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Let $k$ be field and consider the power series $A=k[[x,y]]$. What is the simplest way (in the sense of using the least "heavy" theorems) to show that $\operatorname{dim} A=2$, where $\operatorname{dim} A$ means dimension as a ring (Krull dimension)?

Here is my approach: first of all, note that $A$ is a local ring with maximal ideal $m=(x,y)$. Hence $\operatorname{dim} A = \operatorname{ht} (m)$. Now i use the following "heavy" theorem (Matsumura, 13.5): "Let $A$ be a Noetherian ring and $I$ an $r$-generated ideal. Let $P$ be a minimal prime divisor of $I$. Then $\operatorname{ht}(P) \le r$." Apply this theorem to $m$: $m$ is $2$-generated and it is a minimal prime divisor of itself. Hence $\operatorname{ht}(m) \le 2$. Now we have the chain of prime ideals $m \supsetneq (x) \supsetneq 0$, which gives $\operatorname{ht}(m)=2$.

Could you please confirm the validity of the above approach? It seems right to me. Also, any other interesting alternatives?

Manos
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  • Dimension...as what? Vector space, algebra...and over what? – DonAntonio Jun 20 '13 at 14:46
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    Dimension as a ring. I will edit. – Manos Jun 20 '13 at 14:47
  • Do you really consider that theorem "heavy" in the context of dimension calculations? It follows pretty readily from Krull's Height Theorem, which is difficult, but it's the cornerstone of dimension theory. So by your criteria, any dimension proof is necessarily going to be "heavy." – Potato Jun 20 '13 at 17:17
  • When you say "Krull's height theorem", you mean the theorem i am quoting? And yes, you are right. I was wondering if there is any argument without involving dimension theory. In any case, is my argument correct? Any simple answer would do :) – Manos Jun 20 '13 at 17:31
  • Yes, it's right. –  Jun 20 '13 at 18:25

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Yes, your argument is correct.