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Is it possible to have an Abelian group under two different binary operations but the binary operations are not distributive?

Are there broad or powerful theorems of rings that do not involve the familiar numerical operations (+) and (*) in some fundamental way?Are the axioms for abelian group theory independent?A question about groups: may I substitute a binary operation with a function?Question about the definition of a field…Is a ring closed under both operations?Non-commutative or commutative ring or subring with $x^2 = 0$Can a group have a subset that is stable under all automorphisms, but not under inverse?Using a particular definition of a field to argue that $0$ commutes with the other elements in the field under multiplicationExample of a communtative ring with two operations where the identity elements are not distinct?In abstract algebra, what is an intuitive explanation for a field?

2

1

$begingroup$

I am trying to show that if $(R, +)$ is an Abelian group and $(R - {0_R}, cdot)$ is an Abelian group, then $(R, +, cdot)$ is not necessarily a field. Note that $0_R$ is the identity element of $(R, +)$. I know that a field is a commutative division ring and one of a ring's properties is that $forall a,b in R, ~ acdot (b + c) = a cdot b + acdot c$. Therefore, I am trying to come up with a set and two binary operations that satisfy the first property, but together do not form a field.

So far, I have come up with a group over polynomials with $+$ being normal addition and $cdot$ being composition, but then $(R - {0_R})$ is not commutative. I would appreciate any help/guidance.

Thanks.

group-theory ring-theory field-theory

share|cite|improve this question

asked 4 hours ago

sepehr78sepehr78

675

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Composition isn't invertible either.
  $endgroup$
  – jgon
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Let $R$ be any six-element set, and put any Abelian group structures you like on $R$ and $R-{0}$.
  $endgroup$
  – Lord Shark the Unknown
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're working too hard. Just literally take any random abelian group structures at all and they almost certainly will not be distributive.
  $endgroup$
  – Eric Wofsey
  15 mins ago
  

  
  
  
  

add a comment | 

2

1

$begingroup$

I am trying to show that if $(R, +)$ is an Abelian group and $(R - {0_R}, cdot)$ is an Abelian group, then $(R, +, cdot)$ is not necessarily a field. Note that $0_R$ is the identity element of $(R, +)$. I know that a field is a commutative division ring and one of a ring's properties is that $forall a,b in R, ~ acdot (b + c) = a cdot b + acdot c$. Therefore, I am trying to come up with a set and two binary operations that satisfy the first property, but together do not form a field.

So far, I have come up with a group over polynomials with $+$ being normal addition and $cdot$ being composition, but then $(R - {0_R})$ is not commutative. I would appreciate any help/guidance.

Thanks.

group-theory ring-theory field-theory

share|cite|improve this question

asked 4 hours ago

sepehr78sepehr78

675

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Composition isn't invertible either.
  $endgroup$
  – jgon
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Let $R$ be any six-element set, and put any Abelian group structures you like on $R$ and $R-{0}$.
  $endgroup$
  – Lord Shark the Unknown
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're working too hard. Just literally take any random abelian group structures at all and they almost certainly will not be distributive.
  $endgroup$
  – Eric Wofsey
  15 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

I am trying to show that if $(R, +)$ is an Abelian group and $(R - {0_R}, cdot)$ is an Abelian group, then $(R, +, cdot)$ is not necessarily a field. Note that $0_R$ is the identity element of $(R, +)$. I know that a field is a commutative division ring and one of a ring's properties is that $forall a,b in R, ~ acdot (b + c) = a cdot b + acdot c$. Therefore, I am trying to come up with a set and two binary operations that satisfy the first property, but together do not form a field.

So far, I have come up with a group over polynomials with $+$ being normal addition and $cdot$ being composition, but then $(R - {0_R})$ is not commutative. I would appreciate any help/guidance.

Thanks.

group-theory ring-theory field-theory

share|cite|improve this question

asked 4 hours ago

sepehr78sepehr78

675

$endgroup$

share|cite|improve this question

asked 4 hours ago

sepehr78sepehr78

675

share|cite|improve this question

asked 4 hours ago

sepehr78sepehr78

675

asked 4 hours ago

sepehr78sepehr78

675

asked 4 hours ago

sepehr78sepehr78

675

2 Answers
2

active

oldest

votes

3

$begingroup$

Here is a concrete example, inspired by LStU:

The set is ${0,1,2,3,4,5}$. Addition is just addition mod $6$.

Multiplication is defined by
$$
acdot b = left{ begin{array}{cl} 0& a=0 \ 0 & b=0 \
1 & a = b= 5 \
5 & a=5 wedge b in [1,4]\
5 & b=5 wedge a in [1,4]\
ab pmod{5}& mbox{otherwise}end{array} right.
$$
or as a table
$$
begin{array}{c|cccccc} cdot&0&1&2&3&4&5 \ hline
0 & 0&0&0&0&0&0 \
1 & 0&1&2&3&4&5 \
2 & 0&2&4&1&3&5 \
3 & 0&3&1&4&2&5 \
4 & 0&4&3&2&1&5 \
5 & 5&5&5&5&5&1
end{array}
$$
The group properties, as well as commutativity, are easily checked.

Now consider $$ (1+4)cdot 5 = 5cdot 5 = 1 \
1cdot 5 + 4 cdot 5 = 5+5 = 4 neq 1
$$

share|cite|improve this answer

answered 4 hours ago

Mark FischlerMark Fischler

33.4k12452

$endgroup$

add a comment | 

4

$begingroup$

As @LordSharktheUnknown implicitly points out, if you just take a finite set with non-prime-power order (six is the first such integer $ge 2$) and put any group structures you want, it will have to work, because finite fields have prime-power order.

But just to be clear, you can also do it with infinite sets. Pretty much anything you try will work, provided you let loose a bit. Take $R = mathbb{Z}$, with $+$ being regular addition. Let $S = mathbb{Z}setminus {0}$, let $phi:S to R$ be the bijection which shifts negative numbers up by one and is constant on positive numbers. Now define $acdot b = phi^{-1}(phi(a)+phi(b))$. We're just relabeling $S$ to be $mathbb{Z}$ again and then doing regular addition. Now, doing addition first, we have
$$-2cdot(1+1) = -2cdot 2 = phi^{-1}(-1+2) = 1,$$
but distributing first, we have
$$-2cdot(1+1) = -2cdot 1 + -2cdot 1 = phi^{-1}(-1+1) + phi^{-1}(-1+1) = -1+(-1) = -2.$$

In terms of guidance, you should expect that you'll need to do something perverse like this, because most of the examples you'll think of where two binary operations already exist are rings, where distributivity necessarily holds.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

cspruncsprun

2,00829

$endgroup$

add a comment | 

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3

$begingroup$

Here is a concrete example, inspired by LStU:

The set is ${0,1,2,3,4,5}$. Addition is just addition mod $6$.

Multiplication is defined by
$$
acdot b = left{ begin{array}{cl} 0& a=0 \ 0 & b=0 \
1 & a = b= 5 \
5 & a=5 wedge b in [1,4]\
5 & b=5 wedge a in [1,4]\
ab pmod{5}& mbox{otherwise}end{array} right.
$$
or as a table
$$
begin{array}{c|cccccc} cdot&0&1&2&3&4&5 \ hline
0 & 0&0&0&0&0&0 \
1 & 0&1&2&3&4&5 \
2 & 0&2&4&1&3&5 \
3 & 0&3&1&4&2&5 \
4 & 0&4&3&2&1&5 \
5 & 5&5&5&5&5&1
end{array}
$$
The group properties, as well as commutativity, are easily checked.

Now consider $$ (1+4)cdot 5 = 5cdot 5 = 1 \
1cdot 5 + 4 cdot 5 = 5+5 = 4 neq 1
$$

share|cite|improve this answer

answered 4 hours ago

Mark FischlerMark Fischler

33.4k12452

$endgroup$

add a comment | 

3

$begingroup$

Here is a concrete example, inspired by LStU:

The set is ${0,1,2,3,4,5}$. Addition is just addition mod $6$.

Multiplication is defined by
$$
acdot b = left{ begin{array}{cl} 0& a=0 \ 0 & b=0 \
1 & a = b= 5 \
5 & a=5 wedge b in [1,4]\
5 & b=5 wedge a in [1,4]\
ab pmod{5}& mbox{otherwise}end{array} right.
$$
or as a table
$$
begin{array}{c|cccccc} cdot&0&1&2&3&4&5 \ hline
0 & 0&0&0&0&0&0 \
1 & 0&1&2&3&4&5 \
2 & 0&2&4&1&3&5 \
3 & 0&3&1&4&2&5 \
4 & 0&4&3&2&1&5 \
5 & 5&5&5&5&5&1
end{array}
$$
The group properties, as well as commutativity, are easily checked.

Now consider $$ (1+4)cdot 5 = 5cdot 5 = 1 \
1cdot 5 + 4 cdot 5 = 5+5 = 4 neq 1
$$

share|cite|improve this answer

answered 4 hours ago

Mark FischlerMark Fischler

33.4k12452

$endgroup$

add a comment | 

$begingroup$

Here is a concrete example, inspired by LStU:

The set is ${0,1,2,3,4,5}$. Addition is just addition mod $6$.

Multiplication is defined by
$$
acdot b = left{ begin{array}{cl} 0& a=0 \ 0 & b=0 \
1 & a = b= 5 \
5 & a=5 wedge b in [1,4]\
5 & b=5 wedge a in [1,4]\
ab pmod{5}& mbox{otherwise}end{array} right.
$$
or as a table
$$
begin{array}{c|cccccc} cdot&0&1&2&3&4&5 \ hline
0 & 0&0&0&0&0&0 \
1 & 0&1&2&3&4&5 \
2 & 0&2&4&1&3&5 \
3 & 0&3&1&4&2&5 \
4 & 0&4&3&2&1&5 \
5 & 5&5&5&5&5&1
end{array}
$$
The group properties, as well as commutativity, are easily checked.

Now consider $$ (1+4)cdot 5 = 5cdot 5 = 1 \
1cdot 5 + 4 cdot 5 = 5+5 = 4 neq 1
$$

share|cite|improve this answer

answered 4 hours ago

Mark FischlerMark Fischler

33.4k12452

$endgroup$

share|cite|improve this answer

answered 4 hours ago

Mark FischlerMark Fischler

33.4k12452

answered 4 hours ago

Mark FischlerMark Fischler

33.4k12452

answered 4 hours ago

Mark FischlerMark Fischler

33.4k12452

4

$begingroup$

As @LordSharktheUnknown implicitly points out, if you just take a finite set with non-prime-power order (six is the first such integer $ge 2$) and put any group structures you want, it will have to work, because finite fields have prime-power order.

But just to be clear, you can also do it with infinite sets. Pretty much anything you try will work, provided you let loose a bit. Take $R = mathbb{Z}$, with $+$ being regular addition. Let $S = mathbb{Z}setminus {0}$, let $phi:S to R$ be the bijection which shifts negative numbers up by one and is constant on positive numbers. Now define $acdot b = phi^{-1}(phi(a)+phi(b))$. We're just relabeling $S$ to be $mathbb{Z}$ again and then doing regular addition. Now, doing addition first, we have
$$-2cdot(1+1) = -2cdot 2 = phi^{-1}(-1+2) = 1,$$
but distributing first, we have
$$-2cdot(1+1) = -2cdot 1 + -2cdot 1 = phi^{-1}(-1+1) + phi^{-1}(-1+1) = -1+(-1) = -2.$$

In terms of guidance, you should expect that you'll need to do something perverse like this, because most of the examples you'll think of where two binary operations already exist are rings, where distributivity necessarily holds.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

cspruncsprun

2,00829

$endgroup$

add a comment | 

4

$begingroup$

As @LordSharktheUnknown implicitly points out, if you just take a finite set with non-prime-power order (six is the first such integer $ge 2$) and put any group structures you want, it will have to work, because finite fields have prime-power order.

But just to be clear, you can also do it with infinite sets. Pretty much anything you try will work, provided you let loose a bit. Take $R = mathbb{Z}$, with $+$ being regular addition. Let $S = mathbb{Z}setminus {0}$, let $phi:S to R$ be the bijection which shifts negative numbers up by one and is constant on positive numbers. Now define $acdot b = phi^{-1}(phi(a)+phi(b))$. We're just relabeling $S$ to be $mathbb{Z}$ again and then doing regular addition. Now, doing addition first, we have
$$-2cdot(1+1) = -2cdot 2 = phi^{-1}(-1+2) = 1,$$
but distributing first, we have
$$-2cdot(1+1) = -2cdot 1 + -2cdot 1 = phi^{-1}(-1+1) + phi^{-1}(-1+1) = -1+(-1) = -2.$$

In terms of guidance, you should expect that you'll need to do something perverse like this, because most of the examples you'll think of where two binary operations already exist are rings, where distributivity necessarily holds.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

cspruncsprun

2,00829

$endgroup$

add a comment | 

$begingroup$

As @LordSharktheUnknown implicitly points out, if you just take a finite set with non-prime-power order (six is the first such integer $ge 2$) and put any group structures you want, it will have to work, because finite fields have prime-power order.

But just to be clear, you can also do it with infinite sets. Pretty much anything you try will work, provided you let loose a bit. Take $R = mathbb{Z}$, with $+$ being regular addition. Let $S = mathbb{Z}setminus {0}$, let $phi:S to R$ be the bijection which shifts negative numbers up by one and is constant on positive numbers. Now define $acdot b = phi^{-1}(phi(a)+phi(b))$. We're just relabeling $S$ to be $mathbb{Z}$ again and then doing regular addition. Now, doing addition first, we have
$$-2cdot(1+1) = -2cdot 2 = phi^{-1}(-1+2) = 1,$$
but distributing first, we have
$$-2cdot(1+1) = -2cdot 1 + -2cdot 1 = phi^{-1}(-1+1) + phi^{-1}(-1+1) = -1+(-1) = -2.$$

In terms of guidance, you should expect that you'll need to do something perverse like this, because most of the examples you'll think of where two binary operations already exist are rings, where distributivity necessarily holds.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

cspruncsprun

2,00829

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

cspruncsprun

2,00829

answered 4 hours ago

cspruncsprun

2,00829

answered 4 hours ago

cspruncsprun

2,00829