# Example of two prime ideals whose intersection of the squares not equal to the square of the intersection

In this topic the OP raised the following question:

Let $R$ be a commutative noetherian ring and $\mathfrak p,\mathfrak q \in \operatorname{Spec}(R)$. Is it true that $(\mathfrak p\cap \mathfrak q)^2=\mathfrak p^2 \cap \mathfrak q^2$?

Obviously, we always have $(\mathfrak p\cap \mathfrak q)^2 \subseteq \mathfrak p^2 \cap \mathfrak q^2$ and there is no reason to think that, in general, the converse holds.

What remained unsolved in that topic is to

Give an example of prime ideals $\mathfrak p,\mathfrak q$ (in a noetherian ring) such that $$(\mathfrak p\cap \mathfrak q)^2 \neq \mathfrak p^2 \cap \mathfrak q^2.$$

Edit. It would be nice to have such an example for $R$ a noetherian integral domain.

• This is quite interesting. There must be some geometric way to think about this that would give an idea for a counterexample (or suggest a proof). I'm trying to think if embedded points/non-Cohen-Macaulay type things do the trick.
– Matt
Jul 19, 2013 at 23:57
• Let $b_1,\cdots,b_s,d_1,\cdots,d_t \in \mathbb{C}^n$. Then the ideals of $\mathbb{C}[x]=\mathbb{C}[x_1,\cdots,x_n]$ defined by $I_1 = (b_1^Tx,\cdots,b_s^Tx), I_2 = (d_1^Tx, \cdots,d_t^Tx)$ are prime. Do they satisfy inequality or equality? I am not sure how to approach this. Jul 20, 2013 at 14:40
• @Manos It's hard to believe that $I_1,I_2$ are prime ideals without any constraint on the vectors $b_i,d_j$.
– user26857
Jul 20, 2013 at 15:47
• @YACP: yes, you are right, let's assume that the set of $b_i$ is linearly independent and the set of $d_j$ is also linearly independent. Then they are prime. Jul 20, 2013 at 16:27

What about the following: Let $R$ be the ring $k[x,y]/(x^2y-y^2x)$. Let us take $\mathfrak p=(x)$ and $\mathfrak q=(y)$. These are prime ideals. Furthermore $\mathfrak p\cap \mathfrak q=(xy)$, so $(\mathfrak p\cap \mathfrak q)^2=(x^2y^2)$. But, $x^2y=y^2x\in \mathfrak p^2\cap \mathfrak q^2.$

Edit. The same idea works for integral domains. Let $R=k[x,y,u,v]/(x^2y-u^2v)$. Then take $\mathfrak p=(x,v)$ and $\mathfrak q=(y,u)$, these are prime ideals. Then $\mathfrak p\cap \mathfrak q=(xy,xu,vy,vu)$, so $(\mathfrak p\cap \mathfrak q)^2$ is generated by elements of degree $4$. But $x^2y=u^2v$ is in $\mathfrak p^2\cap \mathfrak q^2$.

• Ah. So this has nothing to do with "badness" of the ring, but rather the "size." Geometric interpretation: Note the entire example is built on just the xy-axis even though the space is the xy-axis union the extra line $y=x$. The LHS says find functions that vanish along $x$ and $y$, then double the order of vanishing. The RHS says find functions that vanish to at least order $2$ along $x$ and $y$. If you try the obvious thing just on the space that is the $xy$-axis, then you don't have enough functions to produce some single global function different from coming from the LHS.
– Matt
Jul 21, 2013 at 16:10
• (cont). Thus the a "post-obvious" thing to do is to stick the example inside some larger space where you have lots of extra functions and can find one with the desired vanishing properties that doesn't come from the localized example. Thanks!!
– Matt
Jul 21, 2013 at 16:11

How about $R=k[x,y,u,v]/(xy-uv)$. Let $I=(x,y)$, $J=(u,v)$. Clearly $xy=uv\in I^2\cap J^2$, but $xy\notin (I\cap J)^2$.

Edit. Let $R=k[x,y,u,v,a,b,c,d]/(xy+uv-ab-cd)$, $\mathfrak p=(x,y,u,v)$, $\mathfrak q=(a,b,c,d)$. Clearly $t:=xy+uv=ab+cd\in\mathfrak p^2\cap\mathfrak q^2$, but $t\notin (\mathfrak p\cap\mathfrak q)^2$.

• True, but $I, J$ are not prime.
– user18119
Jul 21, 2013 at 16:51
• Even though it doesn't answer the original question, it could be turned into something like: a simpler example with $I^2\cap J^2\ne (I\cap J)^2$ for non necessarily prime ideals.
– user18119
Jul 22, 2013 at 10:32
• @QiL8, I somehow had ideals in mind, I had read it as prime ideals, but i came to thinking about it 2 days later and forgot that the problem asked for prime ideals. Jul 22, 2013 at 13:12
• @QiL8, I think i fixed it, I hope. Jul 22, 2013 at 14:29