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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?

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    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?2012-04-13
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    @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 $S$?2012-04-21
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    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.2012-04-21
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    @ArturoMagidin Oh! I'm sorry, I didn't mean to ignore your comment. My question arose while reading [some random person's notes](http://www.math.columbia.edu/~ums/pdf/Rankeya_Free_Objects_and_Tensor_Products.pdf). Page 3 is where the equality is from.2012-04-21
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    Nowhere on page 3 can I find the statement that "the free group on $S$ is the intersection of all modules containing $S$." There is the statement that if $N$ is a module, and $S$ is a subset of $N$, then $\langle S\rangle$ is the submodule of $N$ obtained by intersecting all submodules of $N$ that contain $S$. And there is the definition of what a "basis for $N$" might mean. But this does not lead to the statement "a free module on $S$ is the intersection of all modules containing $S$". Do you not see why that statement does not make sense? You cannot intersect modules in a vacuum.2012-04-21
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    You cannot intersect modules in a vacuum because the structures don't have to match. The underlying sets may be disjoint (and so you would get a "module" with an empty underlying set). It only makes sense to intersect modules when they are both submodules of some larger module.2012-04-21
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    @ArturoMagidin Indeed! I had already corrected it. Thank you for pointing out this, I had it wrong.2012-04-21
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    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.2012-04-21
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    @ArturoMagidin Ok ok, give me some time, I haven't finished editing. I also need to read again all your comments now that I have understood one mistake. I didn't mean to annoy you. Give me a while to think about this.2012-04-21
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    Take a look at [this answer](http://math.stackexchange.com/a/54334/742). It might help clarify a few things. I'll write an answer to what I think your question *actually is* (nothing to do with free modules) shortly.2012-04-21
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    @ArturoMagidin Thank you. Your penultimate comment might help me resolve my confusion: I thought that $S$ was a basis of $\langle S \rangle$ because $$\langle S \rangle = \{ \sum_{k = 1}^n r_k s_k : r_k \in R, s_k \in S \}$$ Is that wrong? (I know that the module $\langle S \rangle$ generated by $S$ and the free module $F(S) = \oplus_{s \in S} R$ over $S$ are not the same but free just means it has a basis so if $S$ is a basis of $\langle S \rangle$...)2012-04-21
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    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 $r_1=\ldots=r_n=0$.)2012-04-21
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    @ArturoMagidin Yes! Right! You are awesome, thank you very much!! Would you still like to write an answer? If yes, I'll wait and accept, if no, I'll re-accept the one I've got.2012-04-21
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    I'll write an answer sometime later today...2012-04-21
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    @ArturoMagidin Ok, cool. No rush.2012-04-21
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    @ArturoMagidin And if you have time maybe you could have a look at [this](http://math.stackexchange.com/questions/134939/direct-sum-construction), it's somewhat related but I think one answer is incomplete and the second poster is ignoring me it seems...2012-04-21

2 Answers 2

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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

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.

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    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?2012-04-21
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    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. $T \subset \langle S \rangle$.2012-04-22
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    @Arturo: Nothing wrong with your answer, but: Why are you keep repeating what everyone can read in a textbook on algebra? And why is it always to detailed? I hope that, at least, you have fun doing that. But remember that beginners also have to start thinking by themselves. Then it is not good to elaborate every tiny argument.2012-04-22
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    @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.2012-04-22
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    This is a misunderstanding. I don't have a problem at all. I am just curious and wanted to share my opinion.2012-04-22
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    @ClarkKent: I argued using the intersection because that uses the definition of $\langle S\rangle$ as an intersection; it's a way of emphasizing why the intersection has the properties that we want it to have.2012-04-22
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    @Martin: It seems clear that the OP is having trouble following what he finds in textbooks and on-line notes, where things are not quite as detailed... hence the tiny details.2012-04-22
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    @ClarkKent, please be civil. This is **not** "your thread" not is it anyone's.2012-04-23
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    @MarianoSuárez-Alvarez I asked a mathematical question and a random person provides their unsolicited opinion about a third person which has nothing to do with anything asked in the question. I thought this site was about mathematics not about opinions.2012-04-24
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    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.*2012-04-24
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    @Clark: Your reaction seemed to me to be somewhat out of proportion. It was, IMHO, a valid question: why not simply say "Look at textbook `x`, page `y`"? I gave my reason.2012-04-25
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    @ArturoMagidin Just wanted to give my thanks about this answer because its really complete and makes the definitions extremely clear. Keep it up man!2014-09-17
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    @ArturoMagidin Thanks for all the details, I really appreciate also your connection with vector spaces. By the way, do you have a reference for your last theorem?2015-04-27
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    Thank is very helpful. Thank you.2016-07-02
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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.

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    I think I made a mistake. See edit in question. And thank you for the hint.2012-04-13
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    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).2012-04-13
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    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.2012-04-21
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    @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.2012-04-21