Infinite direct sum of abelian groups

While reading about graded rings, I read that a graded ring $R$ is an infinite direct sum of abelian groups $\displaystyle R=\bigoplus_{i \in \mathbb Z} A_i$ together with a bilinear map $A_i\oplus A_j\longrightarrow A_{i+j}$.

I know that in the finite case we have that the direct sum of two abelian groups $A_1\oplus A_2$ is an abelian group and the addition is defined as $(a_1,a_2)+(a_1',a_2')=(a_1+a_1',a_2+a_2')$ but what if we take an infinite direct sum $\bigoplus_{i \in \mathbb Z} A_i$ of abelian groups, does the infinite case give an abelian group and what an element in this infinite direct sum looks like and how do we add two elements together?

and my last question is that is it always possible to construct a bilinear multiplication $A_i\oplus A_j\longrightarrow A_{i+j}$ especially when the $A_i$ are not subgroups of the same group but rather they are completely disctinct abelian groups put together into an infinite direct sum. Thank you for your clarification !!

• Do you know the definition of what a (possibly infinite) direct sum is? If not, read it, and perhaps you'll be able to answer your own question! – Pedro Tamaroff Jun 4 '16 at 15:56
• yes I know the finite case and I gave it in my post but can't figure out what an infinite case should be – palio Jun 4 '16 at 15:58

It is simply the subgroup of all elements $(a_i)$ in the product $\displaystyle\prod_{i\in\mathbf Z}A_i$ such that all $a_i$ are $0$ but a finite number.
You may think of the construction of the ring of polynomials with coefficients in $R$; it is the $R^{(\mathbf N)}$ of sequences of coefficients almost all $0$.
• $R[X]=\displaystyle\bigoplus_{i\in\mathbf N}RX^i$. – Bernard Jun 4 '16 at 16:26
• No it's the set of monomials of degree $i$, plus $0$. – Bernard Jun 4 '16 at 16:43
• @palio: The Wikipedia information is only about finite direct sums: in the category of abelian groups, it is the same as a finite product. For infinite direct sums, one has to add the condition that all components but a finite number are $0$. – Bernard Jun 4 '16 at 18:18
• For your last comment: no, the sum of the $A_i$s is indeed the whole of $R[X]$, but it is not a direct sum, since $A_i\subset A_{i+1}$. – Bernard Jun 4 '16 at 18:22