Prove that: Every Q - vector space V≠0 is not free over subring Z⊂Q

Question

Prove that:  Every Q - vector space V≠0 is not free over subring Z⊂Q

Jack Maxson · Accepted Answer

Step 1
Dietrich Burde essentially already proved the result. Here are some more details.
First we show the claim for finite vector spaces.
So let V be a finite-dimensional

Q

vector space,

n = \dim_{Q} V

.
Suppose V was free as a

Z

-module, then

Q \subset V = Q^{n}

is a submodule of a free

Z

-module hence it is itself free.
But

Q

is not free as a

Z

-module, since any two rationals are linearly dependent and clearly

Q

is not generated by one rational as a

Z

-module.
To deal with the infinite dimensional case it should suffice to write V as the union of its finite-dimensional subspaces.

Buck Henry · Accepted Answer

Step 1  Every Q -vector space V is divisible as an abelian group: for every  x∈V  and every nonzero integer n, there exists  y∈V such that ny=x.  Claim. No nonzero free abelian group is divisible.  Indeed, if G is a divisible abelian group, then the only homomorphism  G→Z  is the trivial one, so Z cannot be a homomorphic image of G.  But Z is a homomorphic image of every nonzero free abelian group.

alenahelenash · Accepted Answer

Step 1Consider a Q -vector spaceV≠0and suppose, toward a contradiction that it is free over Z, say with basis B. Consider any basis vectorb∈B.Since V is a Q -vector space, we can consider the element 12b of V, and we can express it as a Z -linear combination∑inibiof some elements bi of B (possibly including b itself as one of the bi′s).So we have a linear dependence relationb−∑i2nibi=0among the basis vectors b and bi, in which the basis vector b has coefficient 1 (if b is not among the bi′s) or1−2ni (if b=bi).In any case, the coefficient of b is odd and, in particular, not 0. That contradicts the fact that a basis B has to be linearly independent.

Prove that: Every \mathbb{Q} - vector space V\ne0 is not free

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