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.


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 $L$.


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