# Hypersurfaces in P^n: Always One Polynomial?

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?

## 1 Answer

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 $f = \prod_i f_i$.

• 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. – Cass Dec 15 '13 at 7:26
• Hmm, I'm assuming when you said permitting $Z$ reduced, you meant $Z$ non-reduced? – Dori Bejleri Dec 15 '13 at 7:28
• Ugh, yes. Thank you. – Cass Dec 15 '13 at 7:28
• 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. – Dori Bejleri Dec 15 '13 at 7:29
• Yes it indeed doesn't work for non-reduced. I edited it. – Dori Bejleri Dec 15 '13 at 7:39