It is a well know fact that if $L$ is a Galois Extension of Field $K$ of degree $n$ then we have :
$$ L \otimes_K L \simeq L^n $$
I am trying to get insight on what happens when L is separable but not normal
Thus I am considering $\mathbb{Q}[\alpha]$ with $\alpha = \sqrt[3]{2}$ and :
$$ \mathbb{Q}[\alpha] \otimes_\mathbb{Q} \mathbb{Q}[\alpha] $$ $$ \simeq \mathbb{Q}[\alpha][X]/(X^3 - 2)$$ $$ \simeq \mathbb{Q}[\alpha][X]/(X-\alpha) \times \mathbb{Q}[\alpha][X]/(X^2+\alpha X + \alpha^2) $$ $$ \simeq \mathbb{Q}[\alpha] \times \mathbb{Q}[\alpha,\omega] $$
is a product of $\mathbb{Q}[\alpha]$ and $M=\mathbb{Q}[\alpha,\omega]$ a splitting field of $X^3-2$.
Now, contrary to the Galois case, I guess we cannot say that every element of the Galois Group corresponds bijectively to a maximal ideal of the tensor product, since the Galois group only has one element here and the tensor product has more than one maximal ideal.
But can we still deduce something from the set of maximal ideals ?
Does the fact that we only get one copy of $\mathbb{Q}[\alpha]$ correspond to the fact the Galois Group only has one element ?
Do the other Maximal Ideals tell us something about $\mathbb{Q}$-homomorphisms of $Q[\alpha]$ into an algebraic closure ?
Ok thanks, this was my understanding. In this case it looks to me as if there are 3 possibilities for the compositum each corresponding to a $Q$-homomorphism of $L$ into an algebraic closure with the last two being isomorphic : $\mathbb{Q}[\alpha]=\mathbb{Q}[\alpha].\mathbb{Q}[\alpha]$ corresponding to the identity, $\mathbb{Q}[\omega,\alpha]=\mathbb{Q}[\alpha].\mathbb{Q}[\omega\alpha]$ and $\mathbb{Q}[\omega,\alpha]=\mathbb{Q}[\alpha].\mathbb{Q}[\omega^2\alpha]$ corresponding to the two other non trivial homomorphisms. Is that correct ?