Let $X$ consist of two distinct points. Prove $k[X] \simeq k \oplus k$.

Question

Let $X$ consist of two distinct points, $a$ and $b$. Prove $k[X] \simeq k \oplus k$.

First, I attempted to construct the isomorphism as follows:

$$\phi:k[X] \rightarrow k \oplus k$$ $$ak_1+bk_2 \mapsto (k_1,k_2)$$

I thought this would work, but then it dawned on me that I don't think, in general, every element of $k[X]$ can be written this way. (Right?)

I would appreciate any tips!

Answer 1

$X= a \cup b=V(x-a) \cup V(x-b)=V((x-a) \cap(x-b))$. By the chinese remainder theorem, we have an isomorphism $$k[x]/((x-a)(x-b)) \to k[x]/(x-a) \times k[x]/(x-b).$$