I was reviewing the definition of a vector space recently, and it occurred to me that one could allow for only scalar multiplication by the integers and still satisfy all of the requirements of a vector space.

Take for example the set of all ordered pairs of integers. Allow for scalar multiplication over the integers and componentwise vector addition as usual. It seems to me that this is a perfectly well-defined vector space.

The integers do not form a Field, which begs the question: Is there any reason that the "field" over which a vector space is defined must be a mathematical Field? If so, what is wrong with the vector field I attempted to define above? If not, what are the requirements for the scalars? (For instance, do they have to be a Group - Abelian or otherwise?)

  • 2
    $\begingroup$ Not mentioned in the other answers yet: A vector space over the integers is properly called a "$\mathbb{Z}$-module," and every such module is equivalent to an abelian group, in your case the group $\mathbb{Z}^2$ with addition defined componentwise. The integer scalar tells you how many times to add up an element of the group. $\endgroup$ Jul 2, 2017 at 22:10
  • $\begingroup$ So basically whether a structure of this sort is a vector space or a module depends on whether its scalars form a ring or a field. So it's largely a definition thing, then? Vector space have scalar fields, and modules have scalar rings. $\endgroup$
    – Geoffrey
    Jul 2, 2017 at 22:21
  • $\begingroup$ +1 for a great question, thanks for posting! $\endgroup$
    – Rax Adaam
    Nov 15, 2020 at 19:47

3 Answers 3


If you pick the scalars from a general ring instead of insisting on a field (in particular, $\mathbb Z$ is a ring), you get a structure known as a module rather than a vector space.

Modules behave like vector spaces in certain respects, but there are also points where they are not at all as well-behaved as vector spaces. For example, a module does not necessarily have a basis, or even a well-defined dimension. This makes matrices less useful for understanding modules than they are for vector spaces. (You can still have matrices with entries in a ring; they just don't tell you everything about linear maps between the modules anymore).

  • $\begingroup$ So, just to be clear, what I'm hearing is that if it has the right properties and its scalars form a Field, then it is called a vector space. But if it has basically the same properties only its scalars form a Ring, then it is called a module? Do I have that right? $\endgroup$
    – Geoffrey
    Jul 2, 2017 at 22:15
  • $\begingroup$ @Geoffrey: Correct -- the axioms are identical. $\endgroup$ Jul 2, 2017 at 22:20
  • $\begingroup$ @Geoffrey basically, if you it's not a field, theorems about vector spaces don't apply $\endgroup$
    – PyRulez
    Jul 3, 2017 at 12:52

These things are studied: they are called modules over the ring instead of vector spaces.

The main difference is that the elements of general modules do not allow a lot of the geometric intuition we have for vector spaces, so we still retain the traditional term "vector space" because it is still a useful term.

So, modules over fields (and also noncommutative fields) are called vector spaces.


While the other answers (and comments) implicitly address the question stated in the title of the OP, I thought it may be useful to include an explicit answer, as well.

Does the “field” over which a vector space is defined have to be a Field?

Yes, a vector space is defined over a field; i.e. if the scalars do not refer to a field, the resulting object is not by definition a vector space.

For completeness: as pointed out in the other answers and comments, there are objects with analogous definitions, in the case that the scalars belong to a ring (as in the example provided in the OP), and the other axioms are met, the resulting object is called a "module." As the subsection on modules on the Wikipedia vector space page says (emphasis added):

Modules are to rings what vector spaces are to fields: the same axioms, applied to a ring R instead of a field F, yield modules. The theory of modules, compared to that of vector spaces, is complicated by the presence of ring elements that do not have multiplicative inverses. For example, modules need not have bases, as the Z-module (that is, abelian group) Z/2Z shows; those modules that do (including all vector spaces) are known as free modules. Nevertheless, a vector space can be compactly defined as a module over a ring which is a field, with the elements being called vectors. Some authors use the term vector space to mean modules over a division ring.[105] The algebro-geometric interpretation of commutative rings via their spectrum allows the development of concepts such as locally free modules, the algebraic counterpart to vector bundles.


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