Let $\theta : \mathbb{C} → \mathbb{R}$ be a homomorphism. Prove that $\theta(x) = 0$ for all $x \in \mathbb{C}$.

Question

Let $\theta: \mathbb{C} \to \mathbb{R}$ be a homomorphism. Prove that $\theta(x) = 0$ for all $x\in \mathbb{C}$.

All I really know is the following:

Let $a + bi, c + di\in \mathbb{C}$.

I don't know what $θ(a +bi)$ equals. Am I meant to give a rule for the function? If so the most natural rule is $θ(a +bi)$ = a

Thus $θ((a+bi) + (c+di))$ = a + b and $θ((a+bi)(c+di))$ = ac.

^Is what I even wrote correct? If so, where do I go from here?

Answer 1

Since $\theta$ is a homomorphism, we have $ 0 = \theta(0) = \theta(i^2 + 1^2) = \theta(i^2) + \theta(1^2) = \theta(i)^2 + \theta(1)^2. $ But then, since $\theta(i), \theta(1) \in \mathbb{R}$ we must have $\theta(i) = \theta(1) = 0.$

Thus, $\theta(x1+yi) = \theta(x) \theta(1) + \theta(y) \theta(i) = \theta(x)0 + \theta(y)0 = 0.$

Answer 2

You have $θ(1)^2 = θ(1^2) = θ(1)$. There are only two elements $z ∈ ℂ$ with $z^2 = z$, namely $0$ and $1$. (Consider $z^2 - z = z(1 -z)$ and use that $ℂ$ is a domain to see that there are no others.)

• If $θ(1) = 1$, then $θ(\mathrm i)^2 = θ(\mathrm i^2) = θ(-1) = -θ(1) = -1$. Hm.
• If $θ(1) = 0$, then for all $x ∈ ℂ$, $θ(x) = θ(x·1) = …$.