3

In his first chapter of "The Arithmetic of Elliptic Curves", Silverman develops some necessary background in algebraic geometry. He works over ground fields $K$, which are perfect but not necessarily algebraically closed. For example, we get the coordinate ring of $V/K$: $$ K[V] = \frac{K[X_1, \dots, X_n]}{I(V/K)}. $$

I am then wondering why Silverman defines the maximal ideal $M_P = \{f \in \bar{K}[V] : f(P) = 0\}$ and the corresponding local ring $\bar{K}[V]_{M_P}$ only for algebraically closed fields?

It seems to me that (for $P$ a $K$-rational point) we could define a maximal ideal $m_P = \{f \in K[V] : f(P) = 0\}$ and a corresponding local ring $K[V]_{m_P}$ — although I am not sure if this is still a DVR?

Noa Even
  • 2,801
H. Löw
  • 161

1 Answers1

2

Yes, all of this still works for non-algebraically-closed fields. One doesn't even need $P$ to be $K$-rational: it's perfectly fine to think about the local ring of $\operatorname{Spec} \Bbb R[x]$ at the point $(x^2+1)$, for instance.

As for whether the local ring is a DVR or not, this is only the case when $V$ is of dimension one - in general, the local ring of a point has dimension equal to the codimension of the closure of the point in the variety. So for $K[V]_{I(P)}$ to have dimension one, $\overline{\{P\}}$ should be a codimension one subvariety of $V$, which if $P$ is a closed point can only happen when $V$ is of dimension one.

KReiser
  • 65,137
  • at the maximal ideal $(x^2+1) = {f \in K[V] : f(i) = 0}$ (in elliptic curves we want a group law on the points so it is weird to call a maximal ideal a point) – reuns Aug 07 '19 at 18:17