Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.