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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1922.0. "2 elementary ring algebra problems" by EVTSG8::ESANU (Au temps pour moi) Mon Dec 19 1994 13:09

A. If R is a unitary ring such that  x^3 = x  and  2x = 0  for every  x 
in R, then  R  is commutative.

B. If R is a ring such that  x^3 = x  and  3x = 0  for every  x 
in R, then  R  is commutative.

(Mathematical folklore).

Mihai.
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1922.1HANNAH::OSMANsee HANNAH::IGLOO$:[OSMAN]ERIC.VT240Fri Dec 30 1994 17:5674
Well, I'll give part A a try, even though I don't know what a unitary ring
is, and aren't sure I remember what a ring is.  I remember there were rings,
fields, and groups.

Maybe as I go I'll remember stuff !

We're given that for every x in R,

	x^3 = x
	2x=0

Writing these longer,

	x*x*x = x

	x+x=0

So, I must assume that rings have two operators associated with them, "*"
and "+".

Then there's that 0.  I'll assume that the element 0 is some existing element
in R (must it be unique ?) such that for every x,

	x+0=x

Anyway, we want to show that given those 2 statements, that for every x and
y in R, it is true that

	x*y = y*x

But hmm.  Maybe we're supposed to prove

	x+y = y+x

Or maybe we can prove *both* ???

Well, let's start with

	x*y

What can we do with it ?  We can apply the cube premiss:

	x*x*x*y

What else can we do ?  We could apply it to y as well:

	x*x*x*y*y*y

Maybe we should apply it to the middle (replace x*y with its cube):

	x*x*x*y*x*y*x*y*y*y

Actually, we were sloppy in that step, since we really started with

	(x*x*x)*(y*y*y)

and in order to cube the middle, we must be sure we're allowed to regroup as

	(x*x)*(x*y)*(y*y)

This, of course, presupposes associativity, that is, for every x and y, and
z,

	(x*y)*z = x*(y*z)

Is this a given quality of rings ?

Oh well, maybe I'll wait for some answers to my elementary questions before
proceeding.

Thanks.

/Eric