# On rings and semirings [closed]

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.

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## closed as not a real question by Emil Jeřábek, Martin Brandenburg, Mark Sapir, S. Carnahan♦Jan 14 '12 at 6:46

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

I wouldn't call "allowing infinite sums" to what you define. It's not even a set if you don't bound $I$. –  Fernando Muro Jan 13 '12 at 9:24
I am not sure if the thing is well-defined at all. For instance, let $a,b\in R$. Do you then identify $a+b$ (addition in $R$) with the formal infinite sum $\sum_{i=1}^{\infty }a_i$, where $a_1=a$, $a_2=b$ and $a_i=0$ for $i>2$? If you want this and other "obvious" relations to hold, then I think you always have $\hat{R}=0$. –  Robert Kucharczyk Jan 13 '12 at 13:21
If you want $\sum_{i=0}^{\infty} 1$ to be $1 + \sum_{i=1}^{\infty} 1$ (which is a rather reasonable thing for sums to satisfy - how else would you compute sums?), but also want $\sum_{i=0}^{\infty} 1$ to be $\sum_{i=1}^{\infty} 1$ (by a substitution of $i\mapsto i-1$, since it is a bijection), you get $0=1$. Welcome to Hilbert's hotel. –  darij grinberg Jan 13 '12 at 13:38
I don't see here any definition of multiplication that was approved by the proposer of the question. S. Carnahan mentioned a definition, but he seems quite doubtful about it, and it seems to have gotten no response from Ali Bleybel. So I would answer Question 1 in the negative. You can't have a ring or a semi-ring without a definition of multiplication. –  Andreas Blass Jan 13 '12 at 14:23
Okay, we've had a lot of interaction with the questioner, but there doesn't seem to be any clarification forthcoming. I'm closing until the question is edited to a meaningful state. –  S. Carnahan Jan 14 '12 at 6:46

## 2 Answers

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.

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Thank you for your very nice answer! Maybe it will fit my purposes. –  user16974 Jan 14 '12 at 10:07
Does anyone have the paper "On limits in complete semirings" in pdf? I don't have access to Springer. –  user16974 Jan 14 '12 at 10:25

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.)

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The unit is $1+0+0+...$, not $0+1+0+...$ (you have been working with $\lambda$-rings for too long...). –  darij grinberg Jan 13 '12 at 14:33
What you wrote here is perfectly OK, modulo Darij's comment, but it goes way beyond the original question, where $I$ was just a countable set. You've taken a specific countable set $\mathbb N$, and you've made essential use of its additive structure in defining multiplication. You could proceed similarly if $I$ is any (additive) semigroup in which each element can be expressed as a sum of two others in only finitely many ways. But I don't see anything similarly intelligent to do when $I$ is, as in the original question, merely a countable set. –  Andreas Blass Jan 13 '12 at 14:39
I intended to restrict to well ordered I, so what Martin does here is OK for me. –  user16974 Jan 13 '12 at 15:48
Is there a possible way to make multiplication more symmetrical? It seems there isn't. –  user16974 Jan 13 '12 at 16:19
@Ali: Martin's approach doesn't work for general well-ordered countable $I$; he really uses that it's $\mathbb N$. –  Andreas Blass Jan 13 '12 at 18:34