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Let $F$ be a field and $R$ a commutative ring where $R$ is not the zero ring. Suppose $\varphi : F \rightarrow R$ is a ring homomorphism. Show that $\varphi$ is injective.

Any help would be appreciated. I feel pretty lost on this.

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    $\begingroup$ The kernel is an ideal of $F$. Now, what are the only ideals of $F$? $\endgroup$ – Crostul Mar 24 '17 at 0:17
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$\varphi$ is zero homomorphism, or it is injective:

Assume that $\varphi$ is not injective. Let $0\neq x\in F$, such that $\varphi (x)=0$. Since $F$ is a field, $x$ has an inverse. So $$1=\varphi (1)=\varphi (xx^{-1}) =\varphi (x)\varphi (x)^{-1}=0.$$

Hence $\varphi$ is zero homomorphism.

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The kernel of a ring homomorphism is an ideal. A field has exactly two ideals, the unit ideal and the zero ideal. So either $\ker{\varphi}$ is the zero ideal or $\ker{\varphi}$ is the unit ideal, $F$ itself. If $\ker{\varphi}$ is the zero ideal, then $\varphi$ is injective (a general result you should know). If $\ker{\varphi}$ is the unit ideal $F$, then every element of $F$ is sent to $0_R$. However, since $R$ is not the zero ring, $0_R\neq 1_R$. But by the definition of a ring homomorphism, $\varphi(1_F)=\varphi(1_R)$. Therefore $\ker{\varphi}$ cannot be the unit ideal, so $\varphi$ is injective.

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Do you know or can you show that $\phi(0_F) = 0_R$? (EDIT: see the comment below, this isn't actually sufficient. You have to show $\phi(a) = 0_R \implies a = 0_F.$)

Once you have that, say $a,b \in F$. To show that $\phi$ is an injection, go to the definition. You just need to show that $\phi(a) = \phi(b) \implies a=b$. So let's suppose $\phi(a) = \phi(b)$. Now use the definition of a ring homomorphism to show $\phi(a-b) = 0$. Now if you can logically piece this together with the first statement, you've got it.

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  • $\begingroup$ That won't do it-one needs to show that $\phi(a) = 0_R \implies a = 0_F$. $\endgroup$ – David Wheeler Mar 24 '17 at 11:05
  • $\begingroup$ Oh, you're right. Thanks. I'll edit. $\endgroup$ – user2697520 Mar 24 '17 at 11:18

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