Adjoining a root to a field that already has that element

Question

I know that one can take an irreducible polynomial $f$ over a field $K$ and make an extension containing a root of $f$ by the construction $K[x]/(f)$. What happens if $f$ is in fact reducible? For example, what does $\mathbb Q(\sqrt2)[x]/(x^2 - 2)$ look like?

Apologies for the vague question. I guess I'm asking what kind of structure this is (for one, I can't tell if it's still a field). What parts of it depend on the nature of $f$ or $K$?

Answer 1

Not sure how satisfactory this answer is, but in general for commutative rings $k \subseteq K$ and a polynomial $f \in k[x]$, one has $$K[x]/(f) \cong K \otimes_k k[x]/(f)$$ as a tensor product of $k$-algebras. So for your example this implies $$\mathbb{Q}(\sqrt2)[x]/(x^2-2) \cong \mathbb{Q}(\sqrt2) \otimes_{\mathbb{Q}} \mathbb{Q}(\sqrt2)$$ which has zero divisors: $$(\sqrt{2} \otimes 1 + 1 \otimes \sqrt{2})(\sqrt{2} \otimes 1 - 1 \otimes \sqrt{2}) = 2 \otimes 1 - 1 \otimes 2 = 0.$$

Answer 2

It won't be an integral domain since $(x - \sqrt{2})(x+\sqrt{2}) = x^2-2 = 0$ in the quotient. You'll end up with four square roots of $2$.