Let R a ring with subring $\mathbb{C}$. Show there is no ring homomorphism $\phi:R\to\mathbb{R}$.

My attempt: every ring homomorphism from a field to a non-zero ring is injective. So we can consider the restriction of $\phi$ to $\mathbb{C}$.

If $\phi(i)=r$, then $r^4=1\implies r^2=\pm1$.
$r^3=-r\implies r(r^2-1)=0\implies r=0 $ or $r=\pm 1$. In any case, $\phi$ is not injective. QED

Is this right?

  • $\begingroup$ But why should $\phi$ be injective? If $R=\mathbb C \times \mathbb R$ then we can just project onto the second factor. Yes, it is not injective on $\mathbb C$. $\endgroup$ – lulu Sep 15 '17 at 0:08
  • $\begingroup$ Well, the previous question was: show that any ring homomorphism $k\to R$ from a field to a non-zero ring is injective, so since the subring is also a field, then the restriction of $\phi$ must be injective, and I (think) I showed that whatever it is, it cannot be injective? $\endgroup$ – George Sep 15 '17 at 0:13
  • $\begingroup$ My error. I keep forgetting that the inclusion map for direct product of rings is not a ring homomorphism. That is, in my example, the obvious inclusion of $\mathbb C$ into the direct product is not a ring homomorphism. Thus it is not a counterexample to your claim. $\endgroup$ – lulu Sep 15 '17 at 11:29

You were on the right track, but you went a little off course.

Starting as you did, let $r=\phi(i)$.

Since $\phi$ is a ring homomorphism, we must have $\phi(1) = 1$, hence $$i^2+1 = 0 \implies \phi(i^2+1) = 0 \implies r^2 + 1 = 0$$ contradiction, since for any $r \in \mathbb{R}, r^2+1 > 0$.

Hence there is no such homomorphism.

  • 1
    $\begingroup$ Complex numbers were a mistake $\endgroup$ – George Sep 15 '17 at 0:15
  • 3
    $\begingroup$ @George: I don't follow. $\endgroup$ – quasi Sep 15 '17 at 0:17

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