# For ring homomorphism $\phi:R\to S$, prove that $\phi(R)$ is a subring of $S$. [duplicate]

This question already has an answer here:

Could someone please verify whether my solution is correct?

For ring homomorphism $\phi:R\to S$, prove that $\phi(R)$ is a subring of $S$.

Subset/nonempty: Since $\phi(0_{R})=0_{S}$, then $\phi(R)\subset S$ and $\phi(R)\neq \varnothing$.

Closure under subtraction: Let $\phi(r),\phi(s)\in \phi(R)$. Since $R$ is a ring, $r-s\in R$ and $\phi(r-s)\in S$. Then $\phi(r-s)=\phi(r)-\phi(s)\in \phi(R)$.

Closure under multiplication: Since $rs\in R$ and $\phi(rs)\in S$, then $\phi(rs)=\phi(r)\phi(s)\in \phi(R)$.

I am not sure what to say for closure, whether it should be stated that $\phi(rs)\in S$, etc.

## marked as duplicate by Eevee Trainer, Lord Shark the Unknown, mrtaurho, Cesareo, MickGMar 22 at 10:55

• Same question here, with an answer. We have $\phi(rs)=\phi(r)\phi(s)\in S$. – Dietrich Burde Mar 27 '18 at 21:29
• @DietrichBurde Thank you. The answer defines $s_{1}=\phi(a_{1})$ and $s_{2}=\phi(a_{2})$, then later states that $\phi(s_{1}-s_{2})\in \phi(R)$. I am very confused about this part... – numericalorange Mar 29 '18 at 19:36
$$\phi(R)\subseteq S$$. To show it's a subring, we use the subring test: closure under subtraction and multiplication.
Pick $$s_1,s_2\in\phi(R)$$. There will be $$a_1,a_2\in R:s_1=\phi(a_1),s_2=\phi(a_2)$$. But then $$s_1-s_2=\phi(a_1)-\phi(a_2)=\phi(a_1-a_2)\in\phi(R)$$, so closure under subtraction is done.
Pick $$s_1,s_2\in\phi(R)$$. Again, we have the above $$a_1,a_2$$. But then $$s_1s_2=\phi(a_1)\phi(a_2)=\phi(a_1a_2)\in\phi(R)$$, and we're done.