Franklin at 21:03, 18 February 2008

2008-02-18T21:03:39Z

New page

If <math>n\neq m</math> then <math>\mathbb{Z}^n\not\cong\mathbb{Z}^m</math>.

Proof

Assume that <math>f:\mathbb{Z}^n\rightarrow\mathbb{Z}^m</math> is an isomorphism. Let <math>A</math> be the matrix of <math>f</math> in the canonical basis and <math>M</math> the maximum of the absolute values of the entries of <math>A</math>. If we evaluate <math>f</math> in all the vectors of <math>\mathbb{Z}^n</math> who's entries have absolute values less than or equal to <math>r</math> (there are <math>(2r)^n</math> of such elements) then we get elements who's entries have absolute value less than or equal to <math>nrM</math> (there are <math>(2rnM)^m</math> of such elements in <math>\mathbb{Z}^m</math>). Since <math>f</math> is injective we must have <math>(2r)^n\leq(2rnM)^m</math> for every <math>r</math>. Replacing <math>f</math> by its inverse if necessary we can assume that <math>n>m</math> but if this is the case the inequality above can not be true for arbitrarily large values of <math>r</math>.

0708-1300/fact - Revision history

Franklin at 21:03, 18 February 2008