# 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 $$\mathbb{Z}_{5}$$?

• Thanks for checking out my problem Brian! Its a mocking test Im trying to solve for studying purposes, I dont know where this problem is taken from. The notation abuse is mine, it appears in the exam as $M_{2 \times 2}$ @BrianMoehring
– Sok
Commented May 14, 2021 at 21:47

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.

• Thanks!! But Im wondering how you came up with this idea? @user26857
– Sok
Commented May 14, 2021 at 23:39
• @Sok I've learned from basic algebra lectures that the field of complex numbers is isomorphic to $\mathbb R[X]/(X^2+1)$ and in the same time with the ring of matrices $\begin{pmatrix} a & b \\-b & a \end{pmatrix}$ with $a,b\in\mathbb R$. The last isomorphism is just the Chinese Remainder Theorem for the ideals $(X-2)$ and $(X+2)$ whose product is noting but the ideal $(X^2+1)$. Commented May 14, 2021 at 23:41

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.

• Thanks! I already got the other direction by my own. Btw, Im pretty sure that $(0,5), (5,0)$ are pairs such that $a^{2} + b^{2}=0$ in $\mathbb{Z}_{5}$ isnt?
– Sok
Commented May 14, 2021 at 23:30
• 5 is equivalent to 0. Commented May 14, 2021 at 23:34
• Sure! Got it! My mistake
– Sok
Commented May 14, 2021 at 23:36
• Do you have an idea in order to compute the ideals of $A$?? @DougM
– Sok
Commented May 15, 2021 at 0:18
• Sorry but how we get from multiplying elements from matrix in $A$ and $R$ to mutiply couples of elements of $\mathbb{Z}_{5}$? @Doug_M
– Sok
Commented May 16, 2021 at 18:58