I would like to show that the homogeneous ideal of any point $x$ in projective space $\mathbb{P}^n(k)$, where $k$ is an algebraic closed field, is a maximal homogeneous ideal in the polynomial ring $k[X_0, \dots, X_n]$.

By maximal homogeneous ideal I mean a homogeneous ideal in the polynomial ring that is properly included in the irrelevant ideal $(X_0, \dots, X_n)$, and that is maximal among all homogeneous ideals that also are properly included in the irrelevant ideal.

I have already found, that if $x=(x_0: \dots: x_n)$ with $x_i = 1$, I can write its homogeneous ideal as $(X_0 - x_0 X_i, \dots, X_n - x_n X_i)$. I think the next step should be to show that this is indeed a maximal homogeneous ideal, but I just don't know how to do this, and I would be grateful for some help. Is it possible to check it directly by showing that any homogeneous ideal containing properly $(X_0 - x_0 X_i, \dots, X_n - x_n X_i)$ has to be the the irrelevant ideal $(X_0, \dots, X_n)$? Is there some general criterion for checking this kind of maximality, similar to the maximal ideals ($R/I$ is a field $\Leftrightarrow $ I is a maximal ideal in the ring $R$)?

Also, I was wondering whether, given any homogeneous ideal properly included in the irrelevant ideal, there always exists a maximal homogeneous ideal containing it. Since we have this with ideals, I was thinking it was in this situation also true, and started proving this using Zorn's Lemma, but got stuck at some point because of some technical problem. Is this statement true?

Thanks very much in a advance!

  • $\begingroup$ I don't understand: between $I$ and the irrelevant ideal it is supposed that there is no other homogeneous ideal. However, $(0)\subsetneq(Y^2)\subsetneq(Y)$. $\endgroup$ – user26857 Sep 14 '16 at 16:37
  • $\begingroup$ Ok, yes, it is strange, so I guess this result just doesn't hold after all. I will reformulate my question. $\endgroup$ – Thomas Delerue Sep 15 '16 at 8:10
  • $\begingroup$ 1. The homogeneous ideal of a point as you have found it is not a maximal homogeneous ideal for the same reason I've pointed out in my first comment: $K[X_i]$ has plenty of homogeneous ideals. 2. The characterization of maximal homogeneous ideals is the following: $I$ is maximal iff $R/I\simeq K[Y,Y^{-1}]$. 3. The polynomial ring over a field is noetherian, so every non-empty family of ideals has a maximal element. $\endgroup$ – user26857 Sep 16 '16 at 2:50

The maximal homogeneous relevant ideals of $k[X_0, \dots, X_n]$ are exactly the ideals of the form $$\mathfrak m_a=\langle a_iX_j-a_jX_i\vert i,j=0\cdots n\rangle $$ where $a=[a_0:a_1:\cdots:a_n]\in \mathbb P^n(k) $.
The correspondence $a \leftrightarrow \mathfrak m_a$ is the link between the classical points of projective space $\mathbb P^n(k)$ and the closed points of projective space seen as the scheme $\mathbb P^n_k=\operatorname {Proj} (k[X_0, \dots, X_n])$.

  • $\begingroup$ Dear Georges, do you consider somehow that the maximal relevant ideals are prime? $\endgroup$ – user26857 Sep 16 '16 at 16:27
  • $\begingroup$ Ok, let me ask in other way: Dear Georges, can you elaborate on the form of maximal homogeneous relevant ideals as you described it in your answer? Thanks. $\endgroup$ – user26857 Sep 21 '16 at 3:04
  • $\begingroup$ Dear @user26857: the ideals which are maximal with respect to proper inclusion in the irrelevant ideal $\langle X_0, \dots, X_n \rangle$ are not maximal in the usual sense in the ring $k[X_0, \dots, X_n]$. However they are prime (in the usual sense) in that ring, because they are the ideals of lines in the affine space $\mathbb A^{n+1}_k=\operatorname {Spec (k[X_0, \dots, X_n])}$ (Notice carefully that one of the $a_j$'s is nonzero since these $a_j$'s are the homogeneous coordinates of a point in projective space) $\endgroup$ – Georges Elencwajg Sep 21 '16 at 7:12
  • $\begingroup$ Dear Georges, the first part is pretty clear. I'll think (algebraically) about their prime-ness. Thanks. $\endgroup$ – user26857 Sep 21 '16 at 7:14
  • $\begingroup$ Dear Georges, could you please elaborate why $\mathfrak{m}_a$ is maximal among homogenous relevant ideals, and why every maximal homogenous relevant ideal is of this form? $\endgroup$ – Layer Cake May 1 '17 at 21:12

Thanks very much for your help and your answers! However I am still confused, because it seems to me that the ideal $m_a = (a_iX_j - a_jX_i \vert i,j=0, \dots,n)$ is the same as $(X_0 - \tilde{a}_0 X_i, \dots, X_n - \tilde{a}_n X_i)$, supposing $$a = (a_0:\dots:a_n) = (\tilde{a}_0:\dots:\tilde{a}_{i-1}:1:\tilde{a}_{i+1}:\dots:\tilde{a}_n) \in \mathbb{P}^n(k),\ \tilde{a}_i=1.$$ On the other hand we have $k[X_0, \dots, X_n]/(X_0 - \tilde{a}_0 X_i, \dots, X_n - \tilde{a}_n X_i) \cong k[X_i]$, and so by the remark of user26857, $(X_0 - \tilde{a}_0 X_i, \dots, X_n - \tilde{a}_n X_i)=m_a$ cannot be a maximal homogeneous ideal. Am I missing something? Maybe we don't use the same definition of maximal homogeneous ideal? In any case this correspondence is what I was looking for. I wanted to obtain a subset of $\mathrm{Proj} \, k[X_0, \dots, X_n]$, that is isomorphic to $\mathbb{P}^n(k)$ as a locally ringed space. Because in the affine case this subset is the set $\mathrm{Spm} \,k[X_1, \dots, X_n]$ of all maximal ideals of $k[X_1, \dots, X_n]$, I thought it would be similar in the projective case if I adapted the definition of maximality, hence if I considered the subset of all maximal homogeneous ideals in $k[X_0, \dots, X_n]$.

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    $\begingroup$ Dear Thomas, indeed the ideal $\mathfrak m_a$ is not maximal in the usual sense: it is maximal in the set of ideals strictly included in the irrelevant ideal. See our exchange with user26857 in the comments to my answer. $\endgroup$ – Georges Elencwajg Sep 21 '16 at 7:23
  • $\begingroup$ Dear Georges, thank you for your answer. Unfortunately I still do not understand why the ideal $m_a$, as you wrote it, is a maximal homogeneous ideal. $\endgroup$ – Thomas Delerue Sep 21 '16 at 15:01
  • $\begingroup$ Because the point $a$ is a minimal variety in the projective space $\mathbb{P}^n(k)$. $\endgroup$ – Georges Elencwajg Sep 21 '16 at 18:35

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