# Isomorphic varieties

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.

• @BenjaLim It seems easier to me to view it as a corollary of Bezout's theorem. Commented Jul 3, 2013 at 13:05
• @BenjaLim I am not sure why I have to do that. My argument is just that if $\phi:W \to W'$ was an isomorphism, then $\phi(V)$ and $\phi(V')$ should be disjoint, but 1-dim subvarieties of W' can't be disjoint. Commented Jul 3, 2013 at 13:10

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 $H_i$.
• If one wishes to distinguish between $W$ and $W'$, it is sufficient to do it over a single field, say $\mathbb{Q}$ :-) Commented Jul 3, 2013 at 13:21
• Surely you mean $\mathbb{C}$, because the singular homology is not what you say it is otherwise... Commented Jul 3, 2013 at 14:02
• Shall we agree to use homotopy groups? Then we don't need to worry about coefficients :-) Note that in my answer I used the ring $\mathbb{Z}$. Commented Jul 3, 2013 at 14:13