# Basis of a module

I know that not all modules have bases, and those that do are called free modules. I know that all vector spaces have bases, and that a module $M$ over $R$ becomes a vector space if $R$ is a division ring. So my question is, why is it that $R$ being a division ring allows $M$ to have a basis?

Thanks for any replies.

• To find a basis, take a generating set X. If it is not linearly independent, then there are numbers ai so that Sum aixi = 0. Solve for x1 to get x1 = Sum( (1/a1)*aixi, i > 1). Now x2,.... are still a spanning set and are closer to being linearly independent. Repeat this until you get a basis. The only thing that can go wrong: what if (1/a1) doesn't exist. The key step is being able to divide. Hence division rings. – Jack Schmidt Apr 30 '14 at 20:18
• @JackSchmidt Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Mar 18 '15 at 21:25

To find a basis for $M$, take a generating set $X = \{{\bf x_i} \} \subset M$. If it is not linearly independent, then there are numbers $a_i \subset R$ so that $\sum_i a_i\, {\bf x}_i = {\bf 0}$. Solve for ${\bf x}_1$ to get $${\bf x}_1 = -\sum_{i > 1} (a_1)^{-1} a_i\, {\bf x}_i.$$
Now ${\bf x}_2, \dots$ are still a spanning set and are closer to being linearly independent. Repeat this until you get a basis. The only thing that can go wrong: what if $(a_1)^{-1}$ doesn't exist. The key step is being able to divide. Hence division rings.
[Additional comments from tparker: If the set $X$ is only one element too large, then the equation above indicates how to express an arbitrary element ${\bf x}_1$ as a linear combination of elements of the basis $\{ {\bf x}_2, \dots, {\bf x}_n \}$. If it's $k>1$ elements too large, then you need to keep track of a system of $k$ equations of the form above. Once you've gotten the original set down to a basis $B = \{ {\bf x}_{k+1}, \dots, {\bf x}_n \} \subset X$, you need to recursively plug each equation into the previous one in reverse order, expanding out each ${\bf x}_i$ in terms of the basis elements in $B$, starting with ${\bf x}_k$.]