# 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?