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In article <11482497.1161538502599.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
2Pac <poinsotlaurent@xxxxxxxx> wrote:
>I know that it exists an algebraic structure (K,+,.) where (K,+) is a
>loop and the nonzero elements of K equipped with "." is also a loop
>(this is called a "double-loop"). Are there such finite or infinite
>(but preferably finite for my concerns) double-loops with a (left or
>right) distribution of . over + which are not quasifields (i.e. (K,+)
>is not associative) ? In this case can we find examples where + is
>noncommutative ?
>
>Thank for the help and the references you could give me,
Assume a structure satisfies all the axioms for a ring with unity
except perhaps for commutativity of addition. Then the commutativity
of addition will follow. Explicitly, let (R,+,*) be a set with two
binary operations such that for all a,b,c in R:
(i) (a+b)+c = a+(b+c)
(ii) There exists 0 in R such that for all a a+0=0+a = a.
(iii) For all a in R there exists b in R such that a+b=b+a=0.
(iv) (a*b)*c = a*(b*c)
(v) a*(b+c) = (a*b) + (a*c)
(vi) (a+b)*c = (a*c) + (b*c)
(vii) There exists an element 1 in R such that for all a in R,
1*a=a*1=a.
If you drop (vii), then you can have a structure with noncommutative
addition: just pick your favorite nonabelian group G, and define
multiplication to always be the identity element for the group; this
will satisfy axioms (i)-(vi) with a non-commutative "+".
Let a and b be arbitrary elements of R. We want to prove that
a+b=b+a.
Now, since R satisfies (vii), then we have that
(1+1)*(a+b) = [(1+1)*a] + [(1+1)*b] by (v)
= [1*a + 1*a] + [1*b + 1*b] by (vi)
= [a+a]+[b+b] (by (vii)).
On the other hand,
(1+1)*(a+b) = [1*(a+b)] + [1*(a+b)] by (vi)
= [a+b] + [a+b] by (vii).
Thus, we have
a + a + b + b = a + b + a + b (applying (i) to drop parenthesis).
Call this (*)
Now, let a' be the element guaranteed for a by axiom (iii), and b' for
b. That is, a'+a=a+a' = 0, and b'+b=b+b' = 0.
Adding a' on the left we have
a'+(a+a+b+b) = a' + (a+b+a+b) by (*)
(a'+a) + (a + b + b) = (a'+a) + (b + a + b) by (i)
0 + (a + b + b) = 0 + (b + a + b) by (iii)
a + b + b = b + a + b by (ii)
Call this (**)
Now add b' on the right to get
(a+b+b)+b' = (b+a+b) + b' by (**)
(a+b)+(b+b') = (b+a) + (b+b') by (i)
(a+b) + 0 = (b+a) + 0 by (iii)
(a+b) = (b+a) by (ii)
Thus, a+b=b+a, and + must be commutative if the other six axioms are
present.
--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
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