Elementary proof that projective space is a quotient - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T21:47:25Z http://mathoverflow.net/feeds/question/16146 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16146/elementary-proof-that-projective-space-is-a-quotient Elementary proof that projective space is a quotient Andrew Critch 2010-02-23T11:54:33Z 2010-02-23T19:16:21Z <p>Fix an algebraically closed <b>&dagger;</b> ground field $k$ of any characteristic. I want to use the classical definition of projective $n$-space $\mathbb{P}^n$ as set quotient of $\mathbb{A}^{n+1}\setminus 0$ by the action of $k^*$, and for its Zariski topology the definition that closed sets are the "zero sets" of homogeneous polynomials in $n+1$ variables considered as functions from $\mathbb{P}^n$ to {$0,1$}.</p> <p>I'm not satisfied with my understanding of why this is the same as the quotient topology from the map <br>$\mathbb{A}^{n+1}\setminus 0\to \mathbb{P}^{n}$. This means proving the map is surjective, continuous, and that a set with closed pre-image is closed, the first two properties being clear. The last part is the least trivial, meaning that:</p> <blockquote> <p>If a set $S\subseteq \mathbb{P}^n$ has closed preimage $\widehat{S}$ in $\mathbb{A}^{n+1}\setminus 0$, i.e. $\widehat{S}$ is cut out by some polynomials $f_1,\ldots,f_r$, then $S$ is closed in $\mathbb{P}^n$, i.e. $\widehat{S}$ is cut out by some <i>homogeneous</i> polynomials $g_1,\ldots,g_s$.</p> </blockquote> <p>I'm bad at algebra, so I can't seem to start doing anything here that doesn't look ugly/hard... I really want to "see" what's going on and avoid just quoting results about invariants without understanding how they work in this "toy" example.</p> <p>Thanks for the help, anyone!</p> <p><b>&dagger; Follow up:</b> the algebraic closed hypothesis is not necessary, since David/Kevin's proof works for any infinite field, and for a finite field our spaces (as defined here) are discrete so the result is trivial.</p> http://mathoverflow.net/questions/16146/elementary-proof-that-projective-space-is-a-quotient/16147#16147 Answer by David Speyer for Elementary proof that projective space is a quotient David Speyer 2010-02-23T12:00:39Z 2010-02-23T12:00:39Z <p>Let $f$ be a polynomial which vanishes on $\hat{S}$. Write $f=\sum f_i$, where $f_i$ is homogenous of degree $i$. The set $\hat{S}$ is homogenous so, for any $\lambda \in k^*$, the polynomial $f(\lambda \cdot x) = \sum \lambda^i f_i$ also vanishes on $f$.</p> <p>Since $k$ is infinite, we can find more equations of the form $\sum \lambda^i f_i=0$ than there are terms in $f$. Taking linear combinations, we deduce that each $f_i$ individually vanishes on $\hat{S}$.</p> http://mathoverflow.net/questions/16146/elementary-proof-that-projective-space-is-a-quotient/16148#16148 Answer by Kevin Buzzard for Elementary proof that projective space is a quotient Kevin Buzzard 2010-02-23T12:04:08Z 2010-02-23T12:04:08Z <p>Look at the subspace of $\mathbf{A}^{n+1}$ cut out by your polynomials. This set is invariant under the diagonal action of $k^\times$. So the functions that vanish on it will be an ideal $I$ (the radical of the ideal generated by your $f_i$) which is invariant under this action. But because $k$ is infinite, it's now relatively easy to check that if $f\in I$ and if you write $f$ as a sum of homogeneous pieces, then each homogeneous piece will be in $I$ too (the point is that the homogeneous pieces scale differently under $k^\times$ so if the degree of f is $d$ then you just use $d+1$ elements of $k^\times$ and then put a linear combination of them together and use the fact that a certain Van der Monde determinant isn't zero). Hence $I$ is generated by homogeneous elements and now you're home, unless I missed something.</p>