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Let $k$ be a field, not necessarily algebraically closed. Then how would you show that the extension $k[x] \subset k[x,y]$ does or does not satisfy Going-Up?

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up vote 5 down vote accepted

No, that extension does not satisfy Going-Up.

Indeed the associated morphism $f:\mathbb A^2_k \to \mathbb A^1_k$ is the first projection and is well known not to be a closed map:
the closed hyperbola $V(xy -1)\subset \mathbb A^2_k $ projects to $\mathbb A^1_k \setminus V(x)=\mathbb A^1_k \setminus \lbrace O \rbrace \subset \mathbb A^1_k $, which is not closed in $\mathbb A^1_k$

What has this got to do with Going-Up? A lot! Atiyah-Macdonald prove in (Exercise 6.11, page 79) that a morphism of rings $A\to B$ with $B$ noetherian satisfies Going-Up if and only if the associated morphism of affine schemes $Spec(B)\to Spec(A)$ is closed.
This proves that your inclusion$k[x] \to k[x,y]$ does not satisfy Going-Up.

The direct way Alternatively, we can atttack the problem frontally.
Consider the inclusion of prime ideals $(0)\subsetneq (x)$ in $k[x]$ and the prime ideal $(xy-1)$ in $k[x,y]$ above $(0)$ (= whose intersection with $k[x]$ is $(0)$).
Going-Up does not hold because there is no prime ideal ${\mathfrak q}\subset k[x,y]$ with both $(xy-1) \subset {\mathfrak q}$ and ${\mathfrak q} \cap k[x]=(x)$ [since then we would have $xy-1,x\in {\mathfrak q}$, so that $-1\in {\mathfrak q}$: this is impossible for a prime ideal].

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Thank you for the direct way. While I am aware of the closed map criterion, I was not sure how to show it does not hold. – Rankeya Oct 11 '11 at 0:18

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