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I tried to solve this question but not sure that my solution was the right one. Help me please with this question:

Find $\dim X\times \mathbb{P}^{2}$, where $X=\left \{ w_{0}^{3}=w_1(w_{1}^{2}-w_{2}^{2}) \right \}\subseteq \mathbb{P}_{w}^{2}$

Thanks!

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    What have you tried so far? Isn't $dim(X \times Y)=dim(X)+dim(Y)$? So aren't you really only asking what is $dim(X)$?2012-08-21
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    Yes, you are right2012-08-21
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    @hayd, can you give intuition for how to prove $dim(x \times y) = dim(x) + dim(y)$ for generic linear spaces $x$ and $y$?2013-10-26
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    @spencer What's the definition of times?2013-10-27

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I can tell you what the dimension of $X$ is: By the principal ideal theorem, the ideal generated by $f:=w_0^3-w_1^3 + w_2^2w_1$ has height one, so the codimension of $X$ in $\mathbb{P}^2$ is also equal to one (It cannot have height $0$ because $f$ is not a unit). $\mathbb{P}^2$ having dimension $2$, this yields $\dim(X)=1$. Much in line with haydoni's comment, I would then argue that $\dim(X\times\mathbb{P}^2)=\dim(X)+\dim(\mathbb{P}^2)=1+ 2 = 3$.

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    Dear rattle, you mean $\dim(X\times\mathbb{P}^2)=\dim(X)+\dim(\mathbb{P}^2)=1+2=3$.2012-08-21
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    Oh my god. Thanks. Edited =)2012-08-21
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    @ rattle Thanks a lot! my solution was the same.2012-08-21