Let $X=\operatorname{Spec}\mathbb{C}[x,y,t]/(xy-t)$, $Y=\operatorname{Spec}K[x,y]/(xy-t)\rightarrow \operatorname{Spec}K$ and $Z=\operatorname{Spec}R[x,y]/(xy-t)\rightarrow \operatorname{Spec}R$, where $K$ is the rational function field with variable $t$, and $R$ is the Laurent series ring of variable $t$.

Question: What are the points in $Y$ and $Z$?

I know that points of $X$ are the follows. The generic point $\xi$ is given by the ideal $(0)$, and corresponding to $V=\{ xy=t\}\subset \mathbb{C}^{3}$. Any point of $V$ is a closed point of $X$. We also some points in the middle, for example one corresponding to curve $\{ xy=t\}$ with a fixed $t\neq0$, or $\{x=0, t=0\}$ or $\{y=0, t=0\}$ or $\{xa=t, y=a\}$ etc.

What's the analog description of $Y$ and $Z$?

  • $\begingroup$ Thanks zcn for the answer, and thanks user26857 for the explanation. I know how to do that now. Sorry, new guy in town. $\endgroup$ – user180625 Oct 13 '14 at 9:18

If $S$ is any ring, $u$ is a unit in $S$, $x, y$ are indeterminates over $S$, then $S[x,y]/(xy-u)$ is isomorphic to the localization $S[x,x^{-1}]$, i.e. Laurent polynomials over $S$. The points of $\text{Spec}(S[x,x^{-1}])$ correspond precisely to the prime ideals of $S[x]$ not containing $x$.

Now suppose $S$ is a field. Then $S[x]$ is a PID, and prime ideals of $S[x]$ are given by $(f(x))$, where $f$ is an irreducible polynomial over $S$, along with $(0)$; moreover the only prime ideal containing $x$ is $(x)$ itself. So $\dim S[x,x^{-1}] = 1$, the generic point is $(0)$, and every other point is closed.

This is enough to cover both instances: for $Y$, $K = \mathbb{C}(t)$ is a field where $t$ is a unit, and for $Z$, $R = \mathbb{C}[[t, t^{-1}]]$ is also a field where $t$ is invertible (note that $R$ is the localization of $\mathbb{C}[[t]]$ at the multiplicative set $\{t^n\}$, but $\mathbb{C}[[t]]$ is local with maximal ideal generated by $t$). Explicitly,

$$Y \leftrightarrow \{(f) \mid f \in K[x] \text{ irreducible over } K, f \ne x\} \cup \{(0)\}$$ $$Z \leftrightarrow \{(f) \mid f \in R[x] \text{ irreducible over } R, f \ne x\} \cup \{(0)\}$$


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