# A module without a basis

Let’s start by recalling some background about modules.

Suppose that $$R$$ is a ring and $$1_R$$ is its multiplicative identity. A left $$R$$-module $$M$$ consists of an abelian group $$(M, +)$$ and an operation $$R \times M \rightarrow M$$ such that for all $$r, s \in R$$ and $$x, y \in M$$, we have:

1. $$r \cdot (x+y)= r \cdot x + r \cdot y$$ ($$\cdot$$ is left-distributive over $$+$$)
2. $$(r +s) \cdot x= r \cdot x + s \cdot x$$ ($$\cdot$$ is right-distributive over $$+$$)
3. $$(rs) \cdot x= r \cdot (s \cdot x)$$
4. $$1_R \cdot x= x$$

$$+$$ is the symbol for addition in both $$R$$ and $$M$$.
If $$K$$ is a field, $$M$$ is $$K$$-vector space. It is well known that a vector space $$V$$ is having a basis, i.e. a subset of linearly independent vectors that spans $$V$$.
Unlike for a vector space, a module doesn’t always have a basis. Continue reading A module without a basis