# If $f : \mathbb{C} → \mathbb{R}$ is a ring homomorphism, then $f$ must be trivial

My problem is

Show that if $f : \mathbb{C} → \mathbb{R}$ is a ring homomorphism, then $f$ must be trivial, i.e. $f(a) = 0$ for all $a ∈ \mathbb{C}$.

Let $\phi$ be the ring homorphism. I think I should start with the fact that $ker(\phi)$ is an ideal and as the only ideals in $\mathbb{C}$ are $\{0\}$ and $\mathbb{C}$ (because $\mathbb{C}$ is a field) then $ker(\phi)=\{0\}$ or $\mathbb{C}$.

If $ker(\phi)=\mathbb{C}$ we have that $\phi$ is the trivial homomorphism.

But if $ker(\phi)=\{0\}$ this gives that $\phi$ is one-to-one. I don't know what to do here. Is it known that there exists no one-to-one ring homomorphism from $\mathbb{C}$ to $\mathbb{R}$? How would I show this?

If it is not the trivial ring, then $i$ must be mapped to something. So suppose $f(i) = x \in \mathbb{R}$. Then you have $f(i^2) = f(-1) = -f(1) = -1$ since $f$ is a homomorphism. But then, $f(i^2) = f(i)*f(i) = x^2 > 0$.

• Carefully noting that $f(1)^2=f(1)$ implies $f(1)=1$ because $\Bbb R$ is a domain, and not suggesting that $f$ is unital a priori, since that would not jive with allowing $f$ to be zero... +1 – rschwieb Oct 28 '15 at 3:59

Hint: Suppose that $\ker\phi=\{0\}$, then $i$ must be mapped to something nonzero under $f$, say $x$. Then you have that $f(-1) = f(i^2) = f(i)^2 = x^2$.

• Did you intend $\text{ker}\phi = \lbrace 0 \rbrace$? Because it could be the case that the kernel is all of $\mathbb{C}$, in which case $i$ is not necessarily mapped to something nonzero. – 211792 Oct 28 '15 at 0:58
• Yeah I made a typo. Thanks for catching that. – Cameron Williams Oct 28 '15 at 0:59

An answer which perhaps gives intuition on the problem.

Suppose $$f:\mathbb{C}\rightarrow \mathbb{R}$$ is a ring homomorphism. If we can show that the kernel is the entirety of $$\mathbb{C}$$ then we show that $$f$$ is trivial. (because for any $$a\in \mathbb{C}=Ker(f)$$, $$f(a)=0$$)

Then we have that

$$Ker(f) = \{a+bi \quad | f(a+bi) = 0\}=\{a+bi \quad | f(a)+f(b)f(i)=0 \}=\{a+bi \quad |f(a)f(i)^4+f(b)f(i) \}=\{a+bi\quad |f(i)(f(a)f(i)^3+f(b))=0\}$$

Now we will show that $$f(i)=0$$

First observe that $$f(i)^3 = f(i^3)=f(-i)=-f(i)$$ or $$f(i)( f(i)^2+1 )= 0$$. Notice that since $$f(i)$$ is in $$\mathbb{R}$$, then there is no element where $$f(i)^2+1=0$$. Which forces $$f(i)=0$$

Going back to the kernel, we sub this in to get

$$Ker(f) = \{a+bi \quad | 0(f(a)f(i)^3+f(b))=0\}=\{a+bi \quad | 0=0 \}=\mathbb{C}$$

In other words, the kernel of $$f$$ is the entire $$\mathbb{C}$$, and hence $$f$$ is trivial.