3
$\begingroup$

Consider the homomorphism $f : \mathbb{C}[X] \to \mathbb{C}[X]$, $X \mapsto X^2$. It induced a morphism of affine schemes $\operatorname{spec} f : \mathbb{A} \to \mathbb{A}$ which topologically is the identity function. How am I meant to think about $\operatorname{spec} f$ geometrically?

EDIT: As pointed out in the comments, $spec \; f$ is not infact topologically the identity, it does exactly what it should. It sends $a$ to $a^2$...

  • 8
    am I missing something? Why is it topologically identity? $(x-a)$ pulls back to $(x-a^2)$. And this is also how you can think about it geometrically: it's the map $z \to z^2$ from $\mathbb{A}^1$ to $\mathbb{A}^1$.2011-09-01
  • 0
    you are absolutely right....2011-09-01
  • 1
    @Soarer: Perhaps you'd like to post that as an answer?2011-09-26
  • 0
    @Zev, done. (random characters)2011-09-29

1 Answers 1

4

$(x−a)$ pulls back to $(x−a^2)$. This is also how you can think about it geometrically: it's the map $z \to z^2$ from $\mathbb{A}^1$ to $\mathbb{A}^1$.