Let's denote the free $R$-module generated by a set $S \subset M$ where $M$ is a module by $\langle S \rangle$. Then this is by definition the intersection of all modules $N \subset M$ containing $S$.

I'd like to show that $$\langle S \rangle = \{ \sum_{k = 1}^n r_k s_k : r_k \in R, s_k \in S \}$$

The inclusion $\supset$ is obvious: if $x = \sum_{k = 1}^n r_k s_k $ then clearly $x$ is in every module $M$ with $S \subset M$. Hence $x \in \langle S \rangle$.

Now for some reason the other inclusion is not obvious to me. Let $x \in \langle S \rangle$. Then $x$ is in every module $N \subset M$ with $S \subset N$. How do I proceed from here?

Edit I made a mistake when I wrote this question. The module generated by $S$ is only free if $S$ is finite. Right?

  • $\begingroup$ Your first paragraph makes no sense to me. If $\langle S\rangle$ is the free $R$-module generated by the set $S$, then how is this "by definition the intersection of all modules $M$ containing $S$"? And intersection done how? I can give you two modules that contain $S$, and whose underlying sets intersect in nothing except $S$. The notation is also lousy notation, since given a module $M$ and a subset $X\subseteq M$, then $\langle X\rangle$ is the smallest submodule of $M$ that contains $X$ (which is the intersection of all submodules of $M$ that contain $X$). Are you missing stuff? $\endgroup$ – Arturo Magidin Apr 13 '12 at 18:20
  • $\begingroup$ @ArturoMagidin I missed this comment for some reason! Sorry for the late response! I probably am missing stuff. For example: what if $S$ isn't countable? Doesn't the equality above assume countability of $S$? $\endgroup$ – Rudy the Reindeer Apr 21 '12 at 20:17
  • $\begingroup$ You seem to be ignoring my comment and instead continue to push through, ignoring the problems I raised. It's not a question of whether $S$ is finite, countable, or uncountable. What you wrote makes no sense as written, regardless of the size of $S$. So while you seem to think that you are missing some subtleties further down in the argument, what I'm pointing out is that you can't even get started the way you wrote things. $\endgroup$ – Arturo Magidin Apr 21 '12 at 20:25
  • 1
    $\begingroup$ Your edit still doesn't make sense. You cannot declare the submodule generated by $S$ to be free. Do you understand what "free module" means? You have no warrant for asserting that $\langle S\rangle$ is free, regardless of the size of $S$. There is a big difference between "the submodule of $M$ generated by a subset $S\subseteq M$" and "the free module on a set $S$." You are mixing two very different concepts. $\endgroup$ – Arturo Magidin Apr 21 '12 at 20:39
  • 1
    $\begingroup$ Yes, it's wrong. A basis has to have two properties: (i) it must generate the module; and (ii) it must not have any nontrivial relations. While $S$ certainly generates $\langle S\rangle$, in general it does not satisfy (ii). (By "satisfiies no nontrivial relations", we mean that if $s_1,\ldots,s_n\in S$ are distinct elements of $S$, and $r_1,\ldots,r_n\in R$ are elements of $R$ such that $r_1s_1+\cdots+r_ns_n=0$, then $r_1=\ldots=r_n=0$.) $\endgroup$ – Arturo Magidin Apr 21 '12 at 21:38

You seem to be confusing the notions of free module, basis, generating set, and submodule generated by a set.

Submodule generated by a subset, and generating sets

Let $M$ be a given module, and let $S\subseteq M$ be a subset of $M$. We define the submodule of $M$ generated by $S$, denoted by $\langle S\rangle$, as the smallest submodule of $M$ that contains $S$. Equivalently, since the intersection of an arbitrary family of submodules of $M$ is again a submodule of $M$, we can define $$\langle S\rangle = \bigcap_{N\leq M, S\subseteq N} N,$$ where the intersection is over all submodules of $M$ that contains $S$.

Note that his definition is taking place in the context of a given module $M$. We are always staying inside the module $M$. The intersection is the intersection of all submodules of $M$ that contain $S$. We are only considering elements and modules that are contained in $M$ and have structure compatible with $M$.

Now, the above is the "top-down" description of the submodule generated by $M$. As usual in this situation (see link above), we also want a "bottoms up" description of $\langle S\rangle$. That is provided by the following, which is, I believe, what you are actually trying to prove:

Theorem. Let $M$ be a $R$-module, and let $S\subseteq M$ be a subset. Then $\langle S\rangle$ consists exactly of all elements of $M$ of the form $$\sum_{i=1}^n r_is_i,$$ for some $s_1,\ldots,s_n\in S$ and $r_1,\ldots,r_n\in R$.

Proof. Let $T$ be the set of all such elements of $M$.

Note that if $N$ is any submodule of $M$ such that $S\subseteq N$, then we will necessarily have $T\subseteq N$; indeed, let $x\in T$. Then there exist $s_1,\ldots,s_n\in S$ and $r_1,\ldots,r_n\in R$ such that $x = r_1s_1+\cdots + r_ns_n$. Since $S\subseteq N$ by assumption, we have $s_1,\ldots,s_n\in N$. Since $N$ is a submodule, we must also have $r_1s_1+\cdots + r_ns_n\in N$. Thus, $x\in N$, and we have shown that $T\subseteq N$.

That means that $T$ is contained in $\langle S\rangle$: because $\langle S\rangle$ is an intersection, and $T$ is contained in every set that is being intersected. So $T\subseteq \langle S\rangle$.

In order to show that $\langle S\rangle\subseteq T$, it is enough to show that the set $T$ is a submodule of $M$ that contains $S$. Because if this is indeed the case, then $T$ will be one of the sets being intersected in the definition of $\langle S\rangle$, and therefore $\langle S\rangle$ will necessarily be contained in $T$.

So we want to show that $T$ is a submodule of $M$. It is contained in $M$ because $M$ is a module, $S\subseteq M$, so whenever $s_1,\ldots,s_n\in S$ and $r_1,\ldots,r_n\in R$, we will have $r_1s_1+\cdots+r_ns_n\in M$. $T$ is not empty, since the sum with no summands (when $n=0$) is, by definition, equal to $0$, the identity element of $M$. (If $S\neq\varnothing$, then you can also conclude that $T\neq\varnothing$ by taking $n=1$, $s_1\in S$ arbitrary, and considering the sum with $r_1=0_R$, the zero of $R$).

Now suppose that $x$ and $y$ are elements of $T$, and that $r\in R$. We want to show that $x-ry\in T$. If we can do that, this will prove that $T$ is a submodule of $M$. Since $x\in T$, there exist $s_1,\ldots,s_n\in S$ and $r_1,\ldots,r_n\in R$ such that $x=r_1s_1+\cdots+r_ns_n$. Since $y\in T$, there exist $s'_1,\ldots,s'_m\in S$ and $r'_1,\ldots,r'_m\in R$ such that $y=r'_1s'_1+\cdots + r'_ms'_m$. The only thing we know about the sets $\{s_1,\ldots,s_n\}$ and $\{s'_1,\ldots,s'_m\}$ is that they are contained in $S$; they could be equal, disjoint, one could be contained in the other, or none of the above. Note that the definition of $T$ just says that there are some elements of $S$ that have a certain property, it places no conditions on those elements (not even them being distinct!) other than being contained in $S$.

Now we have: $$\begin{align*} x-ry &= \Bigl( r_1s_1+\cdots+r_ns_n\Bigr) -r\Bigl( r'_1s'_1+\cdots + r'_ms'_m\Bigr)\\ &= r_1s_1 + \cdots + r_ns_n + (-rr'_1)s'_1 + \cdots + (-rr'_m)s'_m. \end{align*}$$ This is an element of $T$: we have $s_1,\ldots,s_n,s'_1,\ldots,s'_m$ are elements of $S$, $r_1,\ldots, r_n, -rr'_1,\ldots,-rr'_m$ are elements of $R$, and we are taking the corresponding sum of multiples of elements of $S$. Thus, $x-ry\in T$ when $x,y\in T$ and $r\in R$. So $T$ is a submodule of $M$.

Finally, we need to show that $S\subseteq T$. Indeed, if $s\in S$, then taking $n=1$, $r_1=1_R$, and $s_1=s$, we get $s = r_1s_1\in T$.

So $T$ is a submodule of $M$ that contains $S$. Therefore, we have: $$\langle S\rangle = \bigcap_{N\leq M, S\subseteq N} N = T\cap\bigcap_{N\leq M, S\subseteq N} N \subseteq T,$$ so $\langle S\rangle \subseteq T$. Since we already have shown that $T\subseteq \langle S\rangle$, we conclude that $T=\langle S\rangle$, as desired. $\Box$

A few things to notice:

  • The definition of $T$ places no restrictions on the size of $S$. We do not require the sum expression to "use" all elements of $S$, just some. If $S$ is finite, we could describe every element of $T$ in terms of all elements of $S$, but we don't have to.

  • The definition of $T$ places no restrictions on the elements $s_1,\ldots,s_n$ that are used to express an element of $T$, except that they must be elements o. They could be the same element, repeated $n$ times, or all distinct; or some repeated, or none.

Now: given a module $M$ and a subset $S\subseteq M$, we say that $S$ generates $M$ if and only if $\langle S\rangle = M$.

If $M$ is a module, and $S\subseteq M$, then $S$ always generates $\langle S\rangle$. Every module has a generating set; in fact, many. We can always take $S=M$; generating sets are not, in general, unique: if $S$ generates $M$, and $S\subseteq S'\subseteq M$, then $S'$ also generates $M$. And there may not be a "smallest" generating set, or even "minimal" generating sets. Sometimes there are, sometimes there are not.


Bases are special types of generating sets.

Let $M$ be a module. A subset $B\subseteq M$ of $M$ is a basis for $M$ if and only if it satisfies two conditions:

  1. $B$ generates $M$; that is, $\langle B\rangle = M$; and
  2. $B$ is "independent": if $b_1,\ldots,b_n$ are any finite set of pairwise distinct elements of $B$, and $r_1,\ldots,r_n$ are elements of $R$ such that $$r_1b_1+\cdots+r_nb_n = 0,$$ then $r_1=r_2=\cdots=r_n=0$.

Again, note that this is all taking place in the context of a given module. There are several equivalent ways of saying $B$ is a basis; for example, you can say that $B$ is a basis if and only if every element of $M$ can be expressed uniquely as a sum of nonzero multiples of finitely many elements of $B$. A basis may or may not exist for any given module $M$.

Free modules and free generating sets

Let $X$ be a set. A module $M$ is said to be a free module on $X$ if and only if $X\subseteq M$ and $M$ satisfies the following condition:

  • Given any module $N$ and any (set-theoretic; that is, not necessarily a module homomorphism) function $f\colon X\to M$, there exists a unique moudle homorphism $\mathcal{F}\colon M\to N$ such that $\mathcal{F}(x) = f(x)$ for all $x\in X$.

We have the following result (which is where, I think, you got confused):

Theorem. Let $M$ be a module, and let $X\subseteq M$. Then $M$ is a free module on $X$ if and only if $X$ is a basis for $M$.

Now, given any set $X$ (finite, countable, or infinite, doesn't matter), there always is a "free module on $X$". One can construct it as the set of all formal sums of multiples of elements of $X$, or as a direct sum, or any other number of ways. But here we start with a set, and we construct a module (as opposed ot the notion of "submodule generated", where we start with a module).

Finally, we say that a module $M$ is free if there exists $X\subseteq M$ such that $X$ is a basis for $M$ (equivalently, such that $M$ is free on $X$).

Also, given a module $M$ and a subset $X$, if $X$ is independent, then by the theorem above we can conclude that $\langle X\rangle$ is a free module on $X$; this may or may not be equal to all of $M$. But again, here we start with a module, not with a set.

  • $\begingroup$ Is someone seriously upvoting this less than six minutes after I posted it? I would expect to take longer than that to read it! Again, I appreciate the vote of confidence, but shouldn't one wait to upvote at least until one has read the whole thing? $\endgroup$ – Arturo Magidin Apr 21 '12 at 23:24
  • $\begingroup$ For the first inclusion, $T \subset \langle S \rangle$, do we have to argue using the intersection? How about: Let $x = \sum_{k=1}^n s_k r_k$ for $s_k \in S$. Then $s_k \in \langle S \rangle$. Then since $\langle S \rangle$ is a module, $\sum_{k=1}^n s_k r_k = x \in \langle S \rangle$, i.e. $T \subset \langle S \rangle$. $\endgroup$ – Rudy the Reindeer Apr 22 '12 at 8:22
  • 1
    $\begingroup$ @MartinBrandenburg Do you have a problem with Arturo? If so, keep it out of my threads please. I don't want my threads to be polluted with opinionated argumentative off-topic spats. Thank you. $\endgroup$ – Rudy the Reindeer Apr 22 '12 at 9:10
  • 1
    $\begingroup$ @ClarkKent, please be civil. This is not "your thread" not is it anyone's. $\endgroup$ – Mariano Suárez-Álvarez Apr 23 '12 at 15:56
  • 2
    $\begingroup$ Dear Clark: Nothing justifies being uncivil. There is no gentle way to read your «pollute with opinionated argumentative off-topic spats» and such a description surely might appear to some to be out of proportion Martin's one comment on Arturo's pedagogical method. Independently of what others write, say, or do in this site, nothing justifies being uncivil. $\endgroup$ – Mariano Suárez-Álvarez Apr 24 '12 at 23:16

Hint: Show that $\left\{\sum^n_{k=1}r_ks_k:r_k\in R,s_k\in S\right\}$ is an $R$-module containing $S$. Then if $x\in\langle S\rangle$, it must also be in this set.

  • $\begingroup$ I think I made a mistake. See edit in question. And thank you for the hint. $\endgroup$ – Rudy the Reindeer Apr 13 '12 at 11:16
  • $\begingroup$ OK - in answer to the new question, no I don't think $S$ needs to be finite, and in fact the same description holds (in that you still take finite linear combinations of elements of $S$, but they can be arbitrarily long). $\endgroup$ – mdp Apr 13 '12 at 11:55
  • $\begingroup$ I removed the accept and upvoted instead because I realised there are a few other things that I don't understand that Arturo helped me clear up. He is going to write an answer and to keep the balance I'll accept that but won't upvote. $\endgroup$ – Rudy the Reindeer Apr 21 '12 at 21:54
  • $\begingroup$ @ClarkKent: You are allowed to both upvote and accept; and there is no balance to be kept. Upvotes for Matt don't count against me or vice-versa. $\endgroup$ – Arturo Magidin Apr 21 '12 at 23:18

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.