Is f(x) Idempotent for Any Matrix B in M2(R)?

[DanielThrice]

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?

 

[micromass]

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??

 

[DanielThrice]

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?

 

[micromass]

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

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

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

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

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