Q (the rationals) not free abelian yet iso to Z^2?

Thread starter TopCat
Start date Nov 25, 2011

Nov 25, 2011

TopCat

So I just proved that Q (the rationals) is not free abelian since it must have an empty basis. But Q is isomorphic to Z^2 = ƩZ which is equivalent to Q having a nonempty basis. I must be wrong about the isomorphism but I don't see why.

Last edited: Nov 25, 2011

Physics news on Phys.org

Google claims its latest quantum algorithm can outperform supercomputers on a real-world task
Electrohydrodynamics pump and machine learning enable portable high-performance excimer laser
Controlled atomic defects in nickelate films narrow down explanations of superconductivity emergence

Nov 25, 2011

mathwonk

Science Advisor

Homework Helper

2024 Award

there is a map ZxZ*-->Q but it isn't injective.

Nov 25, 2011

TopCat

D'oh! Thanks, mathwonk.

Thread 'When are Markov Matrices also Martingales?'

Are there known conditions under which a Markov Chain is also a Martingale? I know only that the only Random Walk that is a Martingale is the symmetric one, i.e., p= 1-p =1/2.

View full post »

Thread 'The colorful world of ##2\times 2## complex matrices'

The world of 2\times 2 complex matrices is very colorful. They form a Banach-algebra, they act on spinors, they contain the quaternions, SU(2), su(2), SL(2,\mathbb C), sl(2,\mathbb C). Furthermore, with the determinant as Euclidean or pseudo-Euclidean norm, isu(2) is a 3-dimensional Euclidean space, \mathbb RI\oplus isu(2) is a Minkowski space with signature (1,3), i\mathbb RI\oplus su(2) is a Minkowski space with signature (3,1), SU(2) is the double cover of SO(3), sl(2,\mathbb C) is the...

View full post »