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If $V$ is affine variety and $W$ is not a affine varieties, then $V\times W$ is an affine variety?

this is just from

Prove that if $V$ and $W$ are affine varieties, then $V \times W$ is an affine variety.

loose argument using their notation. do both $V$ and $W$ be affine varieties?

I think that only one needs to be. If $V\times W$ to be an affine it needs to be defined by polynomials $h_1 , \dots, h_r $. I think that $r=s+t$ where $V$ has $s$ polynomials and $W$ has $t$ polynomials

In order for $V\times W$ to be an affine variety it needs to have a set of polynomials that take in $n+m$ vector. so $V \times W$ is the product of the polynomials def by $V \times W$

is my reasoning wrong? is there something misunderstood with def ?? Things get weirder with higher and higher dimensions

Adam Higgins
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Tiger Blood
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1 Answers1

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Take $V$ to be a point, and $W=\mathbb{P}^1$. $W$ is not affine and $V$ is affine. The product variety $V\times W$ is clearly isomorphic to $\mathbb{P}^1$, so is not affine.

I think one error in your logic is that $V\times W$ need not globally be cut out by polynomials, as you claim. Your logic only applies on an affine neighborhood in $V\times W$.

Andrew
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