I am reading a proof of Bezout theorem for algebraic curves and I encountered this morsphism that the author says is an isomorphism by the chinese remainder theorem: $$\tau:k[[x]][y]/(f_a,g_a) \rightarrow\prod_{i=1}^ n k[[x,y]]/(f_{p_i},g_{p_i})$$
defined by $\tau(h) = (h_{p_1},\dots, h_{p_n})$, where the subindex ${p_i}$ means that if $h \in k[[x,y]]$ then $h_{p_i}(x,y)=h(x+a,y+p_i)$ and $f_{a}(x,y)=f(x+a,y)$, the same for $g_a$. The morphism $\tau$ maps injectively from $k[[x]][y]$ to $k[[x,y]]$. However, I can't see why this is true.
More context about the question:
$k[[x,y]]$ is the ring of formal series in two variables.
The text proves a proposition about two polynomials $f$ and $g$ in $k[x,y]$ with no factors in common and all their zeros in common with first coordinate fixed as $a$ are of the form $(a,p_i)$.
I thought that I could show that the ideals $(f_{p_i},g_{p_i})$ are pairwise coprime and their intersection or product is $(f_a,g_a)$ then $$\prod_{i=1}^ n k[[x,y]]/(f_{p_i},g_{p_i})\cong k[[x,y]]/(f_a,g_a)$$ but I cant prove neither of these statements. Other thing I could do is to show that $$\prod_{i=1}^ n k[[x]][y]/(f_{p_i},g_{p_i})\cong k[[x]][y]/(f_a,g_a)$$ again using the chinese remainder theorem, but I dont see the coprimality between these ideals or that $(f_a,g_a)=\prod_{i=1}^ n(f_{p_i},g_{p_i})$ or $(f_a,g_a)=\bigcap_{i=1}^ n(f_{p_i},g_{p_i})$ as ideals.