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

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) .
