Modules

• Dec 4th 2011, 09:02 AM
jcir2826
Modules
Given a set X, prove that there exists a free R-module F with a basis B for which there is a
bijection ϕ : B → X.
• Dec 4th 2011, 09:34 AM
Drexel28
Re: Modules
Quote:

Originally Posted by jcir2826
Given a set X, prove that there exists a free R-module F with a basis B for which there is a
bijection ϕ : B → X.

Consider the free module $\displaystyle R^{\oplus X}$ (i.e. $\displaystyle X$-fold coproduct). This has a natural basis of the form $\displaystyle \{e_x:x\in X\}$ where $\displaystyle e_x$ is the tuple with $\displaystyle 1$ in the $\displaystyle x^{\text{th}}$ coordinate and zero elsewhere. I think the rest should be obvious.
• Dec 4th 2011, 07:54 PM
jcir2826
Re: Modules
So I am just missing to show that there is a bijection from this natural basis to the free module direct summand R and X?
• Dec 4th 2011, 08:19 PM
Drexel28
Re: Modules
Quote:

Originally Posted by jcir2826
So I am just missing to show that there is a bijection from this natural basis to the free module direct summand R and X?

The basis is $\displaystyle \{e_x:x\in X\}$ and isn't $\displaystyle x\mapsto e_x$ a bijection?