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Let $R$ be a commutative, unital ring. Define $$ R[\mathbf{t}]= \left\{ \begin{bmatrix} \mathbf{w} & \mathbf{z} \\ -\mathbf{z} & \mathbf{w}-\mathbf{z} \end{bmatrix}\in R_2^2\;\middle|\; \mathbf{w},\mathbf{z}\in R \right\}. $$ Show that $R[\mathbf{t}]$ forms a commutative, unital ring under the usual matrix addition and multiplication.

I'm struggling with proving that rings of matrices maintain characteristics of the ring that their elements are in. This is an example problem I've found that I can't quite figure out where to start at. If anyone could give an example solution for me to examine and question I would be very grateful!

egreg
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Se yaJ Me
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    Check that the sum, product and scalar product of matrices of that form result in more matrices of that form. Check the identity element is of that form, and then check $AB=BA$ for general matrices $A$ and $B$ of the given form. Start writing! – anon Feb 04 '14 at 21:08
  • @anon thanks! I'll give this a try in a moment! – Se yaJ Me Feb 04 '14 at 21:17
  • By the way, this "exercise" comes from considering the commutative ring $R[x]/(x^2-x+1)$ and writing it down very complicated. – Martin Brandenburg Feb 05 '14 at 00:18

2 Answers2

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Since $R$ is a unital ring, there are two obvious elements around to play with: the additive identity $0$ and the multiplicative identity $1$. Given that, it may be helpful to note that there are two distinguished elements of $R[t]$: $$I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\ (w=1,z=0),\qquad J=\begin{bmatrix}0&1\\-1&-1\end{bmatrix}(w=0,z=1).$$ Moreover, we can write any element $A\in R[t]$ as $A=wI+zJ$ for some $w,z\in R$ with the usual matrix operations, and conversely by definition any element of $R[t]$ has this form.

Can you see how to use this decomposition to prove the statements @anon mentioned?

HINTS BELOW:

$wI+w'I=(w+w')I$...

$IJ=JI=J$

$J^2=J-I$

Sean Clark
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You need to start at the definition of ring. You just need to show that all the ring axioms indeed hold.

You can take two generic elements of the set and show that their sum lands back in the set, and their product lands back in the set. Here are two generic elements:

$X=\begin{bmatrix}a&b\\-b&a-b\end{bmatrix}$, and $Y=\begin{bmatrix}c&d\\-d&c-d\end{bmatrix}$

Adding these two:

$\begin{bmatrix}a&b\\-b&a-b\end{bmatrix}+\begin{bmatrix}c&d\\-d&c-d\end{bmatrix}=\begin{bmatrix}a+c&b+d\\-b-d&a+c-b-d\end{bmatrix}=\begin{bmatrix}a+c&b+d\\-(b+d)&(a+c)-(b+d)\end{bmatrix}$.

See? The last matrix is in the set! Do the same thing for the product $XY$.

There are still more axioms to be checked, namely distributivity and associativity. You don't have to do matrix computations for these, though. Argue that associativity is inherited from the ring containing it. Argue that things distribute inside this ring because they distribute in the big ring.

Finally, can you see why the zero matrix is in the set? And can you see what the identity is? These are the last two things you need to confirm for the set to be a ring with unity under this multiplication and addition.

rschwieb
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