Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

As a follow-up to this question, I am trying to prove the following:

If $F$ be a free $R$-module on a set $S$ via the function $i:S \longrightarrow F$ then $i$ is necessarily injective.

Attempted Proof: This means that for every $R$-module $X$ and every function $f:S \longrightarrow X$ there exists a unique $R$-linear function $\tilde{f}:F \longrightarrow X$ such that $\tilde{f} \circ i = f$. So, consider the function $f:S \longrightarrow X$ defined by $f(x) := 0_X \;\forall\; x \in S$ where $0_X$ denotes the additive identity in $X$. Then, by definition of a free module,

$$ \tilde{f} \circ i = f \implies \tilde{f} \circ i = 0_X $$ But, this means that $\tilde{f}$ is a left-inverse for $i$ and $i$ has a left-inverse if and only if it is injective. Therefore, $i$ must be injective.

So, my question is, does this proof work?

share|improve this question
1  
No, if the composition is the zero function, then this does not mean that $\tilde{f}$ is a left inverse of anything; you would need the composition to be the identity function to get that conclusion. –  Arturo Magidin Jan 14 '12 at 22:26
1  
Note that if you have a composition $\overline{f}\circ i$, you want to show that the composition is one-to-one in order to conclude that $i$ is one-to-one. But if the composition is the zero function, then the composition is certainly not one-to-one, you cannot conclude anything about the first function being applied. –  Arturo Magidin Jan 14 '12 at 22:41
add comment

1 Answer

up vote 2 down vote accepted

To show that $i$ is injective, consider $R$ as a module over itself. Then $R$ is cyclic, generated by $1_R$.

Now, let $s,t\in S$ such that $i(s)=i(t)$. Define $f\colon S\to R$ by $$f(u) = \left\{\begin{array}{ll} 1_R &\text{if }u=s,\\ 0 &\text{if }u\neq s. \end{array}\right.$$ Then $f$ extends to a module homomorphism $\overline{f}\colon F\to M$. Then $$1_R = f(s) = \overline{f}(i(s)) = \overline{f}(i(t)) = f(t).$$

(If your rings don't have identity, then pick any $x\in R$, $x\neq 0$, and consider ideal generated by $x$, $\{nx + rx\mid n\in\mathbb{Z}, r\in R\}$; then map $s$ to $x$ and everything else to $0$).

share|improve this answer
add comment

Your Answer

 
discard

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.