Let $R, S$ be commutative rings, $f : R \rightarrow S$ an epimorphism, I an ideal of R. Show that $f(I)$ is an ideal of $S$.
As far as I understand, I need to show 4 things:
1) $0_s \in f(I)$
2) $sf(i) \in f(I), f(i)s \in f(I) \ \forall s \in S \ \forall f(i) \in f(I)$
and the subgroup properties
3) $f(a) + f(b) \in f(I) \ \forall a,b \in I$
4) $f(a)^{-1} \in f(I) \ \forall a \in I$
that's what I have so far:
1) $I$ ideal of R $\Rightarrow 0_R \in I \Rightarrow_{f(0_R) = 0_S} 0_S \in f(I)$
2)
$f$ surjective $\Rightarrow \exists s' \in R : f(s') = s$. $sf(i) = f(s')f(i) = f(\underbrace{s'i}_{\in I}) \in f(I)$.
(and the other direction similar).
3) $f(a) + f(b) = f(\underbrace{a + b}_{\in I}) \in f(I)$
4) I tried a few approaches, but didn't succeed in this part. Can you help me to show why the inverse element exists?
Is my approach correct?
$a^{-1} \in I \Rightarrow f(a^{-1})=f(a)^{-1} \in f(I)$
EDIT:(as Simon suggested)
There is no need for inverse to show ideal