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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?

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  • $\begingroup$ Your title does not make sense: what you mean is «The image of a ring homomorphism is a subring». $\endgroup$ – Mariano Suárez-Álvarez May 20 '16 at 3:08
  • $\begingroup$ Why is $\phi$ written to the right of the argument?! $\endgroup$ – YiFan Mar 22 at 10:51
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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.

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