Finite Dimensional Rings Containing Algebraically Closed Fields (2.9.47)

William Fulton Algebraic Curves, chapter 2, section 9, exercise 47:

Let $R$ be a commutative ring sharing an identity with $k⊆R$, an algebraically closed field. Show that $R$ is isomorphic to a finite direct product of local rings.

Proof: Let $r_1,...,r_n$ be a basis for $R$ over $k$. Then $k[x_1,...,x_n] / I ≅ R$ for some ideal $I$ via the natural identification $x_i ↦ r_i$. We see that distinct (surjective) morphisms $R → k$ preserving $k$ thus correspond to maximal ideals in $k[x_1,...,x_n]$ containing $I$, which themselves correspond to points in the locus of $I$. It follows that if we can prove there are only finitely many such morphisms, then Proposition 6 of this same section will imply the result. To this end, we show that the image of $r_i$ in $k$ under a morphism $φ$ may be among only finitely many elements; indeed, this follows by considering that $1,r_i,r_i^2,...$ are linearly dependent over $k$, and thus $φ(r_i)$ must be a root of the polynomial of $r_i$ over $k$ induced by this dependence.$~\square$