2

I am reading Silverman's 'The Arithmetic of Elliptic Curves'. enter image description here

It says $\mathbb{P}^n(K)=\{[x_0,\cdots,x_n]\in\mathbb{P}^n:\text{ all }x_i\in K\}$, and then a remark says: if $P=[x_0,\cdots,x_n]\in\mathbb{P}^n(K)$, it does not follow that each $x_i\in K$.

I am confused by the remark because clearly in the definition it says all $x_i\in K$

Stéphane Jaouen
  • 2,954
  • 5
  • 21
Eulerian
  • 21
  • 2
  • 4
    The author probably means that the equivalence class must have some representative where all coordinates belongs to $K$. It doesn't mean that all representatives satisfy that. So $[x_0,...,x_n]\in\mathbb{P}^n(K)$ means that there is some $0\ne\lambda\in\overline{K}$ such that $\lambda x_0, \lambda x_1,...,\lambda x_n$ all belong to $K$. – Mark Mar 31 '24 at 13:00
  • If you want to learn what a projective space is, we could start by taking some special cases before we get into the general definition of Silverman. – Stéphane Jaouen Mar 31 '24 at 13:05
  • What is $\mathbb A$ in Silverman's definition ? I've never seen such a convoluted definition for such a simple notion – Stéphane Jaouen Mar 31 '24 at 13:09
  • 1
    To follow up on the comment by @Mark here is an example. If $ K $ is the field of rational numbers, so that $ \bar K $ is (I presume) the field of algebraic complex numbers, then $ [ 0 , \sqrt 2 , 2 \sqrt 2 ] \in \mathbb P ^ 2 ( K ) $, because $ [ 0 , \sqrt 2 , 2 \sqrt 2 ] = [ 0 , 1 , 2 ] $, and $ 0 , 1 , 2 \in K $, even though $ \sqrt 2 , 2 \sqrt 2 \notin K $. However, if we pick either $ \sqrt 2 $ or $ 2 \sqrt 2 $ (but not $ 0 $) and divide everything by that, then we get a representative with coordinates in $ K $: either $ [ 0 , 1 , 2 ] $ or $ [ 0 , 1 / 2 , 1 ] $. – Toby Bartels Mar 31 '24 at 13:18
  • By the way, it seems weird (as @StéphaneJaouen noticed) to define $ \mathbb P ^ n ( K ) $ as a subset of $ \mathbb P ^ n ( \bar K ) $, since the definition of $ \mathbb P ^ n ( K ) $ could proceed exactly like the definition of $ \mathbb P ^ n ( \bar K ) $ (which Silverman abbreviates as $ \mathbb P ^ n $) by starting from $ \mathbb A ^ { n + 1 } ( K ) $ instead of $ \mathbb A ^ { n + 1 } ( \bar K ) $ (which I presume is what Silverman means by $ \mathbb A ^ { n + 1 } $). But I guess it works. – Toby Bartels Mar 31 '24 at 13:21

0 Answers0