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The question is: suppose that $\phi:\mathbb{Z}[X] \to \mathbb{Q}$ is a ring homomorphism sending $X \to q/p$ in its lowest term. Show that $n/m \in \mathbb{Q}$ is in image of $\phi$ if every prime divisor of $m$ divides $p$.

I couldn't really think about any effective approach in an entire hour: what I can see from the info given is that $p^k$ will be a multiple of $m$ for sufficiently large $k$, but that doesn't seem to lead anywhere.

Also the problem didn't specify how the homomorphism deals with constant polynomials. But clearly, if the map just sent everything to $0$ then above theorem is false, which probably implies that the homomorphism is an identity on integers (?)

I might try to prove the contrapositive of the result, but then it doesn't seem to simplify the problem (if not make it worse).

I seem to fail at grasping one critical observation...

Any hints appreciated

user160738
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  • Are $q,p$ fixed in advance in the definition of $\phi$, or do they depend on the element being operated on? – Mario Carneiro Feb 17 '15 at 00:45
  • The map sending everything to zero is not a ring homomorphism. To be a ring homomorphism, $\mathbb Z$ must by fixed by $\phi$, and the homomorphism is determined by where $X$ is sent. If $X$ is sent to $0$, then consider $q=0, p=1$. – Dustan Levenstein Feb 17 '15 at 00:48
  • @MarioCarneiro $p,q$ are fixed (given) in the definition, as far as I see. For instance one example would be $X\to 4/5$ where $q=4, p=5$ – user160738 Feb 17 '15 at 00:50
  • @DustanLevenstein Oh right, that clarifies the problem. But why is $0$ map not an homomorphism? I cannot see any of conditions failing... Also, if $X$ is sent to $0$, is the "lowest term" defined to be $q=0,p=1$? – user160738 Feb 17 '15 at 00:55
  • A homomorphism is required to carry the multiplicative identity to the multiplicative identity, so $\phi(1)$ must be $1$. Lowest terms means no common factors other than $\pm 1$. – Dustan Levenstein Feb 17 '15 at 00:58

3 Answers3

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**Hint 1: ** I am assuming that the homo-morphism is unitary, that is $\phi(1)=1$. This is often a condition which is understood in problems.

Prove by induction that $\phi(n)=n$ for all $n \in \mathbb Z$.

Hint 2: Use the above to show that if $P(X) =a_nX^n+...+a_1x+a_0$ then

$$\phi(P)=a_k \frac{p^k}{q^k}+...+a_1\frac{p}{q}+a_0= $\frac{a_kp^k+...+a_1q^{k-1}p+a_0q^k}{q^k}$$$

The claim follows from here:

If $\frac{m}{n}$ is in the image of $\phi$ then it is obtained after canceling from a fraction of the form $\frac{a_kp^k+...+a_1q^{k-1}p+a_0q^k}{q^k}$. Therefore $n$ is a divisor of $ q^k$, and the condition on primes is easy to deduce from here.

The converse is a bit trickier. Assume that a fraction $\frac{m}{n}$ has the given condition. Show that this implies that $n|q^k$ for some $k$. Chose such $k$ to be the smallest possible, and use the fact that $gcd(p,q)=1$ to show that there exists some $a_k$ such that $\frac{m}{n}-a_k\frac{p^k}{q^k}$ can be reduced to a fraction of the form $\frac{*}{q^{k-1}}$. Use induction by $k$.

N. S.
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  • Could you give a bit more hint as to why the last result help us to show the converse? I used Bezout's lemma to show that there are $a_k, b_k$ such that $a_k(p^k/q^k)+b_k=1/q^k$ but I'm not very sure what to do after that. I see that I haven't used the fact that $k$ is the smallest one... and it is easy to show that we can reduce $m/n - qa_k(p^k/q^k)$ into some fraction with denominator $q^{k-1}$... but thats rather trivial. – user160738 Feb 17 '15 at 01:40
  • @user160738 After bringing it to the same denominator $\frac{m}{n}-a_k\frac{p^k}{q^k}$ has the form $\frac{s-a_kp^k}{q^k}$, where $s,k, q$ and $p$ are fixed. Now, pick $a_k$ so that $$s \equiv a_kp^k \pmod{q}$$ This is possible as $p^k$ has an inverse mod $q$.... Also not that you can probably get a shorter solution by solving the equation $\pmod{q^k}$... – N. S. Feb 17 '15 at 01:45
  • Nice, but sorry if it sounds stupid, but what does the fact that $m/n-a_k (p^k/q^k)$ can be written in the form $*/(q^{k-1}$ imply? in particular, what do you mean by induction by k? – user160738 Feb 17 '15 at 02:04
  • @user160738 Now repeat... You can find $a_{k-1}$ so that this difference $-a_{k-1}(p^{k-1}/q^{k-1})$ is a fraction with the denominator $q^{k-2}$ and repeat again and again... – N. S. Feb 17 '15 at 04:24
  • Got it! Thanks a lot – user160738 Feb 18 '15 at 17:38
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Hint: By Bézout's identity, there exist $a$ and $b$ so that $ap^k+bq^k = 1$, since $p$ and $q$ are assumed to be in lowest terms. This can be rewritten as $a+b(q/p)^k=1/q^k$. Now interpret this identity in terms of $\phi$.

Dustan Levenstein
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Hint $\ (a,b)=\!\!\!\overset{\rm\quad\ Bezout}{1\,\Rightarrow\,} 1 = ma+nb\,\Rightarrow\, \frac{1}b = m\, \frac{a}b + n\,$ so $\ \frac{a}b \in R\,\Rightarrow\,\frac{1}b\in R.\,$ Since $\,b\,$ is a unit so too are all its (prime) divisors. $\ $ QED

Remark $ $ Generally every ring between a Bezout domain and its fraction field is a localization, i.e. it can be generated by adjoining inverses of elements. See here for much further discussion, including literature references.

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Bill Dubuque
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