Projection $\operatorname{Spec}(K[x,y]) \rightarrow \operatorname{Spec}(K[x])$ not closed Let $K$ be a field. I am trying to show that the map $\operatorname{Spec}(K[x,y]) \rightarrow \operatorname{Spec}(K[x])$ induced by the inclusion $K[x] \hookrightarrow K[x,y]$ is not closed (topologically). My "geometric" idea was to show that the image of $V(xy-1) \subseteq \operatorname{Spec}(K[x,y])$ is exactly $\operatorname{Spec}(K[x]) \setminus \lbrace 0 \rbrace$ which is not open. This comes from the intuition that this map "is the projection $\mathbb{A}_K^2 \rightarrow \mathbb{A}_K$", but I do not know how to show this rigurously.
 A: You're on the right track.
To look at the image of $V(xy-1)$ under the projection, consider the composite morphism $\operatorname{Spec} k[x,y]/(xy-1) \to \operatorname{Spec} k[x,y] \to \operatorname{Spec} k[x]$, which corresponds to the ring morphism $k[x]\to k[x,y]/(xy-1)$ by $x\mapsto x$. To show that this map is not closed, we can show that it has nonclosed image.
The closed subsets of $\operatorname{Spec} k[x]$ are the empty set, the entire space, and finite unions of closed points. Since $k[x,y]/(xy-1)$ is integral, $\operatorname{Spec} k[x,y]/(xy-1)$ is irreducible; combining this with the fact that the image of an irreducible topological space is irreducible, we see that the if the image of $\operatorname{Spec} k[x,y]/(xy-1)$ in $\operatorname{Spec} k[x]$ is closed, it can only be the whole space or a single closed point.
It cannot be the whole space: there is nothing that maps to $(x)$, as $k[x,y]/(xy-1)\otimes_{k[x]} k[x]/(x) = 0$.
It cannot be a single closed point: the generic point is in the image (look at the zero ideal).
Thus this map is not closed.
A: Denote the induced function $\operatorname{Spec}(K[x,y]) \to \operatorname{Spec}(K[x])$ by $\pi$. Two steps:

*

*What is $\pi\big((xy-1)K[x,y]\big)$? This should tell you immediately what the image of $V(xy-1)$ must be if it is to be closed.

*Why is $xK[x]$ (the maximal ideal which corresponds to the point $0$ in $\mathbb A_K$) not in the image $\pi\big(V(xy-1)\big)$? This is actually really easy: what is $x K[x,y] + (xy-1)K[x,y]$?

