1. ## Matrix Ring

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?

2. Well, call the matrix M. Then you want that $(Mx+B)^2=Mx+B$ in $M_2(\mathbb{R})[x]$. So just perform the calculation and see what you 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 $M^2x^2+(MB + BM)x + B^2 = Mx+B$? (Put $B=\left(\begin{array}{cc}a&b\\c&d\end{array}\right )$ and see what happens)

3. So for this to be true, M^2 would have to equal M/x, and B would have to be it's own multiplicative inverse right?

4. Originally Posted by DanielThrice
So for this to be true, M^2 would have to equal M/x, and B would have to be it's own multiplicative inverse right?
No. You know what $M^2$ is, you can just work it out (as you know what M is). And if B was its own multiplicative inverse than B^2=I, but you need $B^2=B$.

Basically, you have $\left(\begin{array}{cc}0 & 1\\0&0\end{array}\right)^2+\left(\left(\begin{arra y}{cc}0 & 1\\0&0\end{array}\right)B + B\left(\begin{array}{cc}0 & 1\\0&0\end{array}\right)\right)x + B^2 = \left(\begin{array}{cc}0 & 1\\0&0\end{array}\right)x+B$.

Now, if $B=\left(\begin{array}{cc}a & b\\c&d\end{array}\right)$ then you basically have to solve the equation,

$\left(\begin{array}{cc}0 & 1\\0&0\end{array}\right)^2+\left(\left(\begin{arra y}{cc}0 & 1\\0&0\end{array}\right)\left(\begin{array}{cc}a & b\\c&d\end{array}\right) + \left(\begin{array}{cc}a & b\\c&d\end{array}\right)\left(\begin{array}{cc}0 & 1\\0&0\end{array}\right)\right)x + \left(\begin{array}{cc}a & b\\c&d\end{array}\right)^2$ $= \left(\begin{array}{cc}0 & 1\\0&0\end{array}\right)x+\left(\begin{array}{cc}a & b\\c&d\end{array}\right)$ for a, b, c and d. So do this.