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Let $L$ be a finitely generated free module over a commutative ring $A$. Let $e_1, \dots, e_n$ be a basis of $L$. Let $x_1,\dots,x_m$ be generators of $L$. Then $m \ge n$? If $m = n$, then $x_1,\dots,x_m$ is a basis of $L$?

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If $m<n$, then these generators give you a surjective endomorphism $L\to L$ which is necessarily injective, contradiction. – user26857 Nov 20 '12 at 16:10
Note that this would not be a contradiction if $A$ was not commutative. – Gregor Bruns Nov 20 '12 at 16:15
up vote 5 down vote accepted

If $x_1,\ldots,x_m$ generate $L$, then you get a surjective $A$-module map $A^m\rightarrow L$. Tensoring with $k(\mathfrak{m})=A/\mathfrak{m}$, $\mathfrak{m}$ a maximal ideal, gives you a surjection from an $m$-dimensional $k(\mathfrak{m})$-vector space to an $n$-dimensional $k(\mathfrak{m})$-vector space, so $m\geq n$.

If $n=m$, then you get a surjective endomorphism $L\rightarrow L$, and any surjective endomorphism of a finite $A$-module is injective. So in this case the elements form a basis.

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+1 This is a nice answer. However, the first assertion can be proved without using axiom of choice as YACP's comment shows. – Makoto Kato Nov 20 '12 at 16:21
@MakotoKato, I'm not so sure if the uniqueness of rank for commutative rings can be proved without choice. The proof I know uses it. So it may not be a drawback here. – Gregor Bruns Nov 20 '12 at 16:25
@GregorBruns Please see YACP's comment or my answer. – Makoto Kato Nov 20 '12 at 16:30

I would like to prove the first assertion without using axiom of choice. Suppose $m < n$. Then $\bigwedge^n L = 0$. This is a contradiction because $\bigwedge^n L$ is a free module of rank $1$.

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This is pretty nice. – user18119 Nov 20 '12 at 21:17
@MakotoKato +1 Nice answer, great idea to consider exterior powers!! – user38268 Dec 8 '12 at 5:52
Thanks. It can also be proved in essentially the same but a bit more elementary way using the module of alternating forms $Alt^n(L,A)$. – Makoto Kato Dec 8 '12 at 14:47

Your last question can be answered using a nice fact that I learnt from Atiyah - Macdonald. Suppose we have $x_1,\ldots,x_n$ that generate $L \cong A^n$. We now recall the following facts:

  1. Localisation commutes with finite direct sums

  2. If $M,N$ are $A$ - modules then $\phi : M \to N$ is injective iff for all maximal ideals $\mathfrak{m} \in A$ the induced map $\phi_\mathfrak{m} : M_{\mathfrak{m}} \to N_{\mathfrak{m}}$ on localisation is injective.

Using these it is enough to assume that $A$ is a local ring with maximal ideal $\mathfrak{m}$. Now define a map $\phi : A^n \to A^n$ by $\phi(x_i) = e_i$ where $e_i$ are the canonical basis vectors of $A^n$. Then $\phi$ is surjective and we have a ses $$0 \longrightarrow \ker \phi \longrightarrow A^n \stackrel{\phi}{\longrightarrow} A^n \longrightarrow 0$$

which upon tensoring with $A/\mathfrak{m} = k$ gives that $$0 \longrightarrow \ker \phi \otimes_A k \longrightarrow A^n \otimes_A k\stackrel{\phi \otimes 1}{\longrightarrow} A^n\otimes_A k \longrightarrow 0.$$

Rank - nullity implies that $\ker \phi \otimes_A k =0$. But now $\ker \phi \otimes_A k \cong \ker\phi / \mathfrak{m} \ker\phi$ which implies that $\ker \phi = \mathfrak{m}\ker \phi$. We know that $\ker \phi$ is finitely generated and $A$ is local by assumption. The hypotheses of Nakayama's Lemma are now satisfied and applying it shows that $\ker \phi = 0$ and hence $\phi$ is an isomorphism. Hence $x_1,\ldots,x_n$ are a basis for $A^n$.

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+1 This is a nice answer. – Makoto Kato Dec 8 '12 at 14:36

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