Re: Surjective morphism (Q_+, *) to (Q, +)
- From: "jsher" <josh.sher@xxxxxxxxx>
- Date: 10 Jul 2006 11:38:29 -0700
Off the top of my head I think the answer is no.
Sketch of proof:
Suppose there was a functional that did this. Call it F.
Then F(r*w)=F(r)+F(w) for every r and w in Q.
Prove that this implies that F=C*Ln (where C is a constant). (Hint this
is clearly true if F was continuous since you can extend the map from R
to R instead of just Q to Q since Q is dense in R)
And Since Ln doesn't map Q to Q we have a contradiction.
Now if there was a non-continous functional that did this. This would
be essentially a map from P:Z to Z such that F is the composition of
the maps below:
Q(+) <--- Z
^
|
Q(*)----> Z
Where the map from Q to Z and from Z to Q was some ordering of the
rational numbers. I think the goal is again to prove that this
functional would have to be the log function...
You might also consider looking at the representations of the two
groups in question (Q, +) and (Q,*).
eugene wrote:
Does there existsa surjectiv morhism between groups (Q_+, *) to (Q, +),
here Q_+ menas the set of positiive rationals ?
My supposion is that it exists, but it seems i can't construct it. Log
maps the first group onto (R, +), the problem would be solved if one
could construct surjective morphism from (R, +) onto (Q, +)... I tried
also idea with ord_a(x) for some natural a. This would map the (Q_+, *)
onto
(Z,+) and once again, how to proceed ... ?
Thanks
.
- References:
- Surjective morphism (Q_+, *) to (Q, +)
- From: eugene
- Surjective morphism (Q_+, *) to (Q, +)
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