Let $R= \oplus_{n \ge 0} R_n$ be a graded Noetherian commutative ring and suppose $R_0$ is Artinian.

Do all maximal homogeneous ideals of $R$ have the same height ?

Let $R_{>0}$ be the ideal generated by elements of positive degree. It's easy to see that the maximal homogeneous ideals of $R$ are exactly the ideals $\mathfrak{p} \oplus R_{>0}$ where $\mathfrak{p}$ runs through the prime ideals of $R_0$.

So the question is trivially true if $R_0$ is a field. I wonder, if this generalizes to the case when $R_0$ is Artinian, but I have no idea how this could be proved.

*Edit 2:* For the sake of completeness, let me add that $R$ is of finite Krull dimension. In fact, the dimension equals the order of the pole at $t=1$ of the Poincaré series $p_R(t) = \sum_{n\ge 0} \ell(R_n)t^n$ defined via the length of $R_n$ as (finitely generated) $R_0$-module.

*Edit:* If $R=A[x_1,...,x_n]$ is a polynomial ring with $A$ Artinian, then the maximal homogeneous ideals of $R$ are $\mathfrak{p}R + (x_1,...,x_n)$ that admit a prime ideal chain
$$\mathfrak{p}R \varsubsetneqq \mathfrak{p}R + (x_1) \varsubsetneqq ... \varsubsetneqq
\mathfrak{p}R + (x_1,...,x_n)$$
of length $n+1$. Since $\dim(R) = n$, all max. homoegenous ideals have height $n$. Thus the question can be answered affirmatively in this case.

**Edit 3:** a-fortiori's example already contains the whole story: Since $A := R_0$ is Artinian, we can decompose $A$ into a finite product of local rings $(A^i,\mathfrak{m}^i)$. This yields a decomposition $R = \prod_i R^i$ with graded rings $R^i = A^iR$ having $R^i_0 = A^i$. Since the product is finite, it's easy to see that the maximal ideals of $A$ are just the ideals
$$\mathfrak{m} = A^1 \times \cdots \times \mathfrak{m}^i \times \cdots \times A^r.$$
Let $M^i = \mathfrak{m}^i \oplus R^i_{>0} \trianglelefteq R^i$ be the homogeneous max. ideal belonging to $\mathfrak{m}^i$ and let $M \trianglelefteq R$ be the homogeneous max. ideal belonging to $\mathfrak{m}$. Then

$$M = \mathfrak{m} \oplus R_{>0} = R^1 \times \cdots \times M^i\times \cdots\times R^r.$$

Hence the height of $M$ equals the height of $M^i$. Since $R^i_0 = A^i$ is local, $M^i$ is the unique homogeneous max. ideal of $R^i$. So its height equals the dimension of $R^i$. Thus we have shown:

The homogeneous maximal ideals of $R$ are given by the ideals $M$ and the height of $M$ is the dimension of $R^i$. In particular, all homogenous maximal ideals of $R$ have the same height, iff all $R^i$ have the same dimension.

Remark: This explains the polynomial ring example above, since there $R^i = A^i[x_1,...,x_n]$ has dimension $n$ for all $i$.

equidimensional). But this is false; consider the ring $R=k[x,y_1, ..., y_n]/ (xy_1, xy_2, ..., xy_n)$, where $k$ is any field and $n\geq 2$ an integer. The minimal primes are $(x)$ and $(y_1, ..., y_n)$, and if $M$ is the homogenous max ideal, we have height $M/(x) = n$ but height $M/(y_1, ..., y_n) = 1$ – Neil Epstein Dec 16 '11 at 11:56