Can anyone explain me why the Krull dimension of the ring $K[t^{-1},t]$, where $K$ is a field, is $1$?
I know I should show some efforts here, in MSE, but I actually couldn't find any clue about that.
Any hints or answer would be highly appreciated.
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2 Answers
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$K[t^{-1},t] = K[t]_t$, meaning $K[t]$ localized at $t$. Hence the prime ideals in $K[t^{-1},t]$ are in one-to-one correspondence with the prime ideals in $K[t]$ which avoid the multiplicative set $\{t^n; n \in \mathbb{N}\}$. It now follows easily that $\dim(K[t^{-1},t]) \leq \dim(K[t]) = 1$.
Also $0 \subset (t-1)$ is a chain of prime ideals in $K[t^{-1},t]$, so that $\dim(K[t^{-1},t]) \geq 1$. The conclusion follows.
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$\begingroup$ @ Malkoun, thanks for your answer, but one thing is not clear to me, how $K[t,t^{-1}] = K[t]_{t}$ $\endgroup$– R. SinghMay 18, 2017 at 11:39
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1$\begingroup$ yes, well I mean the localization with multiplicative subset $\{1, t, t^2,...\}$ (not to be confused with the localization at a prime ideal). So in that localization, the elements are of the form $p(t)/t^n$, for some $n \in \mathbb{N}$, where $p(t) \in K[t]$. Hence, the elements in the localization consists of polynomials in $t^{-1}$ with coefficients in the ring $K[t]$. Hence $K[t^{-1},t] = K[t]_t$. $\endgroup$– MalkounMay 18, 2017 at 11:44
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$\begingroup$ Also: being the localization of a PID, it is a PID, hence $1$-dimensional. But that presupposes being confident about this fact of PIDs. $\endgroup$– rschwiebMay 30, 2017 at 20:31
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$\begingroup$ @Malkoun Right, I thought it was more than obvious it was a nonfield PID , but it should be mentioned explicitly. $\endgroup$– rschwiebMay 30, 2017 at 22:06
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For the irreducible polynomial $f(y,t) = yt-1$$\in$ $K[y,t]$ , Krull dimension of the quotient ring $K[y,t]/(f(y,t)$ is $1$ so from the isomorphism $ K[t^{-1},t] \rightarrow K[y,t]/(f(y,t)) $ , Krull dimension of the ring $ K[t^{-1},t]$ is $1$.
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$\begingroup$ How do you know that "Krull dimension of the quotient ring $K[y,t]/(f(y,t)$ is $1$"? $\endgroup$ Jan 9, 2020 at 14:43