Rings with 'non-harmless' zero-divisors
$begingroup$
The following excerpt is from pp. 246–247 of Paolo Aluffi's Algebra: Chapter 0:
1.2. Prime and irreducible elements. Let $R$ be a (commutative) ring [with $1$], and let $a,bin R$. We say that $a$ divides $b$, or that $a$ is a divisor of $b$, or that $b$ is a multiple of $a$, if $bin(a)$, that is
$$
(exists cin R), quad b = ac.
$$
We use the notation $a mid b$.
Two elements $a,b$ are associates if $(a) = (b)$, that is, if $amid b$ and $bmid a$.
Lemma 1.5. Let $a,b$ be nonzero elements of an integral domain $R$. Then $a$ and $b$ are associates if and only if $a = ub$, for $u$ a unit in $R$.
[Proof omitted.]
Incidentally, here the reader sees why it is convenient to restrict our attention to integral domains. This argument really shows that if $(a) = (b) ne (0)$ in an integral domain, and $b = ca$, then $c$ is necessarily a unit. Away from the comfortable environment of integral domains, even such harmless-looking statements may fail: in $Bbb Z/6Bbb Z$, the classes $[2]_6,[4]_6$ of $2$ and $4$ are associates according to our definition, and $[4]_6 = [2]_6cdot[2]_6$, yet $[2]_6$ is not a unit. However, $[4]_6 = [5]_6cdot [2]_6$ and $[5]_6$ is a unit, so this is not a counterexample to Lemma 1.5. In fact, Lemma 1.5 may fail over rings with 'non-harmless' zero-divisors (yes, there is such a notion) [emphasis added].
Since at this point, Aluffi does not say what such rings are called, I was hoping someone might know what type of rings Aluffi is referring to. (And hopefully provide a little context as to why they are interesting!)
abstract-algebra ring-theory
asked Jan 7 at 1:44
AOrtizAOrtiz
10.5k21341
$endgroup$
add a comment |
$begingroup$
The following excerpt is from pp. 246–247 of Paolo Aluffi's Algebra: Chapter 0:
1.2. Prime and irreducible elements. Let $R$ be a (commutative) ring [with $1$], and let $a,bin R$. We say that $a$ divides $b$, or that $a$ is a divisor of $b$, or that $b$ is a multiple of $a$, if $bin(a)$, that is
$$
(exists cin R), quad b = ac.
$$
We use the notation $a mid b$.
Two elements $a,b$ are associates if $(a) = (b)$, that is, if $amid b$ and $bmid a$.
Lemma 1.5. Let $a,b$ be nonzero elements of an integral domain $R$. Then $a$ and $b$ are associates if and only if $a = ub$, for $u$ a unit in $R$.
[Proof omitted.]
Incidentally, here the reader sees why it is convenient to restrict our attention to integral domains. This argument really shows that if $(a) = (b) ne (0)$ in an integral domain, and $b = ca$, then $c$ is necessarily a unit. Away from the comfortable environment of integral domains, even such harmless-looking statements may fail: in $Bbb Z/6Bbb Z$, the classes $[2]_6,[4]_6$ of $2$ and $4$ are associates according to our definition, and $[4]_6 = [2]_6cdot[2]_6$, yet $[2]_6$ is not a unit. However, $[4]_6 = [5]_6cdot [2]_6$ and $[5]_6$ is a unit, so this is not a counterexample to Lemma 1.5. In fact, Lemma 1.5 may fail over rings with 'non-harmless' zero-divisors (yes, there is such a notion) [emphasis added].
Since at this point, Aluffi does not say what such rings are called, I was hoping someone might know what type of rings Aluffi is referring to. (And hopefully provide a little context as to why they are interesting!)
abstract-algebra ring-theory
asked Jan 7 at 1:44
AOrtizAOrtiz
10.5k21341
$endgroup$
$begingroup$
maybe there is a misprint. it should read "if $(a)=(b) ne 0$ and $a = bc$ then $c$ is a unit.
$endgroup$
– David Holden
Jan 7 at 2:09
$begingroup$
@DavidHolden: Thank you for the catch.
$endgroup$
– AOrtiz
Jan 7 at 2:10
1
$begingroup$
You may find helpful the papers I cite here which discuss generalizations of "associate" and related notions to non-domains.
$endgroup$
– Bill Dubuque
Jan 7 at 2:12
3
$begingroup$
You can find definitions of "harmless" zero-divisors here and here
$endgroup$
– Bill Dubuque
Jan 7 at 2:15
$begingroup$
A zero-divisor $z$ is a harmless zero-divisor if $1-z$ is a unit.
$endgroup$
– lhf
Jan 7 at 10:34
add a comment |
$begingroup$
The following excerpt is from pp. 246–247 of Paolo Aluffi's Algebra: Chapter 0:
1.2. Prime and irreducible elements. Let $R$ be a (commutative) ring [with $1$], and let $a,bin R$. We say that $a$ divides $b$, or that $a$ is a divisor of $b$, or that $b$ is a multiple of $a$, if $bin(a)$, that is
$$
(exists cin R), quad b = ac.
$$
We use the notation $a mid b$.
Two elements $a,b$ are associates if $(a) = (b)$, that is, if $amid b$ and $bmid a$.
Lemma 1.5. Let $a,b$ be nonzero elements of an integral domain $R$. Then $a$ and $b$ are associates if and only if $a = ub$, for $u$ a unit in $R$.
[Proof omitted.]
Incidentally, here the reader sees why it is convenient to restrict our attention to integral domains. This argument really shows that if $(a) = (b) ne (0)$ in an integral domain, and $b = ca$, then $c$ is necessarily a unit. Away from the comfortable environment of integral domains, even such harmless-looking statements may fail: in $Bbb Z/6Bbb Z$, the classes $[2]_6,[4]_6$ of $2$ and $4$ are associates according to our definition, and $[4]_6 = [2]_6cdot[2]_6$, yet $[2]_6$ is not a unit. However, $[4]_6 = [5]_6cdot [2]_6$ and $[5]_6$ is a unit, so this is not a counterexample to Lemma 1.5. In fact, Lemma 1.5 may fail over rings with 'non-harmless' zero-divisors (yes, there is such a notion) [emphasis added].
Since at this point, Aluffi does not say what such rings are called, I was hoping someone might know what type of rings Aluffi is referring to. (And hopefully provide a little context as to why they are interesting!)
abstract-algebra ring-theory
asked Jan 7 at 1:44
AOrtizAOrtiz
10.5k21341
$endgroup$
The following excerpt is from pp. 246–247 of Paolo Aluffi's Algebra: Chapter 0:
1.2. Prime and irreducible elements. Let $R$ be a (commutative) ring [with $1$], and let $a,bin R$. We say that $a$ divides $b$, or that $a$ is a divisor of $b$, or that $b$ is a multiple of $a$, if $bin(a)$, that is
$$
(exists cin R), quad b = ac.
$$
We use the notation $a mid b$.
Two elements $a,b$ are associates if $(a) = (b)$, that is, if $amid b$ and $bmid a$.
Lemma 1.5. Let $a,b$ be nonzero elements of an integral domain $R$. Then $a$ and $b$ are associates if and only if $a = ub$, for $u$ a unit in $R$.
[Proof omitted.]
Incidentally, here the reader sees why it is convenient to restrict our attention to integral domains. This argument really shows that if $(a) = (b) ne (0)$ in an integral domain, and $b = ca$, then $c$ is necessarily a unit. Away from the comfortable environment of integral domains, even such harmless-looking statements may fail: in $Bbb Z/6Bbb Z$, the classes $[2]_6,[4]_6$ of $2$ and $4$ are associates according to our definition, and $[4]_6 = [2]_6cdot[2]_6$, yet $[2]_6$ is not a unit. However, $[4]_6 = [5]_6cdot [2]_6$ and $[5]_6$ is a unit, so this is not a counterexample to Lemma 1.5. In fact, Lemma 1.5 may fail over rings with 'non-harmless' zero-divisors (yes, there is such a notion) [emphasis added].
Since at this point, Aluffi does not say what such rings are called, I was hoping someone might know what type of rings Aluffi is referring to. (And hopefully provide a little context as to why they are interesting!)
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Jan 7 at 1:44
AOrtizAOrtiz
10.5k21341
asked Jan 7 at 1:44
AOrtizAOrtiz
10.5k21341
edited Jan 7 at 2:09
AOrtiz
asked Jan 7 at 1:44
AOrtizAOrtiz
10.5k21341
asked Jan 7 at 1:44
AOrtizAOrtiz
10.5k21341
asked Jan 7 at 1:44
AOrtizAOrtiz
10.5k21341
10.5k21341
$begingroup$
maybe there is a misprint. it should read "if $(a)=(b) ne 0$ and $a = bc$ then $c$ is a unit.
$endgroup$
– David Holden
Jan 7 at 2:09
$begingroup$
@DavidHolden: Thank you for the catch.
$endgroup$
– AOrtiz
Jan 7 at 2:10
1
$begingroup$
You may find helpful the papers I cite here which discuss generalizations of "associate" and related notions to non-domains.
$endgroup$
– Bill Dubuque
Jan 7 at 2:12
3
$begingroup$
You can find definitions of "harmless" zero-divisors here and here
$endgroup$
– Bill Dubuque
Jan 7 at 2:15
$begingroup$
A zero-divisor $z$ is a harmless zero-divisor if $1-z$ is a unit.
$endgroup$
– lhf
Jan 7 at 10:34
add a comment |
$begingroup$
maybe there is a misprint. it should read "if $(a)=(b) ne 0$ and $a = bc$ then $c$ is a unit.
$endgroup$
– David Holden
Jan 7 at 2:09
$begingroup$
@DavidHolden: Thank you for the catch.
$endgroup$
– AOrtiz
Jan 7 at 2:10
1
$begingroup$
You may find helpful the papers I cite here which discuss generalizations of "associate" and related notions to non-domains.
$endgroup$
– Bill Dubuque
Jan 7 at 2:12
3
$begingroup$
You can find definitions of "harmless" zero-divisors here and here
$endgroup$
– Bill Dubuque
Jan 7 at 2:15
$begingroup$
A zero-divisor $z$ is a harmless zero-divisor if $1-z$ is a unit.
$endgroup$
– lhf
Jan 7 at 10:34
$begingroup$
maybe there is a misprint. it should read "if $(a)=(b) ne 0$ and $a = bc$ then $c$ is a unit.
$endgroup$
– David Holden
Jan 7 at 2:09
$begingroup$
maybe there is a misprint. it should read "if $(a)=(b) ne 0$ and $a = bc$ then $c$ is a unit.
$endgroup$
– David Holden
Jan 7 at 2:09
$begingroup$
@DavidHolden: Thank you for the catch.
$endgroup$
– AOrtiz
Jan 7 at 2:10
$begingroup$
@DavidHolden: Thank you for the catch.
$endgroup$
– AOrtiz
Jan 7 at 2:10
1
1
$begingroup$
You may find helpful the papers I cite here which discuss generalizations of "associate" and related notions to non-domains.
$endgroup$
– Bill Dubuque
Jan 7 at 2:12
$begingroup$
You may find helpful the papers I cite here which discuss generalizations of "associate" and related notions to non-domains.
$endgroup$
– Bill Dubuque
Jan 7 at 2:12
3
3
$begingroup$
You can find definitions of "harmless" zero-divisors here and here
$endgroup$
– Bill Dubuque
Jan 7 at 2:15
$begingroup$
You can find definitions of "harmless" zero-divisors here and here
$endgroup$
– Bill Dubuque
Jan 7 at 2:15
$begingroup$
A zero-divisor $z$ is a harmless zero-divisor if $1-z$ is a unit.
$endgroup$
– lhf
Jan 7 at 10:34
$begingroup$
A zero-divisor $z$ is a harmless zero-divisor if $1-z$ is a unit.
$endgroup$
– lhf
Jan 7 at 10:34
add a comment |
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$begingroup$
maybe there is a misprint. it should read "if $(a)=(b) ne 0$ and $a = bc$ then $c$ is a unit.
$endgroup$
– David Holden
Jan 7 at 2:09
$begingroup$
@DavidHolden: Thank you for the catch.
$endgroup$
– AOrtiz
Jan 7 at 2:10
1
$begingroup$
You may find helpful the papers I cite here which discuss generalizations of "associate" and related notions to non-domains.
$endgroup$
– Bill Dubuque
Jan 7 at 2:12
3
$begingroup$
You can find definitions of "harmless" zero-divisors here and here
$endgroup$
– Bill Dubuque
Jan 7 at 2:15
$begingroup$
A zero-divisor $z$ is a harmless zero-divisor if $1-z$ is a unit.
$endgroup$
– lhf
Jan 7 at 10:34