# If $R$ is a ring s.t. $(R,+)$ is finitely generated and $P$ is a maximal ideal then $R/P$ is a finite field

Let $R$ be a commutative unitary ring and suppose that the abelian group $(R,+)$ is finitely generated. Let's also $P$ be a maximal ideal of $R$.

Then $R/P$ is a finite field.

Well, the fact that the quotient is a field is obvious. The problem is that I have to show it is a finite field. I do not know how to start: I think that we have to use some tools from the classification of modules over PID (the hypotesis about the additive group is quite strong).

I found similar questions here and here but I think my question is (much) easier, though I don't manage to prove it.

What do you think about? Have you got any suggestions? Thanks in advance.

-
You are correct that your case is easier than both of the linked questions, and that you should use the classification of finitely generated abelian groups. – Jack Schmidt Aug 13 '12 at 16:00

As abelian groups, both $\,R\,,\,P\,$ are f.g. and thus the abelian group $\,R/P\,$ is f.g....but this is also a field so if it had an element of additive infinite order then it'd contain an isomorphic copy of $\,\Bbb Z\,$ and thus also of $\,\Bbb Q\,$, which of course is impossible as the last one is not a f.g. abelian group. (of course, if an abelian group is f.g. then so is any subgroup)
The last statement is not obvious, by the way (and is false with "abelian" removed); it needs the fact that $\mathbb{Z}$ is Noetherian. – Qiaochu Yuan Aug 13 '12 at 16:13
Thanks for your answer. So $(\mathbb Q,+)$ is not finitely generated. I do trust you, but I'd like to prove this. Suppose there are $\frac{r_1}{s_1}, \ldots ,\frac{r_n}{s_n}$ generators. Then if we take a prime $p$ that doesn't divide any $s_i$ we cannot write $\frac{1}{p}$ as a linear combination (over $\mathbb Z$) of the generators and this is absurd. Am I right? Is this proof correct? Thanks a lot for your kind help. – Romeo Aug 13 '12 at 16:18
@Romeo, your proof is right, and you can try also to prove the following: any subgroup of the rationals generated by two (2) elements is cyclic, and then generalize inductively to any finite generating set. Since $\,\Bbb Q\,$ is not cyclic this wraps up the matter...and, btw, you can also deduce that $\,\Bbb Q\,$ is not a free (abelian, of course) group. – DonAntonio Aug 13 '12 at 16:28