Let $\phi : R\to S $ be a ring homomorphism.
Definition:
$Im\left ( \phi \right )=\left ( R \right )\phi=\left \{ \left ( a \right )\phi \mid a \in R \right \}$
Prove that $Im\left ( \phi \right )$is a subring of S
By the subring test, we need to show that the $Im\left ( \phi \right )$ is closed under multiplication and subtraction.
Suppose $\left ( a_{1} \right )\phi,\left ( a_{2} \right )\phi \in Im\left ( \phi \right )$.
Thus, $s_{1}=\left ( a_{1} \right )\phi$ and $s_{2}=\left ( a_{2} \right )\phi \in Im\left ( \phi \right )$
$\left ( a_{1} \right )\phi$ + (-$\left ( a_{2} \right ))\phi= s_{1}-s_{2}$ but S is a ring so $s_{1}-s_{2} \in$ S
Is this correct?
Best Answer
A little changes needed.Suppose $\left ( a_{1} \right )\phi,\left ( a_{2} \right )\phi \in Im\left ( \phi \right )$. Thus, $s_{1}:=\left ( a_{1} \right )\phi$ and $s_{2}:=\left ( a_{2} \right )\phi $.
Now $\left ( a_{1} \right )\phi + (-\left ( a_{2} \right ))\phi= s_1-s_2$ but S is a ring so $s_{1}-s_{2} \in S,$ Hence $s_1-s_2\in Im \phi.$
One also should check in similar way the multiplication.