Let $k$ be a field and let $a\in k$ be a non zero element. Consider the quotient ring $A = k[x,y]/(xy)$. If $f\in k[x,y]$ we denote by $\overline{f}$ its image in $A$. Now consider the quotient ring $B = A/(\overline{x+y-a})$. I want to show that $B$ is isomorphic, as a ring, to $k\times k$.
Here is my attempt: Consider the ring homomorphism $$ k[x,y]\to k\times k, \quad f(x,y)\mapsto (f(a,0),f(0,a)). $$ Its kernel crearly contains $(xy)$ so it induces a homomorphism $$ \varphi':A \to k\times k, \quad \overline{f(x,y)} \mapsto (f(a,0),f(0,a)). $$ Finally, it is clear that the kernel of this homomorphism contains $I=(\overline{x+y-a})$ so it induces a ring homomorphism $$ \varphi:B\to k\times k, \quad \overline{f(x,y)}+I \mapsto (f(a,0),f(0,a)). $$ It is easy to see that $\varphi$ is surjective. So I need to show that this is injective or, equivalently, that $\ker(\varphi')=I$. Here is where I'm stuck.
I also tried by defining $$ \eta:k\times k\to B, \quad (r,s) \mapsto \overline{a^{-1}(rx+sy)}. $$ This can be shown (by using the defining relations of $B$) to be a ring homomorphism and we have that $\varphi\circ\eta$ is the identity map. Thus we can try by proving that $\eta\circ \varphi$ is the identity map. I'm stuck here too.
Another attempt is the following: Intuitively, the relation $\overline{x}= \overline{a} - \overline{y}$ tells me that we can conjecture the isomorphism $$ B\cong k[x]/(x(x-a)), $$ by first defining $$ k[x,y]\to k[x]/(x(x-a)), \quad f(x,y)\mapsto f(x,a-x), $$ which clearly induces a homomorphism $$ A\to k[x]/(x(x-a)), $$ which in turn induces $$ \theta: B\to k[x]/(x(x-a)). $$ Now the aim is to prove that this last map is a ring isomorphism. If done, then as $(x)$ and $(x-a)$ are comaximal ideals, the Chinese Remainder Theorem implies that $$ k[x]/(x(x-a)) \cong \frac{k[x]}{(x)}\times \frac{k[x]}{(x-a)}\cong k\times k $$ and we are done. But I don't know how to show that $\theta$ is an isomorphism.
Thank you in advance