# Generators of a finitely generated free module over a commutative ring

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 $m ## 3 Answers If$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$-module is injective. So in this case the elements form a basis. -  – Makoto Kato Nov 20 '12 at 16:13 +1 This is a nice answer. However, the first assertion can be proved without using axiom of choice as YACP's comment shows. – Makoto Kato Nov 20 '12 at 16:21 @MakotoKato, I'm not so sure if the uniqueness of rank for commutative rings can be proved without choice. The proof I know uses it. So it may not be a drawback here. – Gregor Bruns Nov 20 '12 at 16:25 @GregorBruns Please see YACP's comment or my answer. – Makoto Kato Nov 20 '12 at 16:30 I would like to prove the first assertion without using axiom of choice. Suppose$m < n$. Then$\bigwedge^n L = 0$. This is a contradiction because$\bigwedge^n L$is a free module of rank$1$. - This is pretty nice. – QiL'8 Nov 20 '12 at 21:17 @MakotoKato +1 Nice answer, great idea to consider exterior powers!! – BenjaLim 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