Surjective homomorphism from a Noetherian ring to another ring Let $f:A\to B$ be a surjective ring homomorphism and assume $A$ is Noetherian.
We want to show that $B$ is also Noetherian.
This question was similarly asked, but I wanted to get some clarification on some differences:
$\textbf{Proof Sketch}$
Take any ideal in $B$, say $J$, and its preimage, say $J'\in A$ and show it's an ideal (I know this is true from a previous result I proved)
Since the $J'$ lives in $A$, it is finitely generated, say $\langle a_1,...,a_l\rangle=J'$
We then show $\langle f(a_1),...,f(a_l)\rangle=J$ Thus every ideal of $B$ is finitely generated
Questions:
1) Does this work?
2) Technically by definition, we define Noetherian rings to be commutative, is that necessary for $B$ here? (Or should I assume $B$ is commutative?) 
3) Did we ever need the homomorphism to be surjective?
 A: *

*Yes, this works, although you're sweeping something under the rug (namely, in your statement "We then show $\langle f(a_1),\dots, f(a_\ell)\rangle = J$").

*If $A$ is commutative and admits a surjection to $B$, $B$ must also be commutative: any $b,b'\in B$ are images of some $a,a'\in A$, and so
$$
b b' = f(a) f(a') = f(a a') = f(a' a) = f(a') f(a) = b' b.
$$

*Surjectivity is definitely needed. $\Bbb Z$ is noetherian, and admits a ring map to any other ring $R$ with unity, even if $R$ is nonnoetherian. This is what you swept under the rug, as I mentioned earlier. If $f : A\to B$ is not surjective, you might not have $f(J') = J$: in general, given a map of sets $g : X\to Y$ and a subset $S\subseteq Y$, you have $f(f^{-1}(S))\subseteq S$, but not necessarily equality. For an example, consider the map
\begin{align*}
g : \{\ast\}&\to\Bbb Z\\
\ast&\mapsto 0.
\end{align*}
Then $g^{-1}(\Bbb Z) = \{\ast\}$, but $g(\{\ast\}) = \{0\}\subsetneq\Bbb Z$. Although this is only an argument for maps of sets, the same phenomenon can happen with ring maps.

