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Why is commutativity optional in multiplication for rings?

Why is ring addition commutative?Why can't the Polynomial Ring be a Field?Why Does Induction Prove Multiplication is Commutative?In a Commutative Ring, is Addition Necessarily Commutative?Is there a theory of “rings” with partially defined multiplication?Why do they need commutativity in the proof?What is a semiring without commutativity requirement for addition?Does commutativity of $+$ on $mathbb{N}$ imply the associativity of $+$ on $mathbb{N}$?PID which is not a Prüfer domainAbout the hypotheses for a ring to contain a maximal idealAre rings closed under addition and multiplication?

4

1

$begingroup$

More precisely, why is it that all rings are required by the axioms to have commutativity in addition, but are not held to the same axiom regarding multiplication? I know that we have commutative and non-commutative rings depending on whether or not they are commutative in multiplication, but I am wondering why it is that the axioms were defined that way, providing us with this option.

I am using this list of axioms, from David Sharpe’s Rings and factorization:

abstract-algebra ring-theory commutative-algebra

share|cite|improve this question

edited 2 hours ago

Eevee Trainer

7,05321337

asked 2 hours ago

KusaKusa

323

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For what it's worth, we also have the notion of a "commutative ring," one in which multiplication does commute. (en.wikipedia.org/wiki/Commutative_ring) Of course I get you're trying to get at why we don't require this, and I don't know how to answer you on that, but I figured it's a point worth bringing up.
  $endgroup$
  – Eevee Trainer
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EeveeTrainer Thank you! I dunno if you missed it, but I did express my awareness of that.
  $endgroup$
  – Kusa
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that matrices are an example where addition is commutative but multiplication is not necessarily
  $endgroup$
  – J. W. Tanner
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I did miss it, sorry about that Kusa. xD
  $endgroup$
  – Eevee Trainer
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The first rings that were considered were generally commutative, but it soon became apparent that rings with noncommutative multiplication were far too common, starting with matrices and, more generally, endomorphism rings of abelian groups. On the other hand, if you take the definition of ring with unity, but omit the condition that addition is commutative, it turns out that you can prove that the other conditions force commutativity of addition.
  $endgroup$
  – Arturo Magidin
  2 hours ago
  

  
  
  
  

 | 
show 4 more comments

4

1

$begingroup$

More precisely, why is it that all rings are required by the axioms to have commutativity in addition, but are not held to the same axiom regarding multiplication? I know that we have commutative and non-commutative rings depending on whether or not they are commutative in multiplication, but I am wondering why it is that the axioms were defined that way, providing us with this option.

I am using this list of axioms, from David Sharpe’s Rings and factorization:

abstract-algebra ring-theory commutative-algebra

share|cite|improve this question

edited 2 hours ago

Eevee Trainer

7,05321337

asked 2 hours ago

KusaKusa

323

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For what it's worth, we also have the notion of a "commutative ring," one in which multiplication does commute. (en.wikipedia.org/wiki/Commutative_ring) Of course I get you're trying to get at why we don't require this, and I don't know how to answer you on that, but I figured it's a point worth bringing up.
  $endgroup$
  – Eevee Trainer
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EeveeTrainer Thank you! I dunno if you missed it, but I did express my awareness of that.
  $endgroup$
  – Kusa
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that matrices are an example where addition is commutative but multiplication is not necessarily
  $endgroup$
  – J. W. Tanner
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I did miss it, sorry about that Kusa. xD
  $endgroup$
  – Eevee Trainer
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The first rings that were considered were generally commutative, but it soon became apparent that rings with noncommutative multiplication were far too common, starting with matrices and, more generally, endomorphism rings of abelian groups. On the other hand, if you take the definition of ring with unity, but omit the condition that addition is commutative, it turns out that you can prove that the other conditions force commutativity of addition.
  $endgroup$
  – Arturo Magidin
  2 hours ago
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

More precisely, why is it that all rings are required by the axioms to have commutativity in addition, but are not held to the same axiom regarding multiplication? I know that we have commutative and non-commutative rings depending on whether or not they are commutative in multiplication, but I am wondering why it is that the axioms were defined that way, providing us with this option.

I am using this list of axioms, from David Sharpe’s Rings and factorization:

abstract-algebra ring-theory commutative-algebra

share|cite|improve this question

edited 2 hours ago

Eevee Trainer

7,05321337

asked 2 hours ago

KusaKusa

323

$endgroup$

share|cite|improve this question

edited 2 hours ago

Eevee Trainer

7,05321337

asked 2 hours ago

KusaKusa

323

share|cite|improve this question

edited 2 hours ago

Eevee Trainer

7,05321337

asked 2 hours ago

KusaKusa

323

edited 2 hours ago

Eevee Trainer

7,05321337

edited 2 hours ago

Eevee Trainer

7,05321337

asked 2 hours ago

KusaKusa

323

asked 2 hours ago

KusaKusa

323

1 Answer
1

active

oldest

votes

5

$begingroup$

The first rings that were considered were generally commutative; polynomial rings, then from work of Dedekind in number fields. The properties were then abstracted Fraenkel and Noether, who still dealt mostly with commutative rings.

However, it soon became apparent that there were too many instances where commutativity of multiplication did not hold. You had famously the quaternions, of course, but you also had matrices and, more generally, the endomorphism ring of an abelian group (where “multiplication” is composition of functions). So that we have two different, related, notions: commutative rings and noncommutative rings, just like we have noncommutative groups and commutative/abelian groups.

Now, why do this with multiplication and not with addition? Well, if you take your definition of ring above, which includes a unity, but you drop condition (3) (that is, you require everything except you do not require that addition be commutative), it turns out that the other eight axioms force commutativity of addition.

Indeed, suppose you have a structure $(R,+,cdot,0,1)$ that satisfies axioms (1), (2), and (4)-(9) above. I claim that one can deduce (3). Indeed, let $a,bin R$. Then using associativity on the left first, and associativity on the right second, we have
$$begin{align*}
(1+1)(a+b) &= 1(a+b) + 1(a+b) = a+b+a+b\
(1+1)(a+b) &= (1+1)a + (1+1)b = a+a+b+b.
end{align*}$$
From this we get that $a+b+a+b = a+a+b+b$. Now add the inverse of $a$ on the left and the inverse of $b$ on the left on both sides to get
$$begin{align*}
(-a) + a + b + a + b + (-b) &= 0+b+a+0 = b+a\
(-a) + a + a + b + b + (-b) &= 0+a+b+0 = a+b
end{align*}$$
Thus, we conclude that $a+b=b+a$. That is, commutativity of addition is a consequence of the other eight axioms.

The reason we include it is two-fold: one, is that it is much nicer to say that the first few axioms force $(R,+)$ to be a commutative/abelian group. The second is that it is also common to consider rings without unity, and if we do that, then it is no longer true that addition is forced to be commutative. To see this, note that if $(G,cdot)$ is any group with identity element $e_G$, and we define a multiplication on $G$ by letting $a*b=e_G$ for all $a,bin G$, then $(G,cdot,*)$ satisfies axioms (1)-(8) given above. But if the original group is not commutative, then the “addition” in this ring is not commutative. So if we want to consider rings without unity, we do want to explicitly require addition to be commutative.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Arturo MagidinArturo Magidin

264k34588915

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is an entire thread on the 4th paragragh, see Why is ring addition commuattive?
  $endgroup$
  – Bill Dubuque
  1 hour ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

The first rings that were considered were generally commutative; polynomial rings, then from work of Dedekind in number fields. The properties were then abstracted Fraenkel and Noether, who still dealt mostly with commutative rings.

However, it soon became apparent that there were too many instances where commutativity of multiplication did not hold. You had famously the quaternions, of course, but you also had matrices and, more generally, the endomorphism ring of an abelian group (where “multiplication” is composition of functions). So that we have two different, related, notions: commutative rings and noncommutative rings, just like we have noncommutative groups and commutative/abelian groups.

Now, why do this with multiplication and not with addition? Well, if you take your definition of ring above, which includes a unity, but you drop condition (3) (that is, you require everything except you do not require that addition be commutative), it turns out that the other eight axioms force commutativity of addition.

Indeed, suppose you have a structure $(R,+,cdot,0,1)$ that satisfies axioms (1), (2), and (4)-(9) above. I claim that one can deduce (3). Indeed, let $a,bin R$. Then using associativity on the left first, and associativity on the right second, we have
$$begin{align*}
(1+1)(a+b) &= 1(a+b) + 1(a+b) = a+b+a+b\
(1+1)(a+b) &= (1+1)a + (1+1)b = a+a+b+b.
end{align*}$$
From this we get that $a+b+a+b = a+a+b+b$. Now add the inverse of $a$ on the left and the inverse of $b$ on the left on both sides to get
$$begin{align*}
(-a) + a + b + a + b + (-b) &= 0+b+a+0 = b+a\
(-a) + a + a + b + b + (-b) &= 0+a+b+0 = a+b
end{align*}$$
Thus, we conclude that $a+b=b+a$. That is, commutativity of addition is a consequence of the other eight axioms.

The reason we include it is two-fold: one, is that it is much nicer to say that the first few axioms force $(R,+)$ to be a commutative/abelian group. The second is that it is also common to consider rings without unity, and if we do that, then it is no longer true that addition is forced to be commutative. To see this, note that if $(G,cdot)$ is any group with identity element $e_G$, and we define a multiplication on $G$ by letting $a*b=e_G$ for all $a,bin G$, then $(G,cdot,*)$ satisfies axioms (1)-(8) given above. But if the original group is not commutative, then the “addition” in this ring is not commutative. So if we want to consider rings without unity, we do want to explicitly require addition to be commutative.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Arturo MagidinArturo Magidin

264k34588915

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is an entire thread on the 4th paragragh, see Why is ring addition commuattive?
  $endgroup$
  – Bill Dubuque
  1 hour ago
  

  
  
  
  

add a comment | 

5

$begingroup$

The first rings that were considered were generally commutative; polynomial rings, then from work of Dedekind in number fields. The properties were then abstracted Fraenkel and Noether, who still dealt mostly with commutative rings.

However, it soon became apparent that there were too many instances where commutativity of multiplication did not hold. You had famously the quaternions, of course, but you also had matrices and, more generally, the endomorphism ring of an abelian group (where “multiplication” is composition of functions). So that we have two different, related, notions: commutative rings and noncommutative rings, just like we have noncommutative groups and commutative/abelian groups.

Now, why do this with multiplication and not with addition? Well, if you take your definition of ring above, which includes a unity, but you drop condition (3) (that is, you require everything except you do not require that addition be commutative), it turns out that the other eight axioms force commutativity of addition.

Indeed, suppose you have a structure $(R,+,cdot,0,1)$ that satisfies axioms (1), (2), and (4)-(9) above. I claim that one can deduce (3). Indeed, let $a,bin R$. Then using associativity on the left first, and associativity on the right second, we have
$$begin{align*}
(1+1)(a+b) &= 1(a+b) + 1(a+b) = a+b+a+b\
(1+1)(a+b) &= (1+1)a + (1+1)b = a+a+b+b.
end{align*}$$
From this we get that $a+b+a+b = a+a+b+b$. Now add the inverse of $a$ on the left and the inverse of $b$ on the left on both sides to get
$$begin{align*}
(-a) + a + b + a + b + (-b) &= 0+b+a+0 = b+a\
(-a) + a + a + b + b + (-b) &= 0+a+b+0 = a+b
end{align*}$$
Thus, we conclude that $a+b=b+a$. That is, commutativity of addition is a consequence of the other eight axioms.

The reason we include it is two-fold: one, is that it is much nicer to say that the first few axioms force $(R,+)$ to be a commutative/abelian group. The second is that it is also common to consider rings without unity, and if we do that, then it is no longer true that addition is forced to be commutative. To see this, note that if $(G,cdot)$ is any group with identity element $e_G$, and we define a multiplication on $G$ by letting $a*b=e_G$ for all $a,bin G$, then $(G,cdot,*)$ satisfies axioms (1)-(8) given above. But if the original group is not commutative, then the “addition” in this ring is not commutative. So if we want to consider rings without unity, we do want to explicitly require addition to be commutative.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Arturo MagidinArturo Magidin

264k34588915

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is an entire thread on the 4th paragragh, see Why is ring addition commuattive?
  $endgroup$
  – Bill Dubuque
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

The first rings that were considered were generally commutative; polynomial rings, then from work of Dedekind in number fields. The properties were then abstracted Fraenkel and Noether, who still dealt mostly with commutative rings.

However, it soon became apparent that there were too many instances where commutativity of multiplication did not hold. You had famously the quaternions, of course, but you also had matrices and, more generally, the endomorphism ring of an abelian group (where “multiplication” is composition of functions). So that we have two different, related, notions: commutative rings and noncommutative rings, just like we have noncommutative groups and commutative/abelian groups.

Now, why do this with multiplication and not with addition? Well, if you take your definition of ring above, which includes a unity, but you drop condition (3) (that is, you require everything except you do not require that addition be commutative), it turns out that the other eight axioms force commutativity of addition.

Indeed, suppose you have a structure $(R,+,cdot,0,1)$ that satisfies axioms (1), (2), and (4)-(9) above. I claim that one can deduce (3). Indeed, let $a,bin R$. Then using associativity on the left first, and associativity on the right second, we have
$$begin{align*}
(1+1)(a+b) &= 1(a+b) + 1(a+b) = a+b+a+b\
(1+1)(a+b) &= (1+1)a + (1+1)b = a+a+b+b.
end{align*}$$
From this we get that $a+b+a+b = a+a+b+b$. Now add the inverse of $a$ on the left and the inverse of $b$ on the left on both sides to get
$$begin{align*}
(-a) + a + b + a + b + (-b) &= 0+b+a+0 = b+a\
(-a) + a + a + b + b + (-b) &= 0+a+b+0 = a+b
end{align*}$$
Thus, we conclude that $a+b=b+a$. That is, commutativity of addition is a consequence of the other eight axioms.

The reason we include it is two-fold: one, is that it is much nicer to say that the first few axioms force $(R,+)$ to be a commutative/abelian group. The second is that it is also common to consider rings without unity, and if we do that, then it is no longer true that addition is forced to be commutative. To see this, note that if $(G,cdot)$ is any group with identity element $e_G$, and we define a multiplication on $G$ by letting $a*b=e_G$ for all $a,bin G$, then $(G,cdot,*)$ satisfies axioms (1)-(8) given above. But if the original group is not commutative, then the “addition” in this ring is not commutative. So if we want to consider rings without unity, we do want to explicitly require addition to be commutative.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Arturo MagidinArturo Magidin

264k34588915

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Arturo MagidinArturo Magidin

264k34588915

answered 2 hours ago

Arturo MagidinArturo Magidin

264k34588915

answered 2 hours ago

Arturo MagidinArturo Magidin

264k34588915