Suppose $X$ is smooth variety over an algebraically closed field and $Y \rightarrow X$ some finite generically Galois normal cover. If $C \rightarrow X$ is some smooth curve (by which I mean $C$ is a smooth variety but the map may not be smooth) that is generically an embedding, what is the condition to force $(Y \times_X C)_{red}$ to be normal (=smooth)? The fiber product of two normal varieties is usually not normal. Since smoothness and normality are the same for curves, can one write down a simple condition using the Jacobian in this case?

## 1 Answer

I think this is not likely to happen.

Let $Y$ be an arbitrary singular variety and $Y\to X$ an arbitrary morphism to a non-singular variety $X$. Let $C\subset X$ be a (local) complete intersection curve intersecting the image of the singular set of $Y$. Then the pre-image of $C$ on $Y$ is **not** going to be smooth. The reason is that at any point where a Cartier divisor is smooth, the ambient variety has to be smooth as well, so the same follows for a local complete intersection variety.

For a more explicit example, let $Y\subset \mathbb A^3$ be a singular normal surface and $P\in Y$ a singular point. Let $X=\mathbb A^2$ and $Y\to X$ a general projection. Finally, let $C\subset X$ be a smooth curve containing the image of $P$. Then $Y\times_X C$ is a Cartier divisor on $Y$ and hence cannot be smooth at $P\in Y$. Notice that for most choices this will actually be reduced, so taking $(\quad)_{\text{red}}$ makes no difference. I suppose your only chance is if the map is totally ramified along $C$.