The polynomial $x^3+y^4$ is irreducible in $\mathbb C[x, y]$, thus $\mathbb C[x, y]/(x^3+y^4)$ is an integral domain. My question is: Is $\mathbb C[[x, y]]/(x^3+y^4)$ also an integral domain?
Further thoughts: $\mathbb C[[x, y]]/(x^3+y^4)$ is the completion of the excellent reduced local ring $\mathbb C[x, y]_{(x,y)}/(x^3+y^4)$, hence $\mathbb C[[x, y]]/(x^3+y^4)$ is reduced ... being an integral domain, however, is a much stronger property.
Consider $\Bbb C[x,y]/(x^3+y^4)$. Let's first compute the normalization of this. We have $(\frac{x}{y})^{4}+x=0$, so the normalization contains $z:=\frac{x}{y}$. As $z^4=-x$ and $z^3=-y$, we get $\Bbb C[z] \supseteq R$. As a polynomial ring over a field, $\Bbb C[z]$ is integrally closed. Thus $\Bbb C[z]$ is the normalization of $\Bbb C[x,y]/(x^3+y^4)$ and expressing $x$ and $y$ in terms of $z$ as above, we get that $\Bbb C[x,y]/(x^3+y^4) \cong \Bbb C[z^3,z^4]$. Clearly the only prime ideal in $\Bbb C[z]$ lying over $(z^3,z^4)$ with respect to the normalization map $\Bbb C[z^3,z^4] \hookrightarrow \Bbb C[z]$ is $(z)$. Thus we get an induced map of local rings $\Bbb C[z^3,z^4]_{(z^3,z^4)} \hookrightarrow \Bbb C[z]_{(z)}$. By applying [stacks 0394], we get that $\Bbb C[[z]]$ is the $(z^3,z^4)$-adic completion of $\Bbb C[z]_{(z)}$. (It's easy to see that the required condition is satisfied.) But because $\Bbb C[z]_{(z)}$ is a finite $\Bbb C[z^3,z^4]_{(z^3,z^4)}$-module, this completion may be expressed as a tensor product $\Bbb C[z]_{(z)} \otimes_{A} \widehat{A}$, where $A=\Bbb C[z^3,z^4]_{(z^3,z^4)}$. To finish the argument, we use that $A \to \widehat{A}$ is flat, so the injection $A \to \Bbb C[z]_{(z)}$ induces an injection $A \otimes_A \widehat{A} \hookrightarrow \Bbb C[z]_{(z)} \otimes_{A} \widehat{A} \cong \Bbb C[[z]]$. Being isomorphic to a subring of an integral domain, $\widehat{A}$ is also an integral domain. In fact, if we know that $\widehat{A}$ is isomorphic to a subring of $\Bbb C[[z]]$, it's not hard to figure out that $\widehat{A}=\Bbb C[[z^3,z^4]]$.
