Z^* = {1, -1}.

Every nonzero element of Q, R, and C is invertible. (In other words, they are fields.)

(Z/nZ)^* consists of the classes of m with gcd(m,n) = 1.

(Z[X])^* = {1, -1},

(Q[X])^* = Q^* = Q\{0}, and similarly for R[X] and C[X]. To prove these statements you will need to involve the degree. We leave this to the reader.

If R = Q, R or C, and f is a polynomial in R[X] of positive degree, then (R[X]/(f))^* consists of residue classes of those polynomials g in R[X] for which gcd(g,f) = 1.

If R = Z, then it is harder to describe the invertible elements for general f.

(Z+Zi)^* = {1, -1, i, -i}.

If a + bi is invertible, then there exists an element c + di such that

(a + bi)(c + di) = 1.

Using the property |z||w| = |zw| for the absolute value of complex numbers, we infer that

(a² + b²)(c² + d²) = 1.

Since a, b, c, d are integers, we find that the integer a² + b² divides 1. Then the conclusion that a + bi must be one of the four elements 1, -1, i, -i is obvious.

The invertible elements of M_n(R) are those whose determinant is nonzero.

They are the so-called transformation matrices.

H^* = {ae + bi + cj + dk | a² + b² + c² + d² R^*}

The proof is left to the reader!