# Example for an ideal which is not flat (and explicit witness for this fact)

I'm looking for an ideal $\mathfrak{a}$ of an commutative (possibly nice) ring $A$ together with an injective $A$-module homomorphism $M\hookrightarrow N$ such that the induced map $\mathfrak{a}\otimes_A M\to\mathfrak{a}\otimes_A N$ is not injective.

I want to see such a concrete example since at the moment I have a hard time understanding why it is even possible that this happens, probably because I don't have an intuition about what it means to tensorize with an ideal. Is there any good way to think about this? Or maybe the problem is that all ideals of nice rings are flat? Are there criteria for this?

• Or maybe the problem is that all ideals of nice rings are flat? Are there criteria for this? among integral domains, the ones whose ideals are all flat are the Prufer domains. any other domain will do. – rschwieb Jul 12 '15 at 3:05

First of all we are looking for an ideal $\mathfrak a$ which isn't flat. (A simple example is $\mathfrak a=(X,Y)$ in $A=K[X,Y]$.)
It would be nice if $\mathfrak a$ contains an $A$-sequence $a,b$, that is, $a$ is a non-zero divisor on $A$ and $b$ is a non-zero divisor on $A/(a)$. (For instance, in the previous example $a=X$ and $b=Y$.)
Then take $M=N=A/(a)$, and the multiplication by $b$ which is clearly injective.
We thus have a homomorphism $\mathfrak a/a\mathfrak a\to\mathfrak a/a\mathfrak a$ which is given by the multiplication with $b$. However, the last map is not injective since $b\hat a=0$ (why?) while $\hat a\ne0$ (why?).
I believe this works. Consider $A=k[x,y]$, $I=(x,y)$ (note that the ideal correspond to a codimension 2 subset). Consider the map $\varphi:k[x,y]/(x)\rightarrow k[x,y]/(x)$ which is the multiplication by $y$. It is obviously injective.
Now consider $(x,y)\otimes\varphi:(x,y)\otimes_A k[x,y]/(x)\rightarrow (x,y)\otimes_A k[x,y]/(x)$.
The element $x\otimes 1$ is not zero in $(x,y)\otimes_A k[x,y]/(x)=(x,y)/x(x,y)$ but its image is $x\otimes_A y=xy\otimes_A 1=y\otimes_A x=0$ because $x=0$ in $k[y]=k[x,y]/(x)$. Hence $\varphi$ is not injective.