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Let $F$ be a field and define R to be: \begin{bmatrix}a&-b\\b&a\end{bmatrix}

  1. Show that with the usual matrix addition and multiplication, $R$ is a commutative ring with identity.
  2. Is $R$ a field if $F =Q,R,Z_5$?

For the first question is all I have to show is the matrix provided can be reduced to the identy matrix. So for example \begin{bmatrix}a&-b\\b&a\end{bmatrix}+\begin{bmatrix}-a+1&b\\-b&1a+1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix} and the inverse of this matrix is \begin{bmatrix}a/(b^2+a^2)&b/(b^2+a^2)\\-b/(b^2+a^2)&a/(b^2+a^2)\end{bmatrix} which by mutilpication of the orginal matrix gives us the identy matrix. So this matrix is a commutative ring...

I'm new to ring theory so I am not sure how to approach this second question. I know to be a field you have to be an abelian group which I am not sure how to check with matricies...

  • One way is to check that $a+ib\mapsto\begin{pmatrix}a&-b\b&a\end{pmatrix}$ preserves multiplication and addition. Then, in the case of $F$ being $\mathbb{Q}$ or $\mathbb{R}$ you have an isomorphism with the complex numbers, either with rational components or all the complex numbers. – conditionalMethod Nov 03 '19 at 20:15

2 Answers2

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For the first question is all I have to show is the matrix provided can be reduced to the identy matrix.

This makes no sense. What you've shown is that $x + (1 - x) = 1$ which doesn't have anything to do with showing that this set is a ring.

What you need to show is that the set is closed under matrix addition and multiplication (this makes it a ring or more accurately, a subring of the full matrix ring). Then you need to check that multiplication is commutative (which isn't generally the case for matrices. It is clear that the zero matrix and the identity matrix are in this set.

For question 2, write the matrix as $a I + b J$ where

$$ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, J = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}.$$

Notice that $J^2 = -I$. This should make you think of complex numbers. The question is, what happens if your field already has a square root of $-1$?

Trevor Gunn
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Starting with $\Bbb Q$ gives the field $\Bbb Q(i)$, starting with $\Bbb R$ gives the complex numbers $\Bbb C$. The matrix $A(a, b)$ is mapped to $a+ib$. This will be a field isomorphism, see the following duplicates:

Complex number isomorphic to certain $2\times 2$ matrices?

Prove that a mapping from C to M2(R) is injective and a homomorphismm

The field $\Bbb F_5$ has already a square root of $-1$, since $2^2=-1$.

Dietrich Burde
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