Let \(F\) be a field and let \(R\) be the set of matrices \[ \begin{pmatrix}a&-b\\ b&a\end{pmatrix}, \qquad a,b\in F. \] Show that \(R\) is a commutative ring with identity under the usual matrix operations. Determine for which of \[ F=\mathbb Q,\qquad \mathbb C,\qquad \mathbb Z_5,\qquad \mathbb Z_7 \] the ring \(R\) is a field.
Solution
For \(a,b\in F\), write \[ M(a,b)= \begin{pmatrix} a&-b\\ b&a \end{pmatrix}. \]
1. The set \(R\) is closed under addition, additive inverses, and multiplication, and contains the identity matrix.
For \(a,b,c,d\in F\), \[ M(a,b)+M(c,d)=M(a+c,b+d), \] \[ -M(a,b)=M(-a,-b), \] and direct multiplication gives \[ M(a,b)M(c,d) = M(ac-bd,ad+bc). \] Also \[ I=M(1,0). \] Thus \(R\) is a subring of \(M_2(F)\) containing the identity.
2. The ring \(R\) is commutative.
The multiplication formula from step 1 gives \[ M(a,b)M(c,d) = M(ac-bd,ad+bc). \] Since multiplication and addition in \(F\) are commutative, \[ ac-bd=ca-db \] and \[ ad+bc=cb+da. \] Hence \[ M(a,b)M(c,d)=M(c,d)M(a,b). \]
3. A nonzero element \(M(a,b)\) of \(R\) is invertible in \(R\) if and only if\[ a^2+b^2\neq0. \]
The determinant is \[ \det M(a,b)=a^2+b^2. \] Thus invertibility as a matrix requires and is implied by \[ a^2+b^2\neq0. \] When this holds, \[ M(a,b)^{-1} = \frac1{a^2+b^2} \begin{pmatrix} a&b\\ -b&a \end{pmatrix} = M\left( \frac{a}{a^2+b^2}, -\frac{b}{a^2+b^2} \right), \] which belongs to \(R\).
4. The ring \(R\) is a field if and only if \(-1\) is not a square in \(F\).
By step 3, \(R\) fails to be a field exactly when there is a nonzero pair \((a,b)\) such that \[ a^2+b^2=0. \] For such a pair, \(b\neq0\), since \(b=0\) would force \(a=0\). Dividing by \(b^2\) gives \[ \left(\frac ab\right)^2=-1. \] Thus failure of fieldness implies that \(-1\) is a square.
Conversely, if \(c^2=-1\) in \(F\), then \[ M(c,1)\neq0 \] but \[ \det M(c,1)=c^2+1=0. \] Hence \(R\) has a nonzero noninvertible element and is not a field.
5. For \(F=\QQ\), the ring \(R\) is a field.
No rational square equals \(-1\), since every rational square is nonnegative as a real number. Apply step 4.
6. For \(F=\CC\), the ring \(R\) is not a field.
In \(\CC\), \[ i^2=-1. \] Apply step 4.
7. For \(F=\ZZ_5\), the ring \(R\) is not a field.
Modulo \(5\), \[ 2^2=4=-1. \] Apply step 4.
8. For \(F=\ZZ_7\), the ring \(R\) is a field.
The nonzero squares modulo \(7\) are \[ 1^2=6^2=1, \qquad 2^2=5^2=4, \qquad 3^2=4^2=2. \] Thus the set of nonzero squares is \[ \{1,2,4\}, \] which does not contain \[ -1=6. \] Apply step 4.
9. Therefore\[ \boxed{ R\text{ is a field precisely for }F=\QQ\text{ and }F=\ZZ_7. } \]
This follows from steps <1>5–<1>8.
10. Q.E.D.
Steps 1 and 2 prove that \(R\) is a commutative ring with identity, and step 9 gives the requested field classification.