# Morphism smooth over the function field. What does it mean?

Look at the following lines I found in the book "Moriwaki - Arakelov geometry" (beginning of chap. 4):

Let $S$ be a connected Dedekind scheme with function field $K$. Let $\pi:X\to S$ be a projective flat morphism of relative dimension 1. Assume that $X$ is regular and that $\pi$ is smooth over $K$.

I know the definition of a smooth morphism between two schemes, but what is the smoothness over $K$ (the function field of $S$)?

edit: I know the following definition of smoothness from Liu's book:

$f:X\to S$ of finite type is smooth at $x\in X$ if it is flat at $x$ and moreover if $X_{f(x)}\to\text {Spec } k(f(x))$ is smooth at $x$. Then, a scheme $Y$ over a field $k$ is smooth at $y\in Y$ if the points in the base change $Y_{\overline k}$ lying over $y$ are regular.

Explicitly, you have the generic point $Spec(K)\to S$, so you can take the pull-back $\pi_K: X_K = X\times_S Spec(K)\to Spec(K)$, and you ask that this is a smooth morphism.
• It's simply because $\pi_K$ is what I call the generic fiber. In general if $s: Spec(L)\to S$ is a point then the fiber of $\pi$ over $s$ is $X_L:= X\times_S Spec(L)$, equiped with its structural morphism $\pi_L: X_L\to Spec(L)$. So saying the fiber is smooth is by definition saying that $\pi_L$ is a smooth morphism. – Captain Lama Apr 17 '16 at 20:46