# Confusion on the direct sum of abelian groups and the definition of graded ring.

A graded ring is a ring $$S$$ together with a family $$(S_{d})_{d\geq 0}$$ of subgroups of the additive group of $$S$$, such that $$S=\bigoplus_{d\geq 0}S_{d}$$ and $$S_{e}S_{d}\subseteq S_{e+d}$$ for all $$e,d\geq 0$$.

I am having (and seems like always having) trouble to understand the part $$S=\bigoplus_{d\geq 0}S_{d}$$ as abelian groups.

Does this mean the following?

Every $$s\in S$$ can be written as $$s=(x_{0},x_{1},x_{2},\cdots)$$ where $$x_{i}=x_{i}(s)$$ is uniquely determinted by $$s$$, $$x_{d}\in S_{d}$$ for each $$d$$, and only finitely number of $$x_{d}$$ are nonzero.

If this is correct, why would many resources I find consider $$s=\sum_{d\geq 0}x_{d}$$, where $$x_{d}=x_{d}(s)$$ is uniquely determined by $$s$$, $$x_{d}\in S_{d}$$ and only finitely finitely number of $$x_{d}$$ are nonzero?

Are these two things the (essentially) same? If so, then I have trouble making them compatible with the ring multiplication.

For example, if we let $$s,r\in S$$, then we can write $$s=(x_{0},x_{1},x_{2},\cdots)$$ and $$r=(y_{0},y_{1},y_{2},\cdots)$$, so that $$sr=(x_{0}y_{0},x_{1}y_{1},x_{2}y_{2},\cdots),$$ so if the above two notions are equivalently, $$sr$$ should correspond to a sum $$\sum_{d\geq 0}x_{d}y_{d}.$$ However, if we write $$s=\sum_{d\geq 0}x_{d}$$ and $$r=\sum_{d\geq 0}y_{d}$$, then $$sr\neq \sum_{d\geq 0}x_{d}y_{d}.$$ Right?

This confuses me a lot. For example, in this post, Definition of graded rings, the argument is writing $$1=\sum_{d\geq 0}x_{d}$$ and then consider $$x_{0}=x_{0}1=\sum_{d\geq 0}x_{d},$$ which will give $$x_{0}=x_{0}x_{d}$$ for all $$d\geq 0$$.

However, if we do this analogously to vector notions, then we write $$1=(x_{0},x_{1},\cdots, )$$ so $$(x_{0},0,0,\cdots)=x_{0}=1x_{0}=(x_{0}x_{0}, x_{1}x_{0},\cdots),$$ which only gives us $$x_{0}=x_{0}x_{0}, x_{0}x_{d}=0\ \text{for all}\ d>0.$$

So in my view I do not think these notions coincide. Am I missing something here?

In "An introduction to the theory of groups", Rotman gives Lemma 10.4 in page 310:

Let $$\{A_{k},k\in K\}$$ be a family of subgroups of a group $$G$$, where $$K$$ is an index set. Then, the following are equivalent:

1. $$G\cong\bigoplus_{k\in K}A_{k}$$;

2. Every $$g\in G$$ has a unique expression of the form $$g=\sum_{k\in K}a_{k}$$ where $$a_{k}\in A_{k}$$, $$k$$ are distinct and $$a_{k}\neq 0$$ for only finitely many $$k$$.

3. $$G=\langle \bigcup_{k\in K}A_{k}\rangle$$ and for each $$j\in K$$, $$A_{j}\cap\langle\bigcup_{k\neq j}A_{k}\rangle=0$$.

So can I just consider $$\bigoplus_{d\geq 0}S_{d}=\langle \bigcup_{d\geq 0}S_{d}\rangle$$ such that $$S_{d}\cap \langle \bigcup_{e\neq d}S_{d}\rangle=0$$? Or For instance, the ideal $$S_{+}:=\bigoplus_{d>0}S_{d}$$ is just $$\langle\bigcup_{d>0}S_{d}\rangle$$ with the similar trivial intersection?

Yes, they are essentially the same. You may be seeing the distinction between an "external direct sum" and "internal direct sum", with you using the external sum notation, and the authors you are seeing utilizing the internal notation. Identify an element $$x_d\in S_d$$ with the image of $$S_d$$ in the (external) direct sum (i.e., the tuple whose $$d$$th entry is $$x_d$$, and every other entry is $$0$$).

The reason you are having trouble with the multiplication is that the direct sum is not a direct sum of rings, it is a direct sum of groups. I don't know if you've ever seen the construction of a polynomial ring that is done with formal direct sums, but it is the same idea here. You don't multiply the tuples entry-by-entry. Instead, you are supposed to have a multiplication defined on the entire direct sum which satisfies that $$S_rS_k\subseteq S_{r+k}$$.

For instance, let me expand on the construction of a polynomial ring alluded to above. What "is" the ring $$\mathbb{Z}[x]$$?

Well, I am going to define $$\mathbb{Z}[x]$$ as follows: its underlying set is the abelian group direct sum $$\bigoplus_{k=0}^{\infty} \mathbb{Z}=\mathbb{Z}\oplus\mathbb{Z}\oplus\cdots.$$ Sum is done as usual: entry by entry.

Now I'm going to define a multiplication; it uses the multiplication I know from $$\mathbb{Z}$$, but it is not the multiplication entry by entry:

Given two elements $$(a_0,a_1,a_2,\ldots)$$ and $$(b_0,b_1,b_2,\ldots)$$ of the direct sum, I will define their product as $$(a_0,a_1,a_2,\ldots)\times(b_0,b_1,b_2,\ldots) = (c_0,c_1,c_2,\ldots)$$ where for each nonnegative integer $$k$$, $$c_k = a_0b_k + a_1b_{k-1}+\cdots+a_kb_0 = \sum_{j=0}^k a_jb_{k-j},$$ where the products and sum on the right hand side are the usual integer products.

We then define "$$x$$" to be the tuple $$(0,1,0,0,\ldots)$$, and for each $$m\in\mathbb{Z}$$, wee identify $$m$$ with the tuple $$(m,0,0,\ldots)$$. Then we prove that every element of $$\mathbb{Z}[x]$$ can be written uniquely as $$a_0+a_1x+\cdots+a_nx^n\quad \text{or as}\quad 0$$ where $$a_n\neq0$$, and the sums and products on the left hand side are the ones we've defined, not the ones from $$\mathbb{Z}$$.

Note this makes $$\mathbb{Z}[x]$$ into a ring (in fact a graded ring). But even though the elements are "really" tuples, we do not define the product entry-by-entry.

Rotman's definition is again trying to get at the fact that it is equivalent to describe an external direct sum as it is to describe an internal direct sum as subgroups of a group satisfying certain properties.

• Thank you so much for the detailed answer and I am really sorry for the late reply. Let me summarize your points and see if I understand them. So these two views (writing as a vector and writing as a sum) are essentially the same, the key point here is the definition of the multiplication. If we view an element $s\in S$ as a vector $s=(x_{0},x_{1},\cdots)$, then the multiplication of two elements $s,e\in S$ is what you defined above $se=(u_{0},u_{1},\cdots)$ where $u_{k}=\sum_{i=0}^{k}x_{i}y_{k-i},$ where $x_{i}y_{k-i}$ is the ring multiplication. Jan 22 at 4:50
• However, if we view $s\in S$ as $s=\sum_{d\geq 0}x_{d}$, then for two elements $s,e\in S$, the product of them is just the ring multiplication $se=(\sum_{d\geq 0}x_{d})(\sum_{d\geq 0}y_{d})$, and since both sums are finite, I believe that after some kind of rearrangement, $se$ can finally be written as $\sum_{k\geq 0}\sum_{i=0}^{k}x_{i}y_{k-i}$. And thus, if we want to go from here back to the vector view, we can do this, because then $se=(u_{0},u_{1},\cdots,)$ if we identity each summand as a coordinate in the vector. Jan 22 at 4:52
• I am understanding it correctly? So in the future, I will just consider an element $x\in S=\bigoplus_{d\geq 0}S_{d}$ as $x=\sum_{d\geq 0}x_{d}$ with $x_{d}\in S_{d}$, finitely many of them nonzero, unique. And if I want to prove that some element $y$ belongs to $S$, I will just need to prove that $y$ can be written as a finite sum of elements from the $S_{d}$, right? In this way, it is more like $S$ is a "sum" of elements from itself, much better than considering vectors every time. Jan 22 at 5:02
• @JacobsonRadical I think you are understanding correctly. Jan 23 at 4:26
• Great. Thank you so much for this great answer! Jan 23 at 18:15

I believe your confusion comes from the definition of the multiplication map for the direct product of rings. Both definitions of direct sum are right and coincide module isomorphism, but the multiplication is very different in the case of a graded ring: If $$S=\bigoplus_{g\geq 0} S_d$$ is a graded ring, and if you consider its elements as sequences that is, you see the direct sum as an external direct sum), then if $$r=(x_0,x_1,\dots)$$ and $$r=(y_0,y_1,\dots)$$ are elements in $$S$$ then its multiplication is given by "polynomial multiplication", that is $$rs = (u_0,u_1,\dots)$$ where $$u_i = \sum_{j=0}^{i} x_j y_{i-j}.$$ As both sequences are eventually zero, this multiplication is well defined. With this definition of the multiplication in the case of external direct sum, it coincides with multiplication in the case of internal direct sum.

• @ArturoMagidin Oops! Sorry for that, I know the difference, I just had a "lapsus brutus" (or "lapsus linguis" if you prefer). Jan 21 at 20:53