2
$\begingroup$

In a free $R$-module $M$, a set $\{a_1,...,a_k\} \subset M$ is linearly independent if and only if whenever $\sum r_ia_i=0$ with $c_i \in R$, it follows that for every $i$, $c_i=0$.

Now suppose the set $A=\{a_1,...,a_k\}$ is linearly independent but doesn't span $M$. If an element $\alpha \in M$ is not in the span of $A$, then there do not exist $c_i$'s in $R$ such that $\sum r_ia_i=\alpha$. Does this imply that $A \bigcup \{\alpha\}$ is linearly independent?

I know that in linear algebra over a field this is true, but it is conceivable that over a general ring, it could be possible that some element $\alpha$ wouldn't be in the span of $A$, though $r\alpha$ is for some $r \in R$. Are there times when this occurs, or is it always the case that non spanning implies linear independence? If it does occur, are there necessary and sufficient conditions for it to occur?

$\endgroup$
2
$\begingroup$

No. Take $\{2\}$ and 7 in $\Bbb Z$. Suppose that your condition holds: $A$ is a ring such that given any $A$-module $M$ and any linearly independent subset $S$ of $M$ and $m$ not in the span of $M$, then $S\cup m$ is linearly independent. Assume that $M$ is nonzero. Since the empty set $\varnothing$ is always linearly independent, the hypothesis implies there exists $m\in M$ that is linearly independent. If $m$ spans $M$, we're done and $M$ is free with basis $\{m\}$. Else we can continue this process, using Zorn's lemma: consider a chain of linearly independent subsets $S_1\subseteq S_2\subseteq\cdots$. Then their union is linearly independent, since linear independence is afforded by finitely many elements (prove this!). By Zorn there exists a maximal linearly independent subset $S$ of $M$. This must be a basis, i.e. it must also span $M$, for if it doesn't, we have $m$ not in the span of this linearly independent subset. By your hypothesis, adjoining this element to $S$ gives a larger linearly independent subset over $S$, and this contradicts the fact that $S$ was chosen to be maximal. It follows that $\langle S\rangle =M$; and so $M$ is free with basis $S$. This proves every $A$-module is free.

$\endgroup$
  • $\begingroup$ Simple enough. Are there any general conditions under which this occurs or does not occur? $\endgroup$ – leibnewtz Dec 3 '15 at 1:31
  • 1
    $\begingroup$ @leibnewtz For any ring $R$ we have that $\{a\}$ is $R$-linearly independent iff $a$ is a nonzerodivisor; on the other hand any two elements $a,b\in R$ are automatically $R$-linearly dependent ($(-b)a+ab=0$). $\endgroup$ – Matemáticos Chibchas Dec 3 '15 at 1:35
  • $\begingroup$ What about when we have $M=R^n$, for $n \geq 2$? $\endgroup$ – leibnewtz Dec 3 '15 at 1:37
  • $\begingroup$ @PedroTamaroff I'm a bit confused. Why do you say every module admits a basis? $\endgroup$ – leibnewtz Dec 3 '15 at 1:38
  • 1
    $\begingroup$ @leibnewtz: Matemáticos Chibchas's comment shows that if $R$ is commutative and satisties your condition, then $R$ is a field. $\endgroup$ – Eric Wofsey Dec 3 '15 at 2:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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