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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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    $\begingroup$ I wouldn't call "allowing infinite sums" to what you define. It's not even a set if you don't bound $I$. $\endgroup$ Jan 13, 2012 at 9:24
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    $\begingroup$ 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 $\hat{R}=0$. $\endgroup$ Jan 13, 2012 at 13:21
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    $\begingroup$ 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. $\endgroup$ Jan 13, 2012 at 13:38
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    $\begingroup$ 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. $\endgroup$ Jan 13, 2012 at 14:23
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    $\begingroup$ 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. $\endgroup$
    – S. Carnahan
    Jan 14, 2012 at 6:46

1 Answer 1

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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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  • $\begingroup$ Thank you for your very nice answer! Maybe it will fit my purposes. $\endgroup$
    – user16974
    Jan 14, 2012 at 10:07
  • $\begingroup$ Does anyone have the paper "On limits in complete semirings" in pdf? I don't have access to Springer. $\endgroup$
    – user16974
    Jan 14, 2012 at 10:25