Let $f: R \longrightarrow S$ be a surjective homomorphism of commutative rings with identity. Let $I$ be an ideal of $R$ which contains $Ker(f)$. Show that $I$ is an irreducible ideal of $R$ if and only if $I^e$ is an irreducible ideal of $S$.
(Where $I^e$ is extension ideal of $I$ and $I$ is irreducible if and only if $I\ \neq R$ and, whenever $I=I_1 \cap I_2$, with $I_1$ and $I_2$ ideals of $R$, then $I=I_1$ or $I=I_2$ ).
I could not show any of the implications, I tried to show directly by definition, but I do not know if this is the way.
While in general $f(I)$ need not be an ideal in $S$, here you are assuming that $f$ is surjective. This means that the extension ideal $I^e$ actually is equal to $f(I)$.
Indeed, since $f$ is a group homomorphism, we certainly have that $f(I)$ is a subgroup of $S$. To show it is an ideal, you just need to show that if $a\in f(I)$ and $s\in S$, then $sa\in f(I)$. But since $a\in f(I)$, there exists $x\in I$ such that $f(x)=a$; and since $f$ is onto, there exists $r\in R$ such that $f(r)=s$. Therefore, $$ sa = f(r)f(x) = f(rx)\in f(I)$$ since $x\in I$ and $r\in R$ implies $rx\in I$. Thus, $I^e=f(I)$.
At this point, the condition should trivially follow by applying the Lattice Isomorphism Theorem a few times, as this yields an inclusion-preserving, join and meet preserving, correspondence between ideals of $f(R)=S$ and ideals of $R$ that contain $\mathrm{ker}(f)$, of which $I$ is one.