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.
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 $1$.