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.
First of all i believe that what the authors really meant was the more precise "non-negatively graded" instead of "positively graded".
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.
