# Finitely generated free $A$-module is isomorphic to $A^n$

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...

• @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 :( – Bryan Shih Feb 1 '17 at 11:03
• Sorry I got confused :|, see answer. – Hermès Feb 1 '17 at 12:20

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.

• 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" ? – Christopher.L Nov 18 '18 at 14:30
• 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? – Christopher.L Nov 18 '18 at 14:59
• @Christopher.L I would have to know what your definition of direct sum is to begin. – rschwieb Nov 18 '18 at 16:14
• @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. – rschwieb Nov 18 '18 at 19:21
• 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? – Christopher.L Nov 18 '18 at 23:52