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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 $dim(X)$? –  hayd Aug 21 '12 at 11:31
    
Yes, you are right –  Mushka Aug 21 '12 at 17:28
    
@hayd, can you give intuition for how to prove $dim(x \times y) = dim(x) + dim(y)$ for generic linear spaces $x$ and $y$? –  spencerlyon2 Oct 26 '13 at 23:13
    
@spencer What's the definition of times? –  hayd Oct 27 '13 at 15:50

1 Answer 1

up vote 5 down vote accepted

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 $\dim(X\times\mathbb{P}^2)=\dim(X)+\dim(\mathbb{P}^2)=1+2=3$. –  Robert Auffarth Aug 21 '12 at 13:29
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Oh my god. Thanks. Edited =) –  Jesko Hüttenhain Aug 21 '12 at 13:31
    
@ rattle Thanks a lot! my solution was the same. –  Mushka Aug 21 '12 at 17:30

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