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I have a question about the definition on this page.

I feel like conditions 1 and 2 imply 3, because a group homomorphism maps identity to identity.

Am I missing something? The proof of the lemma on that page uses condition 3 to assert that $1_F$ is not in the kernel of $\psi$, but I fail to see why this is not already implied by the first two conditions.


Finally, I would like to prove the lemma without the use of ideals. Could someone give me a hint as to how to show that the kernel of $\psi$ is trivial?

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Consider the map $f:\Bbb R\to\Bbb R$ such that $f(x)=0$ for all $x\in\Bbb R$: $f$ satisfies (1) and (2) but not (3).

To prove that if $\psi:F\to K$ is a field homomorphism, then $\ker\psi$ is trivial, let $x\in F\setminus\{0_F\}$; we want to show that $\psi(x)\ne 0_K$. Suppose that it is, and consider $\psi(x\cdot x^{-1})$; on the one hand it’s $\psi(1_F)=1_K\ne 0_K$, but on the other hand it’s ... ?

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  • $\begingroup$ Thanks! I have a few more questions: 1. Are there other simple counterexamples for $\mathbb{R}\rightarrow\mathbb{R}$? What about for other fields? 2. I see that the first part of (3) is necessary, but what about the second part of (3)? I ask this because I have seen another definition of field homomorphism that omits it; moreover, this counterexample does not violate it. $\endgroup$ – angryavian Jul 8 '13 at 9:02
  • $\begingroup$ @blf: You’re welcome. The second part of (3) follows from (1): $\psi(0_F)=\psi(0_F+0_F)=\psi(0_F)+\psi(0_F)$, and the only solution to $x+x=x$ in $K$ is $x=0_K$. For the first question, note that (1) implies that $\psi(1_F)$ satisfies $x^2=x$, or $x(x-1_K)=0_K$; this has only the solutions $x=0_K$ and $x=1_K$, so either $\psi$ is a field homomorphism, or it’s like my example and collapses everything to $0_K$. $\endgroup$ – Brian M. Scott Jul 8 '13 at 9:07
  • $\begingroup$ Ah ok, so it is the existence of $0_K$ as an "extra" solution to $x^2 = x$ that makes it more complicated than an abelian group under multiplication. Thanks for the great explanations! $\endgroup$ – angryavian Jul 8 '13 at 9:20
  • $\begingroup$ @blf: You’re welcome! $\endgroup$ – Brian M. Scott Jul 8 '13 at 9:21
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supposing that $\psi:F\to K$ has the property $\psi(a+_F b) = \psi(a)+_K \psi(b)$, consider $\psi(0_F) = \psi(0_F+_F 0_F) = \psi(0_F)+_K \psi(0_F)$, hence, $\psi(0_F) = 0_K$... That's what I see...

But as for this map $f(x) = 0$ for all $x\in\mathbb{R}$, it doesn't have the property $f(1) = 1$... or I don't unerstand something...

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The kernel is trivial because if $f(a) = 0$ for some nonzero $a$, then $f(1) = f(aa^{-1}) = f(a)f(a^{-1}) = 0$ contradicting 3.

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