$V(I)$ consists of a finite set of points if and only $k[x_1,\ldots,x_n]/I$ has Krull dimension zero

Question

I am trying to prove the following:

Let $I\subset k[x_1, ..., x_n]$ be an ideal. Show that $V(I)$ consists of a finite set of points if and only if $k[x_1,..., x_n]/I$, seen as k-space vector, is an algebra of finite dimension.

To do this, I would like to use the following result:

$k[x_1,\ldots,x_n]/I$ has Krull dimension zero $\iff$ it is finite-dimensional as a $k$-vector space.

Here is where several questions arise:

Question 1: why "$V(I)$ consists of a finite set of points if and only $k[x_1,\ldots,x_n]/I$ has Krull dimension zero"?

If the above is true, it is sufficient to prove then that: $k[x_1,\ldots,x_n]/I$ has Krull dimension zero $\iff$ it is finite-dimensional as a $k$-vector space.

Edit: In everything, k is an algebraically closed field.

Answer 1

NB: for an ideal $I\subset R$, we assume $V(I)$ to be closed subscheme of $\operatorname{Spec} R$ given by the closed immersion $\operatorname{Spec} R/I \to \operatorname{Spec} R$ which is Spec of the natural quotient map $R\to R/I$. If you're working with $V(I)$ being points of $k^n$ satisfying blah blah blah then you should make $k=\overline{k}$ here and consider adopting the schematic viewpoint.

Krull dimension zero implies that every prime ideal is maximal, and furthermore that any prime ideal is a minimal prime ideal. As noetherian rings have finitely many minimal primes and $k[x_1,\cdots,x_n]/I$ is noetherian, this shows that Krull dimension zero implies finitely many points.

For the reverse implication, note that finitely many points implies that up to nilpotents, $k[x_1,\cdots,x_n]/I$ is a finite product of fields, which has krull dimension zero. Since nilpotents don't change the krull dimension, this proves the claim.

Answer 2

Note: Here I take $V(I)$ to simply mean the zero locus of the set of polynomials in $I$.

First of all, the assertion you are trying to prove is false. Take the ideal $I = (x^2 + y^2 +1) \subset \mathbb{R}[x,y]$, then $V(I)$ is empty but $\mathbb{R}[x,y]/I $ is not a finite dimensional vector space over $\mathbb{R}$. The problem is that over $\mathbb{C}$, the polynomial $x^2 + y^2 +1$ has infinitely many zeroes.

Second of all, you can prove the following correct claim using no more than Hilbert's Nullstellensatz.

Correct claim: $$\mathrm{dim}_k k[x_1, \dots, x_n]/I < \infty \iff |V_K(I)| < \infty \text{ for every } k \subset K.$$

Proof. If part: Let $\bar{k}$ be the algebraic closure of $k$. If $V_{\bar{k}}(I)$ is empty then by the Nullstellensatz we have that $I$ is the unit ideal and the asserion is trivial. If not, we have $V_{\bar{k}}(I) = \{ a_1 , \dots, a_m \}, $ where $a_j = (a_{1j}, \dots, a_{nj}) \in \bar{k}^n$. Now let $f_{ij} \in k[x_i]$ be the minimal polynomial of $a_{ij}$. Then $f_i := f_{i1} \dots f_{im}$ vanishes at every point of $V_{\bar{k}}(I)$, so $f_i \in I(V_{\bar{k}}(I))$ and by the Nullstellensatz we have that $f_i \in \sqrt I \cap k[x_i]$. Therefore $I \cap k[x_i] \neq \{0\} $ for all $i$. Now using the division algorithm one can show that $k[x_1, \dots, x_n]/I$ is finite. I leave the details up to you.

Only if part: $k[x_1,\dots,x_n]/I$ is finite dimensional over $k$, so any $1,x_i, x_i^2,\dots $ must have a linear relationship between them, so $I \cap k[x_i] \neq \{0\}$ for all $i$. Now pick a nonzero polynomial $g$ in $K[x_i] \cap I$. Then $V_K(I)$ is bounded by the product of the set of finite solutions of each such $g$ in $K[x_i]$.

Answer 3

$k$ a field and $R = k[x_1,\ldots,x_n]/I$.

It is an infinite dimensional $k$-algebra iff there is some $x_j \in R$ which is not algebraic over $k$.

(in a $n$-dimensional $k$-algebra, there is a non-trivial $k$-linear relation between $1,x_j,x_j^2,\ldots,x_j^n$, conversely if all the $x_j$ are algebraic then $R$ is finite dimensional)

• If every element in $R$ is algebraic over $k$ then $V(I) = \{ a \in \overline{k}^n, \forall f \in I, f(a)=0\}$ is finite, since for $a \in V(I), a_j$ is one of the finitely many roots of the $k$-minimal polynomial of $x_j \in R$.

• Otherwise take $x_j$ non-algebraic over $k$ and make iteratively an ideal $J$ from the $x_i$ which are not algebraic over $k[x_j]$, then $R/J$ is an algebraic extension of $k[x_j]$.

  $k[x_j]$ has infinitely many maximal ideals, only finitely of them become $(1)$ in $R/J$, the others extend to distinct prime (thus maximal) ideals of $R/J$ and they give rise to distinct points in $V(I)$.