Is $\mathfrak{a}\otimes_{k[x_1,x_2]}A\to A$ injective? I am considering the algebra $A$ generated by four elements over $k$ (some field), subject to the relations
$$x_1^2=x_3^2\quad x_2^2=x_4^2\quad x_1x_2=x_3x_4\quad x_1x_3=x_2x_4$$
and I'm considering it as a module over the polynomial ring $k[x_1,x_2]$ and I would like to know if the following is true: define a map
$$\mathfrak{a}\otimes_{k[x_1,x_2]}A\to A\quad f\otimes g\mapsto fg$$
with $\mathfrak{a}\subseteq k[x_1,x_2]$ an ideal. Is this map injective? I tried using the relations to minimise the possible powers of $x_3$ and $x_4$. That is, to write a generic element of $A$ as
$$g=x_3g_1(x_1)+x_3g_2(x_2)+x_4g_3(x_2)+g_4(x_1,y_2)$$
with $g_i$ being polynomials, in the hopes that multiplying this by some element of $k[x_1,x_2]$ would reveal that this cannot be zero, unless $f=0$ or $g=0$. However, the fact that $x_2x_4=x_1x_3$ does not make the desired result obvious. Furthermore, I don't think I'm utilising the properties of the tensor product here. Could someone suggest a better approach to this problem? I would appreciate a hint in the right direction.
 A: Consider the case $\mathfrak{a} = \langle x_1^2 - x_2^2 \rangle$.  Then since $\mathfrak{a}$ is a principal ideal of an integral domain, it is isomorphic as a module to $k[x_1, x_2]$, and we have $\mathfrak{a} \otimes{k[x_1, x_2]} A \simeq A$.  Under this identification, $(x_1^2 - x_2^2) \otimes x_1$ corresponds to $x_1 \ne 0$.
On the other hand, under the multiplication map $\mathfrak{a} \otimes_{k[x_1,x_2]} A \to A$, we have $(x_1^2 - x_2^2) \otimes x_1 \mapsto x_1 (x_1^2 - x_2^2) = 0$.  (I found this last relationship using Macaulay 2 in the case $k = \mathbb{Q}$.  It looks like there should probably be a more elementary argument using the relations, but I was too lazy to track down the details of that.  There might also be some questions of whether the eventual argument would work for example if $k$ has characteristic 2; but again, my suspicion would be that some elementary argument transforming $x_1^3$ into $x_1 x_2^2$ using the difference of monomial relations should work over fields of any characteristic.)

As a brief indication of what led me to considering the ideal $\langle x_1^2 - x_2^2 \rangle$ in particular: let us consider the case $k = \mathbb{C}$ and find the set of (closed) points of the closed subset of $\mathbb{A}^4_k$ corresponding to $A$.  Here, from the first two relations, we have $x_3 = \pm x_1$ and $x_4 = \pm x_2$.  Then $x_1 x_2 = x_3 x_4$ gives that the signs in both cases must be the same (or can be chosen to be the same in case one of $x_1, x_2$ is zero); and $x_1 x_3 = x_2 x_4$ then implies that $x_1^2 = x_2^2$.  We thus get that the closed subset is $\{ (t, t, t, t) \} \cup \{ (t, t, -t, -t) \} \cup \{ (t, -t, -t, t) \} \cup \{ (t, -t, t, -t) \}$.  Thus, the image of the projection onto $\mathbb{A}_k^2$ via the first two coordinates is $\{ (t, t) \} \cup \{ (t, -t) \}$ which is precisely the zero set of $x_1^2 - x_2^2$.
