A finite dimensional extension of a field is direct product of local rings, the Fulton way. Theorem. If $k$ is a field, $R$ a ring containing $k$ as a subring and $R$ finite dimensional over $k$, then $R$ is isomorphic to a direct product of local rings.
I know the usual proof using the Chinese remainder theorem, but I want to understand the proof by Fulton in his 'Algebraic curves', only valid for algebraically closed $k$. Fulton shows in an entertaining way that, if $k$ is algebraically closed and if $I$ is an ideal in $k[X_1, \ldots, X_n]$ with finite vanishing set $V(I)=\left \{  P \in k^n \mid F(P)=0, 
 \forall F\in I \right \}$ then $k[X_1, \ldots, X_n]/I$ is isomorphic to a direct product of local rings. Then he claims that the above Theorem can be deduced. I know how to get $R$ as a quotient $k[X_1, \ldots, X_n]/I$, using the fact that $R$ is finite dimensional over $k$, but I don't know how to assure that $I$ has a finite vanishing set. Any ideas?
 A: I worked out the excellent but succinct comment of reuns.
By hypothesis, $k$ is an algebraically closed field, $R$ is a ring with
$k \subset R$, and $R$ is a vector space over $k$ of finite dimension $n$,
with basis $\{ r_1, \ldots, r_n \}$. Since $R$ is a vector space,
$r_i \cdot r_j = \sum_{ k = 1 }^{ n } a_{i,j,k} \cdot r_k$ with unique
coefficients $a_{i,j,k} \in K$. That is,
$R \simeq k[X_1, \ldots, X_n] \mathbin{/} I$, where $I$ is generated by the
corresponding relations
$X_i \cdot X_j - \sum_{ k = 1 }^{ n } a_{i,j,k} \cdot X_k$.
Also, since $R$ is a vector space over $k$ of dimension $n$, for each $i$, the
elements of $\{ 1, r_i, \ldots, r_i^n \}$ are linearly dependent, such that a
unique monic polynomial $f_i(X_i)$ of minimal degree (at most $n$) exists with
$f_i(r_i) = 0$. These minimal polynomials can be naturally embedded in
$k[X_1, \ldots, X_n]$, and so $f_i \in I$ (because $I$ comprises, by
construction, all relations pertaining to the basis elements $r_i$).
For each $P = (u_1, \ldots, u_n) \in V(I)$, $φ(X_i) = u_i$ defines a unique
ring homomorphism $\phi$ from $k[X_1, \ldots, X_n]$ to $k$. Also, for each
$g \in I$: $\phi(g(X_1, \ldots, X_n)) = g(\phi(X_1), \ldots, \phi(X_n)) =
g(u_1, \ldots, u_n) = 0$ (since $P \in V(I)$ and $g \in I$). Thus,
$I \subset \ker(\phi)$, such that $\phi$ induces an homomorphism from
$k[X_1, \ldots, X_n] \mathbin{/} I$ to $k$.
For each homomorphism $\phi$ from $k[X_1, \ldots, X_n] \mathbin{/} I$ to $k$,
for each $g \in I$:
$g(\phi(X_1), \ldots, \phi(X_n)) = \phi(g(X_1, \ldots, X_n)) = \phi(0) = 0$,
that is, $(\phi(X_1), \ldots, \phi(X_n)) \in V(I)$. In particular, since
$f_i \in I$, $f_i(\phi(X_i)) = 0$, and thus $\phi(X_i)$ can assume only
finitely many values (because each such value must be a root of $f_i$), such
that only finitely many such homomorphisms exist.
Hence, there is a one-to-one correspondence between $V(I)$ and the
homomorphisms from $k[X_1, \ldots, X_n] \mathbin{/} I$ to $k$, and since there
are finitely many such homomorphisms, $V(I)$ is a finite set. Furthermore,
$V(I)$ is not empty, because otherwise $I = k[X_1, \ldots, X_n]$ (by the
Nullstellensatz, since $k$ is algebraically closed) and thus $R$ is the
zero-ring, which is excluded by the hypothesis $k \subset R$. Therefore, the
theorem from Fulton's book mentioned in the question applies, such that
$k[X_1, \ldots, X_n] \mathbin{/} I$ (isomorphic to $R$) is direct product of
local rings.
