Let $k$ be an algebraically closed field of characteristic not 2 or 3. Consider the map $\text{spec}(k[t])\rightarrow \text{spec}(k[x,y]/(y^2-x^3)$ given by the ring map $x\mapsto t^2$ and $y\mapsto t^3$. (This is just the normalizalition of the nodal cubic). Then consider the map from $\text{spec}(k[\![t]\!])\rightarrow \text{spec}(k[x,y]/(y^2-x^3)$ defined similarly by the ring map $x\mapsto t^2$ and $y\mapsto t^3$. What is the fiber product $\text{spec}(k[\![t]\!]) \times_{\text{spec}(k[x,y]/(y^2-x^3)} \text{spec}(k[t])$?
I believe it should be $\text{spec}(k[\![t]\!][s]/(st,s^2)) $, because the if we consider the inclusion of the origin into $\text{spec}(k[x,y]/(y^2-x^3)$ then the pullback along the normalization is $\text{spec}(k[s]/(s^2))$, but I am unsure how to compute this using the tensor product. I believe one can do it by carefully using the universal property of tensor products to construct a map $k[\![t]\!][s]/(st,s^2)) \rightarrow k[\![t]\!] \otimes_{(k[x,y]/(y^2-x^3)}k[t]$.
Best Answer
I think your quotient ideal may be a little too big. The correct ring may be calculated as follows (note that $k[s]\otimes_{k[x,y]/(x^2-y^3)}k[t]$ is finite over $k[s]$): \begin{align*} k[[s]]\otimes_{k[x,y]/(x^2-y^3)}k[t] &\cong k[[s]]\otimes_{k[s]}k[s,t]/(s^2-t^2,s^3-t^3) \\ &\cong k[[s]]\otimes_{k[s]}k[s,t]/(s^2-t^2,(s+t)(s^2-t^2)-st(s-t)) \\ &\cong k[[s]]\otimes_{k[s]}k[s,t]/((s+t)(s-t),st(s-t)) \\ &\cong k[[s]]\otimes_{k[s],s=(u+v)/2}k[u,v]/(uv,u(u^2-v^2)/4) \\ &\cong k[[s]]\otimes_{k[s],s=u+v}k[u,v]/(uv,u^3), \end{align*} where $u:=s-t,v:=s+t$. It is clear that for any $N\geq 3$, the following equality of ideals holds in $k[u,v]/(uv,u^3)$: $$((u+v)^N) = (v^N) = (u,v)^N.$$ This implies that $$ k[[s]]\otimes_{k[x,y]/(x^2-y^3)}k[t]\cong k[[s]]\otimes_{k[s],s=u+v}k[u,v]/(uv,u^3) \cong k[[u,v]]/(uv,u^3). $$