First problem: I have set of matrices \begin{pmatrix} x & y \\ ry & x \end{pmatrix} where, $x,y \in R$, $R$ is a ring, and $r$ is a fixed element from $R$. I need to proof that this set is a ring with respect to matrix multiplication and addition.

My attempt:

  1. I need to show that it is an abelian group with respect to addition.

1.1) Associativity is quite simple

1.2)I need to determine the Identity element. The identity element here is $$\begin{pmatrix} x & y \\ ry & x \end{pmatrix} + \begin{pmatrix} e & e \\ re & e \end{pmatrix}= \begin{pmatrix} x & y \\ ry & x \end{pmatrix} $$

where $e$ is identity element from $R$

1.3) Need to determine the inverse element

$$\begin{pmatrix} x & y \\ ry & x \end{pmatrix} + \begin{pmatrix} -x & -y \\ r(e-y) & -x \end{pmatrix} = \begin{pmatrix} e & e \\ re & e \end{pmatrix}$$ 1.4) the group with respect to + is abelian

  1. To proof that multiplication is distributive with respect to addition we need just doing matrix multiplication, nothing special here.

Am I missing something in the first case?

Second problem: I have another set of matrices: $$\frac{1}{2} \begin{pmatrix} x & y \\ ay & x \end{pmatrix}$$

where $a \in Z$ and not divisible by square of prime, and $x,y \in Z$ have the same parity. How these conditions effect the proof? I don't see any differences with proof if $x,y,a \in Z$.

  • 1
    $\begingroup$ Did you check that sums and products of matrices of the form $\left(\begin{smallmatrix} x&y\\ry&x\end{smallmatrix}\right)$ are again of this form? $\endgroup$
    – Christoph
    Commented Oct 31, 2020 at 17:23
  • $\begingroup$ @Christoph yes, this is the second part, and it is very long to write it here, I wonder did I make mistake when I determine this set as an abelian group? $\endgroup$
    – John G.
    Commented Oct 31, 2020 at 17:26

1 Answer 1


Your first proof seems to be correct (You could add that is clossed under addition and multiplication).

For the second problem, you have that: $$\frac{1}{2} \begin{pmatrix} x & y\\ ay & x \end{pmatrix}\frac{1}{2} \begin{pmatrix} x' & y'\\ ay' & x' \end{pmatrix}=\frac{1}{2} \begin{pmatrix} \frac{xx'+ayy'}{2} & \frac{xy'+yx'}{2}\\ a\frac{xy'+yx'}{2} & \frac{xx'+ayy'}{2} \end{pmatrix} $$

You must check that $\frac{xx'+ayy'}{2}\in\mathbb{Z}$ and $\frac{xy'+yx'}{2}\in \mathbb{Z}$, so for that, you need to use that $a$ is not divisible by square of prime, and $x$, $y$ have the same parity.


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