Do commutative rings without unity have the IBN property?
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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.
ac.commutative-algebra
asked 2 hours ago
M.G.
2,67522537
add a comment |Â
up vote
1
down vote
favorite
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.
ac.commutative-algebra
asked 2 hours ago
M.G.
2,67522537
1
Simply adjoint a unit to $R$ to obtain $R_1$ (e.g., $R_1 = R[x]/(r x - r : r in R)$). Then an isomorphism $R^m cong R^n$ automatically extends to an isomorphism $R_1^m cong R_1^n$, and you can use the IBN property for $R_1$.
– LSpice
2 hours ago
5
The answer is "not necessarily". For a counterexample, let $R = oplus^omega mathbb Z$ with zero multiplication.
– Keith Kearnes
2 hours ago
3
It should be pointed out that $R^n$ doesn't have a basis either when $R$ is non-unital.
– Benjamin Steinberg
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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.
ac.commutative-algebra
asked 2 hours ago
M.G.
2,67522537
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.
ac.commutative-algebra
ac.commutative-algebra
asked 2 hours ago
M.G.
2,67522537
asked 2 hours ago
M.G.
2,67522537
edited 2 hours ago
asked 2 hours ago
M.G.
2,67522537
asked 2 hours ago
M.G.
2,67522537
asked 2 hours ago
M.G.
2,67522537
2,67522537
1
Simply adjoint a unit to $R$ to obtain $R_1$ (e.g., $R_1 = R[x]/(r x - r : r in R)$). Then an isomorphism $R^m cong R^n$ automatically extends to an isomorphism $R_1^m cong R_1^n$, and you can use the IBN property for $R_1$.
– LSpice
2 hours ago
5
The answer is "not necessarily". For a counterexample, let $R = oplus^omega mathbb Z$ with zero multiplication.
– Keith Kearnes
2 hours ago
3
It should be pointed out that $R^n$ doesn't have a basis either when $R$ is non-unital.
– Benjamin Steinberg
1 hour ago
add a comment |Â
1
Simply adjoint a unit to $R$ to obtain $R_1$ (e.g., $R_1 = R[x]/(r x - r : r in R)$). Then an isomorphism $R^m cong R^n$ automatically extends to an isomorphism $R_1^m cong R_1^n$, and you can use the IBN property for $R_1$.
– LSpice
2 hours ago
5
The answer is "not necessarily". For a counterexample, let $R = oplus^omega mathbb Z$ with zero multiplication.
– Keith Kearnes
2 hours ago
3
It should be pointed out that $R^n$ doesn't have a basis either when $R$ is non-unital.
– Benjamin Steinberg
1 hour ago
1
1
Simply adjoint a unit to $R$ to obtain $R_1$ (e.g., $R_1 = R[x]/(r x - r : r in R)$). Then an isomorphism $R^m cong R^n$ automatically extends to an isomorphism $R_1^m cong R_1^n$, and you can use the IBN property for $R_1$.
– LSpice
2 hours ago
Simply adjoint a unit to $R$ to obtain $R_1$ (e.g., $R_1 = R[x]/(r x - r : r in R)$). Then an isomorphism $R^m cong R^n$ automatically extends to an isomorphism $R_1^m cong R_1^n$, and you can use the IBN property for $R_1$.
– LSpice
2 hours ago
5
5
The answer is "not necessarily". For a counterexample, let $R = oplus^omega mathbb Z$ with zero multiplication.
– Keith Kearnes
2 hours ago
The answer is "not necessarily". For a counterexample, let $R = oplus^omega mathbb Z$ with zero multiplication.
– Keith Kearnes
2 hours ago
3
3
It should be pointed out that $R^n$ doesn't have a basis either when $R$ is non-unital.
– Benjamin Steinberg
1 hour ago
It should be pointed out that $R^n$ doesn't have a basis either when $R$ is non-unital.
– Benjamin Steinberg
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
(I will write my comment as an answer.)
The answer is "not necessarily". For a counterexample, let $R=oplus^omega mathbb Z$
with zero multiplication.
However let me add this: when $R$ is nonunital, $R^n$ is not the $n$-generated free $R$-module.
answered 26 mins ago
Keith Kearnes
5,65412742
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
(I will write my comment as an answer.)
The answer is "not necessarily". For a counterexample, let $R=oplus^omega mathbb Z$
with zero multiplication.
However let me add this: when $R$ is nonunital, $R^n$ is not the $n$-generated free $R$-module.
answered 26 mins ago
Keith Kearnes
5,65412742
add a comment |Â
up vote
4
down vote
(I will write my comment as an answer.)
The answer is "not necessarily". For a counterexample, let $R=oplus^omega mathbb Z$
with zero multiplication.
However let me add this: when $R$ is nonunital, $R^n$ is not the $n$-generated free $R$-module.
answered 26 mins ago
Keith Kearnes
5,65412742
add a comment |Â
up vote
4
down vote
up vote
4
down vote
(I will write my comment as an answer.)
The answer is "not necessarily". For a counterexample, let $R=oplus^omega mathbb Z$
with zero multiplication.
However let me add this: when $R$ is nonunital, $R^n$ is not the $n$-generated free $R$-module.
answered 26 mins ago
Keith Kearnes
5,65412742
(I will write my comment as an answer.)
The answer is "not necessarily". For a counterexample, let $R=oplus^omega mathbb Z$
with zero multiplication.
However let me add this: when $R$ is nonunital, $R^n$ is not the $n$-generated free $R$-module.
answered 26 mins ago
Keith Kearnes
5,65412742
edited 20 mins ago
answered 26 mins ago
Keith Kearnes
5,65412742
answered 26 mins ago
Keith Kearnes
5,65412742
answered 26 mins ago
Keith Kearnes
5,65412742
5,65412742
add a comment |Â
add a comment |Â
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1
Simply adjoint a unit to $R$ to obtain $R_1$ (e.g., $R_1 = R[x]/(r x - r : r in R)$). Then an isomorphism $R^m cong R^n$ automatically extends to an isomorphism $R_1^m cong R_1^n$, and you can use the IBN property for $R_1$.
– LSpice
2 hours ago
5
The answer is "not necessarily". For a counterexample, let $R = oplus^omega mathbb Z$ with zero multiplication.
– Keith Kearnes
2 hours ago
3
It should be pointed out that $R^n$ doesn't have a basis either when $R$ is non-unital.
– Benjamin Steinberg
1 hour ago