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Free R-modules...

by Artusartos
Tags: free, rmodules
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Artusartos
#1
Jan13-13, 01:35 PM
P: 247
From my textbook:

A free R-module is "A left R-module F is called a free left R-module if F is isomorphic to a direct sum of copies of R..."

I know that another definition of an R-module a module with a basis...but I don't know how to connect that definition with this one. Also, what does "copies of R" mean?

Thanks in advance
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algebrat
#2
Jan21-13, 02:51 AM
P: 428
[itex][/itex]
Quote Quote by Artusartos View Post
From my textbook:

A free R-module is "A left R-module F is called a free left R-module if F is isomorphic to a direct sum of copies of R..."
Can we think of it as
$$F\cong\prod_{\alpha\in J}R_\alpha$$
This is the underlying abelian group (analogous to vectors in vector space), and it looks like there is a natural way to multiply on the left by elements of R (analogous to scalars in a vector space).

I know that another definition of an R-module a module with a basis...but I don't know how to connect that definition with this one.
For F above, a basis could be elements like (1,0,...,0), (0,1,0,...,0) and so on. Notice each coordinate alone looks like R.

So that seems to suggest that a left R-module does indeed have a basis. Now let's consider if we think a left R-module with a basis is a free module. Uh, never mind, I'll leave that for someone else


Also, what does "copies of R" mean?
The product above consists of copies of R.


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