Re: Free modules over rings without unity ?

From: Hagen <knaf@xxxxxxxxxxx>
Date: Wed, 28 Jan 2009 15:07:28 EST

Suppose R is a ring with*out* unity (1), and
consider the direct sum of finitely many copies of R,
i.e. R^n.

Is R^n a free R-module?

In this case, I don't see what a natural choice of
basis would be (even when n = 1)!

What if we suppose that R is, say, an integral
domain?

Are there *any* rings without unity for which R^n
*is* free?

Yes. You should take a look at the free-ideal-rings.

H
.

References:
- Free modules over rings without unity ?
  - From: qsymmetry

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