Currently I'm reading going down theorem from Atiyah and Macdonald's Commutative Algebra text. I'm trying to find a counterexample if the conditions of the theorem are not satisfied. There is a hint in my class notes about counterexample but I'm unable to complete this. Hint goes as follows:

Let $R= \dfrac{\mathbb R[x,y,z]}{(y^2-x^2-x^3)}$. Let $S$ be the integral closure of $R$ in its quotient field. Show that going down theorem does not hold between $R$ and $S$.

I've computed the ring $S$ and this turns out to be the ring $S=R[\frac {y}{x}]$. Any hints/ideas to "guess" the chain of prime ideals for which GD theorem does not hold.


Note that $R\simeq\mathbb R[t^2-1,t^3-t,z]$ by $x\mapsto t^2-1$, $y\mapsto t^3-t$, $z\mapsto z$, and $S=\mathbb R[t,z]$.

Now I change the notation and set $S=k[t]$, and $R=k[t^2-1,t^3-t]$. (Note that $R=\{f\in k[t]:f(-1)=f(1)\}$.) Moreover, $S[z]$ is the integral closure of $R[z]$ since $S$ is the integral closure of $R$. To show that Going-Down fails for $R[z]\subset S[z]$ consider the prime ideals $\mathfrak p=(z-t-1)\cap R[z]$ and $Q=(t-1,z)\subset S[z]$.

We show that $Q$ lies over $\mathfrak q=(t^2-1,t^3-t,z)\supseteq\mathfrak p$, but $S[z]$ has no prime in $Q$ lying over $\mathfrak p$.

Observe that we also have $(t+1,z)\cap R[z]=\mathfrak q$, so $\mathfrak p\subseteq\mathfrak q$. Since $z\in\mathfrak q\setminus\mathfrak p$ the inclusion is strict. Suppose there is $P\subset Q$ lying over $\mathfrak p$. Then $P$ is principal generated by an irreducible polynomial, say $f$, and $f(1,0)=0$. We have $(f)\cap R[z]=(z-t-1)\cap R[z]$. Since $(t^2-1)(z-t-1)\in(z-t-1)\cap R[z]$ it follows that $(t^2-1)(z-t-1)\in(f)$, so $f\mid(t^2-1)(z-t-1)$. From $f(1,0)=0$ we get the only possibility $f=t-1$. Now $(t^2-1)(t-1)\in(f)\cap R[z]$, so $(t^2-1)(t-1)\in(z-t-1)$, a contradiction.

  • $\begingroup$ Everything seems fine to me. I've some doubts: $1.$ How did you "guess" that $R=\{f\in k[t]:f(-1)=f(1)\}$ although I don't think that you have used this in your solution. $2.$ Is there any reason for choosing $p$ and $Q$ in this particular way? $\endgroup$ – Arpit Kansal Nov 14 '15 at 11:24
  • 1
    $\begingroup$ 1. Actually I've used that description of $R$ for few times, but I guess one can skip it. (I've just borrowed it from the other examples on the failure of GD.) 2. Well, here is where the geometry can help us, and I gave you a link to some notes where this is explained. Algebraically there are two prime (maximal) ideals $Q_1,Q_2$ in $S[z]$ lying over the same prime $\mathfrak q$ in $R[z]$, and $\mathfrak q$ contains another prime $\mathfrak p$ which comes from $S[z]$ uniquely, that is, only one prime $P$ from $S[z]$ lies over $\mathfrak p$, and $P\subset Q_2$, but $P\not\subset Q_1$.) $\endgroup$ – user26857 Nov 14 '15 at 11:45

I followed the sprit of B. Heinzer, http://www.math.purdue.edu/~heinzer/teaching/math557/gdown.pdf.

Let $k$ be an (algebraically closed) field of characteristic zero, and $R = k[x,y,z]/(y^2-x^2-x^3)$. We are going to show that the going down property fails between $R$ and $S$ its integral closure of $R$.

Let $K = Q(R)$ be the field of fractions of $R$. Since $\frac{y}{x} \in K$ satisfies the equation $T^2 - 1 - x = 0$, $R[\frac{y}{x}] \subseteq S$. Notice that $R[ \frac{y}{x} ] \cong k[x,y,z,U]/ (xU- y, U^2 -1 -x) \cong k[z,U]$. Hence $S = R[\frac{y}{x}]$.

We compute $J$ the Jacobian ideal of $R$: $J = I_1 ([ -2x -3x^2 \quad 2y \quad 0]) = (2x+3x^2 , 2y)R = ( 2x+3x^2, y)R$. Since $R$ is a complete intersection, the non-normal locus is determined by $J$.

Let $Q = (\frac{y}{x} - z)S$. Then $y-xz , z^2 -1 -x \in Q \cap R$. If $J \subseteq Q \cap R$, then $y \in Q \cap R$. Then the height of $Q \cap R$ is at least $2$, and this is a contradiction. Therefore, $p = Q \cap R$ does not contain $J$. Hence $R_p$ is a DVR (so it is integrally closed), and we have $S \subseteq R_p = S_p$, i.e., $Q$ is the only prime lying over $p$. In fact, $p = (y-xz, z^2 - 1-x)R$. (Check that $R/ (y-xz, z^2-1-x)$ is isomorphic to $k[z]$.)

We are going to construct a maximal ideal $M$ of $S$ such that $p \in M$, but $Q \nsubseteq M$. Let $M = (x,y,z+1, \frac{y}{x}-1)$. It is easy to show that $Q \nsubseteq M$, but $p \subseteq M$. Notice that $M \cap R$ is a maximal ideal, so $p_1= M \cap R \supsetneq p$. But there is no prime in $M$ which contracts to $p$. (Recall that $Q$ is the only prime lying over $p$, and $Q \nsubseteq M$.) Hence the going-down property fails.

  • The argument can to modified to work with non algebraically closed field $k$. The only place we use the assumption is using the Jacobian ideal to say $R_p$ is a DVR. This can be easily checked: Since $x$ are not in $p$, $R_p = R[\frac{1}{x}]_p$. But in $R[\frac{1}{x}]$, $p R[\frac{1}{x}] = (y-xz, z^2 -1 -x)R[\frac{1}{x}] = (\frac{y}{x}-z, z^2 - 1 -x) R[\frac{1}{x}] = (\frac{y}{x} - z)R[\frac{1}{x}]$.

  • I would like comment that an assumption on the characteristic of $k$ is necessary ($2 \neq 0$) in order to show the (non)-containment $Q \nsubseteq M$.

  • $\begingroup$ I'm trying to prove $R/(y-xz,z^2-1-x)$ is isomorphic to $k[z]$. Consider the map $x \mapsto z^2-1, y \mapsto z^3 -z, z \mapsto z$. clearly $p$ is in kernal but how to prove this is exactly the kernal? $\endgroup$ – user270331 Jan 23 '18 at 5:39
  • $\begingroup$ Is your $R = k[x,y,z]$? Then first go modulo $y-xz$, to replace $y$ by $xz$, and you can do the same for $x$ with the other relation. $\endgroup$ – Youngsu Jan 30 '18 at 17:11
  • $\begingroup$ No, $R= k[x,y,z]/(y^2-x^2-x^3)$. $\endgroup$ – user270331 Jan 30 '18 at 19:22
  • $\begingroup$ You will have the same conclusion since $y^2-x^2-x^3 \in (y-xz, z^2-1-x)$. $\endgroup$ – Youngsu Jan 31 '18 at 16:29

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