Affine variety consisting $(\mathbb Z,\mathbb Z)$ should be equal to whole affine space.

Asked Jan 26 '22 at 20:11

Active Jan 26 '22 at 20:11

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In this question I was searching to find vanishing ideal of $$\{(n,n)| n\in\mathbb Z\}\subset A^2(\mathbb C)$$

I thought that vanishing ideal is $$\left<\prod_{n\in\mathbb Z}(x-n)(y-n)\right>$$

But it is not clearly a polynomial in $\mathbb C[x,y]$

My guts say that the vanishing ideal should be $\emptyset\subset \mathbb C[x,y]$

but having a bit hard time where to start and how to show it actually.

asked Jan 26 '22 at 20:11

User not found

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The vanishing ideal is $(x - y)$. – abacaba Jan 26 '22 at 20:16
but how you prove it – User not found Jan 26 '22 at 20:18
By the transform $(x,y) \to (x - y, y)$ it suffice to show that the vanishing ideal of $(0, Z)$ is $(x)$. If a polynomial $f(x,y)$ vanish on $(0, Z)$ then write $f(x,y) = xg(x,y) + h(y)$. We have $h(Z) = 0$ so $h = 0$, thus $f \in (x)$. – abacaba Jan 26 '22 at 20:21
but it also vanishes on (z,z) for every z\inC – User not found Jan 26 '22 at 20:21
Yes, because the Zariski closure of ${(n,n): n \in \mathbb{Z}}$ is ${(n,n): n \in \mathbb{C}}$. – abacaba Jan 26 '22 at 20:22

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