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M.G.
M.G.
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Let $R$ be a commutative rng, i.e. a commutative ring without an identity element.

Does $R$ still have the Invariant Basis Number (IBN) property?

Recall that a ring is said to have the IBN property if $R^m \cong R^n \Rightarrow m=n$.

All commutative rings have the IBN property, but the standard proof I know makes an essential use of the existence of a maximal ideal by Zorn's Lemma and passing to the residue field, which depends on the presence of a unit.

Let $R$ be a commutative rng, i.e. a commutative ring without an identity element.

Does $R$ still have the Invariant Basis Number (IBN) property?

Recall that a ring is said to have the IBN property if $R^m \cong R^n \Rightarrow m=n$.

All commutative rings have the IBN property, but the standard proof I know makes an essential use of the existence of a maximal ideal by Zorn's Lemma, which depends on the presence of a unit.

Let $R$ be a commutative rng, i.e. a commutative ring without an identity element.

Does $R$ still have the Invariant Basis Number (IBN) property?

Recall that a ring is said to have the IBN property if $R^m \cong R^n \Rightarrow m=n$.

All commutative rings have the IBN property, but the standard proof I know makes an essential use of the existence of a maximal ideal by Zorn's Lemma and passing to the residue field, which depends on the presence of a unit.

Source Link
M.G.
M.G.
  • 7.1k
  • 3
  • 46
  • 60

Do commutative rings without unity have the IBN property?

Let $R$ be a commutative rng, i.e. a commutative ring without an identity element.

Does $R$ still have the Invariant Basis Number (IBN) property?

Recall that a ring is said to have the IBN property if $R^m \cong R^n \Rightarrow m=n$.

All commutative rings have the IBN property, but the standard proof I know makes an essential use of the existence of a maximal ideal by Zorn's Lemma, which depends on the presence of a unit.