Commutative property of ring addition












23












$begingroup$


I have a simple question answer to which would help me more deeply understand the concept of (non)commutative structures. Let's take for example (our teacher's definition of) a ring:



Let $Rneq emptyset$ be a set, let $oplus:Atimes A to A$ and $bullet :Atimes A to A$ be binary operations. Moreover, let $(R, oplus)$ be a commutative group, $(R, bullet)$ be a monoid and following property holds for all $a, b, cin R$:
$$abullet(boplus c) = (abullet b)oplus(a bullet c)$$
$$(boplus c)bullet a = (bbullet a)oplus(c bullet a)$$
Then ordered triple $mathbf R = (R, oplus, bullet mathbf)$ is called a (unitary) ring.



Moreover, we call ring $mathbf R$ commutative iff $(R, bullet)$ is a commutative monoid. Commutativity of a ring is always a matter of its multiplicative operation because the additive operation is always assumed to be commutative.



Could anyone explain me the bold part? Why do we even in non-commutative rings (and fields etc.) assume the addition to be always commutative? Is there some serious reason? Would it make any trouble? Or studying of structures with non-commutative addition just doesn't give us anything new so we can take addition as commutative simply because of our comfort?










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  • $begingroup$
    Possible duplicate of Why is ring addition commutative?
    $endgroup$
    – Santropedro
    Mar 16 '17 at 14:26
















23












$begingroup$


I have a simple question answer to which would help me more deeply understand the concept of (non)commutative structures. Let's take for example (our teacher's definition of) a ring:



Let $Rneq emptyset$ be a set, let $oplus:Atimes A to A$ and $bullet :Atimes A to A$ be binary operations. Moreover, let $(R, oplus)$ be a commutative group, $(R, bullet)$ be a monoid and following property holds for all $a, b, cin R$:
$$abullet(boplus c) = (abullet b)oplus(a bullet c)$$
$$(boplus c)bullet a = (bbullet a)oplus(c bullet a)$$
Then ordered triple $mathbf R = (R, oplus, bullet mathbf)$ is called a (unitary) ring.



Moreover, we call ring $mathbf R$ commutative iff $(R, bullet)$ is a commutative monoid. Commutativity of a ring is always a matter of its multiplicative operation because the additive operation is always assumed to be commutative.



Could anyone explain me the bold part? Why do we even in non-commutative rings (and fields etc.) assume the addition to be always commutative? Is there some serious reason? Would it make any trouble? Or studying of structures with non-commutative addition just doesn't give us anything new so we can take addition as commutative simply because of our comfort?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Possible duplicate of Why is ring addition commutative?
    $endgroup$
    – Santropedro
    Mar 16 '17 at 14:26














23












23








23


7



$begingroup$


I have a simple question answer to which would help me more deeply understand the concept of (non)commutative structures. Let's take for example (our teacher's definition of) a ring:



Let $Rneq emptyset$ be a set, let $oplus:Atimes A to A$ and $bullet :Atimes A to A$ be binary operations. Moreover, let $(R, oplus)$ be a commutative group, $(R, bullet)$ be a monoid and following property holds for all $a, b, cin R$:
$$abullet(boplus c) = (abullet b)oplus(a bullet c)$$
$$(boplus c)bullet a = (bbullet a)oplus(c bullet a)$$
Then ordered triple $mathbf R = (R, oplus, bullet mathbf)$ is called a (unitary) ring.



Moreover, we call ring $mathbf R$ commutative iff $(R, bullet)$ is a commutative monoid. Commutativity of a ring is always a matter of its multiplicative operation because the additive operation is always assumed to be commutative.



Could anyone explain me the bold part? Why do we even in non-commutative rings (and fields etc.) assume the addition to be always commutative? Is there some serious reason? Would it make any trouble? Or studying of structures with non-commutative addition just doesn't give us anything new so we can take addition as commutative simply because of our comfort?










share|cite|improve this question











$endgroup$




I have a simple question answer to which would help me more deeply understand the concept of (non)commutative structures. Let's take for example (our teacher's definition of) a ring:



Let $Rneq emptyset$ be a set, let $oplus:Atimes A to A$ and $bullet :Atimes A to A$ be binary operations. Moreover, let $(R, oplus)$ be a commutative group, $(R, bullet)$ be a monoid and following property holds for all $a, b, cin R$:
$$abullet(boplus c) = (abullet b)oplus(a bullet c)$$
$$(boplus c)bullet a = (bbullet a)oplus(c bullet a)$$
Then ordered triple $mathbf R = (R, oplus, bullet mathbf)$ is called a (unitary) ring.



Moreover, we call ring $mathbf R$ commutative iff $(R, bullet)$ is a commutative monoid. Commutativity of a ring is always a matter of its multiplicative operation because the additive operation is always assumed to be commutative.



Could anyone explain me the bold part? Why do we even in non-commutative rings (and fields etc.) assume the addition to be always commutative? Is there some serious reason? Would it make any trouble? Or studying of structures with non-commutative addition just doesn't give us anything new so we can take addition as commutative simply because of our comfort?







abstract-algebra commutative-algebra ring-theory noncommutative-algebra






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edited Mar 30 '13 at 11:05









Orat

2,86021231




2,86021231










asked Mar 30 '13 at 10:32









JeyekomonJeyekomon

710923




710923












  • $begingroup$
    Possible duplicate of Why is ring addition commutative?
    $endgroup$
    – Santropedro
    Mar 16 '17 at 14:26


















  • $begingroup$
    Possible duplicate of Why is ring addition commutative?
    $endgroup$
    – Santropedro
    Mar 16 '17 at 14:26
















$begingroup$
Possible duplicate of Why is ring addition commutative?
$endgroup$
– Santropedro
Mar 16 '17 at 14:26




$begingroup$
Possible duplicate of Why is ring addition commutative?
$endgroup$
– Santropedro
Mar 16 '17 at 14:26










3 Answers
3






active

oldest

votes


















22












$begingroup$

Perhaps the comment refers to the fact that in order to generalize rings to structures with noncommutative addiiton, one cannot simply delete the axiom that addition is commutative, since, in fact, other axioms force addition to be commutative (Hankel, 1867 [1]). The proof is simple: apply both the left and right
distributive law in different order to the term $rm:(1!+!1)(x!+!y),:$ viz.



$$rm (1!+!1)(x!+!y) = bigglbrace begin{eqnarray}rm (1!+!1)x!+!(1!+!1)y, =, x ,+, color{#C00}{x!+!y} ,+, y\
rm 1(x!+!y)!+1(x!+!y), =, x, +, color{#0A0}{y!+!x}, +, yend{eqnarray}biggrbrace:Rightarrow: color{#C00}{x!+!y},=,color{#0A0}{y!+!x} by cancel x,y$$



Thus commutativity of addition, $rm:x+y = y+x,:$ is implied by these axioms:



$(1) *,$ distributes over $rm,+!: x(y+z), =, xy+xz, (y+z)x, =, yx+zx$



$(2) , +,$ is cancellative: $rm x+y, =, x+z:Rightarrow: y=z, y+x, =, z+x:Rightarrow: y=z$



$(3) , +,$ is associative: $rm (x+y)+z, =, x+(y+z)$



$(4) *,$ has a neutral element $rm,1!: 1x = x$



In order to state this result concisely, recall that a SemiRing is
that generalization of a Ring whose additive structure is relaxed
from a commutative Group to merely a SemiGroup, i.e. here the only
hypothesis on addition is that it be associative (so in SemiRings,
unlike Rings, addition need not be commutative, nor need every
element $rm,x,$ have an additive inverse $rm,-x).,$ Now the above result may
be stated as follows: a semiring with $,1,$ and cancellative addition
has commutative addition. Such semirings are simply subsemirings
of rings (as is $rm:Bbb N subset Bbb Z),$ because any commutative cancellative
semigroup embeds canonically into a commutative group, its group
of differences (in precisely the same way $rm,Bbb Z,$ is constructed from $rm,Bbb N,,$
i.e. the additive version of the fraction field construction).



Examples of SemiRings include: $rm,Bbb N;,$ initial segments of cardinals;
distributive lattices (e.g. subsets of a powerset with operations $cup$ and $cap$;
$rm,Bbb R,$ with + being min or max, and $*$ being addition; semigroup semirings
(e.g. formal power series); formal languages with union, concat; etc.
For a nice survey of SemiRings and SemiFields see [2]. See also Near-Rings.



[1] Gerhard Betsch. On the beginnings and development of near-ring theory.
pp. 1-11 in:
Near-rings and near-fields. Proceedings of the conference
held in Fredericton, New Brunswick, July 18-24, 1993. Edited by Yuen Fong,
Howard E. Bell, Wen-Fong Ke, Gordon Mason and Gunter Pilz.
Mathematics and its Applications, 336. Kluwer Academic Publishers Group,
Dordrecht, 1995. x+278 pp. ISBN: 0-7923-3635-6 Zbl review



[2] Hebisch, Udo; Weinert, Hanns Joachim. Semirings and semifields. $ $ pp. 425-462 in: Handbook of algebra. Vol. 1. Edited by M. Hazewinkel.
North-Holland Publishing Co., Amsterdam, 1996. xx+915 pp. ISBN: 0-444-82212-7
Zbl review,
AMS review






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, it's really interesting! Most books I have, talk about the topics just briefly without any deeper and detailed insight. These are the notes which I'd love to have my books full of... Accepted button goes to you.
    $endgroup$
    – Jeyekomon
    Mar 31 '13 at 19:58










  • $begingroup$
    Never knew distribuitive was THAT amazing.
    $endgroup$
    – Santropedro
    Mar 4 '17 at 4:33



















6












$begingroup$

There are so-called near-semirings (http://en.wikipedia.org/wiki/Near-semiring) in which addition is non-commutative.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Of course one can develop a theory in which the addition is not commutative (see Boris' answer and his mentioning the near-semirings).



    Why "rings with non-commutative addition" are a somewhat side story and commutativity of addition is the usual assumption? Simply because the basic and main examples of these rings, those which primarily occur doing mathematics, do have this property.



    I believe that by far most "rings" can be reconducted in a way or another to the ring of matrices over some algebraic structure with commutative addition (commutative rings or division algebras, tipically). Addition of such matrices commutes.






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      3 Answers
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      3 Answers
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      22












      $begingroup$

      Perhaps the comment refers to the fact that in order to generalize rings to structures with noncommutative addiiton, one cannot simply delete the axiom that addition is commutative, since, in fact, other axioms force addition to be commutative (Hankel, 1867 [1]). The proof is simple: apply both the left and right
      distributive law in different order to the term $rm:(1!+!1)(x!+!y),:$ viz.



      $$rm (1!+!1)(x!+!y) = bigglbrace begin{eqnarray}rm (1!+!1)x!+!(1!+!1)y, =, x ,+, color{#C00}{x!+!y} ,+, y\
      rm 1(x!+!y)!+1(x!+!y), =, x, +, color{#0A0}{y!+!x}, +, yend{eqnarray}biggrbrace:Rightarrow: color{#C00}{x!+!y},=,color{#0A0}{y!+!x} by cancel x,y$$



      Thus commutativity of addition, $rm:x+y = y+x,:$ is implied by these axioms:



      $(1) *,$ distributes over $rm,+!: x(y+z), =, xy+xz, (y+z)x, =, yx+zx$



      $(2) , +,$ is cancellative: $rm x+y, =, x+z:Rightarrow: y=z, y+x, =, z+x:Rightarrow: y=z$



      $(3) , +,$ is associative: $rm (x+y)+z, =, x+(y+z)$



      $(4) *,$ has a neutral element $rm,1!: 1x = x$



      In order to state this result concisely, recall that a SemiRing is
      that generalization of a Ring whose additive structure is relaxed
      from a commutative Group to merely a SemiGroup, i.e. here the only
      hypothesis on addition is that it be associative (so in SemiRings,
      unlike Rings, addition need not be commutative, nor need every
      element $rm,x,$ have an additive inverse $rm,-x).,$ Now the above result may
      be stated as follows: a semiring with $,1,$ and cancellative addition
      has commutative addition. Such semirings are simply subsemirings
      of rings (as is $rm:Bbb N subset Bbb Z),$ because any commutative cancellative
      semigroup embeds canonically into a commutative group, its group
      of differences (in precisely the same way $rm,Bbb Z,$ is constructed from $rm,Bbb N,,$
      i.e. the additive version of the fraction field construction).



      Examples of SemiRings include: $rm,Bbb N;,$ initial segments of cardinals;
      distributive lattices (e.g. subsets of a powerset with operations $cup$ and $cap$;
      $rm,Bbb R,$ with + being min or max, and $*$ being addition; semigroup semirings
      (e.g. formal power series); formal languages with union, concat; etc.
      For a nice survey of SemiRings and SemiFields see [2]. See also Near-Rings.



      [1] Gerhard Betsch. On the beginnings and development of near-ring theory.
      pp. 1-11 in:
      Near-rings and near-fields. Proceedings of the conference
      held in Fredericton, New Brunswick, July 18-24, 1993. Edited by Yuen Fong,
      Howard E. Bell, Wen-Fong Ke, Gordon Mason and Gunter Pilz.
      Mathematics and its Applications, 336. Kluwer Academic Publishers Group,
      Dordrecht, 1995. x+278 pp. ISBN: 0-7923-3635-6 Zbl review



      [2] Hebisch, Udo; Weinert, Hanns Joachim. Semirings and semifields. $ $ pp. 425-462 in: Handbook of algebra. Vol. 1. Edited by M. Hazewinkel.
      North-Holland Publishing Co., Amsterdam, 1996. xx+915 pp. ISBN: 0-444-82212-7
      Zbl review,
      AMS review






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thank you, it's really interesting! Most books I have, talk about the topics just briefly without any deeper and detailed insight. These are the notes which I'd love to have my books full of... Accepted button goes to you.
        $endgroup$
        – Jeyekomon
        Mar 31 '13 at 19:58










      • $begingroup$
        Never knew distribuitive was THAT amazing.
        $endgroup$
        – Santropedro
        Mar 4 '17 at 4:33
















      22












      $begingroup$

      Perhaps the comment refers to the fact that in order to generalize rings to structures with noncommutative addiiton, one cannot simply delete the axiom that addition is commutative, since, in fact, other axioms force addition to be commutative (Hankel, 1867 [1]). The proof is simple: apply both the left and right
      distributive law in different order to the term $rm:(1!+!1)(x!+!y),:$ viz.



      $$rm (1!+!1)(x!+!y) = bigglbrace begin{eqnarray}rm (1!+!1)x!+!(1!+!1)y, =, x ,+, color{#C00}{x!+!y} ,+, y\
      rm 1(x!+!y)!+1(x!+!y), =, x, +, color{#0A0}{y!+!x}, +, yend{eqnarray}biggrbrace:Rightarrow: color{#C00}{x!+!y},=,color{#0A0}{y!+!x} by cancel x,y$$



      Thus commutativity of addition, $rm:x+y = y+x,:$ is implied by these axioms:



      $(1) *,$ distributes over $rm,+!: x(y+z), =, xy+xz, (y+z)x, =, yx+zx$



      $(2) , +,$ is cancellative: $rm x+y, =, x+z:Rightarrow: y=z, y+x, =, z+x:Rightarrow: y=z$



      $(3) , +,$ is associative: $rm (x+y)+z, =, x+(y+z)$



      $(4) *,$ has a neutral element $rm,1!: 1x = x$



      In order to state this result concisely, recall that a SemiRing is
      that generalization of a Ring whose additive structure is relaxed
      from a commutative Group to merely a SemiGroup, i.e. here the only
      hypothesis on addition is that it be associative (so in SemiRings,
      unlike Rings, addition need not be commutative, nor need every
      element $rm,x,$ have an additive inverse $rm,-x).,$ Now the above result may
      be stated as follows: a semiring with $,1,$ and cancellative addition
      has commutative addition. Such semirings are simply subsemirings
      of rings (as is $rm:Bbb N subset Bbb Z),$ because any commutative cancellative
      semigroup embeds canonically into a commutative group, its group
      of differences (in precisely the same way $rm,Bbb Z,$ is constructed from $rm,Bbb N,,$
      i.e. the additive version of the fraction field construction).



      Examples of SemiRings include: $rm,Bbb N;,$ initial segments of cardinals;
      distributive lattices (e.g. subsets of a powerset with operations $cup$ and $cap$;
      $rm,Bbb R,$ with + being min or max, and $*$ being addition; semigroup semirings
      (e.g. formal power series); formal languages with union, concat; etc.
      For a nice survey of SemiRings and SemiFields see [2]. See also Near-Rings.



      [1] Gerhard Betsch. On the beginnings and development of near-ring theory.
      pp. 1-11 in:
      Near-rings and near-fields. Proceedings of the conference
      held in Fredericton, New Brunswick, July 18-24, 1993. Edited by Yuen Fong,
      Howard E. Bell, Wen-Fong Ke, Gordon Mason and Gunter Pilz.
      Mathematics and its Applications, 336. Kluwer Academic Publishers Group,
      Dordrecht, 1995. x+278 pp. ISBN: 0-7923-3635-6 Zbl review



      [2] Hebisch, Udo; Weinert, Hanns Joachim. Semirings and semifields. $ $ pp. 425-462 in: Handbook of algebra. Vol. 1. Edited by M. Hazewinkel.
      North-Holland Publishing Co., Amsterdam, 1996. xx+915 pp. ISBN: 0-444-82212-7
      Zbl review,
      AMS review






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thank you, it's really interesting! Most books I have, talk about the topics just briefly without any deeper and detailed insight. These are the notes which I'd love to have my books full of... Accepted button goes to you.
        $endgroup$
        – Jeyekomon
        Mar 31 '13 at 19:58










      • $begingroup$
        Never knew distribuitive was THAT amazing.
        $endgroup$
        – Santropedro
        Mar 4 '17 at 4:33














      22












      22








      22





      $begingroup$

      Perhaps the comment refers to the fact that in order to generalize rings to structures with noncommutative addiiton, one cannot simply delete the axiom that addition is commutative, since, in fact, other axioms force addition to be commutative (Hankel, 1867 [1]). The proof is simple: apply both the left and right
      distributive law in different order to the term $rm:(1!+!1)(x!+!y),:$ viz.



      $$rm (1!+!1)(x!+!y) = bigglbrace begin{eqnarray}rm (1!+!1)x!+!(1!+!1)y, =, x ,+, color{#C00}{x!+!y} ,+, y\
      rm 1(x!+!y)!+1(x!+!y), =, x, +, color{#0A0}{y!+!x}, +, yend{eqnarray}biggrbrace:Rightarrow: color{#C00}{x!+!y},=,color{#0A0}{y!+!x} by cancel x,y$$



      Thus commutativity of addition, $rm:x+y = y+x,:$ is implied by these axioms:



      $(1) *,$ distributes over $rm,+!: x(y+z), =, xy+xz, (y+z)x, =, yx+zx$



      $(2) , +,$ is cancellative: $rm x+y, =, x+z:Rightarrow: y=z, y+x, =, z+x:Rightarrow: y=z$



      $(3) , +,$ is associative: $rm (x+y)+z, =, x+(y+z)$



      $(4) *,$ has a neutral element $rm,1!: 1x = x$



      In order to state this result concisely, recall that a SemiRing is
      that generalization of a Ring whose additive structure is relaxed
      from a commutative Group to merely a SemiGroup, i.e. here the only
      hypothesis on addition is that it be associative (so in SemiRings,
      unlike Rings, addition need not be commutative, nor need every
      element $rm,x,$ have an additive inverse $rm,-x).,$ Now the above result may
      be stated as follows: a semiring with $,1,$ and cancellative addition
      has commutative addition. Such semirings are simply subsemirings
      of rings (as is $rm:Bbb N subset Bbb Z),$ because any commutative cancellative
      semigroup embeds canonically into a commutative group, its group
      of differences (in precisely the same way $rm,Bbb Z,$ is constructed from $rm,Bbb N,,$
      i.e. the additive version of the fraction field construction).



      Examples of SemiRings include: $rm,Bbb N;,$ initial segments of cardinals;
      distributive lattices (e.g. subsets of a powerset with operations $cup$ and $cap$;
      $rm,Bbb R,$ with + being min or max, and $*$ being addition; semigroup semirings
      (e.g. formal power series); formal languages with union, concat; etc.
      For a nice survey of SemiRings and SemiFields see [2]. See also Near-Rings.



      [1] Gerhard Betsch. On the beginnings and development of near-ring theory.
      pp. 1-11 in:
      Near-rings and near-fields. Proceedings of the conference
      held in Fredericton, New Brunswick, July 18-24, 1993. Edited by Yuen Fong,
      Howard E. Bell, Wen-Fong Ke, Gordon Mason and Gunter Pilz.
      Mathematics and its Applications, 336. Kluwer Academic Publishers Group,
      Dordrecht, 1995. x+278 pp. ISBN: 0-7923-3635-6 Zbl review



      [2] Hebisch, Udo; Weinert, Hanns Joachim. Semirings and semifields. $ $ pp. 425-462 in: Handbook of algebra. Vol. 1. Edited by M. Hazewinkel.
      North-Holland Publishing Co., Amsterdam, 1996. xx+915 pp. ISBN: 0-444-82212-7
      Zbl review,
      AMS review






      share|cite|improve this answer











      $endgroup$



      Perhaps the comment refers to the fact that in order to generalize rings to structures with noncommutative addiiton, one cannot simply delete the axiom that addition is commutative, since, in fact, other axioms force addition to be commutative (Hankel, 1867 [1]). The proof is simple: apply both the left and right
      distributive law in different order to the term $rm:(1!+!1)(x!+!y),:$ viz.



      $$rm (1!+!1)(x!+!y) = bigglbrace begin{eqnarray}rm (1!+!1)x!+!(1!+!1)y, =, x ,+, color{#C00}{x!+!y} ,+, y\
      rm 1(x!+!y)!+1(x!+!y), =, x, +, color{#0A0}{y!+!x}, +, yend{eqnarray}biggrbrace:Rightarrow: color{#C00}{x!+!y},=,color{#0A0}{y!+!x} by cancel x,y$$



      Thus commutativity of addition, $rm:x+y = y+x,:$ is implied by these axioms:



      $(1) *,$ distributes over $rm,+!: x(y+z), =, xy+xz, (y+z)x, =, yx+zx$



      $(2) , +,$ is cancellative: $rm x+y, =, x+z:Rightarrow: y=z, y+x, =, z+x:Rightarrow: y=z$



      $(3) , +,$ is associative: $rm (x+y)+z, =, x+(y+z)$



      $(4) *,$ has a neutral element $rm,1!: 1x = x$



      In order to state this result concisely, recall that a SemiRing is
      that generalization of a Ring whose additive structure is relaxed
      from a commutative Group to merely a SemiGroup, i.e. here the only
      hypothesis on addition is that it be associative (so in SemiRings,
      unlike Rings, addition need not be commutative, nor need every
      element $rm,x,$ have an additive inverse $rm,-x).,$ Now the above result may
      be stated as follows: a semiring with $,1,$ and cancellative addition
      has commutative addition. Such semirings are simply subsemirings
      of rings (as is $rm:Bbb N subset Bbb Z),$ because any commutative cancellative
      semigroup embeds canonically into a commutative group, its group
      of differences (in precisely the same way $rm,Bbb Z,$ is constructed from $rm,Bbb N,,$
      i.e. the additive version of the fraction field construction).



      Examples of SemiRings include: $rm,Bbb N;,$ initial segments of cardinals;
      distributive lattices (e.g. subsets of a powerset with operations $cup$ and $cap$;
      $rm,Bbb R,$ with + being min or max, and $*$ being addition; semigroup semirings
      (e.g. formal power series); formal languages with union, concat; etc.
      For a nice survey of SemiRings and SemiFields see [2]. See also Near-Rings.



      [1] Gerhard Betsch. On the beginnings and development of near-ring theory.
      pp. 1-11 in:
      Near-rings and near-fields. Proceedings of the conference
      held in Fredericton, New Brunswick, July 18-24, 1993. Edited by Yuen Fong,
      Howard E. Bell, Wen-Fong Ke, Gordon Mason and Gunter Pilz.
      Mathematics and its Applications, 336. Kluwer Academic Publishers Group,
      Dordrecht, 1995. x+278 pp. ISBN: 0-7923-3635-6 Zbl review



      [2] Hebisch, Udo; Weinert, Hanns Joachim. Semirings and semifields. $ $ pp. 425-462 in: Handbook of algebra. Vol. 1. Edited by M. Hazewinkel.
      North-Holland Publishing Co., Amsterdam, 1996. xx+915 pp. ISBN: 0-444-82212-7
      Zbl review,
      AMS review







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 5 '18 at 17:44









      Bill Dubuque

      212k29195650




      212k29195650










      answered Mar 30 '13 at 18:00









      Math GemsMath Gems

      17k12039




      17k12039












      • $begingroup$
        Thank you, it's really interesting! Most books I have, talk about the topics just briefly without any deeper and detailed insight. These are the notes which I'd love to have my books full of... Accepted button goes to you.
        $endgroup$
        – Jeyekomon
        Mar 31 '13 at 19:58










      • $begingroup$
        Never knew distribuitive was THAT amazing.
        $endgroup$
        – Santropedro
        Mar 4 '17 at 4:33


















      • $begingroup$
        Thank you, it's really interesting! Most books I have, talk about the topics just briefly without any deeper and detailed insight. These are the notes which I'd love to have my books full of... Accepted button goes to you.
        $endgroup$
        – Jeyekomon
        Mar 31 '13 at 19:58










      • $begingroup$
        Never knew distribuitive was THAT amazing.
        $endgroup$
        – Santropedro
        Mar 4 '17 at 4:33
















      $begingroup$
      Thank you, it's really interesting! Most books I have, talk about the topics just briefly without any deeper and detailed insight. These are the notes which I'd love to have my books full of... Accepted button goes to you.
      $endgroup$
      – Jeyekomon
      Mar 31 '13 at 19:58




      $begingroup$
      Thank you, it's really interesting! Most books I have, talk about the topics just briefly without any deeper and detailed insight. These are the notes which I'd love to have my books full of... Accepted button goes to you.
      $endgroup$
      – Jeyekomon
      Mar 31 '13 at 19:58












      $begingroup$
      Never knew distribuitive was THAT amazing.
      $endgroup$
      – Santropedro
      Mar 4 '17 at 4:33




      $begingroup$
      Never knew distribuitive was THAT amazing.
      $endgroup$
      – Santropedro
      Mar 4 '17 at 4:33











      6












      $begingroup$

      There are so-called near-semirings (http://en.wikipedia.org/wiki/Near-semiring) in which addition is non-commutative.






      share|cite|improve this answer









      $endgroup$


















        6












        $begingroup$

        There are so-called near-semirings (http://en.wikipedia.org/wiki/Near-semiring) in which addition is non-commutative.






        share|cite|improve this answer









        $endgroup$
















          6












          6








          6





          $begingroup$

          There are so-called near-semirings (http://en.wikipedia.org/wiki/Near-semiring) in which addition is non-commutative.






          share|cite|improve this answer









          $endgroup$



          There are so-called near-semirings (http://en.wikipedia.org/wiki/Near-semiring) in which addition is non-commutative.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 30 '13 at 10:48









          Boris NovikovBoris Novikov

          16.1k11529




          16.1k11529























              0












              $begingroup$

              Of course one can develop a theory in which the addition is not commutative (see Boris' answer and his mentioning the near-semirings).



              Why "rings with non-commutative addition" are a somewhat side story and commutativity of addition is the usual assumption? Simply because the basic and main examples of these rings, those which primarily occur doing mathematics, do have this property.



              I believe that by far most "rings" can be reconducted in a way or another to the ring of matrices over some algebraic structure with commutative addition (commutative rings or division algebras, tipically). Addition of such matrices commutes.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Of course one can develop a theory in which the addition is not commutative (see Boris' answer and his mentioning the near-semirings).



                Why "rings with non-commutative addition" are a somewhat side story and commutativity of addition is the usual assumption? Simply because the basic and main examples of these rings, those which primarily occur doing mathematics, do have this property.



                I believe that by far most "rings" can be reconducted in a way or another to the ring of matrices over some algebraic structure with commutative addition (commutative rings or division algebras, tipically). Addition of such matrices commutes.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Of course one can develop a theory in which the addition is not commutative (see Boris' answer and his mentioning the near-semirings).



                  Why "rings with non-commutative addition" are a somewhat side story and commutativity of addition is the usual assumption? Simply because the basic and main examples of these rings, those which primarily occur doing mathematics, do have this property.



                  I believe that by far most "rings" can be reconducted in a way or another to the ring of matrices over some algebraic structure with commutative addition (commutative rings or division algebras, tipically). Addition of such matrices commutes.






                  share|cite|improve this answer









                  $endgroup$



                  Of course one can develop a theory in which the addition is not commutative (see Boris' answer and his mentioning the near-semirings).



                  Why "rings with non-commutative addition" are a somewhat side story and commutativity of addition is the usual assumption? Simply because the basic and main examples of these rings, those which primarily occur doing mathematics, do have this property.



                  I believe that by far most "rings" can be reconducted in a way or another to the ring of matrices over some algebraic structure with commutative addition (commutative rings or division algebras, tipically). Addition of such matrices commutes.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 30 '13 at 11:18









                  Andrea MoriAndrea Mori

                  19.9k13466




                  19.9k13466






























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