Let $R$ be a Noetherian positively graded ring and $M$ a finite graded $R$-module. Prove that $\dim M = \sup\{\dim M_p: p\in\operatorname{Supp} M \text{ graded}\}$.

This is the Exercise 1.5.25 in the book: Winfried Bruns and Jürgen Herzog, Cohen-Macaulay Rings, Cambridge University Press, 1998.

I also want to know why there is positively graded ring in the condition.

On to the main problem now. We have that $R = \oplus_{i \ge 0} R_i$ is Noetherian and $M$ finite graded $R$-module. Since $\operatorname{Ann}(M)$ is homogeneous, and by definition $\dim M = \dim R/\operatorname{Ann}(M)$, we may replace $R$ by $\operatorname{Ann}(M)$ and assume that $\dim M = \dim R$.
We will show that for any maximal ideal $m$ of $R$, there exists a homogeneous ideal $q$ such that it has at least the same height as $m$. So let $m$ be a maximal ideal of $R$ and assume that it is not homogeneous. By Theorem 1.5.8b in B&H, we have that $\operatorname{height}m =\operatorname{height}m^* +1$, where $m^*$ is the ideal generated by the homogeneous components of $m$. Now $m^*$ is by definition a homogeneous ideal and it can not be maximal because $m^* \subsetneq m$. Define $m_0 = m \cap R_0$. Now $R/m$ is a field and this implies that $R_0/m_0$ is a field (see proof in the comments), thus $m_0$ is a maximal ideal of $R_0$. Next define $q = m_0\oplus \left( \oplus_{i \ge 1} R_i\right)$. Then $q$ is a maximal ideal of $R$ that is homogeneous and additionally it contains $m^*$. Hence $\operatorname{height} q >\operatorname{height} m^* = \operatorname{height}m -1$ and so $\operatorname{height} q \ge \operatorname{height} m$.
Alternative Proof: Let $p$ be any homogeneous prime ideal that is $^*$maximal. Then every homogeneous element of $R/p$ is invertible, and we are in the situation of Lemma 1.5.7 in Bruns&Herzog. Hence either $R/p=k$ or $R/p=k[t,t^{-1}]$, where $k$ is a field. But $R/p$ is positively graded since $R$ is. Hence it must be the case that $R$ is a field and so $p$ is a maximal ideal of $R$. This shows that every $^*$maximal ideal is maximal. Now let $q$ be such that $\dim M = \operatorname{height} q$. If $q$ is homogeneous then we are done. Otherwise, $\operatorname{height} q^* =\operatorname{height} q - 1$. Since $q^*$ is properly contained in $q$, it is not maximal, and so it is not $^*$maximal. Thus there must exist some homogeneous prime $P$ such that $P$ properly contains $q^*$. This means that $\operatorname{height} P =\operatorname{height} q = \dim M$, and we are once again done.
• Why if $m$ is maximal in $R$ then $m_0=m\cap R_0$ is maximal in $R_0$? (LaTeX tip: use \subsetneq instead of subset\neq in order to get $m^*\subsetneq m$.) – user26857 Feb 10 '14 at 22:25
• @user121097: Take any element $r \in R_0-m_0$. Then this element is nonzero mod $m$, hence it is invertible mod $m$. So there exist $r' \in R$ and $\mu \in m$ such that $rr'+\mu=1$. So the homogeneous part of degree zero of $rr'+\mu$ must be equal to $1$. Now this part is equal to $r r'_0 + \mu_0$, where $r'_0, \mu_0$ are the degree $0$ homogeneous components of $r', \mu$ respectively. Hence we must have $r r'_0 + \mu_0 = 1$. But $r'_0 \in R_0$ and $\mu_0 \in m_0$ and this equation says that $r \in R_0$ is invertible mod $m_0$. You agree? – Manos Feb 11 '14 at 2:31
• Thanks for the nice answer, let's discuss more. @user121097 For the Z-graded rings, the element in R$_1$, R$_2$,..also can have inverse, but the invertible element can not appear in any ideal which is not R, so how does it disturb the main claim? The Lemma 1.5.7 is an extreme case of this type, “totally invertible”. – Strongart Feb 12 '14 at 11:53