Proving Group Isomorphism: Q& & Z[X]

[TimNguyen]

Prove that there exists a group isomorphism between (Q&,*) and (Z[X],+) where Q& is the set of strictly positive rational numbers.

I was thinking of mapping a p_n, being the nth prime in Q& to x^(n-1). Would this work for this case?

 

[matt grime]

commonly you'd write \mathbb{Q}^+, though what is Z[x]? the polynomial ring in one variable with integer coefficients?

 

[TimNguyen]

Yes, Z[z] is a polynomial ring with integer coefficients. Also, I have no idea how to do the fancy graphic stuff so I wouldn't be able to type out "the set of rational numbers".

 

[matt grime]

read the sticky thread on latex. you can click on any 'fancy' picture to see what the generating text is.

 

[matt grime]

Your idea is on the right lines, though you need to specify more than just p_n goes to x^n: where does 1/p_n go? what about 4, where is that sent?

 

[TimNguyen]

Could I just send those to non-variable coefficients in Z[X]?

 

[matt grime]

No, that can't happen. It is a homomorphism. You must send the reciprocal of a fraction to the inverse in Z[x], and you must send the product of rationals to their sum in Z[x]. I'm just saying that you need to show that the map you defined by specifying where p_n goes is indeed a homomorphism.

 

[TimNguyen]

So, beside the mapping that I've already created, I need to make "additions" to the function in order to make it homomorphic?

 

[matt grime]

If you were doing this and were a proven algebraist what you initially said is fine, since, with experience it's 'obvious' how to fill in the gaps, but you're trying to prove that you know what you're doing, so you need to show that that map, extended to any rational number, not just a prime, is a homomorphism. You don't have to make additions, as such, just prove that it is a homomorphism (and thence an isomorphism).
I mean, I know where that map with the proper details filled in sends 4/9, but do you? Rememver you're trying to both understand what's going on and demonstrate that understanding.

 

[TimNguyen]

So basically, all I need to do is for an arbitrary a,b in Q, then find a way to compute it so it will lead to:

f(ab) = f(a) + f(b)?

 

[matt grime]

But, surely you knew 'how to compute it' already...

you know you have to send, say 4=2^2 to x^2+x^2=2x^2, and 8=2^3 to x^2+x^2+x^2=3x^2 in order for it to be a homomorhism, hey, wonder what that is generally... now check it is indeed a homomorphism (and bijective)

 

[TimNguyen]

Thank you once again, I think I have a brief idea.