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?

  1. Suppose that all $n$ roots of $P(t)$ are in $k$. Since all roots of $P(t)$ are distinct we can say that $k[t]/P \simeq \oplus_n k$.
  2. 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.

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    $\begingroup$ $P(t)$ is irreducible otherwise $k[t]/(P(t))$ has some zero divisors. And $P(t)$ has its roots in the splitting field which is unique up to isomorphisms. $\endgroup$ – reuns Jun 11 '17 at 12:57
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    $\begingroup$ Also if $\{\alpha_j\}_{j=1}^n$ are the roots of $P$ then the splitting field is isomorphic to $$k[t_1,\ldots,t_n]/(I), \qquad I = \{ f \in k[t_1,\ldots,t_n], f(\alpha_1,\ldots,\alpha_n) = 0\}$$ which makes it clear the splitting field is unique up to permutation of the roots $\endgroup$ – reuns Jun 11 '17 at 13:15
  • $\begingroup$ @user1952009, OK, it explains why splitting field is unique, but why there can't be some intermediate Galois extension? Suppose that it exists ($\tilde{K}$ such that $\tilde{K}\subset K$), where will be a contradiction? Can I say that Galois extension $\tilde{K}$ must contain all roots of $P$ and if $\tilde{K} \subset K$ it doesn't, so we are done? $\endgroup$ – Hasek Jun 11 '17 at 13:50

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