TITLE: Isomorphic varieties
QUESTION [7 upvotes]: I just want to see if my approach for this problem is fine:

Show $W=\mathbb{P}^1 \times \mathbb{P}^1$ is not isomorphic to $W'=\mathbb{P}^2.$ 

Well $V= \{ [0:1] \} \times \mathbb{P}^1, V' = \{ [1:0] \} \times \mathbb{P}^1$ are closed subvarieties of $W$ each isomorphic to $\mathbb{P}^1$ so each of dimension $1.$ So $W$ has two dimension 1 closed subvarieties that don't intersect, while $W'$ does not (any two projective plane curves intersect) and thus they are not isomorphic. 
Edit: Isomorphic as varieties.

REPLY [2 votes]: The argument seems fine, though you should clarify what class of objects you are working in: algebraic varieties? projective varieties? Note that they are not even homeomorphic (or even homotopy equivalent) as topological spaces, since $H_2(W;\mathbb{Z})=\mathbb{Z}^2$ while $H_2(W';\mathbb{Z})=\mathbb{Z}$.  The same argument can be stated with $\pi_2$ in place of $H_2$; some people think that the groups $\pi_i$ are easier to define than $H_i$.