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Here is a problem I face practicing the theory of rings:

Define $\phi : \mathbb{Z}[t] \to \mathbb{Q}$, a ring homomorphism (it does map $1$ to $1$). I'm trying to show that if $\phi(t)=\frac{u}{v}$ (in lower terms) then we have $\frac{m}{n}\in\operatorname{Im}(\phi)$ (in lower terms again) if and only if every prime factor of $n$ divides $v$ (in lower terms).

Then I'd like to deduce from this that there does not exist any ideal $I$ of $\mathbb{Z}[t]$ such that $\mathbb{Z}[t]/I$ is isomorphic to $\mathbb{Q}$.

I have:

$$\phi(\sum_{i=0}^{N} a_i t^i)=\sum_{i=0}^{N} a_i\phi(t)^i.$$

If $\frac{m}{n} \in\operatorname{Im}(\phi)$ then there are $N \in \mathbb{N}$ and $a_i\in\mathbb{Z}$ ($i=0,1,...,N$) such that $mv^N=n\sum_{i=0}^{N} a_iu^iv^{N-i}$. Hence one of the implications is quite straightforward as if $p$ prime divides $n$ then it divides the left-hand-side and as it does not divide $m$ it divides $v^N$ and so $v$. But for some reason I can't get to the other way around. I've tried to 'find' the coefficients $a_i$ and show they were integers for some $N$ but I couldn't.

Also, I'm not sure at all how to make progress on the deduction...

Could you hint me please? I'd be very grateful, this is getting a little frustrating...

Thank you!

user26857
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Noome
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1 Answers1

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Hint: If your goal is to show that $\Bbb Z[t]/I$ cannot be isomorphic to $\Bbb Q$, then we need not determine the total image of $\phi$: note that if there exists an ideal $I$ such that $\Bbb Z[t]/I\cong\Bbb Q$, then $I$ is the kernel of a surjective ring homomorphism $\phi : \Bbb Z[t]\to\Bbb Q$. So, you want to show this homomorphism isn't surjective. Suppose $\phi(t) = a/b$ (in lowest terms). Let $p$ be a prime not dividing $b$. Is it possible to wind up with $p$ in the denominator of $\phi(f)$ (after reducing to lowest terms) for some $f\in\Bbb Z[t]$?

Edit: As for the other direction: it is easy to see that any element of the form you propose is in the image if you can show that $1/v$ is in the image. I claim that you can find a linear polynomial $n + mt$ such that $\phi(n + mt) = 1/v$. The equation you want to solve is $n + m(u/v) = 1/v$, which is equivalent to $nv + mu = 1$. This is solvable for $n$ and $m$ integers by Bezout's identity, as $u$ and $v$ are relatively prime by assumption.

Stahl
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  • Yes, that's just the contrapositive of what I have shown, right? Meaning I don't need the other way around ! Is it possible to show it though? Thank you very much, I see the deduction now! – Noome Feb 08 '15 at 20:35
  • I'll try to add something about the other direction. (I originally just wanted to emphasize that it wasn't needed to obtain your deduction!) – Stahl Feb 08 '15 at 20:43
  • Yes, it was very good that you did, it avoids confusion! Thank you – Noome Feb 08 '15 at 20:51
  • Wow, would never have thought of that! Thank you very much for the insight! I had tried Bezout's Identity, the form of the statement was calling for it, but this is really neat, thanks – Noome Feb 08 '15 at 21:26
  • Any element of that form is simply an integer multiple of $1/v$. However $\phi\mid_\Bbb{Z} : \Bbb{Z}\to\Bbb{Z}$ is the identity, so just multiply the polynomial that gets sent to $1/v$ by the appropriate integer. – Stahl Feb 09 '15 at 16:53