Why does $\operatorname{Proj}(B)$ not contain ideals containing the irrelevant ideal?

Question

Let $A$ be a ring and let $B = \bigoplus_{d\geq 0} B_d$ be a graded $A$-algebra.In Liu's book Algebraic Geometry and Arithmetic Curves the set $\operatorname{Proj}B$ is defined as the set of all homogeneous prime ideals of $B$ not containing the irrelevant ideal $B_+ =\bigoplus_{d>0} B_d$.

This basically means don't consider any ideals of the form $I+B_+$ where $I$ is a prime ideal of $B_0$, which contains a copy of $A$.Why is this restriction imposed?

Answer 1

Informal reason: think of the case $B=A[x_0,\dots,x_n]$. Then $B_+=(x_0,\dots,x_n)$ but this will certainly not give a point on $\textrm{Proj }B=\mathbb P^n_A$. (The point $(0:\cdots:0)$ is not defined).

More formally, the main reason is, in my opinion, the following fact: for a homogeneous ideal $\mathfrak a\subset B$, one has $Z(\mathfrak a)=\emptyset$ if and only if $\sqrt{\mathfrak a}=B$ or $\sqrt{\mathfrak a}=B_+$.

Having this in mind, there are two news:

1. Good news: $Z(-)$ establishes (by the above fact) a bijective correspondence similar to the one you probably encountered with affine varieties. More precisely, homogeneous radical ideals in $B$ not containing $B_+$ correspond to algebraic sets in $\textrm{Proj }B$. This is only possible thanks to us excluding $B_+$ from the game.
2. Bad news: the existence of the irrelevant ideal can still be detected in our incapability of getting a regular map $\pi:\textrm{Proj }C\to\textrm{Proj }B$ out of a (graded) homomorphism of graded $A$-algebras $q:B\to C$. (We were able to do this in the affine case!) Indeed, if we try to define $\pi(\mathfrak p)=q^{-1}(\mathfrak p)$, the right hand side is an ideal of $B$ which might contain the irrelevant ideal $B_+$ even though $\mathfrak p$ did not contain $C_+$. We do get a morphism $\pi:U\to\textrm{Proj }B$ defined as above where $U\subset \textrm{Proj }C$ is the set of those $\mathfrak p$ such that $q^{-1}(\mathfrak p)\nsupseteq B_+$. In particular, one could observe that, unlike Spec, Proj is not a functor.

Answer 2

One of the reason I can think of is in the case when $B=k[x_1,\ldots,x_n]$ you want the 1-1 inclusion revising correspondence between algebraic sets and homogeneous radical ideals, similar to the one in affine case. $$Y\rightarrow I(Y)$$ with inverse $$\mathfrak{a}\rightarrow Z(\mathfrak{a})$$ and the little twist showing up in projective case is that the $\emptyset$ corresponds to both $B$ and the irrelevant ideal under above relation. So it should be excluded.