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

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 ?

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)

• Thank you, I thought the answer was much more concise. Oct 5, 2014 at 15:56
• What do you mean by embedding ? Oct 7, 2014 at 10:00
• 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: $\;K\hookrightarrow\overline F\;$ . 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 $$K(\alpha)$$.