Let $\phi: R \longrightarrow R'$ be a surjective ring homomorphism and
$I$ an ideal in $R$. Show that $\phi(I) = \{ \phi (r) : r \in I \}$ is
an ideal in $R'$.
So I asked this question a couple of days ago, but with incorrect notation and symbols. What I've been able to gather thus far is that I know I need to show that it is closed under subtraction and that it absorbs products.
Let $r',s'\in \phi(I)$, then $\exists r,s\in R$ such that $\phi(r)=r', \phi(s)=s'$.
$$\phi(r-s)=\phi(r+(-s))=\phi(r)+\phi(-s)=r'+(-s')=r'-s'\in \operatorname{im}(I)$$
I'm not sure if this is the correct way of showing that subtraction holds, and I'm also not sure of how to show the absorption of products. Any and all help is greatly appreciated, thank you!
Best Answer
Clearly we have $\phi(I) \subseteq R'$ and we have $\phi(I) \neq \emptyset$ since $0 \in I$. Now let $a,b \in \phi(I)$. Then there are $a', b' \in I$ such that $\phi(a')= a$ and $\phi(b')= b$. It follows
$$a - b = \phi(a') - \phi(b') = \phi (a' - b') \in \phi(I)$$
since $\phi$ is a homomorphism and $a ' - b' \in I$ since $I$ is an ideal. Now, since $\phi$ is surjective, for every $s \in R'$ exists an element $r \in R$ such that $\phi(r) = s$. It follows
$$sa = \phi(r)\phi(a') = \phi(ra') \in \phi(I) $$
where we again have used that $\phi$ is a homomorphism and the fact that $I$ is an ideal ($ra' \in I$). Analogous one shows $as \in \phi(I)$ and the claim follows.