abstract algebra – For $ Omega $ an algebraic closure of k $, why is $ k[x]/ (f) otimes_k Omega cong Omega ^ m $?

Fix a field k $ and an algebraic closure $ Omega $ of k $. For a monic, irreducible, separable polynomial $ f in k[x]$, let $ f $ to divide $ Omega $ as

$$ f (x) = prod_ {i = 1} ^ m (x – a_i). $$

It was said that the following holds:

$$ frac {k[x]} {(f)} otimes_k Omega con prod_ {i = 1} ^ m frac { Omega[x]} {(x-a_i)} cong Omega ^ m. $$

I can see the Chinese rest theorem here, but what is not clear to me is the resolution of the tensor because I am not yet comfortable working with them. Why is the first isomorphism true?