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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?

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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, $f(i^2) = f(i)*f(i) = x^2 > 0$.

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  • $\begingroup$ 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 $\endgroup$ – rschwieb Oct 28 '15 at 3:59
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Hint: Suppose that $\ker\phi=\{0\}$, then $i$ must be mapped to something nonzero under $f$, say $x$. Then you have that $f(-1) = f(i^2) = f(i)^2 = x^2$.

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    $\begingroup$ 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. $\endgroup$ – 211792 Oct 28 '15 at 0:58
  • $\begingroup$ Yeah I made a typo. Thanks for catching that. $\endgroup$ – Cameron Williams Oct 28 '15 at 0:59
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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.

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