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[parent] cardinalities of bases for modules (Theorem)

Let $ R$ be a ring and $ M$ a left module over $ R$.

Proposition 1   If $ M$ has a finite basis, then all bases for $ M$ are finite.
Proof. Suppose $ A=\lbrace a_1,\ldots, a_n\rbrace$ is a finite basis for $ M$, and $ B$ is another basis for $ M$. Each element in $ A$ can be expressed as a finite linear combination of elements in $ B$. Since $ A$ is finite, only a finite number of elements in $ B$ are needed to express elements of $ A$. Let $ C=\lbrace b_1,\ldots,b_m\rbrace$ be this finite subset (of $ B$). $ C$ is linearly independent because $ B$ is. If $ C\ne B$, pick $ b\in B-C$. Then $ b$ is expressible as a linear combination of elements of $ A$, and subsequently a linear combination of elements of $ C$. This means that $ b=r_1b_1+\cdots +r_mb_m$, or $ 0=-b+r_1b_1+\cdots r_mb_m$, contradicting the linear independence of $ C$. $ \qedsymbol$
Proposition 2   If $ M$ has an infinite basis, then all bases for $ M$ have the same cardinality.
Proof. Suppose $ A$ be a basis for $ M$ with $ \vert A\vert \ge \aleph_0$, the smallest infinite cardinal, and $ B$ is another basis for $ M$. We want to show that $ \vert B\vert=\vert A\vert$. First, notice that $ \vert B\vert\ge \aleph_0$ by the previous proposition. Each element $ a\in A$ can be expressed as a finite linear combination of elements of $ B$, so let $ B_a$ be the collection of these elements. Now, $ B_a$ is uniquely determined by $ a$, as $ B$ is a basis. Also, $ B_a$ is finite. Let
$\displaystyle B'=\bigcup_{a\in A} B_a.$
Since $ A$ spans $ M$, so does $ B'$. If $ B'\ne B$, pick $ b\in B-B'$, so that $ b$ is a linear combination of elements of $ B'$. Moving $ b$ to the other side of the expression and we have expressed 0 as a non-trivial linear combination of elements of $ B$, contradicting the linear independence of $ B$. Therefore $ B'=B$. This means
$\displaystyle \vert B\vert= \left\vert\bigcup_{a\in A} B_a\right\vert \le \aleph_0 \vert A\vert = \vert A\vert.$
Similarly, every element in $ B$ is expressible as a finite linear combination of elements in $ A$, and using the same argument as above,
$\displaystyle \vert A\vert\le \aleph_0 \vert B\vert \le \vert B\vert.$
By Schroeder-Bernstein theorem, the two inequalities can be combined to form the equality $ \vert A\vert=\vert B\vert$. $ \qedsymbol$



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Cross-references: equality, inequalities, Schroeder-Bernstein theorem, argument, expression, side, spans, collection, proposition, cardinal, cardinality, infinite, expressible, linearly independent, subset, number, linear combination, bases, basis, finite, left module, ring
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This is version 5 of cardinalities of bases for modules, born on 2008-06-04, modified 2008-06-05.
Object id is 10654, canonical name is CardinalitiesOfBasesForModules.
Accessed 286 times total.

Classification:
AMS MSC: 15A03 (Linear and multilinear algebra; matrix theory :: Vector spaces, linear dependence, rank)
 13C05 (Commutative rings and algebras :: Theory of modules and ideals :: Structure, classification theorems)
 16D40 (Associative rings and algebras :: Modules, bimodules and ideals :: Free, projective, and flat modules and ideals)
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