Element in a ring mapping one prime to the next

[tomelwood]

Homework Statement

Let {p_n}n>0 be the ordered sequence of primes. Show that there exists a unique element f in the ring R such that f(p_n) = p_n+1 for every n>0 and determine the family I_f of left inverses of f.

Homework Equations

The ring R is defined to be: The ring of all maps f:Q+-->Q+ such that f(rs)=f(r)f(s) for every r,s in Q+. Define the operations + and x in such a way that (f+g)(r) = f(r)g(r) and (f x g)(r) = f(g(r)) for every r in Q+. This is a non commutative unitary ring with zero divisors.

The Attempt at a Solution

Any pointers would be appreciated, as I cannot see how it is possible to map one prime to the next, given any prime, let alone find the unique f in the ring R that will do it.
Sorry that I haven't provided more, but I have literally been banging my head against a wall for ages on this question, without the faintest idea of where to start!

 

[micromass]

First we define this map on the primes. So we define $$f(p_n)=p_{n+1}$$.

We wish to extend this map to the entire of Q. Take q in Q. Then the prime factorization yields that we can write q as

$$q=\prod_{I\subseteq \mathbb{N}}{p_i^{n_i}}$$

For a certain finite set I and $$n_i\in \mathbb{Z}$$

The only possibility to define f(q) is as

$$f(q)=\prod_{I\subseteq \mathbb{N}}{f(p_i)^{n_i}}=\prod_{I\subseteq \mathbb{N}}{p_{i+1}^{n_i}}$$

Try to show for yourself that f, with this definition, satisfies f(ab)=f(a)f(b).

 

[tomelwood]

Thanks so much, this has been most helpful!

So I got f(qr) = ...= PROD (p_i+1)^ni PROD (s_i+`)^ni = f(q)f(r) which is all I need to prove, right?

So now I need to determine the family of left inverses, which I believe is defined to be I_f = {g in R : gf = 1_r}
So the elements in R are these functions f,g etc. So I need to find the set of these g's which 'cancel out' the f's.
So if I do g(q)f(q) = PROD (g(p_i)^ni) PROD (p_i+1^ni) = 1_R
So I need to find this g. How do I do that.
I was tempted to say whilst writing this that g(q)f(q) = by definition of addition in this R = (g+f)(q) but I don't know if that would have helped at all? Thanks.

 

[micromass]

Yes, you've proven correctly that such an f exist. But you still need to determine for f what f(0) and f(1) is. I forgot to mention it.

For the other question, let g be a left inverse of f. g is uniquely determined by how it acts on the primes. So we simply need to know what $$g(p_i)$$ is.

Since g is a left inverse, we must have that $$g(p_{i+1})=p_i$$. This defines g on every prime number except 2. For 2, we may choose what g(2) is.

 

[tomelwood]

OK. Well surely f(0) is 0 as 0=PROD 0 so f(0) = PROD 0 = 0
and 1 = PROD 1 so f(1) = PROD 1 = 1 ? That seems a little too simple to be true though.

So the set of all these g's is I_f = {g in R : g(p_i+1) = p_i, g(2) = 1} ?

 

[micromass]

No, you don't need to map 2 to 1. g(2) can be anything. We can only say that I_f = {g in R : g(p_i+1) = p_i}

For every $$q\in \mathbb{Q}^+$$, we can find a g in I_f: the g that maps 2 to q.