Lemma 9.16.1. Let $F$ be a field. Let $P \in F[x]$ be a nonconstant polynomial. There exists a smallest field extension $E/F$ such that $P$ splits completely over $E$. Moreover, the field extension $E/F$ is normal and unique up to (nonunique) isomorphism.

## 9.16 Splitting fields

The following lemma is a useful tool for constructing normal field extensions.

**Proof.**
Choose an algebraic closure $\overline{F}$. Then we can write $P = c (x - \beta _1) \ldots (x - \beta _ n)$ in $\overline{F}[x]$, see Lemma 9.10.2. Note that $c \in F^*$. Set $E = F(\beta _1, \ldots , \beta _ n)$. Then it is clear that $E$ is minimal with the requirement that $P$ splits completely over $E$.

Next, let $E'$ be another minimal field extension of $F$ such that $P$ splits completely over $E'$. Write $P = c (x - \alpha _1) \ldots (x - \alpha _ n)$ with $c \in F$ and $\alpha _ i \in E'$. Again it follows from minimality that $E' = F(\alpha _1, \ldots , \alpha _ n)$. Moreover, if we pick any $\sigma : E' \to \overline{F}$ (Lemma 9.10.5) then we immediately see that $\sigma (\alpha _ i) = \beta _{\tau (i)}$ for some permutation $\tau : \{ 1, \ldots , n\} \to \{ 1, \ldots , n\} $. Thus $\sigma (E') = E$. This implies that $E'$ is a normal extension of $F$ by Lemma 9.15.5 and that $E \cong E'$ as extensions of $F$ thereby finishing the proof. $\square$

Definition 9.16.2. Let $F$ be a field. Let $P \in F[x]$ be a nonconstant polynomial. The field extension $E/F$ constructed in Lemma 9.16.1 is called the *splitting field of $P$ over $F$*.

Lemma 9.16.3. Let $E/F$ be a finite extension of fields. There exists a unique smallest finite extension $K/E$ such that $K$ is normal over $F$.

**Proof.**
Choose generators $\alpha _1, \ldots , \alpha _ n$ of $E$ over $F$. Let $P_1, \ldots , P_ n$ be the minimal polynomials of $\alpha _1, \ldots , \alpha _ n$ over $F$. Set $P = P_1 \ldots P_ n$. Observe that $(x - \alpha _1) \ldots (x - \alpha _ n)$ divides $P$, since each $(x - \alpha _ i)$ divides $P_ i$. Say $P = (x - \alpha _1) \ldots (x - \alpha _ n)Q$. Let $K/E$ be the splitting field of $P$ over $E$. We claim that $K$ is the splitting field of $P$ over $F$ as well (which implies that $K$ is normal over $F$). This is clear because $K/E$ is generated by the roots of $Q$ over $E$ and $E$ is generated by the roots of $(x - \alpha _1) \ldots (x - \alpha _ n)$ over $F$, hence $K$ is generated by the roots of $P$ over $F$.

Uniqueness. Suppose that $K'/E$ is a second smallest extension such that $K'/F$ is normal. Choose an algebraic closure $\overline{F}$ and an embedding $\sigma _0 : E \to \overline{F}$. By Lemma 9.10.5 we can extend $\sigma _0$ to $\sigma : K \to \overline{F}$ and $\sigma ' : K' \to \overline{F}$. By Lemma 9.15.3 we see that $\sigma (K) \cap \sigma '(K')$ is normal over $F$. By minimality we conclude that $\sigma (K) = \sigma (K')$. Thus $\sigma \circ (\sigma ')^{-1} : K' \to K$ gives an isomorphism of extensions of $E$. $\square$

Definition 9.16.4. Let $E/F$ be a finite extension of fields. The field extension $K/E$ constructed in Lemma 9.16.3 is called the *normal closure $E$ over $F$*.

One can construct the normal closure inside any given normal extension.

Lemma 9.16.5. Let $L/K$ be an algebraic normal extension.

If $L/M/K$ is a subextension with $M/K$ finite, then there exists a tower $L/M'/M/K$ with $M'/K$ finite and normal.

If $L/M'/M/K$ is a tower with $M/K$ normal and $M'/M$ finite, then there exists a tower $L/M''/M'/M/K$ with $M''/M$ finite and $M''/K$ normal.

**Proof.**
Proof of (1). Let $M'$ be the smallest subextension of $L/K$ containing $M$ which is normal over $K$. By Lemma 9.16.3 this is the normal closure of $M/K$ and is finite over $K$.

Proof of (2). Let $\alpha _1, \ldots , \alpha _ n \in M'$ generate $M'$ over $M$. Let $P_1, \ldots , P_ n$ be the minimal polynomials of $\alpha _1, \ldots , \alpha _ n$ over $K$. Let $\alpha _{i, j}$ be the roots of $P_ i$ in $L$. Let $M'' = M(\alpha _{i, j})$. It follows from Lemma 9.15.6 (applied with the set of generators $M \cup \{ \alpha _{i, j}\} $) that $M''$ is normal over $K$. $\square$

The following lemma can sometimes be used to prove properties of the normal closure.

Lemma 9.16.6. Let $L/K$ be a finite extension. Let $M/L$ be the normal closure of $L$ over $K$. Then there is a surjective map

of $K$-algebras where the number of tensors can be taken $[L : K]_ s \leq [L : K]$.

**Proof.**
Choose an algebraic closure $\overline{K}$ of $K$. Set $n = [L : K]_ s = |\mathop{\mathrm{Mor}}\nolimits _ K(L, \overline{K})|$ with equality by Lemma 9.14.8. Say $\mathop{\mathrm{Mor}}\nolimits _ K(L, \overline{K}) = \{ \sigma _1, \ldots , \sigma _ n\} $. Let $M' \subset \overline{K}$ be the $K$-subalgebra generated by $\sigma _ i(L)$, $i = 1, \ldots , n$. It follows from Lemma 9.15.5 that $M'$ is normal over $K$ and that it is the smallest normal subextension of $\overline{K}$ containing $\sigma _1(L)$. By uniqueness of normal closure we have $M \cong M'$. Finally, there is a surjective map

and note that $n \leq [L : K]$ by definition. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #4189 by Weixiao Lu on

Comment #4383 by Johan on