Counter-example to the Krull intersection theorem In this overflow post some counter-examples to the Krull intersection theorem are given. I'm interested in the counter-example given by Dustin Cartwright. He takes $$R = \mathbb{Q}[x,z,y_1, y_2, \dots]/ (x - y_1 z, x - y_2 z^2, x - y_3z^3, ...)$$ and $I = (\overline{z})$. Then he claims that $\cap_{j \in \mathbb{N}} I^j = (\overline{x})$, but $(\overline{z})(\overline{x}) \neq (\overline{x})$, where $\overline{x}$ and $\overline{z}$ are the images of $x$ and $z$ in the quotient.
I would like to know how to prove both of these last statements. The inclusion $\cap_{j \in \mathbb{N}} I^j \supseteq (\overline{x})$ is clear, but I don't see how to get the reverse inclusion. I tried writing letting $f \in \mathbb{Q}[x,z, y_1, y_2\dots]$ be a polynomial such that $\overline{f} \in \cap_{j \in \mathbb{N}} I^j$ and using the fact that $\overline{f y_j} \in (\overline{x})$ for every $j \in \mathbb{N}$, but I didn't manage to get anything from this. Similarly, I couldn't sort out why $(\overline{z})(\overline{x}) \neq (\overline{x})$ is true.
 A: Let $A=\mathbb{Q}[x,z,y_1,y_2,\dots]$ and consider $J:=(x-y_1z,x-y_2z^2,x-y_3z^3,\dots)$ as an ideal of $A$. Pulling back along the projection map $A\to R$ turns the desired fact about the ideal $I<R$ into a statement about the ideal $(J+Az)<A$; namely, what you wish to show for your first question is that $$\bigcap_{k\in\mathbb{N}}\left[J+(J+Az)^k\right]\subseteq J+Ax.$$ (Personally, I think things become clearer in this reframing, since $A$ is a polynomial ring over a field and hence must more intuitive and familiar than the ring $R$. I will thus prove the statements in the context of $A$ rather than $R$.) Now, first note that $$J+Ax=(x,y_1z,y_2z^2,y_3z^3,\dots)$$ and that $J+Az=(x,z)$, and thus that $$(J+Az)^k=(x,z)^k=(x^k,x^{k-1}z,\dots,xz^{k-1},z^k)$$ for each $k\in\mathbb{N}$. In particular, $J+(J+Az)^k=(x,y_1z,\dots,y_{k-1}z^{k-1},z^k)$. Now suppose that $f$ lies in the intersection on the left hand side. To show $f\in J+Ax$, it suffices to show that every monomial term of $f$ is either divisible by $x$ or divisible by $y_nz^n$ for some $n\in\mathbb{N}$. Suppose therefore for contradiction that this is not the case; then there is a monomial term in $f$ of form $y_{m_1}\dots y_{m_l}z^n$, where $m_i>n$ for each $i$, the $m_i$ are not necessarily distinct, and one of $n$ and $l$ may be zero. But this means that $f\notin\left[J+(J+Az)^{n+1}\right]$, since every generator of that ideal is a monomial and either divisible by $x$, divisible by $y_m$ for some $m\leqslant n$, or divisible by $z^{n+1}$. Thus $f$ cannot lie in the intersection on the left, and this gives the desired contradiction.

Now let's pull back the second statement about the ideals $(\overline{x}),(\overline{z})<R$ into a statement about the ideals $J+Ax,J+Az<A$. What you wish to show is that $$J+(J+Ax)(J+Az)\neq J+Ax.$$ It suffices to show that $x\notin J+(J+Ax)(J+Az)$, so recall that $J+Az=(x,z)$ and that $J+Ax=(x,y_1z,y_2z^2,y_3z^3,\dots)$. In particular, we have
\begin{align}
(J+Ax)(J+Az)=(x^2,xy_1z,xy_2z^2,\dots xz,y_1z^2,y_2z^3,\dots).
\end{align} In particular, every element of $(J+Ax)(J+Az)$ has degree at least $2$. Also, every monomial term appearing in each generator of $J$ has degree at least $1$. Thus, if some $A$-linear combination of the generators of $J$ and the generators of $(J+Ax)(J+Az)$ were equal to $x$, the coefficient of at least one of the generators of $J$ would have to have a non-zero constant term from $\mathbb{Q}$. Suppose for contradiction that there is such an $A$-linear combination, and let $x-y_{n}z^{n}$ be one of the generators of $J$ whose coefficient from $A$ has a non-zero constant term; call this constant term $\lambda\in\mathbb{Q}^\times$.
Note that $\lambda y_nz^n$ cannot be a monomial term in any element of $(J+Ax)(J+Az)$, since every generator of that ideal is a monomial and is either divisible by $x$, divisible by some $y_m$ where $m\neq n$, or divisible by $z^{n+1}$. On the other hand, if $J'$ is the ideal generated by all of the generators of $J$ except $x-y_nz^n$, then no element of $J'$ can have a term $\lambda y_nz^n$ either, since every monomial term of every generator of $J'$ is either divisible by $x$ or divisible by some $y_m$ where $m\neq n$. Thus $\lambda y_nz^n$ must appear as a monomial term in any $A$-linear combination of the generators of $J$ and $(J+Ax)(J+Az)$ for which the $A$-coefficient of $x-y_nz^n$ has constant term $\lambda$. In particular, no such $A$-linear combination can be equal to $x$ if $\lambda\neq 0$, thus giving the desired contradiction.
