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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 $i$.

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You might want to add some condition to exclude trivial counterexamples like $k[X,Y]\times k[Z]$ where $X,Y,Z$ have degree $1$. –  user2035 Dec 16 '11 at 5:45
    
Nice counterexample, thanks. Maybe a positive result can be obtained, if one requires $R_0$ to be indecomposabel. But the counterexample iis sufficient for me. Won't you post it as answer, I'll accept it. –  Ralph Dec 16 '11 at 9:35
    
In fact, the answer is yes, if $R_0$ is in addition indecomposable. For, since an indecomposable Artinian ring is local, $R_0$ has a unique max. ideal and hence $R$ has a unique homogeneous max. ideal. –  Ralph Dec 16 '11 at 9:51
    
A related problem that comes up in these situations is equidimensionality. One might naively think that in this situation, all maximal chains of homogeneous primes ending in the homog. maximal ideal have the same length (such a ring would be called 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

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