Suppose $R$ is commutative, and let $M$ be a free $R$-module of finite rank. Prove that if $A$ and $B$ are bases of $M$, then $|A| = |B|$.

Question

I can't really figure out how to prove this theorem.

Note that if $M$ is free with basis $A$ then $M \simeq R^{|A|}$, so it suffices to show that $R^n \simeq R^k$ implies $n = k$. — Jim, Mar 24 '15 at 07:37

manthanomen · Answer 1 · 2015-03-24T08:08:11.107

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There is a result (Hungerford, p. 186) which states that if $f : R \to S$ is a surjective ring homomorphism such that any two bases of a free $S$-module have the same cardinality, then any two bases of a free $R$-module have the same cardinality. If you know this theorem, all you need to do is construct a surjective homomorphism from an arbitrary commutative ring $R$ to a field (since any two bases of a vector space have the same cardinality). Just take the projection $R \to R/M$ where $M$ is a maximal ideal of $R$.

edited Mar 24 '15 at 08:08

answered Mar 24 '15 at 07:41

manthanomen

3,403

Can I know where I can find the theorem? Or what is the theorem called? – user10024395 Mar 24 '15 at 07:50
All of this is proved in pages 185-186 of Hungerford's Algebra. There may be a quicker way which doesn't use this result, but I'm not sure. – manthanomen Mar 24 '15 at 08:03
$R$ is not necessarily unitary, so there might be no maximal ideal. – MooS Mar 24 '15 at 08:42

Suppose $R$ is commutative, and let $M$ be a free $R$-module of finite rank. Prove that if $A$ and $B$ are bases of $M$, then $|A| = |B|$.

1 Answers1