Let $z$ be a complex number so that $z = a + bi$ for some $a, b \in \mathbb{R}$. Recall that the complex conjugate $\bar z$ of $z$ is defined as $a - bi$. Let $\alpha : \mathbb{C} -> \mathbb{C}$ be the complex conjugation map taking $z$ to $\bar z$ for all $z \in \mathbb{C}$. Prove that $\alpha$ is a ring isomorphism.
Since I want to show that $\alpha$ is a ring isomorphism. I need to show that $\alpha$ is a bijective ring homomorphism.
To first show $\alpha$ is a ring homomorphism, I need to show that if I pick two elements in $\alpha$ and show that multiplication and addition is respected.
pf. Suppose $f, g \in \alpha$. So $f$ = $(a + bi)$ and $g$ = $(c + di)$. Hence, $fg = ac - bd + (ad + bc)i$. Then, $f + g = (a + bi) + (c + di) = (a + c) + (b + d)i$.
To show that this is a ring isomorphism, I need to show one-to-oneness and onto. I know to check one-to-one I need to show that if I pick two random elements, a, b I need to show $f(a) = f(b)$. I am unsure how to show this with my mapping $\alpha : \mathbb{C} -> \mathbb{C}$. Likewise, I am unsure how to show that every output has one distinct input (onto) with my mapping $\alpha : \mathbb{C} -> \mathbb{C}$.
Am I on the right track, or am I missing something?
Any feedback would be appreciated!
You are on the right track!
Injective: if $\alpha(a+bi)=\alpha(c+di)$ then $a-bi=c-di$, which implies $a=c$ and $b=d$.
Surjective: if $a+bi\in \mathbf{C}$, then $\alpha(\underbrace{a-bi}_{\in\mathbf{C}})=a+bi$.