Why there exist prime homogeneous ideals containing the Irrelevant ideal?

Question

The notion of irrelevant ideal is something the doesn't bother me long time now, though the last day I came across it again and realised something that didn't now. That simply I don't understand how the whole $\textbf{Proj}$ construction works.

So my question is:

In Wiki article about $Proj$ sonstruction says: "Define the set $Proj(S)$ to be the set of all homogeneous prime ideals that do not contain the irrelevant ideal". Now since the irrelevant ideal is a maximal ideal (in the usual sense regardless of the grading part/structure), how is possible for a homogeneous prime ideal to contain the irrelevant ideal? Apparently I'm missing something, so can you please help me out and make more clear how does the $Proj$ construction works?

Thank you!

Answer 1

First, let's fix some notation. Let $S=\bigoplus_{i=0}^\infty S_i$ be a graded ring with grading given by $S_i$ for $i\in\mathbb{Z}_{\geq0}$. Let $S_+=\bigoplus_{i=1}^{\infty} S_i$ be the irrelevant ideal of the ring $S$.

First, a caution: $S_+$ is not necessarily maximal. Consider $S=\mathbb{Z}[x_1,x_2,\cdots,x_n]$ with each $x_i$ of degree one. Then $S_+=(x_1,\cdots,x_n)$ and $S/S_+\cong \mathbb{Z}$ which is not a field, so $S_+$ is not maximal.

Secondly, it is certainly possible for a homogeneous prime ideal to contain $S_+$, the irrelevant ideal. Consider $S=\mathbb{Z}[x_1,\cdots,x_n]$ as before and let $\mathfrak{p}=(2,x_1,\cdots,x_n)\subset S$. Then $S/\mathfrak{p}=\mathbb{F}_2$, so $\mathfrak{p}$ is prime, homogeneous, and even maximal while properly containing $S_+$. Even further, $S_+$ is always prime and homogeneous (if $pq$ is positive degree, at least one of $p$ or $q$ must be of positive degree, and $S_+$ is trivially equal to $\bigoplus_{i=0}^\infty \left(S_+\cap S_i\right)$) so there is always at least one prime homogeneous ideal containing $S_+$, $S_+$ itself.

I think some of your confusion may stem from the definition of maximal ideal. A maximal ideal $\mathfrak{a}$ is an ideal which is maximal by inclusion among all proper ideals. This means if any proper ideal $\mathfrak{b}$ contains $\mathfrak{a}$, we must have $\mathfrak{b}=\mathfrak{a}$. This does not prohibit ideals containing maximal ideals, it just says that they must be equal.