2

I read that "smoothness is a geometric property" meaning that if $k\subset K$ is a field extension and $X_k$ is a scheme over $k$ then $X_k$ is smooth over $k$ if and only if $X_K = X_k \times_{Spec(k)}Spec(K)$ is smooth over $K$.

I really do not understand the meaning of this. For instance, it seems to me that we could easily find a polynomial $F\in \mathbb{R}[x,y,z]_d$ whose partial derivatives do not vanish on any point with real coordinates but vanish on some point with complex coordinates. So $X_\mathbb{R} := \{F=0\}\subset\mathbb{P}^2_{\mathbb{R}}$ would be a smooth curve over $\mathbb{R}$ but $X_{\mathbb{C}}\subset\mathbb{P}^2_{\mathbb{C}}$ would be a singular curve over $\mathbb{C}$.

What am I missing here?

Thank you very much.

Puzzled
  • 848

1 Answers1

2

This is directly from the definition: a morphism $f:X\to S$ of schemes is smooth iff it is flat, locally of finite presentation, and the geometric fibers are regular - that is, for any $s\in S$ and any geometric point $\overline{s}\to s$, the fiber product $X\times_S \overline{s}$ is regular.

Your example with subschemes of $\Bbb P^2_{\Bbb R}$ and $\Bbb P^2_{\Bbb C}$ reveals the source of your misunderstanding: the scheme $\Bbb P^2_{\Bbb R}$ has complex points! For instance, the homogeneous ideal $(x^2+y^2)$ represents the point $[1:\pm i:0]=[\pm i:1:0]$ inside $\Bbb P^2_{\Bbb R}$. So the naive tactic of moving the singularity off to a complex-valued point doesn't work the way you think it does. You may be interested to read Affine space over non algebraically closed field - I think it would help you out with your understanding here.

KReiser
  • 74,746