Talk:Extension Theorem for Distributive Operations

The question (paraphrased) is: "Why is it necessary for $\left({T, *}\right)$ to be an abelian group?"

The stipulation was made in Warner that all elements of $R$ are cancellable for $*$. This is necessary for $\left({T, *}\right)$ to be a group (if it's not then not all elements are invertible). Warner then claims that this is necessary for Extension Theorem for Homomorphisms to be applied:

"By Theorem 20.2 (which is Inverse Completion of Commutative Semigroup is Abelian Group) $\left({T, *}\right)$ is a commutative group. Therefore by Theorem 20.4 (which is Extension Theorem for Homomorphisms, every homomorphism from $\left({R, *}\right)$ into $\left({T, *}\right)$ is the restriction to $R$ of one and only one endomorphism of $\left({T, *}\right)$ and so if $g$ and $h$ are two endomorphisms of $T$ coinciding on $R$ (that is, whose restrictions to $R$ are the same function), then $g = h$."

But I can't see why it is necessary for $\left({T, *}\right)$ to be a group (by default abelian) for this to apply. I'm missing something obvious.

Anyone able to help here? --prime mover (talk) 16:00, 3 May 2015 (UTC)

$(S,\circ)$, the Extension Theorem for Homomorphisms form for $(T,*)$ in this context, is supposed to be a commutative semigroup. Since it moreover is the inverse completion of $(R,*)$, it follows that it must be an abelian group. — Lord_Farin (talk) 16:28, 4 May 2015 (UTC)

It's not that it's a commutative semigroup that's important here, it's that it's a commutative semigroup with all cancellable elements. That was the nature of the first question raised by abcxyz (see the version 24th Jan 2013) where he said "Where is the hypothesis that all elements of $R$ are cancellable for $∗$ used?" The answer to that is that this is the crucial info that allows that $(T,*)$ is an abelian group.

The question is, why is it necessary for $(T,*)$ to be an abelian group in order for Extension Theorem for Homomorphisms to be used in this specific context? --prime mover (talk) 19:06, 4 May 2015 (UTC)

I've done a thorough inspection, but couldn't find where this assumption is used. If anywhere, it is in one of the theorems used, but I couldn't find it in there either -- particularly, the call to Extension Theorem for Homomorphisms seems fine. — Lord_Farin (talk) 09:20, 5 May 2015 (UTC)

Okay, I'll leave it where it is for now, and see whether I can find out what -- if anything -- may need to be done. I can see where this is going: it demonstrates the construction whereby the ring of integers is built up from the additive group of integers -- an abelian group constructed from the natural numbers (a commutative semigroup where all the elements are cancellable). So it's constructing a specific object with goal of establishing that this will in fact be the integers. It is probable that if you don't restrict the domain to all-elements-cancellable, then this construction still applies, but it is possible (I haven't followed it through) that distributivity cannot be rigorously established for the non-cancellable elements; or rather, that as the non-cancellable elements have not been included in the exposition, then it is inaccurate to claim that distributivity holds over the entirety of the target inverse-completion. Maybe. --prime mover (talk) 12:48, 5 May 2015 (UTC)