# Geometric reducedness (integral) versus reducedness (integral)

All the schemes here are over $\mathbb{C}$.

Suppose $X \to Y$ is a morphism of varieties, then the geometric reducedness (integral) of the generic fibre implies the geometric reducedness (integral) of general fibres. This can be found in this answer and Theorem 2.2 of this note. However, the statement is false if just using integral instead of geometric integral (but I don't know what happens about reducedness).

Here is an example which I confused about:

Let ${\rm{Spec}}(\mathbb{C}[x,y]/(y^2 - x)) \to {\rm{Spec}}(\mathbb{C}[x])$ be a 2:1 cover. For general points, the fibres are two distinct points. However, the affine ring of the generic fibre is $F:=\{\frac{h(y)}{g(y^2)} \in \mathbb{C}(y)\}$ where $h,g$ are polynomials. $F$ is certainly an integral domain. My questions are:

(1) With respect to which field, do we talk about the geometric integral/reducedness?

(2) Why $F$ is not geometric integral? (I thought every integral domain over $\mathbb{C}$ is automatically geometric integral by definition)

• Isn't the generic fibre actually a scheme over an extension of $\mathbb{C}$? – Zhen Lin Mar 28 '14 at 14:39
• Yes, it is. Say $K(A) \otimes_A B$ is a scheme over $K(A)$, but at the same time, it is also a scheme over $\mathbb{C}$, so how could I know over which field it is geometrically reduced (integral)? – Li Yutong Mar 28 '14 at 15:53
• I think you're confused as to what you are thinking about the fiber as a variety with respect to. You are supposed to think of the fiber as a variety with respect to the generic point of the base scheme. So, in your case, your fiber is a variety over $\mathbb{C}(x)$, and the fiber is NOT geometrically reduced as a variety in that regard. – Alex Youcis Mar 29 '14 at 11:27
• Do you mean not geometrically integral instead of geometrically reduced? – Li Yutong Mar 29 '14 at 13:46
• I think he meant not always geometrically integral. It is always geometrically reduced because we are in characteristic $0$. – Cantlog Mar 29 '14 at 20:52

As per request, of the op.

The theorem to which you are referring is talking about the scheme with respect to the generic point of the base. So, it means that if $f:X\to Y$ is a morphism of finite type, $Y$ irreducible, and if $f^{-1}(\eta)$ (where $\eta$ is the generic point of $Y$) is geometrically _____ over $K(\eta)$, then there is a neighborhood with the same property.

In your example, let's compute the generic fiber. This is merely

$$\mathbb{C}[x,y]/(y^2-x)\otimes_{\mathbb{C}[x]}\mathbb{C}(x)=\mathbb{C}(x)[y]/(y^2-x)$$

So, it's a degree two extension of $\mathbb{C}(x)$. It's not geometrically integral since

\begin{aligned}\mathbb{C}(x)[y]/(y^2-x)\otimes_{\mathbb{C}(x)}\overline{\mathbb{C}(x)}& \cong \overline{\mathbb{C}(x)}[y]/(y^2-x)\\ &\cong \overline{\mathbb{C}(x)}\times\overline{\mathbb{C}(x)}\end{aligned}

In fact, as I'm sure you can easily generalize, any separable, finite extension of a field is not geometrically integral.