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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.

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  • $\begingroup$ @BenjaLim It seems easier to me to view it as a corollary of Bezout's theorem. $\endgroup$
    – Craig
    Commented Jul 3, 2013 at 13:05
  • $\begingroup$ @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. $\endgroup$
    – Craig
    Commented Jul 3, 2013 at 13:10

1 Answer 1

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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 $H_i$.

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  • $\begingroup$ I have not learned much algebraic topology yet. I mainly want to know if my particular approach is fine, I'm sure there are lots of better ways to do this. $\endgroup$
    – Craig
    Commented Jul 3, 2013 at 13:05
  • $\begingroup$ I should have written "homeomorphic". Yes, singular homology for example. $\endgroup$ Commented Jul 3, 2013 at 13:14
  • $\begingroup$ If one wishes to distinguish between $W$ and $W'$, it is sufficient to do it over a single field, say $\mathbb{Q}$ :-) $\endgroup$ Commented Jul 3, 2013 at 13:21
  • $\begingroup$ Surely you mean $\mathbb{C}$, because the singular homology is not what you say it is otherwise... $\endgroup$
    – Zhen Lin
    Commented Jul 3, 2013 at 14:02
  • $\begingroup$ 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 $\mathbb{Z}$. $\endgroup$ Commented Jul 3, 2013 at 14:13

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