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Let $X$, $Y$, $Z$ be integral schemes of finite type over a field $K$ (i.e., locally affine opens are finitely-generated algebras over $K$). Suppose we have the following condition$\colon$

$f \colon Y \to X$ is finite.

For a dominant morphism $g \colon Z \to X$, suppose that there is NO dominant morphism $Z \to Y$ between schemes. (That is, there is NO embedding ${\cal O}_Y \hookrightarrow {\cal O}_Z$ between structural rings).

Q. Is it possble that $Y \times_X Z$ is not integral while $Y \times_X \eta_Z$ being integral ?

As in the answer below, in the case $Z = Y$, there is an example by normalisation of $X$.

Let $X$, $Y$, $Z$ be integral schemes of finite type over a field $K$ (i.e., locally affine opens are finitely-generated algebras over $K$). Suppose we have the following condition$\colon$

$f \colon Y \to X$ is finite.

We consider a dominant morphism $g \colon Z \to X$.

Q. Suppose that ${\cal O}_Y \cap {\cal O}_Z = {\cal O}_X$. Then is it possble that $Y \times_X Z$ is not integral while $Y \times_X \eta_Z$ being integral ?

As in the answer below in the case $Y = Z$, there is an example as $Y$ being a normalisation of $X$.

Let $X$, $Y$, $Z$ be integral schemes of finite type over a field $K$ (i.e., locally affine opens are finitely-generated algebras over $K$). Suppose we have the following condition$\colon$

$f \colon Y \to X$ is finite.

Q. For a dominant morphism $g \colon Z \to X$, suppose $k(Y) \cap k(Z) = k(X)$. If $Y \times_X Z$ is not integral while $Y \times_X \eta_Z$ being integral, what is the characterisation of such $Z$ ?

Is $Z$ uniquely determined or does $g \colon Z \to X$ factor over some minimal morphsim $g \colon Z_{\mathrm{min}} \to X$?

Let $X$, $Y$, $Z$ be integral schemes of finite type over a field $K$ (i.e., locally affine opens are finitely-generated algebras over $K$). Suppose we have the following condition$\colon$

$f \colon Y \to X$ is finite.

For a dominant morphism $g \colon Z \to X$, suppose that there is NO dominant morphism $Z \to Y$ between schemes. (That is, there is NO embedding ${\cal O}_Y \hookrightarrow {\cal O}_Z$ between structural rings).

Q. Is it possble that $Y \times_X Z$ is not integral while $Y \times_X \eta_Z$ being integral ?

As in the answer below, in the case $Z = Y$, there is an example by normalisation of $X$.

Let $X$, $Y$, $Z$ be integral schemes of finite type over a field $K$ (i.e., locally affine opens are finitely-generated algebras over $K$). Suppose we have the following condition$\colon$

$f \colon Y \to X$ is finite, $g \colon Z \to X$ is dominant (i.e., there is an inclusion ${\cal O}_X \hookrightarrow {\cal O}_Z$).

Consider the fibre product $Y \times_X Z$. Let $\eta_Z \colon= {\mathrm{Spec}}\,k(Z)$ be a generic point of $Z$.

Q. Under the assumption

$k(Y) \cap k(Z) = k(X)$,

What is the example such that $Y \times_X Z$ is not integral ?

Let $X$, $Y$, $Z$ be integral schemes of finite type over a field $K$ (i.e., locally affine opens are finitely-generated algebras over $K$). Suppose we have the following condition$\colon$

$f \colon Y \to X$ is finite.

Q. For a dominant morphism $g \colon Z \to X$, suppose $k(Y) \cap k(Z) = k(X)$. If $Y \times_X Z$ is not integral while $Y \times_X \eta_Z$ being integral, what is the characterisation of such $Z$ ?

Is $Z$ uniquely determined or does $g \colon Z \to X$ factor over some minimal morphsim $g \colon Z_{\mathrm{min}} \to X$?