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Let $k$ be a field. Let $R=k[X,Y],\ A=k[X,Y,\frac{X}{Y}]$. Show that going down doesn't hold for $A/R$.

My approach: I took the map $\phi: \operatorname{Spec} A\to \operatorname{Spec} R,\ q\mapsto q\cap R$ to verify the preimages of $q$ in $R$. Sadly, I have not been able to find a counterexample to my problem. I suspect that one could exploit the fact that $k[Y]\in \operatorname{Spec} R, \ k[Y]\not\in \operatorname{Spec} A$. However, to make this work as a counterexample, I would have to find a prime ideal in $R$ containing $k[Y]$ and a prime ideal in $A$ lying above it. This, however, is impossible due to $k[Y]$ being a maximal ideal. Any help is greatly appreciated!

user26857
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Edward Teach
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    $k[Y]$ (the polynomial ring in $Y$) is not an ideal of either rings as it wouldn't contain $XY$. – daruma Jun 04 '21 at 17:59
  • Yes, of course. You're right. Then I'm all out of ideas :) – Edward Teach Jun 04 '21 at 18:33
  • Which property of the going-down theorem doesn't hold for this extension of rings? – daruma Jun 04 '21 at 18:39
  • The standard property for going-down: Let $p\subseteq p'$ be prime ideals of $R$. And let $q\in Spec A$ s.t. $q\sqcap R=p'\supseteq p$. The going-down property holds if there is such a $q'\in Spec A$ s.t. $q'\sqcap R=p$. Here $\sqcap$ denotes the preimage under $R$. I interpreted it as the intersection with $R$ as written in the question – Edward Teach Jun 04 '21 at 18:42
  • I mean which hypotheses of the going-down theorem isn't satisfied? You need certain things on your rings for this theorem to hold. (Not asking for the going-down property) – daruma Jun 04 '21 at 18:44
  • I am sorry. I am afraid I don't quite understand: The only required property, as far as I am aware, is that $A$ is an $R$-algebra. This should, however, be the case in this exercise. – Edward Teach Jun 04 '21 at 18:50
  • Otherwise, the required hypothesis to disprove is just the property, I guess. I.e. find 4 such ideals so that it doesn't hold. At least, that's how I understand the exercise – Edward Teach Jun 04 '21 at 18:55
  • The going-down property is not true for arbitrary extension of (commutative) rings. If $A$ is an integral extension $R$ and $R$ is integrally closed then $A/R$ satisfies the going-down property. This statement is called the going-down theorem. The above extension is not an integral extension so that will give you a hint as to what you want to work with. – daruma Jun 04 '21 at 18:59
  • Can you give an example of chain of prime ideals in $k[X,Y]$? Do you know what the prime ideals of polynomial rings over fields look like? – daruma Jun 04 '21 at 19:13
  • Yes. I have proved that a while back: As far as I know there are three kinds of prime ideals: The $(0)$ ideal, the principal ideal $(f(X,Y))$ where $f$ is irr. over $R$ as well as the ideal $(f(X),g(X,Y))$ where $f[X]$ is prime over $\mathfrak k[X]$ and $g(X,Y)$ is irr. over $(\mathfrak k[X]/(f(X)))[Y]$. – Edward Teach Jun 04 '21 at 19:16

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You should note that this extension is not integral as $X/Y$ is not integral over $k[X,Y]$. Indeed this element is going to be important in constructing your counterexample. The key observation is $X=(X/Y)\cdot Y$.

Consider the chain of ideals $\mathfrak{p}_1\supset\mathfrak{p}_2$ where $\mathfrak{p}_1=(X,Y), \mathfrak{p}_2=(X)$ in $k[X,Y]$. This is a chain of prime ideals(if you don't know why these are prime ideals, you should check they are by looking at the quotients).

Now, $\mathfrak{q}_1=(X,Y)\subset k[X,Y,X/Y]$ is a prime ideal. Indeed the quotient is $k[X/Y]\cong k[Z]$ which is an integral domain. Note that $\mathfrak{q}_1$ lies over $\mathfrak{p}_1$.

Now I want to find a prime ideal $\frak{q}_2$ lying over $\mathfrak{p}_2=(X)$. Is $(X)\subset k[X,Y,X/Y]$ a prime ideal? Unfortunately it isn't as $X=(X/Y)\cdot Y$. So what is a prime ideal lying over $\mathfrak{p}_1$? It turns out $(X/Y)$ is the only choice. But if we let $\mathfrak{q}_2=(X/Y)$ then $\mathfrak{q}_1\not\supset \mathfrak{q}_2$.

daruma
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    Can we also use Krull's principal ideal theorem? q1 is principal and this is a noetherian domain, so the only prime contained inside q1 is 0, right? – Jack Schmidt Jun 04 '21 at 19:40
  • dear @JackSchmidt in fact, Krull's principal ideal theorem is in my opinion slightly overkill here; see the lemma here for a direct proof that proper principal ideals in a Noetherian domain do not contain any non-zero prime ideals. of course the argument of daruma is quite clean as well (+1) – Atticus Stonestrom Jun 04 '21 at 20:42
  • @daruma thank you very much! I don't know why, but I completely missed the maximal ideal (X,Y). I have thought of the chain (X) and (XY), but forgot that the comma doesn't generate the units. Beautiful proof and thank you once again! – Edward Teach Jun 04 '21 at 21:46
  • Why is $(X/Y)$ the only choice in this case? – Meliodas Jun 04 '21 at 23:32