# Krull dimension of the ring $K[t^{-1},t]$

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.

$K[t^{-1},t] = K[t]_t$, 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.

• @ Malkoun, thanks for your answer, but one thing is not clear to me, how $K[t,t^{-1}] = K[t]_{t}$ May 18, 2017 at 11:39
• 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 $K[t^{-1},t] = K[t]_t$. May 18, 2017 at 11:44
• okk.. I got it.. thank you very much.. May 18, 2017 at 11:49
• Also: being the localization of a PID, it is a PID, hence $1$-dimensional. But that presupposes being confident about this fact of PIDs. May 30, 2017 at 20:31
• @Malkoun Right, I thought it was more than obvious it was a nonfield PID , but it should be mentioned explicitly. May 30, 2017 at 22:06

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 $$1$$.

• How do you know that "Krull dimension of the quotient ring $K[y,t]/(f(y,t)$ is $1$"? Jan 9, 2020 at 14:43