Proposition 1.2.
Let $A$ be a ring $\neq 0$. Then the following are equivalent:
i) $A$ is a field
ii) the only ideals in $A$ are $0$ and $(1)$
iii) every homomorphism of $A$ into a non-zero ring $B$ is injective.
I am confused by how ii) implies iii). Given a homomorphism of $A$ into a non-zero ring $B$, the kernel of the homomorphism is an ideal. So by ii), it has to be $0$ or $(1)$. If it is $0$, $A$ is injective. But why can't it be $(1)$?
Say we have that $\phi:A\rightarrow B$ is a homomorphism and $B$ is a non-zero ring, then we have that $\ker\phi$ is an ideal of $A$, so $\ker\phi=0$ or $A$. $\ker\phi\neq A$ since if $\ker\phi=A$ that would imply that $\phi(1_A)=0_B$, but $\phi(1_A)=1_B$, so this would give us that $1_B=0_B$, but the only ring where $1$ and $0$ are the same is the zero ring (and we assumed that $B$ is a non-zero ring). Thus, we conclude that $\ker\phi=0$, so $A$ is injective.
