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Let $F$ be an algebraically closed field, and $A^n(F)$ be an affine space over it with dimension $n$. For $p \in F[X_1,...,X_n]$, show that:

  • $A^n(F) \setminus V(p)$ is infinite for $n \geq 1$
  • $V(p)$ is infinite for $n \geq 2$.

With $V(p)= \{x=(x_1,...,x_n) \in A^n(F) \mid p(x) = 0\}$.

For the second item, I'm thinking of looking $p$ as an element of $F[X_1,...,X_{n-1}][X_n]$, then supposing that it has finite roots, and that should broke the algebraic closure of $F$, so I think I can handle it.

The real problem for me is the first affirmation. How do I prove it? Can I suppose it's finite then get to an contradiction? Any leads?

KReiser
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Gea5th
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  • It seems you want to add the assumption that $p$ is nonconstant. Because if $p=0$ then $V(p)=\mathbf{A}^n(F)$ so that $\mathbf{A}^n(F)\setminus V(p)=\emptyset$ is not infinite. In the same vein, if $p=1$ (or any nonzero element of $F$) then $V(p)=\emptyset$. – Samuel Jan 16 '22 at 20:29
  • Per the first duplicate link, $\Bbb A^n_F\setminus V(p)$ is isomorphic to a closed subvariety in $\Bbb A^{n+1}_F$, so one can apply the second duplicate to see that it has the cardinality of $F$, which is infinite. – KReiser Jan 17 '22 at 00:52

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