Nilpotent, Idepmpotent, units in a ring

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Homework Help Overview

The discussion revolves around identifying nilpotent elements, idempotents, and units in the quotient rings F[x]/ and F[x]/. Participants are exploring the implications of not having a specified field F while attempting to characterize these elements.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the general forms of elements in the rings and attempt to derive conditions for nilpotent and idempotent elements. Questions arise regarding how to proceed without knowing the specific field, and there are attempts to factor expressions and find conditions on coefficients.

Discussion Status

There is active exploration of methods to identify elements in both cases. Some participants have provided partial calculations and are questioning the validity of their approaches, while others are clarifying the relationships between coefficients. No consensus has been reached, but there is a productive exchange of ideas and methods.

Contextual Notes

Participants note the challenge of working without a specified field, which affects their ability to pinpoint exact elements. There are also discussions about the implications of specific values for coefficients in the context of the equations being analyzed.

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Homework Statement


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


Homework Equations





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.
 
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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.
 
missavvy said:
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

Then (ad + cb)x=0
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

missavvy said:
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?
 

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