Question

Prove that Q regarded as a Se module is not free?

Answer 1

Answer:

Modules are a generalization of the vector spaces of linear algebra in which

the “scalars” are allowed to be from an arbitrary ring, rather than a field.

This rather modest weakening of the axioms is quite far reaching, including,

for example, the theory of rings and ideals and the theory of abelian groups

as special cases.

(1.1) Definition. Let R be an arbitrary ring with identity (not necessarily

commutative).

(1) A left R-module (or left module over R) is an abelian group M together

with a scalar multiplication map

· : R × M → M

that satisfy the following axioms (as is customary we will write am in

place of ·(a, m) for the scalar multiplication of m ∈ M by a ∈ R). In

these axioms, a, b are arbitrary elements of R and m, n are arbitrary

elements of M.

(al)a(m + n) = am + an.

(bl)(a + b)m = am + bm.

(cl) (ab)m = a(bm).

(dl)1m = m.

(2) A right R-module (or right module over R) is an abelian group M

together with a scalar multiplication map

· : M × R → M

that satisfy the following axioms (again a, b are arbitrary elements of

R and m, n are arbitrary elements of M).

(ar)(m + n)a = ma + na.

(br)m(a + b) = ma + mb.