## GRE GR9768 Question- 60

Forum for the GRE subject test in mathematics.
mobius70
Posts: 19
Joined: Thu Jan 24, 2008 1:11 am

### GRE GR9768 Question- 60

If S is a ring with property that s= s^2 (s squared) for each s element of S. Which of following must be true?
I) s+s = 0 for each s element of S
II) (s+t)^2 = s^2+t^2 for each s,t in S
III) S is commutative

A) III ONLY
B) I AND II ONLY
C)I AND III ONLY
D) II AND III ONLY
E) I , II AND III.

can someone please explain why I and III are correct.

actuary99
Posts: 1
Joined: Sun Jan 27, 2008 3:04 pm
since every element has a property s^2 = s ,so every element is a multiplicative identity to itself,we see that s is z (additive inverse ) , the rest is easy

mobius70
Posts: 19
Joined: Thu Jan 24, 2008 1:11 am

I understood that every element is multiplicative identity to itself.
But did not get how does that imply that .. the element is also additive identity to itself.

Also in option III) .. when it says ..S IS COMMUTATIVE .. my confusion is around .. if its saying .. this with respect to addtion or multiplication.

If its addtion then its ok .. since S is supposed to be abelian group in addition .. but if its multiplictaion .. then i don't get it ..??

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
Mobius thank you for this problem. Where did you find it? Here is how I can see the solution.

I. Probably, I'm wrong but I didn't hear that there is such term "the identity to itself". Anyway, it doesn't matter and my solution is next:
s+s=(s+s)^2
s+s=ss+2ss+ss=s+2ss+s
last result implies that
2ss=0
2s=0
s+s=0
s=-s

II. I see you solved it, nevertheless want to show you how I did it. If you have another solution would be happy to see it as well.
(s+t)^2=ss+2st+tt=s^2+st+st+t^2
since st=-st for every s,t
(s+t)^2=s^2+t^2

III. Here you should also be careful with definitions. Ring is always commutative under its operation of addition (as it supposed to be abelian group under addition). So the term "commutative ring" means that its multiplication operation is also commutative. I think we can prove it next way:
(s+t)^2=s+t
s^2+st+ts+t^2=s+t
(s+t)+st+ts=(s+t)
st+ts=0
since st+st=0 (s+s=0 was proved in I.) there should be
st+ts=st+st
ts=st

I think these approaches are not really obvious and should be remembered. Please also correct me if I'm wrong.

mobius70
Posts: 19
Joined: Thu Jan 24, 2008 1:11 am
Hi Lime
To answer your question, this is problem from GRE GR9768 Question- 60.

I went through the solutions and all looked kool to me. Well done!!
I was not able to solve for the III) (COMMUTATIVE PART).
You have done it wonderfully.. this is how i solved for first 2 parts.

I ASSUMED THE ADDITION AND MULTIPLICATION ON THE RING TO BE DEFINED AS THE USUAL ARITHMATIC OPERATIONS.

I) Let s be an element of S.
Since S to be a ring S, has to be abelian in addition operation.
=> for every s in S there exists a -s (minus) such that
s+ (-s) = 0 (0 IS THE ADDITIVE IDENTITY)
As defined s= s^2.
=> -s = (-s)^2 = s^2.
=> s = -s.
=> s+s = 0

II) (s)+(t) = (s^2) + (t^2)
also (s+t) = (s+t)^2
=> (s+t)^2 = s^2+t^2

ONLY ONE POINT I WOULD LIKE TO MENTION, FOR AN EXAM LIKE GRE, IT WOULD BE MUCH EASIER TO USE BINARY RING OF ELEMENTS {0,1} TO SOLVE THIS PROBLEM.
I UNDERSTAND THAT THIS APPROACH IS NO WAY SUBSTITUTE FOR A GENERAL SOLUTION AS PRESENTED BY LIME, BUT GUESS IN GRE YOU WANT TO GET ANSWER IN 2 MINS.

Again thanks lime.

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
Cool! Good idea of working with simple rings and fields when it is appropriate, in order to save the time!