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Show that a non-zero ring $R$ is a field if and only if for any non-zero ring $S$, any unital ring homomorphism from $R$ to $S$ is injective.

I would like to verify my proof, especially the reverse implication.

$\Rightarrow$ Let $S$ be any ring, and $f:R\rightarrow S$ be a ring homomorphism. If $x\in \ker f$ where $x$ is non-zero, then $0= f(x)f(x^{-1}) = f(xx^{-1})=f(1) = 1$ contradiction. Thus $x=0$, so $f$ is injective.

$\Leftarrow$ Since any ring homomorphism is injective, the only ideals of $R$ are $\{0\}$ and $R$. Thus $R$ is a field.

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Yes, this proof looks good to me!

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The proof is almost good, but ($\Rightarrow$) can be simplified and ($\Leftarrow$) is a bit too quick (you're using the fact that every proper ideal is the kernel of a nonzero homomorphism, without mentioning it).

For both arguments, we exploit that a (commutative) ring $R$ is a field if and only if its only ideals are $\{0\}$ and $R$.

($\Rightarrow$) Since $R$ is a field, the kernel of $f\colon R\to S$ is either $\{0\}$ or $R$. Since $1\notin\ker f$, we conclude that $\ker f=\{0\}$ and therefore $f$ is injective.

($\Leftarrow$) Suppose $I$ is a proper ideal of $R$. Then the ring $R/I$ is nonzero and so the canonical homomorphism $\pi\colon R\to R/I$ is injective by assumption. Therefore $I=\{0\}$.

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