Let R b e the ring of omplex numb ers C. Show that the map φ : R → Rgiven by φ(z) = ¯z (where ¯z denotes the omplex onjugate of z) is a ringisomorphism.

To show that the map φ : R → R given by φ(z) = z̄ is a ring isomorphism, we need to show that it is a bijective ring homomorphism.

1. Ring homomorphism: A function φ : R → R is a ring homomorphism if for all a, b in R, the following two conditions hold:

   i. φ(a + b) = φ(a) + φ(b)

   ii. φ(a * b) = φ(a) * φ(b)

   For φ(z) = z̄, both conditions hold because the complex conjugate of a sum or product of two complex numbers is the sum or product of the complex conjugates of the numbers.

2. Injective: A function is injective (or one-to-one) if every element of R maps to a unique element in R. For φ(z) = z̄, every complex number has a unique complex conjugate, so φ is injective.

3. Surjective: A function is surjective (or onto) if every element of R is the image of at least one element in R. For φ(z) = z̄, every complex number is the complex conjugate of its own complex conjugate, so φ is surjective.

Therefore, the map φ : R → R given by φ(z) = z̄ is a ring isomorphism.