Basic properties of ring homomorphisms
$R,\ S$ : 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 $R$ implies $ar,\ ra \in A,\ \forall r \in R,\ \forall a \in A$.
From this, $\phi(ar) = \phi(a)s \in \phi(A),\ \phi(ra) = s\phi(a) \in \phi(A), \forall s \in S, \forall \phi(a) \in \phi(A)$.
Since that $\phi$ is onto, $\forall s \in S$ is obvious.
Hence, $\phi(A)$ is ideal of $S$.
4. $\phi^{-1}(B) = \{r \in R \mid \phi(r) \in B\}$ is ideal of $R$.
Obviously, $\phi^{-1}$ is onto. Then, by 3rd property, $\phi^{-1}(B)$ is ideal of R.
5. If $R$ is commutative, then $\phi(R)$ is commutative.
6. If $R$ has a unity $1$, $S \neq \{0\}$, and $\phi$ is onto, then $\phi(1)$ is the unity of $S$.
7. $\phi$ is an isomorphism if and only if $\phi$ is onto and $Ker(\phi)=\{r \in R \mid \phi(r) =0 \} $ is trivial 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 $R$ implies $ar,\ ra \in A,\ \forall r \in R,\ \forall a \in A$.
From this, $\phi(ar) = \phi(a)s \in \phi(A),\ \phi(ra) = s\phi(a) \in \phi(A), \forall s \in S, \forall \phi(a) \in \phi(A)$.
Since that $\phi$ is onto, $\forall s \in S$ is obvious.
Hence, $\phi(A)$ is ideal of $S$.
4. $\phi^{-1}(B) = \{r \in R \mid \phi(r) \in B\}$ is ideal of $R$.
Obviously, $\phi^{-1}$ is onto. Then, by 3rd property, $\phi^{-1}(B)$ is ideal of R.
5. If $R$ is commutative, then $\phi(R)$ is commutative.
6. If $R$ has a unity $1$, $S \neq \{0\}$, and $\phi$ is onto, then $\phi(1)$ is the unity of $S$.
7. $\phi$ is an isomorphism if and only if $\phi$ is onto and $Ker(\phi)=\{r \in R \mid \phi(r) =0 \} $ is trivial ring.
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