Construction of the monoid of fractions

This question comes from my attempt to solve Exercise 17(b) of Bourbaki, Algebra, Chapter 1, §2.

Let $E$ be a commutative monoid (written multiplicatively) and $S$ a submonoid of $E$. Define on $E\times S$ an equivalence relation by $(a,s)\sim(b,t):\Leftrightarrow$ "$\exists u\in S$ such that $atu=bsu$". Denote the set $(E\times S)/\sim$ by $\overline{E}$ and the class of $(a,s)$ by $a/s$. For any $a\in E$, let $\epsilon(a)=a/e$.

So far so good. The Bourbaki exercise is actually more general (replacing commutativity by weaker assumptions), but, even in this special case, I can't do the next step:

Show that there exists on $\overline{E}$ one and only one monoid structure such that $\epsilon$ is a monoid homomorphism and such that, for all $s\in S$, $\epsilon(s)$ is invertible.

My problem is with the "only one" part. Given a monoid structure $\otimes$ on $\overline{E}$ with those properties, I see no reason why $(a/s)\otimes(b/t)=(ab)/(st)$ or even $(s/e)\otimes(e/s)=e/e$. I couldn't find a counterexample in the case $E=\mathbb{N}$, nor did the search give me any intuition on why the assertion should be true.

This commutative case is treated in the text of Bourbaki's Algebra, but there no mention is made of "only one".

I'm glad for anything that gets me started.

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Closely related: Eckmann-Hilton. – t.b. Jul 12 '11 at 17:51
I'm too lazy to write an answer for that, as it is mostly an exercise in overcoming the extreme tedium of the argument. You may want to have a look at the proof of Theorem (10.6) on pages 300ff in Lam's Lectures on Modules and Rings for a very good summary of the ideas involved. I wouldn't assume commutativity of $E$, as it probably obscures more than it simplifies. – t.b. Jul 13 '11 at 9:16
@Theo: By no means I want to urge you into writing an answer if you don't feel like it. I'm just not sure if you've understood my question. I'm quite familiar with the construction itself (in the commutative case). It's just that "only one" statement I find weird and can't prove or disprove. – Stefan Jul 13 '11 at 10:12
Hint: inverses are unique whenever they exist. – Pete L. Clark Jul 15 '11 at 3:13
Please forgive me for possibly being somewhat obtuse but I'm having trouble understanding how these hints lead to the solution. Could someone elaborate a bit more? – LostInMath Jul 15 '11 at 21:53

I think that the "only one" part does not hold. Usually there is plenty of monoid structures on the set $\bar{E}$ such that $\epsilon$ is a monoid homomorphism and $\epsilon(s),s\in S$ are invertible.

For example, let $E=S=\mathbb{N}$ be the natural numbers with its usual addition. Then $\bar{E}=\mathbb{Z}$ is essentially the integers with its usual addition which is denoted by $+$. Let $\phi$ be any non-identity permutation of the integers which fixes the nonnegative integers. Then $m\oplus n=\phi^{-1}(\phi(m)+\phi(n))$ defines a new monoid structure $\oplus$ on the (set of) integers, which has the two required properties.

First let us check that $\oplus$ is an associative and commutative operation which has $0$ as its neutral element; if $m,n,p \in \mathbb{Z}$, then

$$\begin{eqnarray*} (m\oplus n)\oplus p &=& \phi^{-1}(\phi(m)+\phi(n))\oplus p \\ &=& \phi^{-1}(\phi\phi^{-1}(\phi(m)+\phi(n))+\phi(p)) \\ &=& \phi^{-1}((\phi(m)+\phi(n))+\phi(p)) \\ &=& \phi^{-1}(\phi(m)+(\phi(n)+\phi(p)) \\ &=& \phi^{-1}(\phi(m)+\phi\phi^{-1}(\phi(n)+\phi(p))) \\ &=& m \oplus \phi^{-1}(\phi(n)+\phi(p)) \\ &=& m\oplus (n\oplus p), \end{eqnarray*}$$ $$m\oplus n=\phi^{-1}(\phi(m)+\phi(n))=\phi^{-1}(\phi(n)+\phi(m))=n\oplus m$$ and $$m\oplus 0 = \phi^{-1}(\phi(m)+\phi(0)) = \phi^{-1}(\phi(m)+0)=\phi^{-1}(\phi(m))=m.$$

Then let us check that $\epsilon:(\mathbb{N},+)\to(\mathbb{Z},\oplus),\epsilon(n)=n$ is a monoid homomorphism; if $m,n\in \mathbb{N}$, then $$\epsilon(m) \oplus \epsilon(n) = m \oplus n = \phi^{-1}(\phi(m)+\phi(n)) = \phi^{-1}(m+n)=m+n=\epsilon(m+n);\ \epsilon(0)=0.$$

Each of the elements $\epsilon(n),n\in\mathbb{N}$ is invertible since for each $n\in\mathbb{N}$ $$n\oplus \phi^{-1}(-\phi(n)) = \phi^{-1}(\phi(n)+\phi\phi^{-1}(-\phi(n)))=\phi^{-1}(\phi(n)+(-\phi(n))=\phi^{-1}(0)=0.$$

Lastly note that $\oplus$ is different from $+$ because there exists such a positive integer $m$ that $\phi(-m)\neq -m$, which implies that $m+\phi(-m)\neq 0$ and $$m\oplus(-m)=\phi^{-1}(\phi(m)+\phi(-m))=\phi^{-1}(m+\phi(-m))\neq 0\quad (=m+(-m)).$$

Thank you to Theo Buehler for suggesting a correction to my initial flawed attempt to answer the question.

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Do you mean $m \oplus n = \phi^{-1}(\phi(m) + \phi(n))$? Otherwise $\oplus$ is not associative. – t.b. Jul 14 '11 at 23:32
@Theo: No, actually I meant the nonsense I wrote. You are correct. – LostInMath Jul 14 '11 at 23:46
This looks fine now. What I'd find more interesting than a counterexample would be an additional hypothesis that would ensure that the statement holds. I think something like requiring $\ominus (s,0) = (0,s)$ for the $\oplus$-inverse of $(s,0)$ for each $s \in S$ should be sufficient, but I didn't check carefully. At least it excludes counterexamples of the kind you suggest. – t.b. Jul 17 '11 at 11:36
Yes, it would be interesting to know whether requiring the inverses of the invertible elements to be the same forces the operations to be the same. – LostInMath Jul 17 '11 at 12:39
Thanks a lot. I had already looked at your operation with $\phi$ replaced by the permutation exchanging $-1$ and $-2$, but I stopped thinking in that direction beacuse I was sure to have found a counterexample to associativity. Don't know what went on in my head. – Stefan Jul 17 '11 at 14:09