# Necessity of an assumption in the Going-down Theorem

In the Going-down Theorem (Theorem 5.16 p. 64) in Introduction to Commutative Algebra by Atiyah and MacDonald, we have an integral inclusion $A\subset B$ of domains, $A$ being integrally closed. The assumption that $B$ is a domain is used in the proof, and I suspect that this assumption is necessary, but I haven't been able to prove it.

Here is a way of stating the question:

Let $A\subset B$ be an integral inclusion of (commutative) rings, $A$ being an integrally closed domain. Let $\mathfrak m$ be a maximal ideal of $B$, let $\mathfrak p$ be the contraction of $\mathfrak m$ in $A$, and let $\mathfrak p'$ be a prime ideal of $A$ contained in $\mathfrak p$: $$\begin{matrix} B&\supset&\mathfrak m\\ |&&|\\ A&\supset&\mathfrak p&\supset&\mathfrak p' \end{matrix}$$

Is $\mathfrak p'$ the contraction of a prime contained in $\mathfrak m$?

My suspicion is that the answer is "not necessarily".

Such an answer would show the necessity of the assumption that $B$ is a domain.

[The symbol $\subset$ is used as in Bourbaki, i.e. $X\subset X$ for all set $X$.]

• I don't think is really necessary; for some usual statement see here and for some improvements see here. Jul 21, 2017 at 16:27
• @user26857 - Thanks! As you implicitly suggested, the first example on p. 14 of Mel Hochster's text answers the question. (I haven't been able to access the other article.) (Needless to say that I hope you will post answers again!!!) Jul 21, 2017 at 19:18

Here's a counterexample that shows you can't completely drop the assumption of $B$ being an integral domain.

Let $k$ be a field, and consider the commutative rings $\tilde{A}=k[x_1,x_2]$ and $B=k[x_1]\times k[x_1,x_2]$, where $A\subseteq B$ is the image of $\tilde{A}$ under the injection $f(x_1,x_2)\mapsto(f(x_1,0),f(x_1,x_2))$.

First note that $B$ is not an integral domain since $(x_1,0)\cdot(0,x_1)=0.$ Now consider an arbitrary element $(f(x_1),0)\in B$ and note that $$(f(x_1),0)^2-(f(x_1),f(x_1))\cdot(f(x_1),0)=0,\qquad(f(x_1),f(x_1))\in A.$$ So $(f(x_1),0)$ is integral over $A$. Similarly, for $(0,f(x_1,x_2))\in B$ we see that $$(0,f(x_1,x_2))^2 - (f(x_1,0),f(x_1,x_2))\cdot(0,f(x_1,x_2))=0,\qquad(f(x_1,0),f(x_1,x_2))\in A.$$ So $(0,f(x_1,x_2))$ is integral over $A$. Since the set of elements which are integral over $A$ form a subring of $B$, and since elements of these two forms generate $B$, we see that $B$ is integral over $B$.

Finally, we show that the going-down theorem does not hold in this case. Consider the prime ideal $\mathfrak{p}_1=((0,x_2))\subseteq A$. Note that $\mathfrak{q}_1=(0)\times k[x_1,x_2]$ is a prime ideal in $B$ such that $\mathfrak{q}_1\cap A=\mathfrak{p}_1$. Let $\mathfrak{p}_2=((0,0))\subseteq A$. We claim there is no prime ideal $\mathfrak{q}_2\subseteq\mathfrak{q}_1$ such that $\mathfrak{q}_2\cap A=\mathfrak{p}_2$. This is clear since the prime ideals of $B$ are of the form $\mathfrak{p}\times k[x_1,x_2]$ and $k[x_1]\times\mathfrak{q}$ where $\mathfrak{p}$ is a prime ideal of $k[x_1]$ and $\mathfrak{q}$ is a prime ideal of $k[x_1,x_2]$.

• Thanks! It seems to me your last claim contradicts Proposition 3.16 in Atiyah-MacDonald, proposition which says: "Let $A\to B$ be a ring homomorphism and let $\mathfrak p$ be a prime ideal of $A$. Then $\mathfrak p$ is the contraction of a prime ideal of $B$ if and only if $\mathfrak p^{ec}=\mathfrak p$." What am I missing? Jul 21, 2017 at 18:06
• The last claim regarding the structure of the prime ideals of $B$? I'm pretty sure it's correct, see here: math.stackexchange.com/questions/1216203/… . I'll take a look at the Proposition 3.16 and see how it applies here Jul 21, 2017 at 18:10
• I meant the claim "there is no prime ideal $\mathfrak q_2\subseteq\mathfrak q_1$ such that $\mathfrak q_2\cap A=\mathfrak p_2$". (I should have been more precise. Sorry.) Jul 21, 2017 at 18:16
• I think I made a confusion. Proposition 3.16 shows that $\mathfrak p_2$ is the contraction of some prime, but this prime is not contained in $\mathfrak q_1$. Jul 21, 2017 at 18:29
• Yes, you see can see that $\mathfrak{p}_2$ is the contraction of $\mathfrak{q}_*=k[x_1]\times(x_2)$, but this is certainly not contained in $\mathfrak{q}_1$. Jul 21, 2017 at 18:31