3.2 Splitting Fields, Normal Extension and Separable Extensions

Splitting Fields

Definition

Let $f=f(T)\in K[T]$ with $\mathrm{deg}f=n⩾1$. An extension $F/K$ is called a splitting field over $K$ of $f$, if

• $f$ splits in $F[T]$, namely $f(T)={\prod }_{i=1}^n(T-x_i)$ with $x_i\in F$; and
• $F=K(x_1,\cdots ,x_n)$

Let $S\subseteq K[T]$ be a subset. We say $F/K$ is a splitting field over $K$ of $S$, if all $f\in S$ splits in $F$ and $F$ is generated over $K$ by the roots.

Example. Take irreducible $g\in K[T]$ with $g=T^2+aT+b$ and $\mathrm{deg}g=2$. Let $F:=K[T]/(g(T))$, then $g[T]$ as a polynomial in $F[T]$ splits. Because for any $x\in F$, we have $g(T)=(T-x)(T-b/x)$ and $K(x,b/x)=K(x)=F$. Remark that splitting field is not unique, as $K[x]/(g(x))$ and $K[y]/(g(y))$ are not same as sets.

Theorem

Let $K$ be a field and let $f(x)\in K[x]$ with $\mathrm{deg}f⩾1$. Then there exists a splitting field $F$ of $f$ such that $[F:K]⩽n!$.

\begin{proof} By induction on $n$. When $n=2$, we proved it here. If $f$ spilts in $K$, then $F=K$. Otherwise, there exists $g\in K[x]$ with $\mathrm{deg}g⩾2$, which is an irreducible factor of $f(x)$ and $f(x)=g(x)\tilde{g}(x)$. Consider $K_1=K[T]/(g(T))$, then note that $K\hookrightarrow K_1,a\mapsto \overline{a}$. Consider $g(x)\in K[x]\subseteq K_1[x]$, then $\overline{T}$ is a root of $g(x)$ and so $g(x)=(x-\overline{T})h(x)$ with $h\in K_1[X]$. It follows that $f(x)=(x-\overline{T})h(x)\tilde{g}(x)$. By induction hypothesis, there exists a splitting field $F$ of $h(x)\tilde{g}(x)\in K_1[x]$ over $K_1$ and $[F:K_1]⩽(n-1)!$. Assume that $f_1,\cdots ,f_{n-1}$ are roots of $h(x)\tilde{g}(x)$ and $F=K_1(f_1,\cdots ,f_{n-1})$, then $F=K(\overline{T},f_1,\cdots ,f_{n-1})$ is a splitting field of $f$ and $[F:K]⩽n!$.
\end{proof}

Remark. In general, $[F:K]>n$. For example:

• Consider $F=\mathbb{Q}[s_n]$ where $s_n=e^{2\pi i/n}$. Note that $F$ is a splitting field of $x^n-1$ and $[\mathbb{Q}(s_n):\mathbb{Q}]=n-1$ when $n$ is a prime. Claim that $g(x)=x^{n-1}+x^{n-2}+\cdots +1$ is irreducible when $n$ is prime. Note that $g(x+1)=((x+1{)}^p-1)/(x+1-1)={\sum }_{i⩾1}(\genfrac{}{}{0ex}{}{n}{i})x^{i-1}$ is irreducible by Eisenstein criterion.
• Let $f(x)=x^p-2\in \mathbb{Q}[x]$. Then $F=\mathbb{Q}(2^{1/p},s_p)=\mathbb{Q}(2^{1/p})(s_p)$ is a splitting field over $\mathbb{Q}$ and $[F:\mathbb{Q}]=p(p-1)$ by $(p,p-1)=1$.

Corollary

Let $S\subseteq K[x]$ be a subset, then there exists a splitting field over $K$ of $S$.

\begin{proof} If $S=\{f_1,\cdots ,f_n\}$ is finite, then simple take composite of splitting fields of $f_i$. If $|S|=\infty$, use Zorn’s lemma. \end{proof}

Definition

A splitting field of non-const elements in $k[x]$ is called an algebraic closure of $K$, denote as $\overline{K}$.

Example. $\overline{\mathbb{R}}=\mathbb{C}=\overline{\mathbb{C}}\supseteq \overline{\mathbb{Q}}=\overline{\mathbb{Q}(\sqrt{2})}$. Indeed, if $\overline{K}=K$, then $\overline{\overline{K}}=\overline{K}=K$.

Lemma

Suppose $\sigma :K\stackrel{\sim }{\to }L$ is an isomorphism of fields. Let $S=\{f_i{\}}_{i\in I}\subseteq K[T]$ be a subset of monic poly with $\mathrm{deg}{f}_i⩾1$. Let $\sigma (S)=\{\sigma (f_i){\}}_{i\in I}$. Let $L$ and $M$ be splitting fields of $K$ and $L$, respectively. Then $\sigma$ extends to an isomorphism $\sigma :F\stackrel{\sim }{\to }M$.

Remark. Even $\sigma =\mathrm{id}:K\to K$, its extension is non-trivial. Still the example of $K[x]/(g(x))$ and $K[y]/(g(y))$.

\begin{proof} We only consider the case $|S|<\infty$. Indeed, we can suppose $S=\{f(T)\}$ and do induction on the degree of $f(T)$. Suppose the lemma is true if $\mathrm{deg}f⩽n-1$. Now suppose $\mathrm{deg}f=n$, WLOG $f$ is irreducible (otherwise use induction hypothesis).

Assume that $f(T)={\prod }_{i=1}^n(T-u_i)\in F[T]$ and $F=K(u_1,\cdots ,u_n)$. Also assume that $(\sigma f)(T)={\prod }_{j=1}^n(T-v_j)\in M[T]$ and $M=L(v_1,\cdots ,v_n)$. Choose any $u_i,v_j$. By ^zvyjyw, there exists a chain of isomorphism

$$K[u_1]\simeq K[T]/(f(T))\stackrel{\sigma }{\to }L[T]/(\sigma f(T))\stackrel{\sim }{\to }L[v_1],u_1\mapsto \overline{T}\mapsto \sigma (\overline{T})\mapsto v_1.$$

Denote the composition as ${\sigma }_1:K[u_1]\stackrel{\sim }{\to }L[v_1]$. Note that $f_1(T)=f(T)/(T-u_1)$ and $g_1(T)=f(T)/(T-v_1)$ satisfy ${\sigma }_1{f}_1=g_1$. Then by induction hypothesis, there exists $\tilde{\sigma }:F\stackrel{\sim }{\to }M$ extending ${\sigma }_1$, thus extending $\sigma$. \end{proof}

Example. See here and here.

Corollary

Any two splitting fields of $f(T)$ over $K$ are $K$-isomorphism. Moreover, any two algebra closures over $K$ are $K$-isomorphism.

Normal Extension

Definition

Let $F/K$ be an algebraic extension. We say $F/K$ is normal if any irreducible $f(T)\in K[T]$ such that $f(T)$ has one root in $F$ actually spilts in $F$. Equivalently, for any $u\in F$, $\mathrm{Irr}(u,K)$ spilts in $F$.

Theorem

Suppose $[F:K]<\infty$. TFAE:

• $F/K$ is normal;
• $F$ is a splitting field over $K$ of some $g(T)\in K[T]$.

\begin{proof} i)→ii) If $F={\oplus }_{i=1}^{n}K{e}_i$ with $e_i\in F$, then $F$ is a splitting field of $g(T)={\prod }_{i=1}^n\mathrm{Irr}(e_i,K)$.

ii)→i) Let $u\in F$, and let $f(T)=\mathrm{Irr}(u,K)$. It follows that $f(T)={\prod }_{i=1}^m(T-u_i)$ with $u_i\in \overline{K}$ and $u_1=u$. We aim to show $u_i\in F$. Consider any $u_i$ with $i\ne 1$, we have $\sigma :K[u]\simeq K[T]/(f(T))\simeq K[u_i]$. Note that $F$ is a splitting field of $g(T)$ over $K$. It follows that $F$ is a splitting field of $g(T)$ over $K[u]$ and $F[u_i]=K[u_i](u_1,\cdots ,u_m)$ is a splitting field of $g(T)$ over $K[u_i]$. By ^28f933, $\sigma$ extends to an isomorphism $\tilde{\sigma }:F\to F[u_i]$. Note that $\tilde{\sigma }{|}_K=\mathrm{id}{|}_K$, then $\tilde{\sigma }$ is also an homomorphism of $K$-vector space, namely $\tilde{\sigma }(kx)=k\tilde{\sigma }(x)$. Therefore, ${\dim }_{k}F={\dim }_{k}F[u_i]$. Furthermore, $[F:K]=[F(u_i):F][F:K]$ yields $u_i\in F$.

Alternating proof. See 抽象代数 I, page 116. \end{proof}

Remark. Note that it is possible $A\subseteq B$ as fields but $A\simeq B$. For example, $\mathbb{Q}(x^2)\subseteq \mathbb{Q}(x)$.

Separable Extensions

Definition

• Let $f(T)\in K[T]$ irreducible. We call it is separable if $f(T)={\prod }_{i=1}^n(T-u_i)$ inside $\overline{K}[T]$ and $u_i$ are distinct.
• We say $u\in F$ is separable over $K$, if $\mathrm{Irr}(u,K)$ is separable.
• We say $F/K$ is a separable extension, if for any $u\in F$, $u$ is separable over $K$.

Examples.

• $\mathbb{C}/\mathbb{R}$ is separable.
• $\mathbb{Q}(2^{1/3})/\mathbb{Q}$ is separable. In fact, if $F/K$ is algebraic and $\mathrm{char}=0$, then $F/K$ is separable.
• ${\mathbb{F}}_p(x^{1/p})/{\mathbb{F}}_p(x)$ is not separable, as $\mathrm{Irr}(x^{1/p},{\mathbb{F}}_p(x))=T^p-1=(T-x^{1/p}{)}^p$ has only one root with multiplicity $p$.

Remark. Consider $K\subseteq E\subseteq F$, then

• $F/E$ and $E/K$ are separable yields that $F/K$ is separable, see here;
• $F/E$ and $E/K$ are normal does NOT yield that $F/K$ is normal, see here and $\mathbb{Q}[\sqrt[4]{2}]\supseteq \mathbb{Q}[\sqrt{2}]\supseteq \mathbb{Q}$.

Lemma

Assume that $F/K$ is a field extension, and $u\in F$ is algebraic over $K$. Define $f(T)=\mathrm{Irr}(u,K)$ and $f^{\prime }(T)=d/dTf(T)$. Then TFAE:

• $u$ is separable over $K$;
• $f^{\prime }(u)\ne 0$;
• $f^{\prime }(T)\ne 0$.

\begin{proof} Easy. See page 121 of 抽象代数 I. \end{proof}

Corollary

If $F/K$ is algebraic and $\mathrm{char}=0$, then $F/K$ is separable.

Remark. In above example, $(T^p-x{)}^{\prime }=0$ and so $x^{1/p}$ is not separable.

Theorem of Primitive Element

theorem of primitive element

If $K^{\prime }/K$ is a finite separable extension, then there exists $x\in K^{\prime }$ such that $K^{\prime }=K[x]$.

\begin{proof} If $[K^{\prime }:K]=2$, we only need to take $x\in K^{\prime }\setminus K$. By induction, it remains to show the case of $K^{\prime }=K[\alpha ,\beta ]$. We claim that we can find some $\lambda \in K$ such that $K^{\prime }=K[\alpha +\lambda \beta ]$. If it is true, then $\beta \in K[\gamma ]$ with $\gamma =\alpha +\lambda \beta$. Also, it is equivalent to $w(T)=\mathrm{Irr}(\beta ,K[\gamma ])$ has degree $1$.

By method of indeterminate, suppose $f(T)=\mathrm{Irr}(\alpha ,K)$ and $g(T)=\mathrm{Irr}(\beta ,K)$. Let $h(T)=f(\gamma -\lambda T)\in K[\gamma ][T]$. Then $h(\beta )=0$ and so $\omega (T)\mid h(T)$. Similarly $w(T)\mid g(T)$. If $g(T)$ and $h(T)$ have only one common root $\beta$, then $\mathrm{deg}w⩽1$ and we finish the proof.

The roots of $f(T)$ are $\alpha =\alpha ,{\alpha }_2,\cdots ,{\alpha }_m$, and so the roots of $h(T)$ are $(\gamma -{\alpha }_i)/\lambda$. The roots of $g(T)$ are ${\beta }_1=\beta ,{\beta }_2,\cdots ,{\beta }_n$. If $(\gamma -{\alpha }_i)/\lambda ={\beta }_j$, then $\gamma =({\alpha }_i-{\alpha }_1)/({\beta }_j-{\beta }_1)$. We only need to choose

$$\lambda \notin \left\{\frac{{\alpha }_i-{\alpha }_1}{{\beta }_j-{\beta }_1}:1⩽i⩽m,1⩽j⩽n\right\}$$

and then we finish the proof. \end{proof}