# A finite normal extension is also a splitting field

Let $$E/K$$ be a finite normal extension, then there exists $$p(x)\in K[x]$$ s.t. $$E$$ is the splitting field of $$p(x)$$.

The proof goes as follows: $$E=K(a_1,...,a_n)$$ for some $$a_1,...,a_n\in \Omega$$ where $$\Omega$$ is the algebraic closure of $$K$$. Let $$\mu_i$$ be the minimal polynomial of $$a_i$$ for $$i=1,...,n$$ We can send any $$a_i$$ onto another root of $$\mu _i$$ through an homomorphism that fixes $$K$$.

Then the thesis follows.

I don't understand why we can send any $$a_i$$ onto another root of $$\mu _i$$.

Note: the definition that we use of normal extension is: $$L/K$$ is a normal extension if every homomorphism $$\phi :K\rightarrow \Omega$$ that fixes $$K$$ sends $$L$$ onto itself.

## 1 Answer

Here are the steps of a proof.

• The $$\mu_{i}$$ are irreducible.

• Hence, given any two roots $$a, b$$ of $$\mu_{i}$$, there is a $$K$$-isomorphism (meaning that it restricts to the identity on $$K$$) $$K(a) \to K(b)$$ taking $$a$$ to $$b$$.

• This isomorphism can be extended to an automorphism $$\phi$$ of $$\Omega$$.

• Now since $$L/K$$ is normal, $$\phi$$ has to send $$L$$ to $$L$$.