Pseudo Programmer

$R,\ S$ : ring. $\phi$ : $R \rightarrow S$ be ring homomorphism. Let $A$ be subring of $R$ and $B$ be ideal of $S$. 1. For any $r \in R$ and any positive integer $n$, $\phi(nr) = n\phi(r)$ and $\phi(r^{n}) = (\phi(r))^{n}$ By operation-preserving property of ring homomorphism, following equations must be satisfied. $\phi(r + \dots + r) = \phi(r) + \dots + \phi(r)$ $\phi(r \cdot r \dots r) = \phi(r) \cdot \phi(r) \dots \phi(r)$ 2. $\phi(A) = \{\phi(a) \mid a \in A\}$ is subring of $S$. Since that $A$ is subring of R, $a - b \in A$ and $ab \in A,\ \forall a,\ b \in A$. Hence, $\phi(a - b) = \phi(a) - \phi(b) \in \phi(A)$ and $\phi(ab) = \phi(a)\phi(b) \in \phi(A),\ \forall a,\ b \in A)$. $\therefore \text{By subring test, } \phi(A) \text{ is subring of }S.$ 3. If $A$ is an ideal and $\phi$ is onto $S$, then $\phi(A)$ is an ideal. $\phi$ is onto $S$ implies that $\phi(r) = s ,\ \exists r \in R,\ \forall s \in S$ or equivalently, $\phi(A) = S$. $A$ is ideal of $R$ impl...

$\phi$ : $Z_{n} \rightarrow Z_{n}$, $x \mapsto ax$, where $a^{2} = a$ is ring homomorphism. Let $\phi(1) = a$. Then $a = \phi(1) = \phi(1)\phi(1) = a^{2}$. Let $x \in Z_{n}$. Then $\phi(x) = \phi(1 + \dots + 1) = \phi(1)x = ax$. Hence, our original conjecture is true.