Image of an ideal under a surjective ring homomorphism is an ideal

Question

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!

Answer 1

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.

Answer 2

Here is more correct way for the subtraction:

If $r',s'\in \varphi(I)$, there exist $r, s\in I$ such that $r'=\varphi(r),\quad s'=\varphi(s)$. Then $\;r'-s'=\varphi(r)-\varphi(s)=\varphi(r-s)$, and $r-s\in I $ since $I$ is an ideal in $R$.

Stability by product:

Let $r'$ as above, and $a'\in R'$. As $\varphi$ is surjective, there exists $a\in R$ such tha $a'=\varphi(a)$. Then $$a'r'=\varphi(a)\varphi(r)=\varphi(ar)$$ and $ar\in I$ since, again, $I$ is an ideal in $R$.