# Nilpotent, Idepmpotent, units in a ring

## Homework Statement

Determine the nilpotent, idempotents and units in
a) F[x]/<x2-x>
b) F[x]/<x2>

## The Attempt at a Solution

How do I do this without a specified Field?

For a) the elements in R would be {a+bx: a,b are in F; x2=x}
b) {a+bx: a,b in F; x2 =0}
I know what it means to be idempotent, nilpotent and a unit, I just don't understand how I can pin point exact elements without knowing the field.

For (a), just take an element in F[X]/(X2-X). This always has the form aX+b. Now you need to find for which a and b, we have (aX+b)2=0. This a and b will in general be dependent of the field F (which you do not know in this case), however in this case the answer can be described without knowint the field in question.

So just take aX+b, and calculate it's square.

Is it the same for b) ? They will also be of the form ax +b right?

Yes, (b) will use thesame method!

So I get (ax+b)(ax+b) = a2x2 + 2bax + b2 = ax+b
and since x2 = x here, then a2x + 2bax + b2 = ax + b

Whats the best way to factor this? I have ba together so I'm not sure.
I tried

ax(a+ 2b -1) = b-b^2..

I got 0,1 as idempotents...

...sometimes it's the simple things that end up confusing me.

So I get (ax+b)(ax+b) = a2x2 + 2bax + b2 = ax+b
and since x2 = x here, then a2x + 2bax + b2 = ax + b

OK, this is good, so you have that $$(a^2+2ba)x+b^2=ax+b$$. Now you'll have to find conditions on a and b. So for the above equation to hold, it must be true that

$$a^2+2ba=a$$ and $$b^2=b$$.

Now, can you get conditions on a and b from this?

Ok, I think I got it..
I'm working on b), where h(x)= x^2 for the units:
So then elements are of the form

{ax+b: a,b in F, x^2 =0}
Not sure if it's right, but I took
(ax+b)(cx+d) = 1

and expanded to get
axd + bcx + bd = 1

and bd = 1

So b=d=1 or b=d=-1
-> a=-c

Err, am I going somewhere with this?

Most of it seems good, but

So b=d=1 or b=d=-1

how did you arrive there? Can't b=2 and d=1/2 also be a good pick?