0
$\begingroup$

I'm trying to prove that $\mathbb{P}^1$ is not affine over an algebraically closed field, $k$.

My first thoughts are to observe that, by definition, $\mathbb{P}^1$ is affine if $*: \textrm{Morf}(X,\mathbb{P}^1) \to \textrm{Hom}_k(k[\mathbb{P}^1],k[X])$ is bijective for all algebraic varieties $X$. Setting $X = \mathbb{P}^1$, we have $*:\textrm{Morf}(\mathbb{P}^1,\mathbb{P}^1) \to \textrm{Hom}_k(k,k)$ should be bijective. I think that there is a contradiction here, somehow it doesn't make sense that this can be injective. For instance, if we take the identity morphism, $i$, and create a new morphism, $f$, which sends $\infty$ (where $\mathbb{P}^1 = \mathbb{A}^1 \cup {\infty}$) to an element of $k$, then then $f^* = g^*$.

Is this the right idea?

  • 2
    Hint: What is the ring of global sections? – 2017-01-21
  • 0
    Could you please supply a definition? I don't know that phrase. Is it the same as regular functions? – 2017-01-21
  • 0
    Ok, Well, I think in my notation that is $k[\mathbb{P}^1] = k$. – 2017-01-21

1 Answers 1

1

Your idea is correct. I would pick $X=\Bbb A^1$ though. Then, $k[X]=k[t]$ is the polynomial ring in one variable $t$. Note that there is only one $k$-algebra homomorphism $k\to k[t]$. However, there are at least two morphisms $\Bbb A^1\to \Bbb P^1$, one being $a\mapsto[a:1]$ and the other being $a\mapsto[1:a]$. There are plenty more, but this is sufficient to see that the two sets can not be isomorphic.