Show a ring is idempotent 2

  • #1
My professor gave us this query at the end of class, it contained two parts.
1. Show a ring is idempotent
2. Consider the degree one polynomial f(x) is an element of M2(R)[x] given by
f(x) = [0 1
______0 0]x + B
(so f(x) = the matrix []x + B).
For which B is an element of M2(R), if any, is f(x) idempotent?

I proved part 1:
If a2 = a, then a2 - a = a(a-1) = 0. If a does not equal 0, then a-1 exists in R and we have a-1 = (a-1a)(a-1) = a-1[a(a-1)] = a-10 = 0, so a-1 = 0 and a = 1. Thus 0 and 1 are hte only two idempotent elements in a division ring.

Part 2 I simply have no idea on though. How do I do this with a matrix?
 

Answers and Replies

  • #2
22,089
3,286


1. Show a ring is idempotent
I really fail to understand what you mean with this. In your proof, you seem to prove that a division ring has only two idempotents. This is of course correct. Is that what you mean by this question?

2. Consider the degree one polynomial f(x) is an element of M2(R)[x] given by
f(x) = [0 1
______0 0]x + B
(so f(x) = the matrix []x + B).
For which B is an element of M2(R), if any, is f(x) idempotent?
Well, you only have to calculate f(x)2 and compare it to f(x). For what values of B are they equal??
 
  • #3


This is my progress so far...

Well, call the matrix M. Then you want that (Mx+B)^2=Mx+B in M2R[x][/math]. So we just have to perform the calculation and see what we get,

(Mx+B)^2=MxMx + MxB + BMx+B^2 = M^2x^2+(MB + BM)x + B^2 (the x-terms act like this (commutatively) by definition).

So, what is M^2? What must B be for M^2x^2+(MB + BM)x + B^2 = Mx+B?
 
  • #4
22,089
3,286


Well, you know what M is: it is just the matrix

[tex]M=\left(\begin{array}{cc} 0 & 1\\ 0 & 0\end{array}\right)[/tex]

So you can easily calculate M2. Furthermore, if you denote B by

[tex]B=\left(\begin{array}{cc} a & b\\ c & d\end{array}\right)[/tex]

then it's not hard to calculate MB and BM...
 

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