Quesion- Let $R=\bigoplus_{i\ge 0} R_i$ be a (non-trivial) positively graded commutative Noetherian ring with $1(\not=0)$ of (Krull) dimension $d\ge 0$. Let $S\subset R_0$ be a multiplicative set such that $\dim(S^{-1}R_0)<\dim(R_0$. Then can we claim that $\dim(S^{-1}R)<d$ ?
Here is my attempt when $R$ is an integral domain with a stronger hypothesis: Let $S\subset R_0$ be a multiplicative set such that for any maximal ideal $m\in \text{Spec}(R_0)$, we have $S\cap m\neq \emptyset$. Unfortunately, I'm unable to prove the question and have also been unable to find a counterexample
Let $\dim(R_0)=n$. We prove the following statement:
``Let $S\subset R_0$ be a multiplicative set such that $S\cap m\not=\phi$, for any maximal ideal $m\in Spec(R_0)$. Then the graded domain $S^{-1}R$ does not have a graded maximal ideal $S^{-1} M$ such that $M$ is a maximal ideal in $R$."
by induction on $n$. First we note that if $n=0$, then $R_0$ is a field. In this case the statement is vacuously true.
Now we take $n\ge 1$. If there does not exist such a $S$, then the statement is vacuously true. Hence we assume that such a $S$ exists. With contrary we assume that there exists a graded maximal ideal, say $S^{-1}M$, in $S^{-1}R$, where $M\in Spec(R)$ is a maximal ideal. Now there are two possibilities: either $M$ is a graded maximal ideal or $M$ is not a graded ideal. Now if $M$ is a graded maximal ideal, then it is of the form $m\oplus R^+$, where $m$ is a maximal ideal in $R_0$. Then $S\cap m\not=\phi$ implies that $S\cap M\not=\phi.$ This leads to a contradiction.
Now we assume that $M$ is not a graded maximal ideal. Then we first claim that $m_0:=M\cap R_0\not=\{0\}$. Since $S^{-1}M$ is a graded maximal ideal it is of the form $m'\oplus S^{-1}R^+$, where $m'$ is a maximal ideal in $S^{-1}R_0$. In particular, since $R$ is Noetherian, there exists $s\in S$ such that $sR^+\subset M$. As $S\cap M=\phi$, we must have $R^+\subset M.$ Now suppose assume that $m_0=0$. Let $f\in M$. Then $f=f_0+f_1$, where $f_0\in R_0$ and $f_1\in R^+$. As $R^+\subset M$, we have $f_0\in m_0=\{0\}$. This implies $M=R^+$. But then $R_0\cong R/R^+\cong R/M\cong k$, which is not possible as $\dim(R_0)=n\ge 1$, where $k$ is a field. This establish that $m_0\not=\{0\}.$
Let $a\in m_0\setminus \{0\}$, and let ``bar'' denote going modulo $\langle a \rangle $. Then $\dim(\overline R_0)\le n-1$ and $\overline R$ is a graded domain of dimension $\le d-1$. Let $\eta\in Spec(\overline R_0)$ be a maximal ideal in $\overline R_0$. Then the ideal $\langle\eta, a\rangle $ is also a maximal ideal in $R_0$ as $\overline R_0/\eta\cong R_0/\langle\eta, a\rangle $. Therefore, by our hypothesis $S\cap \langle\eta, a\rangle \not=\phi$. This implies that $\overline S\cap \eta \not=\phi$. We also note that $\overline M$ is a maximal ideal in $\overline R$, as $\overline R/\overline M\cong R/M$. Moreover, as $\overline S^{-1}\overline M \cong \overline{S^{-1} M}$ and $a\in R_0$, the ideal $\overline S^{-1}\overline M$ is a graded maximal ideal in $\overline S^{-1}\overline R$ (recall that $S^{-1}M$ is a graded ideal). But by induction hypothesis there does not exist such an maximal ideal in $\overline R$. This completes the induction step.
It is only remaining to show that $\dim(S^{-1}R)<d$. To prove this we note that for an arbitrary graded ring $B=\bigoplus_{i\ge 0} B_i$, there exists a homogeneous maximal ideal $N$ in $B$ such that $ht(N)=\dim(B).$ In $S^{-1}R$, any graded maximal ideal of height $d$ is a localization of a maximal ideal in $R$. But this can not happen, as it is proved previously. Thus the ring $S^{-1}R$ does not have a graded maximal ideal of height $d$. Hence $\dim(S^{-1}R)<d$. This completes the proof. \qed
