What is a Module? Polynomials & Vector Spaces

[marellasunny]

First,could you please clear this doubt-
A polynomial with rational coefficients does not form a vector space over the real numbers.So,will this set of polynomials be called a module instead of a vector space?

Is my understanding correct?Any thing else I should know about modules?
Thanks

 

[micromass]

No, it will not be called a module. I don't think it will be called anything, since the scalar multiplication is probably not well-defined.

In fact, over the real numbers, a module is exactly the same as a vector space. More generally, a module over a field is exactly the same as a vector space.

Thus the notion of module is only interesting for things that are not fields. What is a module?? Well, if we have a ring (with unity, non necessarily commutative) R, then a left module M is an abelian group equipped with a "scalar multiplication"

\cdot:R\times M\rightarrow M

which satisfies the usual properties:

(r+r^\prime)\cdot m=r\cdot m + r^\prime \cdot m
r\cdot (m+m^\prime)=r\cdot m+r\cdot m^\prime
1\cdot m=m

A right module is essentially the same thing, except that the scalar multiplication is now from the right. Thus we have a map
\cdot: M\times R\rightarrow M

 

[algebrat]

First,could you please clear this doubt-
A polynomial with rational coefficients does not form a vector space over the real numbers.So,will this set of polynomials be called a module instead of a vector space?

Is my understanding correct?Any thing else I should know about modules?
Thanks

Yeah both sets of coefficients are fields, so I think if anything is a module it is a vector space. So let's assume your question is about whether or not it is a vector space over the reals.

So I like to think of vector spaces and modules as two sets, which I like to call the vectors and the scalars. So the set of vectors must be an abelian group. Sure, the set of polynomials with rational coefficients is an abelian group. However, there is a problem with your scalars, because multiplying by an inconvenient scalar (real number; in particular an irrational) will pop you out of your abelian group. So it is not a module, which by the way is pretty much the simplest two set "algebra" considered in "abstract algebra".

Keep in mind their is an "algebra" named "algebra", which is also an algebra with two sets, scalars and vector like I mentioned. An algebra is basically a module where the vectors have a multiplication operation. So groups and rings have one set, while modules, vector spaces and "algebras" have two sets (which I call scalars and vectors). Each set has a different number of operations defined on it, check out the definitions for your self.

So the subject of algebra is the study of various algebras, an example of which is an algebra, a vector space where the vectors have a ring structure themselves.

(So the word algebra is used on three different levels here.)

 

[lavinia]

First,could you please clear this doubt-
A polynomial with rational coefficients does not form a vector space over the real numbers.So,will this set of polynomials be called a module instead of a vector space?

Is my understanding correct?Any thing else I should know about modules?
Thanks

A module over a field is a vector space. This may be taken as the definition of a vector space.

The polynomials with rational coefficients are not a vector space over the reals - with usual multiplication - because a real number times a polynomial with rational coefficients may not be a polynomial with rational coefficients.

If there were some non-standard multiplication - which I do not think there is - at least not one that extends usual multiplication by rationals - then the polynomials with rational coefficients would have the cardinality of the reals which they do not. So there is no possible way to make them into a vector space over the reals,

 

[homeomorphic]

A good example of a module would be if you did linear algebra with integers as the scalars, instead of the real numbers--so, here you have modules taking the place of vector spaces. The scalars no longer form a field because there are no multiplicative inverses.