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Let $R=\Bbb{C}[[X,Y,Z]]$ then I want to compute the Krull dimension of $R$.

My idea was to compute the Samuel function and bring it into a polynomial "form" then we immediately know that the Krull dimension is the degree of the polynomial.

For the Samuel function I need to compute $m^k/m^{k+1}$ for the maximal ideal $m\in R$. I know that the maximal ideal of $R$ is $m=(X,Y,Z)$.

After computing some powers I found out that $$m^j=(X^kY^lZ^{j-k-l})_{0\leq k\leq i,~~0\leq l\leq i-k}$$But now I struggle in finding out what $m^j/m^{j+1}$ is. Can someone help me how do proceed here?

user1294729
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  • Have you seen the answers here? We have Theorem. Let $R$ be a Noetherian ring. Then $$\text{dim},R[X_{1},\cdots,X_{n}]=\text{dim},R[[X_{1}\cdots,X_{n}]]=\text{dim},R+n$$ – Dietrich Burde Jul 05 '22 at 12:13
  • @DietrichBurde the problem is, that we haven't had this theorem so I really need to do it by hand – user1294729 Jul 05 '22 at 12:15
  • No, the answer there gives an elementary proof, i.e., an easy induction. So no problem using it for $R=\Bbb C$. – Dietrich Burde Jul 05 '22 at 12:16
  • $m^k$ is just the power series such that the coefficient of $X^a Y^b Z^c$ is $0$ whenever $a+b+c < k$. And $m^k/m^{k+1}$ is a $\Bbb{C}$-vector space with basis the $X^a Y^b Z^c$ with $a+b+c=k$. – reuns Jul 05 '22 at 12:17
  • @DietrichBurde okey I see – user1294729 Jul 05 '22 at 12:18
  • @reuns why do you see that? – user1294729 Jul 05 '22 at 12:18
  • To see what? ${}{}{}$ – reuns Jul 05 '22 at 12:20
  • @reuns I don't see how we get to the fact that $m^k/m^{k+1}$ is generated as a $\Bbb{C}$ vector space. So I don't see how I can see from my $m$ how to proceed – user1294729 Jul 05 '22 at 12:21
  • Any power series starting at $k$ is a power series starting at $k+1$ plus some monomials at $k$ – reuns Jul 05 '22 at 12:21
  • Also with $m$ maximal in $R$ then $m^k/m^{k+1}$ is a $R/m$ module or vector space, that's a general rule. – reuns Jul 05 '22 at 12:23
  • @reuns I don't get it, maybe I have the wrong picture in mind how $m^k$ looks like. Isn't an element of $m^k$ of the form $p(X,Y,Z)X^kY^lZ^{j-k-l}$ where $p\in \Bbb{C}[[X,Y,Z]]$? – user1294729 Jul 05 '22 at 12:24
  • @reuns I think I got it so $m^k$ is the set of all power series of order $\geq i$ and I mod out by all power series of order $\geq i+1$ and therefore what remains are all power series of degree $\leq i$? – user1294729 Jul 05 '22 at 12:30

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