Question

Let $R$ be a semi-ring (resp. a ring). Let $\hat{R}$ be the structure obtained by "allowing infinite sums" in $R$ so $$\hat{R} := \lbrace\sum_{i \in I} a_i| a_i \in R\rbrace,$$ where $I$ is countable set. So my question is:

1. Is $\hat{R}$ a semiring (resp. ring)?
2. If we assume $R \neq 0$ does it follow that $\hat{R} \neq 0$?

Edit: After being closed, I rewrite the question as follows: Given a semiring $R$. Assume we would like to make sense of infinite sums in $R$, so is there a way to define a sort of "completion" to $R$ which allow to make sense of infinite sums? Thanks.

Answer 1

It was pointed out in the comments that if we mean by "$\sum_i a_i$" just the $I$-tuple $(a_i: i \in I)$, then the family of such tuples can of course be made into a semi ring (e.g., by pointwise addition and multiplication, or Cauchy multiplication if you use the index set $\mathbb N$). But in this case, neither the notation/name "sum" makes sense to me, nor does question 2 make sense: $R^I$ is never empty for nonempty $R$.

So I will reinterpret the question: Which semirings can be extended to complete semirings? By "complete semiring" I mean a semiring $(S,+,\cdot,0,1)$ (with commutative addition) in which all sums $\sum_{i\in I} a_i$ (for all index sets $I$, or at least for all countable $I$) are defined and satisfy reasonable laws, in particular:

• infinite associativity and commutativity
• left distributivity: $c\cdot \sum_i a_i = \sum_i c\cdot a_i$, and similarly right distributivity; in particular, the sum of 0s is 0.
• finite sums agree with addition.

If you also impose the following (natural, I think) condition, called "d-completeness" or "discrete completeness":

• Whenever $\sum_{i=0}^n a_i = a_0$ for all $n\ge 0$, then also $\sum_{n=0}^\infty a_i = a_0$

then it is easy to see that the following are equivalent.

1. The semiring $S$ is a subsemiring of a complete semiring.

2. $S$ can be partially ordered by $a \le b$ iff $\exists x\ a+x=b$.

3. For all $a,b,c$ in $S$, $a+b+c=a$ implies $a+b=a$. (Of course, as Darij Grinberg has pointed out, this is never true in rings.)

The point is that $a+(b+c)+(b+c)+\cdots = a+b + (c+b)+(c+b)+\cdots$. A few details can be found here.
Georg Karner's paper "On limits in complete semirings" (Semigroup Forum 45 (1992), no. 2, 148–165, MR1171841 (93h:16078)) has more information, and uses topological concepts rather than order-theoretic ones.

Answer 2

The comments above assume implicitly that we should add relations to the infinite sums, but this is not necessary at all. Let's restrict to $I = \mathbb{N}$. As an abelian group, we have just the direct product $R^{\mathbb{N}}$. The product is given by convolution: $\sum_n r_n \cdot \sum_n s_n = \sum_n \left(\sum_{p+q=n} r_p \cdot s_q\right)$. The unit is $1+0+0+\dotsc$. This makes $\hat{R}$ a ring. The map $R \to \hat{R}$, $r \mapsto r + 0 + 0 \dotsc$ is an injective ring homomorphism, in particular $R \neq 0 \Rightarrow \hat{R} \neq 0$.

If you identify $r_0 + r_1 + r_2 + \dotsc $ with $r_0 + r_1 X + r_2 X^2 + \dotsc$, you see that $\hat{R}$ equals the ring of formal power series $R[[X]]$. If $R$ is a semiring, the same construction works.

(A few years ago I attended a seminar on compact Lie groups where $\hat{R}$ was used instead of $R[[X]]$, I don't remember why.)