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A ring $R$ (with 1) has IBN property if free $R$-modules have unique rank (e.g., commutative rings). In the same fashion, lets call $R$ a good ring if in every free $R$-module any independent set can be extended to a basis (not every commutative ring has this property). I have two questions:

  1. Is the class of good rings well-known?

  2. Is it true that every Noetherian ring has IBN property ?

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closed as unclear what you're asking by David White, Andrés Caicedo, Ryan Budney, Todd Trimble, Eric Naslund Sep 2 '13 at 3:23

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2. It is well-known that every non-zero left noetherian ring has IBN (Lam's Lectures on modules and rings, Prop. 1.8 and Prop. 1.13) . But how is this connected to your class of good rings? –  Martin Brandenburg Aug 27 '13 at 10:01
The answer must be in Lam, as Martin suggests. Certainly that answers (2), and Lam will give much more general conditions to get IBN. As for (1), what do you mean by "known-known?" –  David White Aug 27 '13 at 17:51
@DavidWhite that was a mistyping. I meant well-known. And as Matrin suggested (2) is the Book. But I am sure that there is nothing in Lam's book concerning to what I am calling good rings. –  user39121 Aug 29 '13 at 18:16
@MartinBrandenburg I think the class of good rings must be contained in the class of rings with IBN property. –  user39121 Aug 29 '13 at 18:18

1 Answer 1

up vote 4 down vote accepted

The Rings you are calling good, are called (left or right) Steinitz. They are characterized as: A ring is (left) right Steinitz if and only if it is (left) right perfect and local. This means that every (left) right Steinitz ring has IBN property. You may like to see the following

[1] Chew, Neggers, On the extension of linearly independent subsets of free modules to bases, Proc. Amer. Math. Soc. 24(1970), 466-470.

[2] Brodskii, Steinitz Modules, Mat. Issled. 7(1972), 14-28, 284.

For (2) as Martin said you can see Lam's Book.

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