Center of a ring matrix

Question

I have the following ring matrix:

Ep=

    a     b
    pc   pu+v

where a,b,c,u,v belongs to Zp where p is a prime

now the center of this matrix is:

Z(Ep)=

    x   0    
    0   py+x

where x,y belongs to Zp

My Question is how the center was calculated?

Thank you so much for the help

Answer 1

Assuming $\mathbb{Z}_p$ means the field with $p$ elements, then your ring is $$ R=\left\{ \begin{bmatrix} a & b \\ 0 & v \end{bmatrix} \;\middle|\;\; a,b,v\in\mathbb{Z}_p \right\} $$ A matrix $\left[\begin{smallmatrix}x&y\\0&z\end{smallmatrix}\right]$ is in the center if $$ \begin{bmatrix} a & b \\ 0 & v \end{bmatrix} \begin{bmatrix}x&y\\0&z\end{bmatrix} = \begin{bmatrix}x&y\\0&z\end{bmatrix} \begin{bmatrix} a & b \\ 0 & v \end{bmatrix} $$ for all $a,b,v\in\mathbb{Z}_p$; in other words, doing the products and comparing each entry, $$ \begin{cases} ax=xa\\ ay+bz=xb+yv\\ vz=zv \end{cases} $$ The first and third equalities hold, so what we must have is $$ bx+(v-a)y-bz=0 $$ for all $a,b,v$.

Now use particular values of $a$, $b$ and $v$ to reach the conclusion.

Setting $b=0$, $v=1$, $a=0$ we get $y=0$.

Setting $a=0$, $v=0$ and $b=1$ we get $x=z$.

The center is the set of scalar matrices.

---

Note that this doesn't use any special property of $\mathbb{Z}_p$, so the argument works for any triangular matrix ring with coefficients in a commutative ring.

If the coefficient ring is not commutative, the first and third equation tell us that $x$ and $z$ must be in the center of the coefficient ring. Setting $b=0$, $v=1$, $a=0$ in the middle one again says $y=0$, while $a=0$, $v=0$ and $b=1$ gives $x=z$. So in this case the center is the set of scalar matrices with coefficients in the center of the coefficient ring.