Let \(R\) be a ring with identity, and let \(G(R)\) denote its set of units.
- Prove that \(G(R)\) is a group under multiplication.
- Let \(R=\ZZ[i]\). Prove that \[ G(R)\cong\ZZ/4\ZZ. \]
Solution
1. \(G(R)\) is a group under multiplication.
Multiplication on \(G(R)\) is associative because multiplication in \(R\) is associative, and the identity \(1\) is a unit. If \(x,y\in G(R)\), then \[ (xy)(y^{-1}x^{-1}) =x(yy^{-1})x^{-1} =1 \] and \[ (y^{-1}x^{-1})(xy) =y^{-1}(x^{-1}x)y =1. \] Thus \(xy\) is a unit, with inverse \(y^{-1}x^{-1}\), so \(G(R)\) is closed under multiplication. Finally, every \(x\in G(R)\) has its two-sided inverse \(x^{-1}\in G(R)\). Hence all group axioms hold.
2. The units of \(\ZZ[i]\) are exactly\[ G(\ZZ[i])=\{1,-1,i,-i\}. \]
For \(z=a+bi\in\ZZ[i]\), define \[ N(z)\coloneqq z\overline z=a^2+b^2. \] The norm is multiplicative. If \(z\) is a unit with inverse \(w\in\ZZ[i]\), then \[ 1=N(1)=N(zw)=N(z)N(w). \] Both \(N(z)\) and \(N(w)\) are nonnegative integers, so \(N(z)=1\). Therefore \[ a^2+b^2=1, \] which gives \(z\in\{1,-1,i,-i\}\). Conversely, each of these four elements is a unit.
3. \(G(\ZZ[i])\cong\boxed{\ZZ/4\ZZ}\).
2. Thus the unit group is cyclic of order \(4\), hence isomorphic to \(\ZZ/4\ZZ\).
4. Q.E.D.
Step 1 proves part 1, and steps <1>2–<1>3 prove part 2.