If we define a "hypersurface in $\mathbb{P}^n_k$" ($k$ algebraically closed) to be an effective Cartier divisor - i.e. a locally principal closed subscheme - hence permitting NONreduced components, is it still true that the (saturated) homogeneous ideal in $k[x_0,...,x_n]$ of a hypersurface in $\mathbb{P}^n_k$ can always be generated by a single element?


Yes. In fact, even when we are defining a hypersurface of degree $r$ in the naive way, we are saying that it is the zero set of a homogeneous polynomial $f$ of degree $r$. But homogeneous polynomials of degree $r$ are not functions on projective space. They are sections of the line bundle $\mathcal{O}(r)$. Thus, even under the usual naive definition, a hypersurface is exactly the Cartier divisor corresponding to section of a line bundle. On $\mathbb{P}^n$, all Cartier divisors come up this way and all line bundles are of the form $\mathcal{O}(r)$. From this we see that every effective Cartier divisor is is the zero set of some homogeneous degree $r$ polynomial, specifically the corresponding section of the line bundle.

When the hypersurface is reduced, we don't have to use the machinery about line bundles or divisors. let $Z$ be the closed subscheme corresponding to a reduced Cartier divisor. Then for each irreducible component $Z_i$, we have a homogeneous prime ideal $\mathfrak{p}_i$. Since $Z_i$ are codimension 1, $\mathfrak{p}_i$ are height one, so applying Hauptidealsatz, we see that $\mathfrak{p}_i = (f_i)$ for some homogeneous polynomial $f_i$. Then $Z$ is cut out by the product $f = \prod_i f_i$.

  • $\begingroup$ I agree with your first paragraph (at least upon a first read) and after another read will accept your answer. However, the second paragraph I'm a bit unsure of. If Z is not reduced, how do you break it up into irreducible components? It seems to me that you would have to take the reduction of Z first, which would mean in the end you have a different scheme. $\endgroup$ – Cass Dec 15 '13 at 7:26
  • $\begingroup$ Hmm, I'm assuming when you said permitting $Z$ reduced, you meant $Z$ non-reduced? $\endgroup$ – Dori Bejleri Dec 15 '13 at 7:28
  • $\begingroup$ Ugh, yes. Thank you. $\endgroup$ – Cass Dec 15 '13 at 7:28
  • $\begingroup$ Ok yeah then I don't think my argument in the second paragraph works in the nonreduced case, but I'll think about it and edit it when I know for sure. The first argument goes through in general though. $\endgroup$ – Dori Bejleri Dec 15 '13 at 7:29
  • $\begingroup$ Yes it indeed doesn't work for non-reduced. I edited it. $\endgroup$ – Dori Bejleri Dec 15 '13 at 7:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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