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**HallsofIvy** A "ring" is an algebraic structure that has two operations. We can think of them as "addition" and "multiplication".

Looking only at the "addition", the set of objects is a "commutative group". That is, if x and y are two members of the set then x+ y= y+ x, there exist an additive identity, "0", such that

x+ 0= x, and, for each x, there exist an additive inverse, y. such that x+ y= 0.

We only require that the "multiplication" be "associative"- x(yz)= (xy)z. There is NO requirement that the multiplication be commutative or that there be a "multiplicative identity" or that every element have a "multiplicative identity".

Here, since you are talking about a subset of "all 2 by 2 matrices", the operations are the usual ones for matrices:

$\displaystyle \begin{pmatrix}a & b \\ 0 & a\end{pmatrix}+ \begin{pmatrix}x & y \\ 0 & x\end{pmatrix}= \begin{pmatrix}a+ x & b+ y \\ 0 & a+ x\end{pmatrix}$

The fact that this operation is associative, has an additive identity, and has additive inverses follows from the fact that we "add matrices" by adding the numbers of corresponding entries in the matrices and that addition of numbers satisfies them.

$\displaystyle \begin{pmatrix}a & b \\ 0 & a \end{pmatrix}\begin{pmatrix}x & y \\ 0 & x\end{pmatrix}= \begin{pmatrix}ax & ay+ bx \\ 0 & ax\end{pmatrix}$.

Showing that multiplication is associative is a bit tedious but

$\displaystyle \left(\begin{pmatrix}a & b \\ 0 & a \end{pmatrix}\begin{pmatrix}x & y \\ 0 & x\end{pmatrix}\right)\begin{pmatrix}p & q \\ 0 & p\end{pmatrix}= \begin{pmatrix}ax & ay+ bx \\ 0 & ax\end{pmatrix}\begin{pmatrix}p & q \\ 0 & p\end{pmatrix}$

$\displaystyle = \begin{pmatrix}axp & axq+ ayp+ bxp) \\ 0 & axp \end{pmatrix}$.

Now calculate $\displaystyle \begin{pmatrix}a & b \\ 0 & \end{pmatrix}\left(\begin{pmatrix} x & y \\ 0 & x\end{pmatrix}\begin{pmatrix}p & q \\ 0 & p\end{pmatrix}\right)$

and show that you get the same thing.