Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Update: I've copied this question over to mathoverflow.net:

http://mathoverflow.net/questions/100755/a-graded-ring-r-is-graded-local-iff-r-0-is-a-local-ring

to see if I get any answers there.


Let $R$ be a $\bf{Z}$-graded ring (with no commutativity assumptions whatsoever). Recall that $R$ is "graded-local" if the following equivalent conditions hold:

  1. $R$ has a unique homogeneous left ideal that is maximal among the proper homogeneous left ideals;

  2. $R$ has a unique homogeneous right ideal that is maximal among the proper homogeneous right ideals;

  3. $R\neq 0$ and the sum of two homogeneous non-units is again a non-unit.

Note, in particular, that (3) tells us that $R_0$, the ring of degree 0 elements, is a local ring.

But it seems to me that the converse is true. That is, it seems to me that a $\bf{Z}$-graded ring $R$ is "graded-local" if and only if $R_0$ is a local ring. Since this seems a bit suspicious, I have come here to find out if this is really the case.

Here is my argument: Let $J^g(R)$ denote the intersection of all the homogeneous left ideals that are maximal among the proper homogeneous left ideals. (This is the "graded Jacobson radical". Note that there should be at least one such "maximal homogeneous left ideal" because $R\neq 0$). As an intersection of proper homogeneous left ideals, $J^g(R$) is a proper homogeneous left ideal. I claim that it is maximal among the proper homogeneous left ideals, and hence is the unique such "maximal homogeneous left ideal".

Consider a homogeneous left ideal $I$ with $J^g(R) \subsetneq I$. Since $I$ is homogeneous there exists a homogeneous element $a\in I$ with $a \not\in J^g(R)$. Since $a$ is not in $J^g(R)$ there exists a maximal homogeneous left ideal $\frak{m}$ with $a \not \in \frak{m}$. Well, $R a + \frak{m}$ is a homogeneous left ideal, so by maximality of $\frak{m}$, $1=ra + m$ for some $r\in R$ and $m \in \frak{m}$. Taking the degree 0 components we have $1=r_0a + m_0$ where $m_0$ is a degree zero element in $\frak{m}$ and $r_0a$ is also homogeneous of degree 0. Thus, since $R_0$ is local, either $m_0$ or $r_0a$ is a unit. But $m_0$ cannot be a unit (since it would contradict the properness of $\frak{m}$). Hence $r_0a$ is a unit. In particular, $r_0a$ is left invertible, and thus $a$ is also left invertible. Thus, $I=R$. We conclude that $J^g(R)$ is a "maximal homogeneous left ideal" and hence is the unique such.

Can you see any problems with this? Is it really true that a $\bf{Z}$-graded ring $R$ is graded-local iff $R_0$ is local?

Update: It is clear that an $\bf{N}$-graded ring $R$ is "graded local" iff $R_0$ is local. (If ${\frak m}_0$ is the unique maximal left ideal of $R_0$ then ${\frak m} := {\frak m}_0 \oplus \bigoplus_{d>0} R_d$ is the unique maximal left ideal of $R$.) But the (easy) argument for $\bf{N}$-graded rings doesn't work for $\bf{Z}$-graded rings.

share|improve this question
add comment

1 Answer

I think there is an elementary proof, but let's check it.

I'm relying on:

  1. There always exist maximal proper homogeneous right ideals (by the usual Zorn's Lemma argument.)

  2. The sum of two homogeneous right ideals is again homogeneous.

Suppose $M$ and $N$ are distinct maximal homogeneous right ideals. Then $M+N=R$, and there exists $m+n=1$ with $m\in M$ and $n\in N$. Because of the grading, the grade zero parts must be such that $m_0+n_0=1$, and because $M$ and $N$ are both proper and homogeneous, neither $m_0$ nor $n_0$ can be units of $R_0$. This implies $R_0$ is not local.

By contrapositive then, we have shown if $R_0$ is local, then $R$ is graded local.

share|improve this answer
    
Dear rschwieb, can we think a polynomial ring $k[x_1,...,x_n]$ as a graded local ring with the graded(homogeneous) maximal ideal $m=(x_1,...,x_n)$? –  Arsenaler Oct 1 '12 at 19:51
    
@msnaber It seems to be the case! If there were a homogenous ideal properly containing that $m$, then it would have to contain an element with a nonzero constant term, which would in turn have to be an element of the homogeneous ideal. Thus, the ideal would be the entire ring. –  rschwieb Oct 1 '12 at 20:01
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.