I am a beginner.
This is an exercise from Hartshorne Chapter 1, 4.5. By his hint, it seems this can be argued that there are two curves in image of Segre embedding that do not intersect with each other while in $P^2$ any two curves intersect.
I feel this solution is very special. I would like to know more. Is there any invariant to detect whether two birational equivalent varieties are iso or not?
Thanks!
One way to see it is to note that $\mathbb{P}^1\times\mathbb{P}^1$ maps onto $\mathbb{P}^1$ (by projection to its first factor) while $\mathbb{P}^2$ does not. In fact, any map $\mathbb{P}^2\to\mathbb{P}^1$ is constant.
[Edit: Based on Georges's comment, I think I should explain why all maps $\mathbb{P}^2\to\mathbb{P}^1$ are constant. If $f:\mathbb{P}^2\to\mathbb{P}^1$ is any non-constant map, then its image is irreducible and has $\dim>0$, hence, it must be dense in $\mathbb{P}^1$. Now in $\mathbb{P}^1$, we can take two points $a\neq b$, and pull them back along $f$. This gives (for most choices of $a$ and $b$) two closed, dimension 1 subvarieties (or curves) in $\mathbb{P}^2$ that do not intersect. This is not possible by Bezout's theorem. You can generalize this argument to show that all maps $\mathbb{P}^n\to X$ with $X$ any variety of dimension $<n$ are constant.]
The surface $\mathbb{P}^1\times\mathbb{P}^1$ has disjoint algebraic curves lying on it, like $\{a\}\times \mathbb{P}^1$ and $\{b\}\times \mathbb{P}^1$ (where $a\neq b\in \mathbb P^1$).
The surface $\mathbb{P}^2$ does not have any disjoint algebraic curves lying on it: this is a very weak form of Bézout's theorem.
The Picard groups of both varieties are not isomorphic (this is more advanced) :
We have $\operatorname {Pic} (\mathbb P^2)\cong \mathbb Z$ whereas $\operatorname {Pic} (\mathbb P^1\times \mathbb P^1)\cong \mathbb Z^2$ (see Ischebeck's article here for a very general result concerning the Picard group of the product of two varieties )
Over $\mathbb C$ the underlying topological spaces of the two varieties are not even homeomorphic.
Indeed, we have $H_2(\mathbb P^2,\mathbb Z)\cong \mathbb Z$ whereas $H_2(\mathbb P^1\times \mathbb P^1,\mathbb Z)\cong \mathbb Z^2$.
The first result is completely standard (Greenberg-Harper, Theorem 19.21) and the second results from Künneth (same book, Corollary 29.11.1).
A different way to prove this is to count points. If $k$ is a field with $q$ elements, then $\mathbb{P}^2_k$ has $q^2 + q + 1$ rational points whereas $\mathbb{P}^1_k\times_k \mathbb{P}^1_k$ has $(q + 1)^2$ rational points, so they cannot be isomorphic. If $k$ is an arbitrary field, then any isomorphism $\mathbb{P}^2_k \to \mathbb{P}^1_k\times_k \mathbb{P}^1_k$ must be defined over a non-empty open locus $U$ of a finite ring extension of $\mathbb{Z}$. But since the residue field at any closed point of $U$ is finite, such an isomorphism cannot exist by the previous argument.
This technique of spreading out a morphism is usually not described in introductory text books on scheme theory. This is a pity, since it is quite elementary and a really good advertisment for considering schemes instead of just varieties. Basically, it hinges on the idea that you need finitely many equations to describe your morphism and to describe these equations, it is enough adjoin finitely many roots and inverting finitely many elements in your base.
For the canonical bundle $K$ on these varieties we have the self-intersection numbers $$K^2_{\mathbb P^1\times \mathbb P^1}=8$$ and $$K^2_{\mathbb P^2}=9$$
The canonical bundles of these varieties both only have the zero section, so these sections cannot be used to prove the non-isomorphism of the varieties.
The anticanonical bundles $K^\ast=\Lambda^2T$ of these varieties however have spaces of global sections of different dimensions $h^0$ over the base field and prove that non-isomorphism:
$$h^0(\mathbb P^2, K^\ast_{\mathbb P^2})=h^0(\mathbb P^2,\mathcal O_{\mathbb P^2}(3))=10$$ whereas
$$h^0(\mathbb P^1\times \mathbb P^1, K^\ast_{{\mathbb P^1\times \mathbb P^1}})=h^0(\mathbb P^1\times \mathbb P^1,\mathcal O_{\mathbb P^1 }(2)\boxtimes \mathcal O_{\mathbb P^1 }(2))=9$$ (Whew, that was close: 10 and 9 !)
Edit: notation
Given the projections $p,q:\mathbb P^1\times \mathbb P^1\to \mathbb P^1$ of $\mathbb P^1\times \mathbb P^1$ onto its two factors, the tensor product $p^\ast\mathcal O_{\mathbb P^1}(a)\otimes_{\mathcal O_{\mathbb P^2}} q^\ast\mathcal O_{\mathbb P^1}(b)$ of the pull-backs $p^\ast\mathcal O_{\mathbb P^1}(a)$ and $q^\ast\mathcal O_{\mathbb P^1}(b)$ is denoted by $\mathcal O_{\mathbb P^1}(a)\boxtimes \mathcal O_{\mathbb P^1}(b)$ or simply by $\mathcal O_{\mathbb P^2}(a,b)$.
This convention is used in other answers too.
We have $H^1(\mathbb P^1\times \mathbb P^1,\mathcal O_{\mathbb P^1\times \mathbb P^1}(0,-2))\cong k$ but for any line bundle $ L$ on $\mathbb P^2$ we have $H^1(\mathbb P^2,L)=0 $
We have for the dimension of the space of sections of the tangent bundle $$h^0(\mathbb P^2, T_{\mathbb P^2})=8$$This follows from the Euler sequence of vector bundles $$ 0\to \mathcal O_{\mathbb P^2}\to \mathcal O_{\mathbb P^2}(1)^{\oplus 3} \to T_{\mathbb P^2}\to 0$$ by taking the long exact sequence of cohomology and remembering that $H^1(\mathbb P^2, \mathcal O_{\mathbb P^2})=0$.
The corresponding result on $\mathbb P^1\times \mathbb P^1$ however is $$h^0(\mathbb P^1\times \mathbb P^1, T_{\mathbb P^1\times \mathbb P^1})=h^0(\mathbb P^1\times \mathbb P^1, \mathcal O_{\mathbb P^1\times \mathbb P^1}(2,2)=9$$
