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

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  • $\begingroup$ 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 $\endgroup$
    – Sok
    Commented May 14, 2021 at 21:47

2 Answers 2

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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.

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  • $\begingroup$ Thanks!! But Im wondering how you came up with this idea? @user26857 $\endgroup$
    – Sok
    Commented May 14, 2021 at 23:39
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    $\begingroup$ @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 $(X^2+1)$. $\endgroup$
    – user26857
    Commented May 14, 2021 at 23:41
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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.

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  • $\begingroup$ 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? $\endgroup$
    – Sok
    Commented May 14, 2021 at 23:30
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    $\begingroup$ 5 is equivalent to 0. $\endgroup$
    – user317176
    Commented May 14, 2021 at 23:34
  • $\begingroup$ Sure! Got it! My mistake $\endgroup$
    – Sok
    Commented May 14, 2021 at 23:36
  • $\begingroup$ Do you have an idea in order to compute the ideals of $A$?? @DougM $\endgroup$
    – Sok
    Commented May 15, 2021 at 0:18
  • $\begingroup$ 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 $\endgroup$
    – Sok
    Commented May 16, 2021 at 18:58

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