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Monday, July 8, 2013

Bases and Cardinality (11.1.12-14a)

Dummit and Foote Abstract Algebra, section 11.1, exercises 12-14a:

MathJax TeX Test Page 12. If F is a countable field and V is an infinite dimensional vector space with basis B, prove |B|=|V|.
13. Prove RnR as vector spaces over Q.
14a. Let A be a basis for the infinite dimensional vector space V over F. Prove VaAF.

Proof: Throughout these exercises we shall assume the fact that countable unions and finite direct products of an infinite set S fix the cardinality.

(12) By the inclusion mapping we clearly see |B||V|. Letting Vi for iN be the set of elements of V whose basis sum includes exactly i nonzero vectors, we have V is the countable union of the Vi, so it suffices to show |Vi|=|B|. We can observe|B|=|B×B||F×B|=|(F×B)1×...×(F×B)i||Vi|so that|B|=|iNB||iNVi|=|V|and now |B|=|V|.

(13) First we prove R2R, so that by induction the proposition easily follows. First, assume the basis A of R is finite. Then R is isomorphic to the direct sum of a finite number of copies of Q by the next exercise, which would imply R is countable, a contradiction. Now, there is a basis B for R2 given by the basis A in each component whose cardinality is equal to |AA| so that |B|=|A|. Linearly extend the homomorphism induced by this bijection to obtain an isomorphism.

(14a) By mapping the coefficients of a vector sum of the basis to the proper coordinates in the direct sum we obtain the evident isomorphism. Now, the direct product is clearly a vector space by closure with componentwise multiplication by scalars of F.

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