Show $K(\alpha)$ is a splitting field of $\text {Irr}(\alpha,K)$ over $K$ $\iff$ $K \subset K(\alpha)$ is a normal extension.

I see that if $K \subset K(\alpha)$ is a normal extension, then $\text {Irr}(\alpha,K)$ splits over $K(\alpha)$ by definition and $K(\alpha)$ is by definition a splitting field of $\text {Irr}(\alpha,K)$ over $K$.

However, I cannot prove the other way: $K(\alpha)$ is a splitting field of $\text {Irr}(\alpha,K)$ over $K$ $\Rightarrow$ $K(\alpha)$ is a normal extension.

I know the equivalent definition of a normal extension $K \subset K(\alpha)$ as: Each irreducible polynomial $p \in K[X]$ that has at least one root in $K(\alpha)$ splits over $K(\alpha)$ into linear factors.

Can someone help me out ?


2 Answers 2


You probably need the following equivalent definition:

Let $\;K/F\;$ be an algebraic field extension and let $\;\overline F\;$ be a (the) fixed algebraic closure field of $\;F\;$ containing $\;K\;$ then

**Def.:**$\;\;\;$ The extension $\;K/F\;$ is normal iff every $\;F$- embedding $\;K\hookrightarrow \overline F\;$ is in fact an automorphism of $\;K\;$.

With this you're practically done since if $\;\alpha\in K\;$ is a root of some irreducible $\;p(x)\in F[x]\;$, then for any other root $\;\beta\;$ of the polynomial $\;p(x)\;$ (which, by the way, is contained in $\;\overline F\;$-- why? --) we have an $\;F$- isomorphism $\;K\ge K(\alpha)\longrightarrow K(\beta)\le \overline F\;$ which can be lifted (why?) to an embedding $\;K\hookrightarrow\overline F\;$ , and by the above definition/theorem, this means that in fact $\;\beta\in K\;$ (fill in details)

  • $\begingroup$ Thank you, I thought the answer was much more concise. $\endgroup$
    – Shuzheng
    Oct 5, 2014 at 15:56
  • $\begingroup$ What do you mean by embedding ? $\endgroup$ Oct 7, 2014 at 10:00
  • $\begingroup$ Embedding = injective homomorphism. Thus, $\;K\;$ is embedded in $\;\overline F\;$ means there's an injective homomorphism $\;K\to\overline F\;$ . Some times this is represented by that hooked arrow: $\;K\hookrightarrow\overline F\;$ . $\endgroup$
    – Timbuc
    Oct 7, 2014 at 10:31

Let $f(x)=\text {Irr}(\alpha,K)(x). $ Given $\sigma:K(\alpha) \hookrightarrow \overline K$ fixing $K$, since $f(\sigma(\alpha))=\sigma(f(\alpha))=0, \sigma(K(\alpha))\subseteq K(\alpha)$. For any $\beta\in K(\alpha)$, let $T=\{\gamma\in K(\alpha)\mid \text{Irr}(\gamma,K)(\beta)=0\}$. Then $K(T)\subseteq K(\alpha)$, and for any $\gamma\in T, \text{Irr}(\sigma(\gamma),K)(\beta)=0$, so $\sigma(\gamma)\in T $ as well, i.e. $\sigma(K(T))\subseteq K(T)$. Since $\sigma$ is a non-trivial field homomorphism, its kernel is an ideal of $K(\alpha)$ viewed as a ring, which must be trivial. By the first isomorphism theorem of rings, we get $\sigma(K(T))\cong K(T)$ and hence $\sigma(K(T))=K(T)$. This means that $\beta\in \text{Im}(\sigma)$, and since $\beta$ was an arbitrary element in $K(\alpha)$, we get $\sigma(K(\alpha))= K(\alpha)$ and thus $\sigma$ is in fact an automorphism.

Lemma: Let $F(\alpha)/F$ be algebraic and $\sigma: F\rightarrow L$. Then $\sigma$ has an extension into $F(\alpha)$ if and only if $\exists \beta\in L$ such that it is a root of $\text{Irr}(\alpha,F^{\sigma})$.

Let $p(x)$ be an irreducible polynomial in $K[x]$ with a root $\xi\in K(\alpha)$. We may assume it is monic and therefore the minimal polynomial of $\xi$ over $K$. If $\eta$ is another root of $p$, then by the previous lemma, $\tau: K\rightarrow K(\eta)$ has an extension into $K(\xi)\subseteq K(\alpha)$. This extension may be viewed as a monomorphism from $K(\alpha)$ into $\overline K$, so it is in fact an automorphism, which means that $\eta\in K(\alpha)$. Therefore, every root of $p$ lies in $K(\alpha)$ and hence $p(x)$ splits in $K(\alpha)$.


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