Compute the zero divisors and ideals of $ \left\{ \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \: : \: a,b \in \mathbb{Z}_{5} \right\}$ Let consider the following subset of  $M_{2\times 2}(\mathbb{Z}_{5})$
$$A:= \left\{ \begin{pmatrix} a & b \\
                             -b & a \end{pmatrix} \: : \: a,b \in \mathbb{Z}_{5} \right\}.$$
(1) Prove  that $A$ is subring of $M_{2\times 2}(\mathbb{Z}_{5})$.
(2) Prove that $\begin{pmatrix} a & b \\
                             -b & a \end{pmatrix}$ is a zero divisor if and only if $a^{2}+b^{2}=0$, and compute all the zero divisors of $A$.
(3) Compute all the ideals of $A$.
I proved (1) straightforward, but the problem arives trying to prove (2). For this I take a matrix in  $A$:  $\begin{pmatrix} a & b \\
                             -b & a \end{pmatrix}$, such that $a^{2}+b^{2}=0$. The I want to show there is another matrix lets say $B=\begin{pmatrix} c & d \\
                             -d & c \end{pmatrix} \neq \begin{pmatrix} 0 & 0 \\
                             0 & 0 \end{pmatrix}$ so that
$$\begin{pmatrix} a & b \\
                             -b & a \end{pmatrix} \begin{pmatrix} c & d \\
                             -d & c \end{pmatrix}=\begin{pmatrix} 0 & 0 \\
                             0 & 0 \end{pmatrix}.$$
I tried to compute all the entries of this matrix product which are all equal to $0$ in $\mathbb{Z}_{5}$. But I dont know how to take the values of this matrix $B$. And not sure how to use the hypothesis $a^{2}+b^{2}=0$ for either two implications. And to compute all the zero divisors of $A$ I guess I should test all the six elements of $\mathbb{Z}_{5}$ satisfying $a^{2}+b^{2}=0$. And for (3) Im run out of ideas.
EDIT: For (2) I´ve just noticed the following pairs $(a,b)$ where $a,b \in \mathbb{Z}_{5}$ satisfy $a^{2}+b^{2}=0$:  $(1,2),(2,1),(1,3), (3,1), (0,0), (2,4), (4,2)$. Let me know if there if this is correct? If correct, are there any other choices for $\mathbb{Z}_{5}$?
 A: Hint. Show that $A\simeq\mathbb Z_5[X]/(X^2+1)\simeq\mathbb Z_5\times\mathbb Z_5$. (You should get $\begin{pmatrix} a & b \\-b & a \end{pmatrix}\mapsto(a+2b,a-2b)$.)
The isomorphism gives all you want, including the fact that the ring $A$ has exactly four ideals.
A: The $(a,b)$ pairs such that $a^2 + b^2 = 0$ in $\mathbb Z_5$ are $(0,0), (1,2), (1,3), (2,1), (2,4), (3,1),(3,4),(4,2), (4,3)$
If $a^2 + b^2 = 0$ then $\begin{bmatrix} a & b\\ -b & a\end{bmatrix}\begin{bmatrix} b & a\\ -a & b\end{bmatrix} = 0$
It is still open to prove the other direction.
The (non-trivial) ideals
If we pick one of the pairs... e.g. (1,2) and start multiplying by all the elements in $A.$  It might be worth noting that this ring is commutative.
(1,2)(1,0) = (1,2)
(1,2)(2,0) = (2,4)
(1,2)(3,0) = (3,1)
(1,2)(4,0) = (4,3)
(1,2)(2,1) = (0,0)
(1,2)(3,1) = (1,2)(2,1) + (1,2)(1,0) = (1,2)
(1,2)(4,1) = (2,4)
(1,2)(0,1) = (3,1)
(1,2)(1,1) = (4,3)
(1,2)(4,2) = (0,0), etc.
It looks likes $\{(0,0),(1,2),(2,4),(3,1),(4,3)\}$ is a non-trivial ideal.
And it should be the case that $\{(0,0),(2,1),(4,2),(1,3),(3,4)\}$ will also form and ideal.  And their union will form an ideal.
And the trivial ideals.
