Proving commutativity

Question

Let $R$ be a ring in which $x^2=x$ for all $x\in R$ where $x^2$ of course denotes $x\cdot x$.

I have done part a but how do you do part b?

@rschwieb: Surely this has been asked $\approx 10$ times on math.SE and it costs 2 seconds to find this via google, but sometimes it is good to give an answer anyway so that the OP doesn't just copy complete solutions. — Martin Brandenburg, Oct 18 '14 at 23:38
Dear @MartinBrandenburg: Given a choice between A) (maybe) preventing a single user from successfully fishing a full answer and B) keeping our body of answers as organized as possible, I feel B is the clear winner. The chances of A being successful already seem incredibly slim in a case like this. Regards. — rschwieb, Oct 19 '14 at 00:03

score 3 · Accepted Answer · answered Oct 18 '14 at 20:15

3

Hint: Insert $x=a+b$ into the identity $x^2=x$.

answered Oct 18 '14 at 20:15

Im supposed to prove that for all x,y in R, xy=yx right. So should I just let x=a+b and let y just be y? – snowman Oct 18 '14 at 20:19
No. Insert $a=x+y$ into $a^2=a$. – Berci Oct 18 '14 at 20:22
The trick is to play with $(a+b)^2$, as the hint says. – FormerMath Oct 18 '14 at 20:29
Are we talking about part b? Cuz I have done part a. – snowman Oct 18 '14 at 20:31
I think the same hint works for part a, but making $b=a$ – FormerMath Oct 18 '14 at 20:34
I really dont know how to do it.... part b that is. I tried considering associativity of addition but got no where – snowman Oct 18 '14 at 20:46
I still can't do it...... there needs to be another element in R introduced to prove multiplicative commutativity but what do you do with thw other element say y in R? – snowman Oct 18 '14 at 21:15

score 0 · Answer 2 · answered Oct 18 '14 at 21:29

0

You want to prove $xy=yx$ for given elements $x,y \in X$. This is equivalent to $xy+yx=0$ by a.

There is nothing we can exploit from $x^2=x$ and $y^2=y$. We have to take some element which relates $x$ and $y$. So what about $x+y$?

We have $(x+y)^2=x+y$. Now expand the left hand side, cancel $x$ and $y$, and you are done.

answered Oct 18 '14 at 21:29

So x^2+2xy+y^2=x+y so x+2xy+y=x+y so 2xy=0 and so does xy+yx=0 so u equate them to get xy=yx? – snowman Oct 18 '14 at 21:40
The binomial formula $(x+y)^2=x^2+2xy+y^2$ does not hold in noncommutative rings. But we have (by the distributive law) $(x+y)^2=x^2+xy+yx+y^2$. – Martin Brandenburg Oct 18 '14 at 21:47
but how did you just get $x^2+xy+yx+y^2$ instead of $x^2+xy+xy+^2$? how did it change from xy+xy to xy+yx? – snowman Oct 18 '14 at 21:50
Please think a bit about it. Calculate $(x+y) \cdot (x+y)$ by hand. Recall the ring axioms. – Martin Brandenburg Oct 18 '14 at 23:36
Ok done it now. – snowman Oct 18 '14 at 23:54

2 Answers2