# Thread: Rings- a^2 = a then R must be commutative

1. ## Rings- a^2 = a then R must be commutative

Hi, just need a hint with this one...

Let R be a ring such that a^2 = a for every a in R. Show that R must be commutative

Oh and also having some trouble with...

Show that $\bb{Z}_{n}$ is a field if and only if n is a prime.

2. Originally Posted by liedora
Let R be a ring such that a^2 = a for every a in R. Show that R must be commutative

Using $(x+x)^2=x+x$ prove that

$x=-x,\quad \forall x\in R$

Using $(a+b)^2=a+b$ prove that

$ba=-ab,\quad \forall a,b \in R$

Conclude.

Fernando Revilla

3. Originally Posted by liedora
Show that $\bb{Z}_{n}$ is a field if and only if n is a prime.
Hint ;

If $n=pq$ then, $[p][q]=[0]$

Fernando Revilla

4. Originally Posted by FernandoRevilla
Hint ;

If $n=pq$ then, $[p][q]=[0]$

Fernando Revilla
...and for the other way, use the Euclidean algorithm...

5. Ok managed to get the first question out, but still having difficulties with the second. Any more hints would be greatly appreciated!

6. Originally Posted by liedora
Ok managed to get the first question out, but still having difficulties with the second. Any more hints would be greatly appreciated!
What have you tried?. At any rate, if you know that a finite integral domain is a field then, this fact can help you.

Fernando Revilla

7. I don't know how much of a stickler your instructor is, but (presumably - I mean I can't imagine you not) you've been working with [tex]\mathbb{Z}_n[\math] quite a bit so you get a lot free in the definition of a field. I would say the only thing you need prove is that [tex]\mathbb{Z}_n[\math] contains mult. inverses (as far as going in the forward direction is concerned - going backwards ". . is prime. . .field" shouldn't be too hard).