Return to Answer

In the example $X=\text{Spec}\ k[u,v]\langle u^2 - v^3\rangle$, the scheme $X^{\text{nor}}\times_X X^{\text{nor}}$ is irreducible but not reduced. In the example $X=\text{Spec}\ k[u,v]/\langle u^2 - v^2(v-1) \rangle$, the scheme $X^{\text{nor}}\times_X X^{\text{nor}}$ is reducible (the extra irreducible components come from pairs of distinct points of $X{\text{nor}}$ that are "separated apart" from branches of singularities of $X$).

As extensions of $\mathcal{O}_X$, the rings $\mathcal{O}_Y$, resp. $\mathcal{O}_Z$, equal, $$\mathcal{O}_Y \cong \mathcal{O}_X[f_1]/\langle f_1^2-f_1,tf_1-te_1\rangle,\ \ \mathcal{O}_Z \cong \mathcal{O}_X[f_3]/\langle f_3^2-f_3,tf_3-te_3\rangle, .$$ For the maximal ideal $\mathfrak{m}_X = \langle te_1,te_2,te_3 \rangle$ in $\mathcal{O}_X$ with residue field $k$, both of the fiber rings $\mathcal{O}_Y/\mathfrak{m}_X\cdot \mathcal{O}_Y$ and $\mathcal{O}_Z/\mathfrak{m}_X\cdot \mathcal{O}_Z$ have dimension $2$ as $k$-vector spaces, generated by $1$ and $f_1,$ resp. generated by $1$ and $f_3.$ Thus, the fiber ring $(\mathcal{O}_Y\otimes_{\mathcal{O}_X}\mathcal{O}_Z)/\mathfrak{m}_X\cdot (\mathcal{O}_Y\otimes_{\mathcal{O}_X}\mathcal{O}_Z)$ has dimension $4$ as a $k$-vector space, generated by $$ 1\otimes 1, \ \ 1\otimes f_3, \ \ f_1\otimes 1, \ \ f_1\otimes f_3.$$ On the other hand, $A/\mathcal{m}_X\cdot A$ has dimension $3$ as a $k$-vector space generated by $$1,\mathcal{e}_1, \mathcal{e}_3.$$ Thus, the kernel of the surjective $k[[t]]$-algebra homomorphism, $$\mathcal{O}_Y\otimes_{\mathcal{O}_X} \mathcal{O}_Z\to A,$$ is an ideal of length $1$. Therefore $\mathcal{O}_Y\otimes_{\mathcal{O}_X}\mathcal{O}_Z$ is not an integral domain.

In the example $X=\text{Spec}\ k[u,v]\langle u^2 - v^3\rangle$, the scheme $X^{\text{nor}}\times_X X^{\text{nor}}$ is irreducible but not reduced. In the example $X=\text{Spec}\ k[u,v]/\langle u^2 - v^2(v-1) \rangle$, the scheme $X^{\text{nor}}\times_X X^{\text{nor}}$ is reducible (the extra irreducible components come from pairs of distinct points of $X^{\text{nor}}$ that are "separated apart" from branches of singularities of $X$).

As extensions of $\mathcal{O}_X$, the rings $\mathcal{O}_Y$, resp. $\mathcal{O}_Z$, equal, $$\mathcal{O}_Y \cong \mathcal{O}_X[f_1]/\langle f_1^2-f_1,tf_1-te_1\rangle,\ \ \mathcal{O}_Z \cong \mathcal{O}_X[f_3]/\langle f_3^2-f_3,tf_3-te_3\rangle.$$ For the maximal ideal $\mathfrak{m}_X = \langle te_1,te_2,te_3 \rangle$ in $\mathcal{O}_X$ with residue field $k$, both of the fiber rings $\mathcal{O}_Y/\mathfrak{m}_X\cdot \mathcal{O}_Y$ and $\mathcal{O}_Z/\mathfrak{m}_X\cdot \mathcal{O}_Z$ have dimension $2$ as $k$-vector spaces, generated by $1$ and $f_1,$ resp. generated by $1$ and $f_3.$ Thus, the fiber ring $(\mathcal{O}_Y\otimes_{\mathcal{O}_X}\mathcal{O}_Z)/\mathfrak{m}_X\cdot (\mathcal{O}_Y\otimes_{\mathcal{O}_X}\mathcal{O}_Z)$ has dimension $4$ as a $k$-vector space, generated by $$ 1\otimes 1, \ \ 1\otimes f_3, \ \ f_1\otimes 1, \ \ f_1\otimes f_3.$$ On the other hand, $A/\mathcal{m}_X\cdot A$ has dimension $3$ as a $k$-vector space generated by $$1,\mathcal{e}_1, \mathcal{e}_3.$$ Thus, the kernel of the surjective $k[[t]]$-algebra homomorphism, $$\mathcal{O}_Y\otimes_{\mathcal{O}_X} \mathcal{O}_Z\to A,$$ is an ideal of length $1$. Therefore $\mathcal{O}_Y\otimes_{\mathcal{O}_X}\mathcal{O}_Z$ is not an integral domain.

Answer to original question. I am just posting my comment as an answer. Let $X$ be an integral scheme that is not normal, e.g., $\text{Spec}\ k[u,v]/\langle u^2 - v^3\rangle$ for a field $k$ of characteristic $\neq 2,3$. By a corollary of the Noether normalization theorem, the normalization morphism $\nu:X^{\text{nor}}\to X$ is a finite morphism. Denote by $U\subset X$ the maximal open subscheme over which $\nu$ is flat, and denote by $C\subset X$ the closed complement of $X$. By openness of the normal locus (should be in Section 24 of Matsumura, "Commutative Ring Theory"), since the generic point $\text{Spec}\ k(X)$ is normal, the open $U$ is dense. Over $U$, the morphism $\nu$ is an isomorphism. For every point $p\in C$, the fiber of $\nu$ over $p$ has length $\ell(p)\geq 2$.

Answer to modified question. All of the following rings will be algebras over the power series ring $k[[t]]$. The schemes below are reduced and equidimensional, but not integral. It is straightforward to find integral schemes that are finite type over $k$ and such that the schemes below are obtained via base change by $\text{Spec}\ \widehat{\mathcal{O}}_{X,p} \to X.$ For instance, this works for the plane curve $X=\text{Spec}\ k[u,v]/\langle v^3-(u^3+u^4)\rangle$ and the "triple point" $p=\langle u,v\rangle.$

Answer to original question without any hypothesis that $\mathcal{O}_X$ equals $\mathcal{O}_Y\cap \mathcal{O}_Z.$ I am just posting my comment as an answer. Let $X$ be an integral scheme that is not normal, e.g., $\text{Spec}\ k[u,v]/\langle u^2 - v^3\rangle$ for a field $k$ of characteristic $\neq 2,3$. By a corollary of the Noether normalization theorem, the normalization morphism $\nu:X^{\text{nor}}\to X$ is a finite morphism. Denote by $U\subset X$ the maximal open subscheme over which $\nu$ is flat, and denote by $C\subset X$ the closed complement of $X$. By openness of the normal locus (should be in Section 24 of Matsumura, "Commutative Ring Theory"), since the generic point $\text{Spec}\ k(X)$ is normal, the open $U$ is dense. Over $U$, the morphism $\nu$ is an isomorphism. For every point $p\in C$, the fiber of $\nu$ over $p$ has length $\ell(p)\geq 2$.

Answer to modified question with hypothesis that $\mathcal{O}_X$ equals $\mathcal{O}_Y\cap \mathcal{O}_Z.$ All of the following rings will be algebras over the power series ring $k[[t]]$. The schemes below are reduced and equidimensional, but not integral. It is straightforward to find integral schemes that are finite type over $k$ and such that the schemes below are obtained via base change by $\text{Spec}\ \widehat{\mathcal{O}}_{X,p} \to X.$ For instance, this works for the plane curve $X=\text{Spec}\ k[u,v]/\langle v^3-(u^3+u^4)\rangle$ and the "triple point" $p=\langle u,v\rangle.$

I am just posting my comment as an answer. Let $X$ be an integral scheme that is not normal, e.g., $\text{Spec}\ k[u,v]/\langle u^2 - v^3\rangle$ for a field $k$ of characteristic $\neq 2,3$. By a corollary of the Noether normalization theorem, the normalization morphism $\nu:X^{\text{nor}}\to X$ is a finite morphism. Denote by $U\subset X$ the maximal open subscheme over which $\nu$ is flat, and denote by $C\subset X$ the closed complement of $X$. By openness of the normal locus (should be in Section 24 of Matsumura, "Commutative Ring Theory"), since the generic point $\text{Spec}\ k(X)$ is normal, the open $U$ is dense. Over $U$, the morphism $\nu$ is an isomorphism. For every point $p\in C$, the fiber of $\nu$ over $p$ has length $\ell(p)\geq 2$.

Answer to modified question. All of the following rings will be algebras over the power series ring $k[[t]]$. The schemes below are reduced and equidimensional, but not integral. It is straightforward to find integral schemes that are finite type over $k$ and such that the schemes below are obtained via base change by $\text{Spec}\ \widehat{\mathcal{O}}_{X,p} \to X.$

Answer to original question. I am just posting my comment as an answer. Let $X$ be an integral scheme that is not normal, e.g., $\text{Spec}\ k[u,v]/\langle u^2 - v^3\rangle$ for a field $k$ of characteristic $\neq 2,3$. By a corollary of the Noether normalization theorem, the normalization morphism $\nu:X^{\text{nor}}\to X$ is a finite morphism. Denote by $U\subset X$ the maximal open subscheme over which $\nu$ is flat, and denote by $C\subset X$ the closed complement of $X$. By openness of the normal locus (should be in Section 24 of Matsumura, "Commutative Ring Theory"), since the generic point $\text{Spec}\ k(X)$ is normal, the open $U$ is dense. Over $U$, the morphism $\nu$ is an isomorphism. For every point $p\in C$, the fiber of $\nu$ over $p$ has length $\ell(p)\geq 2$.

Answer to modified question. All of the following rings will be algebras over the power series ring $k[[t]]$. The schemes below are reduced and equidimensional, but not integral. It is straightforward to find integral schemes that are finite type over $k$ and such that the schemes below are obtained via base change by $\text{Spec}\ \widehat{\mathcal{O}}_{X,p} \to X.$ For instance, this works for the plane curve $X=\text{Spec}\ k[u,v]/\langle v^3-(u^3+u^4)\rangle$ and the "triple point" $p=\langle u,v\rangle.$