I have a curiosity question on a fundamental difference between vector spaces and general modules over rings.

Some of the fundamental facts of linear algebra:

(1) A finitely generated vector space has a basis.

(2) Minimal generating (spanning) sets of a vector space are linearly independent and therefore form a basis.

I recently took a course on modules. One basic example discussed: Let $R = K[x,y]$, where $K$ is a field, and let $I = \langle x,y \rangle $. We consider $I$ as a module over $R$.

$I$ is a finitely generated module, however it is not free (does not contain a basis). This is because the smallest generating set has size $2$, and no matter what generating set you choose, you can write a non-trivial $R-$linear combination of the elements of that set that equals $0$.

If $S = \{ f(x,y), g(x,y) \} $ so that $I = \langle S \rangle $, then $g(x,y)f(x,y) + (-f(x,y))g(x,y) = 0 $. A similar argument can be made for any finite generating set for $I$.

This example shows that those fundamental facts of vector spaces are not necessarily true for modules over general rings.

My question is what is it about the scalars coming from a field that makes these facts true but not so when the scalars come from a general ring? I never got a chance to ask my professor during the class. I tried reading proofs from linear algebra texts but I cannot see where the underlying scalar FIELD makes the difference.

Any clarification would be very helpful.

  • $\begingroup$ Because over a field you can "divide" by scalars. $\endgroup$
    – Randall
    Commented May 17, 2021 at 15:16
  • $\begingroup$ And rings aren't necessarily commutative so any of your proofs that rely on the commutativity of scalars won't work for modules. $\endgroup$
    – John Douma
    Commented May 17, 2021 at 15:19
  • $\begingroup$ @Randall How does this "division" help make the difference? Would the facts hold true of the module were over a division ring (that may or may not be commutative)? $\endgroup$ Commented May 17, 2021 at 15:24
  • $\begingroup$ @J-Q-McNeedHelpson read a few proofs of theorems on vector spaces and dimension and note how many times this is used. $\endgroup$
    – Randall
    Commented May 17, 2021 at 15:26
  • $\begingroup$ @JohnDouma True. The example I gave though is a module over a commutative ring, and the results still do not hold. So the difference somehow lies in the the fact that scalars in a vector space come from a field. $\endgroup$ Commented May 17, 2021 at 15:27

3 Answers 3


Since you asked about the theorem that, in a vector space, a minimal spanning set is linearly independent, it seems worthwhile to look into the proof of that theorem to see how division is used. The standard proof begins by assuming that we have a spanning set that isn't independent, so we have some linear dependence relation involving some of the vectors $v_i$ from that spanning set, say $0=c_1v_1+\dots+c_nv_n$, where the scalar coefficients $c_i$ are not all zero. Then we can show that our spanning set isn't minimal as follows. In our linear dependence relation, transpose one of the terms with non-zero coefficient, say $c_j$, to the left side of the equation, obtaining $-c_jv_j=\dots$, where the $\dots$ on the right side is a linear combination of vectors other than $v_j$ from our spanning set. [So far, this works in modules over any commutative ring.] Then we divide both sides of the equation by $-c_j$ [here we need a field] to express $v_j$ as a linear combination of other vectors from our spanning set. So we could delete $v_j$ from the spanning set and what remains would still span. That contradicts the assumed minimality of our spanning set.

If you don't like commutativity, you could replace "module", "commutative ring", and "field" in the preceding paragraph by "left module", "ring", and "division ring", respectively.

It may be worthwhile for you to return to the example in your question, to see exactly how the non-invertibility of coefficients $f$ and $g$ is essential for that example.


In some sense, there’s not much of a difference. In fact, you’d be accurate if you said that a vector space is a module, except in the specialized case where the ring happens to be a field.

But in another sense, there are important differences. Just for notation, I’ll refer throughout to a vector space $V$ over a field $F$ , and a module $M$ over a ring $R$ .

For example, take a single vector $v \in V$ . The set ${v}$ is linearly independent. But not so (in general) for an element $m \in M$ . For example, $\mathbb{Z}_n$ (as an additive group) is a $\mathbb{Z}$ -module with multiplication defined in the natural way: multiply then reduce modulo $n$ . Thus, no single element in $\mathbb{Z}_n$ is a linearly independent singleton set, because $mn=0$ for any $m \in \mathbb{Z}_n$ .

Relatedly, in vector spaces, a set $S$ is linearly dependent if and only if some member of $S$ is a linear combination of other elements of $S$ . Or in a simpler case where $S$ has two elements, $S$ is linearly dependent if and only if one element is a scalar multiple of the other.

Not true in modules. As an example, take $M=\mathbb{Z}$ as a $\mathbb{Z}$ -module, and let $n$ and $m$ be coprime. Because they’re coprime, neither is a scalar multiple of the other. But $S={n,m}$ is linearly dependent. If you wish to solve $an+bm=0$, then write $a=m$ and $b=−n$.

These results are just two among many that illustrate what, to me, is the main difference between modules and vector spaces. In vector spaces, there’s a concept of dimension. It’s simple, and it’s very useful. There are notions of different types of dimensions in modules, but they’re nowhere near as simple, and nowhere near as useful.

  • $\begingroup$ Isn’t a single vector also linearly dependent if the vector space is over a finite field? (Or more generally, a field with positive characteristic?) $\endgroup$
    – tparker
    Commented May 17, 2021 at 15:30
  • $\begingroup$ Thank you for the response. I am glad to see more examples to highlight the differences between vector spaces and modules. I guess my issue I am still having is when it comes down to looking at vector spaces and modules over a ring as abstractly as possible, what, in the arguments, make those results I listed true form vector spaces but not modules? When I read the proofs, I do not see in the arguments where failure occurs if our scalars come from a ring that is not a field. This may just take some deep thought myself to come to an understanding. $\endgroup$ Commented May 17, 2021 at 15:31
  • $\begingroup$ @tparker: No, because in a field with characteristic $p$, multiplying by $p$ means multiplying by $0$. $\endgroup$
    – Bernard
    Commented May 17, 2021 at 15:32
  • $\begingroup$ @Bernard Ah yes, of course. For some reason I was thinking that $v + v + ... + v$ ($p$ times) counted as a nontrivial linear combination, but of course it doesn't. $\endgroup$
    – tparker
    Commented May 17, 2021 at 22:51

Module is a structure of an abelian group with resspect to "addition"over a ring(not necessary commutative) and vector space is the same over a field. As every field is a ring so every vector space is a module but converse it not true.


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