1

Let $k$ be an algebraically closed field and $\mathbb{P}^n_k$ be the $n-$dimensional projective space over $k$. By Bezout's theorem, for homogeneous polynomials $f_1,f_2,\cdots,f_n$ we have $\#V(f_1,f_2,\cdots,f_n)=\displaystyle{\prod^{n}_{i=1}} \deg f_i$ counted with multiplicity. But what about only $n-1$ polynomials? $V(f_1,f_2,\cdots,f_{n-1})$ seems like a curve, so does it has the same cardinality of $k$? If yes, is it a consequence of the Bezout's theorem?

I was trying to add a polynomial $a_0x_0+a_1x_1+\cdots+a_nx_n=0$ for not identically zero $a_0,a_1,\cdots,a_n$, so the system would always be solvable in $\mathbb{P}^n_k$ by the Bezout's theorem. I hope that for in some sense almost all $a_0,a_1,\cdots,a_n$, different $[a_0,a_1,\cdots,a_n]$ would give different solutions. But at the time being I can get nothing. Any help appreciated.

Jianing Song
  • 1,707
  • In your first paragraph, you've left out the condition that $\dim V(f_1,\cdots,f_n)=0$. This matters: if you take all $f_i$ equal, for instance, you don't get a finite set of points.

    The answer to your general question is yes - any positive-dimensional algebraic set over an algebraically closed field $k$ has cardinality equal to the cardinality of $k$. See the linked duplicate.

    – KReiser Jul 03 '22 at 14:36
  • @KReiser Thanks! – Jianing Song Jul 05 '22 at 13:07

0 Answers0