots,\zeta^{p-1}$ , and so the splitting field of $f$ is $\mathbb{Q}{}[\zeta]$ .

(b) Let $F$ have characteristic $p\neq 0$ , and let $f=X^{p}-X-a\in F[X]$ . If $\alpha$ is one root of $f$ in some extension of $F$ , then the remaining roots are $\alpha+1,...,\alpha+p-1$ , and so the splitting field of $f$ is $F[\alpha]$ .

(c) If $\alpha$ is one root of $X^{n}-a$ , then the remaining roots are all of the form $\zeta\alpha$ , where $\zeta^{n}=1$ . Therefore, $F[\alpha]$ is a splitting field for $X^{n}-a$ if and only if $F$ contains all the $n$ th roots of $1$ (by which we mean that $X^{n}-1$ splits in $F[X]$ ). Note that if $p$ is the characteristic of $F$ , then $X^{p}-1=(X-1)^{p}$ , and so $F$ automatically contains all the $p$ th roots of $1$ .

Proposition 2.7

Let $f\in F[X]$ . Let $E$ be an extension of $F$ generated by the roots of $f$ in $E$ , and let $\Omega$ be an extension of $F$ splitting $f$ .

(a)

There exists an $F$ -homomorphism $\varphi\colon E\rightarrow\Omega$ ; the number of such homomorphisms is at most $[E\colon F]$ , and equals $[E\colon F]$ if $f$ has distinct roots in $\Omega$ .
(b)

If $E$ and $\Omega$ are both splitting fields for $f$ , then every $F$ -homomorphism $E\rightarrow\Omega$ is an isomorphism. In particular, any two splitting fields for $f$ are $F$ -isomorphic.

As $f$ splits in $\Omega[X]$ , $f(X)=a\prod\nolimits_{i=1}^{\deg(f)}(X-\beta_{i})$ with $\beta_{1},\beta_{2},\ldots\in\Omega$ . To say that $f$ has distinct roots in $\Omega$ means that $\beta_{i}\neq\beta_{j}$ if $i\neq j$ .

Proof.

We may suppose that $f$ is monic.

We begin with an observation: let $F$ , $f$ , and $\Omega$ be as in the statement of the proposition, let $L$ be a subfield of $\Omega$ containing $F$ , and let $g$ be a monic factor of $f$ in $L[X]$ ; as $g$ divides $f$ in $\Omega[X]$ , it is a product of certain number of the factors $X-\beta_{i}$ of $f$ in $\Omega[X]$ ; in particular, we see that $g$ splits in $\Omega$ , and that it has distinct roots in $\Omega$ if $f$ does..

(a) By hypothesis, $E=F[\alpha_{1},...,\alpha_{m}]$ with each $\alpha_{i}$ a root of $f(X)$ in $E$ . The minimal polynomial of $\alpha_{1}$ is an irreducible polynomial $f_{1}$ dividing $f$ . From the initial observation with $L=F$ , we see that $f_{1}$ splits in $\Omega$ , and that its roots are distinct if the roots of $f$ are distinct. According to Proposition 2.1, there exists an $F$ -homomorphism $\varphi_{1}\colon F[\alpha_{1}]\rightarrow\Omega$ , and the number of such homomorphisms is at most $[F[\alpha_{1}]\colon F]$ , with equality holding when $f$ has distinct roots in $\Omega$ .

The minimal polynomial of $\alpha_{2}$ over $F[\alpha_{1}]$ is an irreducible factor $f_{2}$ of $f$ in $F[\alpha_{1}][X]$ . On applying the initial observation with $L=\varphi_{1}F[\alpha_{1}]$ and $g=\varphi_{1}f_{2}$ , we see that $\varphi_{1}f_{2}$ splits in $\Omega$ , and that its roots are distinct if the roots of $f$ are distinct. According to Proposition 2.2, each $\varphi_{1}$ extends to a homomorphism $\varphi_{2}\colon F[\alpha_{1},\alpha_{2}]\rightarrow\Omega$ , and the number of extensions is at most $[F[\alpha_{1},\alpha_{2}]\colon F[\alpha_{1}]]$ , with equality holding when $f$ has distinct roots in $\Omega.$

On combining these statements we conclude that there exists an $F$ -homomorphism

\varphi\colon F[\alpha_{1},\alpha_{2}]\rightarrow\nolinebreak\Omega,

and that the number of such homomorphisms is at most $[F[\alpha_{1},\alpha_{2}]\colon F]$ , with equality holding if $f$ has distinct roots in $\Omega.$

After repeating the argument $m$ times, we obtain (a).

(b) Every $F$ -homomorphism $E\rightarrow\Omega$ is injective, and so, if there exists such a homomorphism, then $[E\colon F]\leq[\Omega\colon F]$ . If $E$ and $\Omega$ are both splitting fields for $f$ , then (a) shows that there exist homomorphisms $E\leftrightarrows\Omega$ , and so $[E\colon F]=[\Omega\colon F]$ . It follows that every $F$ -homomorphism $E\rightarrow\Omega$ is an $F$ -isomorphism. _□

Corollary 2.8

Let $E$ and $L$ be extension of $F$ , with $E$ finite over $F$ .

(a)

The number of $F$ -homomorphisms $E\rightarrow L$ is at most $[E\colon F]$ .
(b)

There exists a finite extension $\Omega/L$ and an $F$ -homomorphism $E\rightarrow\Omega.$

Proof.

Write $E=F[\alpha_{1},\ldots,\alpha_{m}]$ , and let $f\in F[X]$ be the product of the minimal polynomials of the $\alpha_{i}$ ; thus $E$ is generated over $F$ by roots of $f$ . Let $\Omega$ be a splitting field for $f$ regarded as an element of $L[X]$ . The proposition shows that there exists an $F$ -homomorphism $E\rightarrow\Omega$ , and the number of such homomorphisms is $\leq[E\colon F]$ . This proves (b), and since an $F$ -homomorphism $E\rightarrow L$ can be regarded as an $F$ -homomorphism $E\rightarrow\Omega$ , it also proves (a). _□

Remark 2.9

(a) Let $E_{1},E_{2},\ldots,E_{m}$ be finite extensions of $F$ , and let $L$ be an extension of $F$ . From the corollary we see that there exists a finite extension $L_{1}/L$ such that $L_{1}$ contains an isomorphic image of $E_{1}$ ; then that there exists a finite extension $L_{2}/L_{1}$ such that $L_{2}$ contains an isomorphic image of $E_{2}$ . On continuing in this fashion, we find that there exists a finite extension $\Omega$ / $L$ such that $\Omega$ contains an isomorphic copy of every $E_{i}$ .

(b) Let $f\in F[X]$ . If $E$ and $E^{\prime}$ are both splitting fields of $f$ , then we know there exists an $F$ -isomorphism $E\rightarrow E^{\prime}$ , but there will in general be no preferred such isomorphism. Error and confusion can result if the fields are simply identified. Also, it makes no sense to speak of “the field $F[\alpha]$ generated by a root of $f$ ” unless $f$ is irreducible (the fields generated by the roots of two different factors are unrelated). Even when $f$ is irreducible, it makes no sense to speak of “the field $F[\alpha,\beta]$ generated by two roots $\alpha,\beta$ of $f$ ” (the extensions of $F[\alpha]$ generated by the roots of two different factors of $f$ in $F[\alpha][X]$ may be very different).