I'm trying to show that if $P(t)$ is a polynomial with $n$ distinct roots, then its splitting field $K$ is a minimal Galois extension, that contains $k[t]/(P)$.
Our definition of Galois extension is the following: Let $[K:k]$ be an algebraic closure of $k$. Then $[K:k]$ is called a Galois extension if $K\otimes_k K\simeq \oplus_n K$ for some $n\in\mathbb{N}$.
I have several thoughts about this problem, but can't combine them for a solution. The key point is that I don't clearly understand from the task where does $P(t)$ have distinct roots -- in a field $k$ or somewhere else?
- Suppose that all $n$ roots of $P(t)$ are in $k$. Since all roots of $P(t)$ are distinct we can say that $k[t]/P \simeq \oplus_n k$.
- Denote $i$-th root of $P(t)$ by $\alpha_i$. Then the map $\alpha_i\mapsto\ (t-\alpha_i)$ defines an isomorphism between $K$ and a subfield in algebraic closure of $k$, generated by all roots of $P(t)$.
Now a sketch of proof is:
Suppose that $P(t)$ has exactly $n$ roots (where?). Splitting field $K$ contains $k[t]/(P)$ by construction. Since $k[t]/P \simeq \oplus_n k$, $K\otimes_k K \simeq \oplus_{n^2} k$. If there exists some Galois extension $\tilde{K}$ such that $\tilde{K}\otimes_k\tilde{K}\simeq\oplus_m k$ for some $m < n^2$, then $k[t]/P\simeq\sqrt{m}<n$, that contradicts the fact №2.
Can anyone explain me a correct proof and comment my mistakes? Thanks in advance.