Given $R=k[x_0,\ldots,x_n]$, and $I$ an homogeneous ideal of $R$, se say that $I$ is an artinian ideal if the Krull dimension of $R/I$ is equal to 0.
In a paper the lecturer states without proving (because it seems obvious to anybody) that $I$ is artinian $\iff$ $H(R/I, j)=0$ for $j \gg 0$ (with $H(R/I,-)$ i denote the Hilbert function).
I've been thinking about this quite a lot, but I unfortunately didn't find any proof of this fact: can someone help me making this more clear? Thanks in advance.
The fact which this boils down to is this:
If $R$ is finite dimensional over $k$ and $P$ is a prime ideal, then $R/P$ is a finite-dimensional integral domain over $k$. If $a \in R/P$ is not $0$, then the map $R/P \to R/P, x \mapsto ax$ is an injective (since $a$ is not a zero-divisor) $k$-linear map, thus as $R/P$ is finite-dimensional, this map is surjective by linear algebra. But this means that there is some $b \in R/P$ such that $ab=1$, thus $a$ is invertible and $R/P$ is a field, so $P$ is maximal.
If $R$ has Krull-dimension $0$, then any prime ideal is minimal and since a Noetherian ring has only finitely many minimal prime ideals, $R$ has finitely many prime ideals. Let $P_1, \dots, P_n$ be the prime ideals. $J:=P_1 \cdot \ldots \cdot P_n=P_1 \cap \ldots \cap P_n$ is the nilradical of $R$, which is nilpotent since it is finitely generated. Let $k \in \Bbb N$ such that $J^k=0$. Then we get with the Chinese remainder theorem $$R = R/0 = R/J^k= R/(P_1^k \cdot \ldots \cdot P_n^k) \cong R/P_1^k \times \ldots \times R/P_n^k$$
We only need to make sure that the rings of the form $R/P_1^i$ with $i \geq 1$ are finite-dimensional.
We can do this by induction. For $i=1$, $R/P_1^1=R/P_1$ is a finitely generated $k$ algebra that is also a field. The fact that this is finite-dimensional over $k$ is the statement of Zariski's lemma.
Suppose that $R/P_1^i$ is finite-dimensional for some $i\geq 1$, then the third isomorphism theorem gives an exact sequence
$0 \to P_1^{i}/P_1^{i+1} \to R/P_1^{i+1} \to R/P_1^{i} \to 0$
$R/P_1^{i}$ is finite-dimensional by the induction hypothesis. Since $P_1 (P_1^{i}/P_1^{i+1}) = 0$, $P_1^{i}/P_1^{i+1}$ the $R$-module structure on $P_1^{i}/P_1^{i+1}$ descends to a $R/P_1$-module structure. Now as $R$ is Noetherian $P_1$ is finitely generated, so $P_1^{i}/P_1^{i+1}$ is a finitely generated $R/P_1$-module. Since $R/P_1$ is a finite extension of $k$, this implies that $P_1^{i}/P_1^{i+1}$ is finite-dimensional over $k$. Thus by exactness, $R/P_1^{i+1}$ is finite-dimensional.