# Ideas of Linear Independence in the context of modules and vector spaces

Let $k$ be a ring and suppose $M$ is a module over $k$. A set $X \subseteq M$ is a minimal generating set if it generates $M$ and no proper subset of $X$ generates $M$.

It is easy to see this means that no element of $X$ can be written as a finite $k$-linear combination of the other elements in $X$. However this does NOT correspond to "linear independence" as is the case for vector spaces. For example if you consider $\mathbb Z_3$ as a $\mathbb Z$-module then $\{ 1 \}$ is a minimal generating set but $$6 \cdot 1 = 3 \cdot 1 = 0$$ but $3 \not= 6$.

However don't these notions coincide when we look at modules over a field, i.e. vector spaces? Why does $\sum \alpha_i x_i = 0 \implies$ every $\alpha_i = 0$ if $\{x_i \}$ is a basis for a space $V$, but it doesn't hold for arbitrary modules?

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Turns out:  $$\sum \alpha_i x_i = 0 \implies \text{ each } \alpha_i \text{ is a non-unit}$$ is equivalent to $X$ being a minimal generating set – Paul Slevin Oct 25 '12 at 17:44

## 2 Answers

If $X$ is a minimal generating set in a vector space or more generally in a module $M$ over a skew-field, then $M$ is linearly independent and therefore a basis of $M$: The reason is that if $\sum_{x} \lambda_x x = 0$ is a non-trivial linear combination, then at least one $\lambda_x$ is non-zero and therefore invertible. But then $x$ is a linear combination of the other elements, a contradiction.

As you have already seen, this does not work over general rings (since non-zero does not imply invertible there). Basically you have answered the question yourself.

By the way, the correct definition of a basis $X$ of a module $M$ is that it is a linearly independent generating system. Equivalently, the canonical map $k^{\oplus X} \to M$ is an isomorphism.

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but why does ax = bx imply a = b if x is an element from a min generator set of a vector space? – Paul Slevin Oct 24 '12 at 16:58
If $a \neq b$ then $a - b$ is invertible (if we work over a skew-field), so $(a-b)x=0$ implies $x=0$, impossible. Perhaps you should Linear Algebra before Module Theory. – Martin Brandenburg Oct 24 '12 at 19:42

For that set of generators $x_i$, $1\leq i\leq n$, let $N$ be the submodule of $M$ that it generates. The linear independence condition basically says that the kernel of the obvious map from a free module rank $n$ onto $N$ is zero, and makes $N$ isomorphic to the free module of rank $N$.

Of course, for nonfree modules, it will then be impossible to find a generating set that is also linearly independent.

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