For $f:\Bbb P^1\to\Bbb P^1$ by $[x:y]\mapsto [x^2:y^2]$, show $f_*\mathcal{O}_{\Bbb P^1}\cong \mathcal{O}\oplus\mathcal{O}(-1)$

Question

I am currently preparing for an exam in algebraic geometry and came across the following exercise:

Let $k$ be an algebraically closed field. Let $f\colon\mathbb{P}_k^1\to \mathbb{P}_k^1$, $[x,y]\to [x^2,y^2]$.

1. Show that $f_*(\mathcal{O}_{\mathbb{P}_k^1})$ is a vector bundle $\mathcal{E}$ on $\mathbb{P}_k^1$, which has rank 2.
2. Show that $\mathcal{E}\cong \mathbb{P}_k^1\oplus\mathbb{P}_k^1(-1)$

Somebody showed me a proof that used that $f_*(\mathcal{O}_{\mathbb{P}_k^1})$ is the universal scheme over $\mathbb{P}_k^1$ containig a square root and some properties about $\text{Spec}$ underlined. But I didn't really understand that and feel that is seems unecessarily complicated. I would like to give a more concrete proof showing first that $f_*(\mathcal{O}_{\mathbb{P}_k^1})(U)$ and $f_*(\mathcal{O}_{\mathbb{P}_k^1})(V)$ are both $\mathcal{O}_{\mathbb{P}_k^1}$-modules of rank 2 where $U\cup V= \mathbb{P}_k^1$ is the standard cover and then gluing those along transition maps (which I think should work and would prove 1. and 2.). However I seem to be confused on what the module $\mathcal{O}_{\mathbb{P}_k^1}$-module strucure on $f_*(\mathcal{O}_{\mathbb{P}_k^1})$ actually is since I somehow can't work it out. I would be thankful if somebody could show me/teach me how such a concrete proof might go since I am lost in defintions!

Answer 1

Let $f \colon (X, \mathcal{O}_X) \longrightarrow (Y,\mathcal{O}_Y)$ be a morphism of locally ringed space and $\mathcal{F}$ be a sheaf of $\mathcal{O}_X$-modules, then a structure of a $\mathcal{O}_Y$-modules of $f_*\mathcal{F}$ is given by a morphism (with some additional conditions on compatibility) $$\mathcal{O}_Y \times f_*\mathcal{F} \longrightarrow f_*\mathcal{F}$$ which is by adjunction equivalent to $$f^*(\mathcal{O}_Y \times f_*\mathcal{F}) \longrightarrow \mathcal{F}.$$ This can be given by $$f^*(\mathcal{O}_Y\times f_*\mathcal{F}) = f^*\mathcal{O}_Y \times f^*f_*\mathcal{F} \longrightarrow \mathcal{O}_X \times \mathcal{F} \longrightarrow \mathcal{F},$$ where $f^*f_* \longrightarrow id$ is unit morphism and the last map is the structural morphism of $\mathcal{F}$. Since $\mathbb{P}_k^1$ has a canonical decomposition $D_+(x),D_+(y)$, we may can just reduce to computations to an affine line. The second part of the problem already suggests what you want. It really reduces your problem to a single morphism $f \colon \mathbb{A}_k^1 \longrightarrow \mathbb{A}^1_k, x \longmapsto x^2$. At the level of global sections, this is given by $k[x] \longrightarrow k[x], x \longmapsto x^2$ and hence it identifies one $k[x]$ with $k[x,t]/(t^2-x) \simeq k[x] \oplus k[x]t$. From this, it implies that $$f_*(\mathcal{O}_{\mathbb{A}_k^1})(\mathbb{A}_k^1) = \mathcal{O}_{\mathbb{A}_k^1}(f^{-1}(\mathbb{A}^1_k)) \simeq (\mathcal{O}_{\mathbb{A}_k^1} \oplus \mathcal{O}_{\mathbb{A}_k^1}(-1))(\mathbb{A}_k^1).$$ The twist $(-1)$ comes from the fact that $t$ is of degree $1$ in $k[x,t]$. Indeed $\mathcal{O}(-1)_0 = 0$ and $\mathcal{O}(-1)_1 =\mathcal{O}$.