Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $K$ be an algebraically closed field. Let $n \ge 0$ be an integer. We consider $K^{n+1}$ as a topological space with Zariski topology. Let $G = K^*$ be the multiplicative group of $K$. Let $X = K^{n+1} - {0}$. Then $G$ acts on $X$. Then the projective space $P^n$ over $K$ is, by definition, the orbit space $X/G$. We regard $P^n$ as a topological space with the quotient topology. Let $Z$ be a closed subset of $P^n$. Then how do you prove that $Z$ is the common zeros of homogeneous polynomials?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

A start: Take a $Z'\subset K^{n+1}$ which is a closed and such that $\forall k\in K^*, z\in Z': kz\in Z'$.

Let $I\subset K[x_1,...,x_{n+1}]$ be the ideal of polynomials $f$ such that $f(z)=0$ for all $z\in Z'$.

Take any polynomial $f\in I$ and write it as the sum $\sum f_i$ where the $f_i$ are homogeneous of degree $i$.

Then for any particular $z_0\in Z'$, define $p(x)=\sum f_i(z_0)x^i$. So $p(x)\in K[x]$. If $f_i(z_0)\neq 0$ for some $i$, then $p(x)$ is a non-zero polynomial, so there exists a $k\in K^*$ such that $p(k)\neq 0$. (You don't need $K$ algebraically closed, just $K$ infinite for this step.)

But $p(k)=\sum f_i(z_0)k^i = \sum f_i(kz_0) = f(kz_0)$. So that would contradict the condition that if $z\in Z'$ then $kz\in Z'$.

So that means that $p(x)$ has to be the zero polynomial, which means that $f_i(z_0)=0$ for all $z_0\in Z'$, which means that $f$ is the sum of homogeneous polynomials $f_i$ such tat $f_i(z)=0$ for all $z\in Z'$.

Any closed $Z\subseteq P^n$ must be the image of such a $Z'\subset K^{n+1}$, so it seems like you are done.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.