Let's fix a field $k$ and an algebraic closure $\Omega$ of $k$. For a monic, irreducible, separable polynomial $f\in k[x]$, let $f$ split over $\Omega$ as
$$f(x) = \prod_{i=1}^m (x - a_i).$$
I have been told that the following holds:
$$\frac{k[x]}{(f)}\otimes_k\Omega\cong \prod_{i=1}^m \frac{\Omega[x]}{(x-a_i)} \cong \Omega^m.$$
I can see the Chinese Remainder Theorem in here, but the piece that is unclear to me is resolving the tensor as I'm not quite comfortable working with them yet. Why is the first isomorphism true?
Tensor with $\Omega$ above $k$ and write the Chinese remainder theorem: $$k[x]/(f)\otimes_k\Omega\simeq \Omega[x]/(f)=\Omega[x]\Big/\Bigl( \textstyle\prod_{i=1}^m (x - a_i)\Bigr)\simeq \prod_{i=1}^m \Omega[x]\big/(x - a_i)\simeq\Omega^m $$ (we call apply it because $f$ is separable, so all its roots in $\Omega$ are distinct, hence the linear factors are pairwise coprime.) '