Any maximal linearly independent subset of free module $M$ has cardinality equal to the cardinality of a basis of $M$.

Question

Let $R$ be integral domain and $M$ be an $R$ module.

I want to prove if $M$ is a free module, then any maximal linearly independent subset of $M$ has cardinality equal to the cardinality of a basis of $M$.

Any maximal linearly independent subset of $M$ must have the same cardinality. So it only remains to prove the cardinality is equal to the cardinality of a basis of $M$. So if $\{m_i\}_{i\in I}$ is a maximal linearly independent subset of $M$, then we need to prove $\{m_i\}_{i\in I}$ is a basis of $M$? But how? Any helps would be appreciated greatly.

Answer 1

The general ideal to show this type of results is to use localisation. Let $N$ be a maximal ideal of $R$. Write $R/N=k$, if $R/N\otimes M=V$ is a $k$-vector space. If $(m_1,..,m_n)$ is a maximal independent, $1\otimes m_1,...,1\otimes m_n$ is a basis of $V$, otherwise there exists $v=\sum u_i\otimes v_i$ which is not in $Vect(1\otimes m_1,..,1\otimes m_n)$. Write $x=u'_1v_1+...+u'_nv_n$ where $u'_i$ is the image of $u_i$ by the quotient map $R\rightarrow R/M$, $(m_1,...,m_n,x)$ is independent.

Answer 2

You already know that any two maximal linearly independent subsets of $M$ have the same cardinality, so it suffices to show that a basis is a maximal linearly independent subset of $M$. Can you do this?

Answer 3

You could refer to the book <<Algebra: chapter 0>>:

page307 ,Lemma 1.2: Let M be an R-module, and let S $\in$ M be a linearly independent subset. Then there exists a maximal linearly independent subset of M containing S.

page311, Proposition 1.15: Let R be an integral domain, and let M be a free R-module; assume that M is generated by S: M = $<S>$. Then S contains a maximal linearly independent subset of M.

These two theorems imply that the basis of a R-module,where R is a integral domain, is a maximal linearly independent subset. But the reverse is not true. The integral domains satisfy the 'IBN (Invariant Basis Number) property'. In particular, any two maximal linearly independent subsets of a free module over an integral domain have the same cardinality. Therefore, the basis has the same cardinality with the maximal linearly independent subset.