Maximal ideals and the projective Nullstellensatz

This is a simple question, but it's one of those things that I've been thinking about so much that I've just kind of lost where I am and need some explicit reference.

One of the main corollaries of the affine Nullstellensatz (i.e. that $\mathbb{I}(\mathbb{V}(I)) = \sqrt{I}$) is that maximal ideals in $R=k[x_0,\ldots,x_n]$ are in bijective correspondence with points in $\mathbb{A}^{n+1}$. In all (but one -- see below) of the literature I've read that talks of the projective Nullstellensatz, there is no mention of what maximal ideals of $R$ correspond to, just what prime ideals in general correspond to. I assumed that, if there were a similar fact, then I would have found it somewhere.

But after some thinking we see that points in projective space correspond to lines through the origin in affine space, which are irreducible affine varieties, but not affine points. Thus points in projective space correspond to (prime) ideals in $R$ coming from lines through the origin.

In the opposite direction, maximal ideals in $R$ correspond to points in affine space, and so can't correspond directly to points in projective space but each point in affine space uniquely determines a point in projective space, since it uniquely determines a line through the origin and that point. So it seems like there should be some maximal ideal/points in projective space correspondence, and one way kind of seems almost clear to me, but the other doesn't so much.

In essence could anybody please explain the correspondence between maximal ideals properly contained in $k[x_0,\ldots,x_n]$ and points of projective space. We know that all maximal ideals are of the form $(x_0-a_0,\ldots,x_n-a_n)$, and these correspond to the point $(a_0,\ldots,a_n)$ in affine space. Is there an equally nice formulation for the projective case?

Here is the one reference that I could find: • We always have a morphism $\mathbb{A}^{n+1}-\{0\}\to\mathbb{P}^n$. Thus, any maximal ideal other than $(x_0,\ldots,x_n)$ gives you a point in $\mathbb{P}^n$, the correspondence is not one-to-one, but still nice. Notice that the multiplicative group acts naturally by multiplying all the $x_i$'s by the same number on $R$ and all maximal ideals in a single orbit goes to the same point in projective space and vice versa. – Mohan Aug 20 '15 at 19:16
• I was just curious where the excerpt above comes from. Thanks! – mwmjp Mar 30 '16 at 14:33
• – Tim Mar 31 '16 at 21:16

You're thinking about this the wrong way. There is a map from $\mathbb{A}^{n+1} - \{0\} \to \mathbb{P}^n$ for any $n$, but this is not the way we realize the co-ordinate ring, and the construction one would get if one tried to blow up points into lines in this quotient would not be the affine variety corresponding to some subset of $\mathbb{P}^n$ but the pullback of the tautological bundle onto this variety.

The real maps to consider are the co-ordinate maps: $\mathbb{A}^n \to \mathbb{P}^n$ that make $\mathbb{P}^n$ a variety. These induce isomorphisms between $k[x_1, \dots, x_n]$ and $\mathcal{O}(\mathbb{P}^n - \{(x_1:\dots : x_{n+1}) | x_i = 0\})$, and this is how you prove the correspondence between maximal ideals that do not contain the trivial homogeneous ideal $R(X)_+$ (which obviously cannot be included) and points in $\mathbb{P}^n$, you use these homeomorphisms.

• So given some maximal ideal $m$ not containing $R(X)_+$, this corresponds to some maximal ideal $\hat{m}$ of $k[x_0,\ldots,x_n]$ not containing $(x_0,\ldots,x_n)$. So we know that there exists some $x_i$ such that $\hat{m}$ doesn't contain $(x_i)$... and then where do we go? – Tim Aug 21 '15 at 17:43
• why can we not use a proof that goes something like this: math.stackexchange.com/questions/1405195/… ? – Tim Aug 21 '15 at 18:15
• Then we take the map $\phi: \mathbb{P}^n - \{x_i = 0\} \to \mathbb{A}^n$ that sends $(a_1: \dots : a_{n+1}) \to (a_1/a_i, \dots, a_{n+1}/a_i)$ which induces an isomorphism of coordinate rings on some open set homeomorphic to $mathbb{A}^n$... And from here it should be very obvious to you. This is exactly how they want you to prove it in that link. – Sempliner Aug 22 '15 at 3:54
• How do we deal with patching the affine patches together though? That is, if some $x$ has $x_i,x_j\neq0$, then how do we know that the maximal ideal given by looking at $\mathbb{P}^n\setminus\{x_i=0\}$ is the same as that given by looking at $\mathbb{P}^n\setminus\{x_j=0\}$? – Tim Aug 22 '15 at 11:39
• Further, the Nullstellensatz tells us that all maximal ideals of the polynomial ring $k[x_0,\ldots,x_n]$ are of the form $(x_0-a_0,\ldots,x_n-a_n)$, but this is not homogeneous. So how can there exist maximal homogeneous ideals? – Tim Aug 22 '15 at 11:44

https://math.stackexchange.com/a/1643498/346324

This answer is for the analogous problem in Vakil but is more explicit than what has been given already and looks similar to the affine situation.

• I don't think this can be count as an answer. It's just a comment. – user26857 Oct 3 '16 at 16:01
• @user26857 I agree. It's a very useful comment, but still a comment... – Tim Oct 3 '16 at 22:01
• I'm trying to say that question above has already been asked and answered in the link that I gave... – Tanner Strunk Oct 4 '16 at 1:25