Let $R$ be a ring $M$ an $R$-module. How can I prove that if

  • $M\cong R^n/N$ for some $n\!\in\!\mathbb{N}$ and some submodule $N\leq R^n$

and if

  • $M\cong R^{(I)}/\langle u_1,\ldots,u_m\rangle$ for some set $I$ and some vectors $u_1,\ldots,u_m\in R^{(I)}$,

then

  • $M\cong R^k/\langle v_1,\ldots,v_l\rangle$ for some $k\!\in\!\mathbb{N}$ and some vectors $v_1,\ldots,v_l\in R^k$ ?
  • Is it even possible for $I$ to be infinite in this case? – tomasz Jul 26 '13 at 13:16
  • @tomasz: A module can be finitely generated and infinitely related, e.g. take $R=K[x_1,x_2,x_3,\ldots]$ and the ideal/submodule $\mathfrak{a}=\langle\langle x_1^1,x_2^2,x_3^3,\ldots\rangle\rangle$ and $M=R/\mathfrak{a}$, which is an finitely generated (by $1$) infinitely related (by $x_i^i$ with $i\in\mathbb{N}$) $R$-module. – Leon Jul 26 '13 at 15:22
  • But I'm asking if it's possible for a finitely generated module to be obtained as a quotient of a free module of infinite rank by a finitely generated submodule. It seems intuitively impossible, as the quotient will still have as a submodule (even a direct summand!) a free module of infinite rank. – tomasz Jul 26 '13 at 18:55
  • @tomasz: Well, by YACP's answer, it is not possible. Your situation implies (by Schanuel's lemma) that $R^{(I)}$ is finitely generated, so $|I|<\infty$. – Leon Jul 26 '13 at 20:37
up vote 4 down vote accepted

Using Schanuel's Lemma we find that $N\oplus R^{(I)}\simeq N'\oplus R^n$, where $N'=\langle u_1,\ldots,u_m\rangle$, so we get that $N\oplus R^{(I)}$ is finitely generated which shows that $N$ is finitely generated.

  • How does "$N\oplus R^{(I)}$ finitely generated" imply "$N$ finitely generated"? – Leon Jul 18 '13 at 3:16
  • 2
    It's trivial that $M_1\oplus M_2/M_2\simeq M_1$ and any factor of a finitely generated module is also finitely generated. – user26857 Jul 18 '13 at 7:16
  • VERY elegant, just what I was looking for, thank you! Related question: does "finitely generated & finitely related $\Rightarrow$ finitely presented" also hold for groups, rings, algebras? One would think so, but I'm worried, since we don't have Shanuel's lemma there. Any alternative (more universal) proof? – Leon Jul 18 '13 at 13:59
  • Any ideas about group presentations $\langle\ldots|\ldots\rangle$ and $R$-algebra presentations $R\langle\ldots|\ldots\rangle$? – Leon Jul 26 '13 at 15:31
  • Dear @LeonLampret, maybe you want to post it as a separate question in order to find better answers. – user26857 Jul 26 '13 at 15:59

Every generating system $E$ of a finitely generated module $M$ contains a finite generating system (namely, look at those generators of $E$ which are needed to generate a finite generating system of $M$). Now assume that there are only finitely many relations between the generators of $E$. But these only use finitely many generators of $E$. It follows that every presentation of $M$ can be adjusted to a finite presentation.

  • Could you please be more specific/explicit. I don't understand the last two sentences. Ok, we have elements $x_1,\ldots,x_n$ that generate $R^{(I)}/\langle u_1,\ldots,x_m\rangle$. Now what? – Leon Jul 17 '13 at 7:27
  • Is it clearer now? – Martin Brandenburg Jul 17 '13 at 7:46
  • 1
    No, it's even more confusing. I agree with the first sentence. Don't understand the second sentence: the number of relations is always infinite (when $R$ is inifnite), since it constitutes a submodule. It would really help if you were more explicit, with actual elements and concrete isomorphisms and submodules given by generators. You can assume known that any module has a presentation, which follows from the universal property of the free module $R^{(I)}$. – Leon Jul 17 '13 at 15:13
  • Of course I mean that finitely many relations suffice (to generate all other relations), as in your question! – Martin Brandenburg Aug 5 '13 at 16:36

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