Could someone please verify whether my solution is okay? I understand this proposition is considered trivial, but I am not sure whether my proof is correct.

For ring homomorphism $\phi:R\to S$, prove that $\phi$ is an isomorphism iff $\phi$ is onto and $ker(\phi)=\left \{0_{R} \right \}$.

Let $\phi$ be an isomorphism. Then $\phi$ is bijective hence injective and surjective. Since $\phi$ is injective, then for the group homomorphism $\phi:(R,+)\to (S,+)$, $ker(\phi)=\left \{0_{R} \right \}$.

Let $\phi$ be onto and $ker(\phi)=\left \{0_{R} \right \}$. Then the ring homomorphism is bijective and so it is an isomorphism.

  • 2
    $\begingroup$ As long as you are free to use the fact that $\phi$ is injective iff $ker(\phi) = \{0\}$ your proof is fine. $\endgroup$ – James Mar 29 '18 at 20:42
  • $\begingroup$ @numericalorange Hint: try to learn from your last question. Nearly the exact same ideas suggested in the comments work. $\endgroup$ – rschwieb Mar 29 '18 at 21:25
  • $\begingroup$ @rschwieb Thank you for replying, but I am unsure what you mean. $\endgroup$ – numericalorange Mar 29 '18 at 23:32
  • $\begingroup$ @numericalorange if you think about it and read the comments, you’ll see. $\endgroup$ – rschwieb Mar 30 '18 at 0:16

The proof works since $\phi$ is injective iff $ker\phi=\{0_R\}$. If you need this fact, it can be proved as follows:

Suppose $\phi$ is injective. Then let $x \in ker\phi$ and consider $\phi(x)=\phi(0).$ Since $\phi$ is a homormorphism $\phi(0_R)=0_S$, and since it is injective $x = 0_R$ and thus $ker\phi=\{0_R\}$.

Now suppose $ker\phi=\{0_R\}$.Let $x,y\in R$ and consider $\phi(x)=\phi(y)$. Multiply by $\phi(x)^{-1}$. Then $$\phi(x)^{-1}\phi(x)=\phi(x)^{-1}\phi(y)=\phi(x^{-1}y).$$ Because $\phi$ is a homomorphism $\phi(x)^{-1}=\phi(x^{-1})$ and $\phi(x^{-1})\phi(x)=\phi(x^{-1}x)=\phi(0_R)=0_S$. Thus also $\phi(x^{-1}y)=0_S$ which means $x^{-1}y \in ker\phi$, and since $ker\phi=\{0_R\}$ it follows $y-x=0$ since $x^{-1}=-x$ in ring $R$. Finally, $x=y$.


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