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I tried to search for a similar problem on the site, but none appear.

Definition 1 : Let $A$ be a commuting ring with identity. If $M = \sum_{i \in I } Ax_i$ then $x_i$ are said to be a set of generators of $M$. An $A$-Module is said to be finitely generated if it has a finite set of generators.

Definition 2 : A free $A$-module is one which is isomoprhic to an $A$-module of the form $\bigoplus_{i \in I } M_i$ where each $M_i \cong A$.

Statement : A finitely generated free $A$-module is therefore isomorphic to $A \oplus \cdots \oplus A$ for some $n > 0 $.

What I attempted: I tried constructing an explicit isomorphism. By Def 1. There exists a minimal finite set $\{x_1, \ldots, x_k \} \subseteq M$ such that it generates $M = \bigoplus_{i \in I } A_i$. Hence, consider $$ \phi: A^k \rightarrow M, \quad (a_1, a_2, \ldots a_k ) \mapsto a_1x_1 + \cdots a_kx_k $$ $\phi$ is a homomorphism, and is surjective by definition 1. Suppose exists non zero $(b_1, \cdots, b_k) \in A^k$ such that $b_1 x_1 + \cdots b_k x_k = 0_M$...

but unlike vector space we cannot replace $x_1$ to contradict minimality...

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    @Hermès, that's where I couldn't prove, as the usual method in vector spaces in reducing a spanning set to a basis require the inverse property of the scalar field :( – 2017-02-01
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    Sorry I got confused :|, see answer. – 2017-02-01

1 Answers 1

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Suppose $M=\bigoplus _{j\in I} A$ is finitely generated by $\{x_i\mid i=1\ldots n\}\subseteq M$. We always keep in mind that the definition of the direct sum means that each element is uniquely expressed as a finite sum of elements from different coordinates.

Consider one of the $x_i$: when expressed as a tuple in $M$, it is nonzero only on finitely many indices in $I$, as are all its $A$-multiples.

It is possible, then, to take the union each of these finite sets over all the generators, and still have a finite set $F$. Clearly if you take a linear combination of the generators, the indices on which the combination is nonzero is contained in $F$. But that means the span of the generators (all of $M$) is contained there, so that $M\subseteq \bigoplus _{j\in F}A$.

This shows why it is not possible for $I$ to be infinite.

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    I'm very new to module theory, and am trying to understand the same thing. Your answer is almost precisely what I looked for; But, would you specify more explicitly what you mean by: "the definition of the direct sum means that each element is uniquely expressed as a finite sum of elements from different coordinates" ? – 2018-11-18
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    ps. And is this enough then? Because then the unique representation of $0\in M$ is just $0=0(x_1,x_2,\dots,x_n)=\phi(0)$, so the map $\phi$ defined by the OP would be injective and thus an isomorphism? So, I guess what I mean is, what and how (a bit more explicitly) does uniqueness of the sum-representation come from? – 2018-11-18
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    @Christopher.L I would have to know what your definition of direct sum is to begin. – 2018-11-18
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    @Christopher.L Knowing that $0$ has a unique representation isn't enough. You need to know that *everything else* has a representation that shares a common finite support within the (potentially infinite) direct sum. – 2018-11-18
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    Excuse my confusion; I'm defining the direct sum $\bigoplus_{i\in I} A$ as a set of sequences $(a_i)_{i\in I}$, with pointwise add/mul. I guess I was also just trying to see the explicit isomorphism $\phi: A^k\to M$, like the OP was trying to do. I guess the $\phi$ def. by the OP is the natural one, but how can I see that it is injective? We cant directly say that the given gen. set $\{x_i\}\subset M$ is independent. I managed to see how 'M free' in def2 -> exist independent gen. set X; But, how do I go from there to $|X|=n$, and $M\cong A^k$? Do I have to show two gen. set are of equal size? – 2018-11-18
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    I guess I understand what you mean by "the definition of the direct sum means that each element is uniquely expressed as a finite sum of elements from different coordinates" though. Just that we can write each element like $\sum_i a_ie_i$, where $e_i=(\delta_{ij})_{j\in I}$. I guess I was trying to see the actual isomorphism and got stuck there along the way. I'm sorry if this is turning out to a separate question, you can refer me to a question that might clear up my confusion if you like, or just leave it I guess. – 2018-11-18
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    @Christopher.L can you see how, in an infinite sum, you would necessarily be able to give an element which is nonzero on any particular position? They cannot all fit into a finite subset of positions. The point of my solution is that the positions on which the generators are nonzero is a finite subset – 2018-11-19
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    Yes, well after some thought I got your argument in the answer, I guess I was just a bit confused, and was looking for more than was actually there. So, I see how we can say that $I$ must be finite, as in there must be an isom. to $A^k$ for some $k$, but not if and why we must have $k=n$, as in $n$ being the number of generators in the given generating set for $M$. Which is what seems to be the case seeing this in other places, and what the OP was trying to construct. – 2018-11-19
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    @Christopher.L That is definitely false when the generating set isn't minimal, and I think it is probably still false for general rings, even when the generating set is minimal. I think it might be true for commutative rings though. IMO, that is above and beyond what is asked... I would consider it unnecessary effort. Thanks for taking the time to explain so clearly what you were looking for. I would like to talk with askers as engaged as you are any day of the week. – 2018-11-19
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    Ok, yes, I'm aware that was beyond what was asked for. I guess this was just what I had in mind when I found the question. And yes, I am self-studying a Commutative Algebra course, so I am only dealing with commutative rings; Sometimes I forget that's not necessarily implied =). Thank you too, (mutatis mutandis) right back at you! – 2018-11-20