Is $\mathbb Q[x,z]$ as a $\mathbb Q[x,y]$-module (with morphism $x\mapsto x$, $y\mapsto xz$) flat?

Question

This was an exercise in my class, please help:

Put $A = {\mathbb Q}[x,y]$ and $B = {\mathbb Q}[x,z]$. Consider the morphism $f \colon A \to B$ of ${\mathbb Q}$-algebras given by $x \mapsto x$, $y \mapsto x z$. Then $B$ is an $A$-module. Is $B$ flat ? [Hint: Consider the inclusion $(x,y) \subset A$.]

My guess is that it isn't flat, by using their hint I found that the map

$(x,y)\otimes_A B\to A\otimes_A B$

sends $x\otimes z - y\otimes 1$ to $0$, so if this was nonzero I would be done. But I have a hard time proving that $x\otimes z - y\otimes 1$ is nonzero.

Answer 1

I don't follow Mindlack's proof, as I'm not sure what their $C$ is. The proof I had in mind when posting the question was the following:

For the justification that $$y \otimes 1 \neq x \otimes z,$$ could use the "truncated Koszul complex" $$A \to A \oplus A \to (x,y) \to 0;$$ here the first map sends 1 to $(y,-x)$ and the second map sends $$(1,0) \mapsto x \text{ and }(0,1) \mapsto y.$$ One can check without too much trouble that this is an exact sequence (actually the first map is also injective) .

Now we use that the tensor product is right exact. Tensoring the sequence over $A$ with $B$ yields $$B \to B \oplus B \to (x,y) \otimes_A B \to 0$$ where first map sends $1$ to $(xz,-x)$, and the second sends $$(1,0) \mapsto x\otimes 1 \text{ and } (0,1)\mapsto y\otimes 1.$$ We want to show that $$y \otimes 1 - x \otimes z$$ does not lift to the first copy of $B$. We choose a lift to $ B \oplus B$, say given by $ (-z,1)$. This is not in the image of the first map because neither side is divisible by $x$ (note that $B$ is a UFD) .

Answer 2

Given what the element is, it is enough to show (by the property of the tensor product) that there is an $A$-bilinear $\beta: (x,y) \times B \rightarrow C$ with $\beta(x,z) \neq \beta(y,1)$.

Take $C=\mathbb{Q}$ where $x$ and $y$ act as zero, and define $\beta(P,Q)=Q(0,1)\frac{\partial P}{\partial y}(0,0)$. This is $\mathbb{Q}$-bilinear, and $\beta(y,1)=1$ by definition, while $\beta(x,z)=0$. Moreover, $\beta(xP,Q) = 0 = \beta(P,xQ)$, $\beta(yP,Q)=0$ because $P \in (x,y)$ already, and $\beta(P,y \cdot Q)=\beta(P,xzQ)=\beta(P,x\cdot (zQ)) = 0$ by the above ($\cdot$ denoting the $A$-actions), so $\beta$ is $A$-bilinear.

Here is another proof with almost no calculation. If $B$ is flat over $A$, then $B/xB$ is flat over $A/(x)=\mathbb{Q}[y]$, and given that $A/(x)$ is a PID, $B/xB$ must be torsion-free. But $B/xB$ is nonzero, yet $y \cdot B/xB=0$.