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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.

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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
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up vote 4 down vote accepted

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)

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You don't state it explicitly, but implicitly you are saying: "If an abelian group is f.g. and infinite, then it must have an element of infinite order." That's the crux of the argument, so better to split it out as its own statement. –  Thomas Andrews Aug 13 '12 at 16:13
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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
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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
    
Of course it is false, and big time, without the "abelian" thing, @QiaochuYuan...that's why it is written there. –  DonAntonio Aug 13 '12 at 16:25
    
@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
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