Well, I know that this is noetherian but I don't really understand how I can get a chain of prime ideals here to find the Krull dimension and assure that it is the supremum.
Asked
Active
Viewed 483 times
2
-
What is $ K{ } $? – Bernard Jun 06 '20 at 19:27
1 Answers
3
Assuming $K$ is a field, $\text{dim}\,K[[X]]=1$. More generally there is the following (Theorem 15.4 from Matsumura):
Theorem. Let $R$ be a noetherian ring. Then $$\text{dim}\,R[X_{1},\cdots,X_{n}]=\text{dim}\,R[[X_{1}\cdots,X_{n}]]=\text{dim}\,R+n$$
The proof for the power series case is as follows. By induction we can assume $n=1$. Any prime ideal of $R[[X]]$ is of the form $\mathfrak{M}=(\mathfrak{m},X)$ with $\mathfrak{m}=\mathfrak{M}\cap R$ a maximal ideal of $R$. Thus $\text{ht}\,\mathfrak{M}=\text{height}\,\mathfrak{m}+1$. Conversely if $\mathfrak{m}$ is a maximal ideal of $R$ then $\text{ht}\,(\mathfrak{m},X)=\text{ht}\,\mathfrak{m}+1$, which shows the equality.
Zeek
- 1,837
-
Given the theorem you cited shouldn't it be $\dim K[[X]]=\dim K+1=2$ (maybe I'm missing the correct notion of $\dim$ here)? – mrtaurho Jun 06 '20 at 20:01
-
2
-
-
@Zeek Quick question, how are we using the fact that $R$ is Noetherian in this proof? – slowspider Mar 27 '22 at 22:04