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Let $S$ be a graded ring with decomposition $S = \bigoplus_{d \geq 1} S_d$, where the $S_d$ are additive abelian groups such that $S_d S_e \subseteq S_{d+e}$ for $e,d \geq 1$. An element in $S_d$ is called a homogeneous element of degree $d$. An ideal $\mathfrak{a}$ of $S$ is defined to be homogenous if $\mathfrak{a} = \bigoplus_{d\geq 1} (\mathfrak{a}\cap S_d)$.


I'm trying to prove that an ideal $\mathfrak{a}$ of $S$ is homogenous if and only if it can be generated by homogeneous elements.

If $\mathfrak{a}$ is a homogeneous ideal, then $\mathfrak{a} = \bigoplus_{d\geq 1} (\mathfrak{a}\cap S_d)$ and it is generated by $\bigcup_{d \geq 1} (\mathfrak{a} \cap S_d)$, which is a set of homogeneous elements.

I'm unsure about the other direction, though. Suppose $\mathfrak{a}$ is an ideal generated by a set $H$ of homogeneous elements in $S$, say $H = \bigcup_{d \geq 0} H \cap S_d$. Then $ \displaystyle \mathfrak{a} = \left\{ \sum_{h_i \in H} r_i h_i \big| \ \mathrm{finitely \ many} \ r_i \in S \ \mathrm{are \ non-zero} \right\}$. Given an element $a \in \mathfrak{a}$, we can thus write $a = \displaystyle \sum_{i=1}^{n} r_i h_i$ for $r_i \in S, \ h_i \in H$. But we can also write $r_i = \displaystyle \sum_{j=1}^{m_{i}} \alpha_{i,j}$ for each $i$, where $\alpha_{i,j} \in S_d$ for some $d$. So we have that $ \displaystyle a = \sum_{i=1}^n \sum_{j=1}^{m_i} \alpha_{i,j}h_i$. But since each $\alpha_{i,j}$ and each $h_i$ are homogeneous, and the $S_i$ are additive abelian groups, we have that $ \displaystyle a = \sum_{i=1}^k t_i$ where $t_i$ are homogeneous.

Why are the $t_i$ necessarily in $\mathfrak{a}$? I apologise if the notation I have used causes difficulty in following what I'm doing or in answering my question. In particular, I realise I have lost some information: we know precisely which of the $S_i$ each $t_j$ is in, and we also know that $k = p \ \mathrm{max}_i \ m_i $, where $p$ is the largest $i$ such that $r_j$ is contained in $S_i$ for some $j$.

Thanks

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Are you sure you don't want to look at $d \geq 0$? Where else is $1$ going to live? –  Dylan Moreland Feb 6 '12 at 1:08
    
Thanks, that was a typo. –  Matt Feb 6 '12 at 1:11
    
@DylanMoreland. Just to check I understand; because of the fact that $S_d S_e \subseteq S_{d+e}$, if $1$ is in $S_i$ then it's also in $S_{2i}$. But this contradicts the definition of a direct sum unless $i = 2i$, i.e. $i = 0$? –  Matt Feb 6 '12 at 1:20
    
Well, a priori it might not be homogeneous at all. But I think anything besides $1 \in S_0$ should cause problems. –  Dylan Moreland Feb 6 '12 at 1:27

1 Answer 1

up vote 4 down vote accepted

I think you're getting lost in your notation. You need to write the given element $a$ as a sum of elements that belong to $\mathfrak a\cap S_d$. But this is already the case, because each $h$ belongs to some $\mathfrak a\cap S_d$.

So here is how I would write this down:

Assume $\mathfrak a$ is generated by homogeneous elements, say $\mathfrak a = \langle \bigcup_{i\geq 1}H_i \rangle$ (*), where $H_i\subseteq S_i$. Then for any $a\in\mathfrak a$, there exist a collection $\{r_h\mid h\in H\}$ (only a finite number of them non-zero) such that

$$ a = \sum_{h\in H} r_h h = \sum_{i\geq 1} \sum_{h\in H_i} r_h h \in \bigoplus_{i\geq 1} (\mathfrak a\cap S_i). $$

I should explain why the "$\in$" holds: we may write each $r_h$ as $r_h = \sum_{j\geq 1} r_{h,j}$, where $r_{h,j}\in S_j$ and $h\in H_i$ so $r_{h,j}h\in S_{i+j}$. But on the other hand, $h\in \mathfrak a$, and thus $r_{h,j}h\in r_{h,j}\mathfrak a\subseteq a$, because $\mathfrak a$ is an ideal. Therefore $r_{h,j}h \in \mathfrak a \cap S_{i+j} $.

Since this holds for any $a\in\mathfrak a$, we have that $$ \bigoplus_{i\geq 1}(\mathfrak a\cap S_i) \subseteq \mathfrak a \subseteq \bigoplus_{i\geq 1}(\mathfrak a\cap S_i), $$ (the first inclusion is trivial, the seconds follows from the above). This means both are equal and $\mathfrak a$ is homogeneous.

(*) With $\langle X\rangle$ I mean: generated as an ideal. So these are all elements of the form $\sum_{x\in X} r_x x$, with $r_x = 0$ for almost all $x$.)

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@Matt: I should point out there was a shortcoming in a previous version of this (already accepted) answer. Let me know if you disagree on anything. –  Myself Feb 6 '12 at 1:47
    
Thanks a lot for your attention to detail; what you have written makes perfect sense to me. –  Matt Feb 6 '12 at 2:21

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