= Solution
Suppose first that every <ideal> of $R$ is finitely generated. For an ascending chain
$$
I_1\subseteq I_2\subseteq\cdots,
$$
the union $I=\bigcup_nI_n$ is an ideal. Its finite generating set lies in one $I_N$, so $I_n=I_N$ for all $n\geq N$. Thus $R$ satisfies the <ascending chain condition> and is a <Noetherian ring>.
Conversely, if an ideal $I$ is not finitely generated, choose $a_1\in I$, and after choosing $a_1,\ldots,a_n$, choose
$$
a_{n+1}\in I\setminus(a_1,\ldots,a_n).
$$
This creates a strictly ascending chain of ideals, contradicting Noetherianity. Hence every ideal is finitely generated.
If $\varphi:R\to S$ is surjective and $J\lhd S$, then $\varphi^{-1}(J)$ is an ideal of $R$. If it is generated by $a_1,\ldots,a_m$, then $J$ is generated by $\varphi(a_1),\ldots,\varphi(a_m)$. Thus every ideal of $S$ is finitely generated, so $S$ is Noetherian.
The <Hilbert basis theorem> states that if $R$ is a commutative Noetherian ring, then $R[X]$ is Noetherian. To prove it, let $I\lhd R[X]$. The leading coefficients of polynomials in $I$ generate an ideal of $R$; choose generators that occur as leading coefficients of $f_1,\ldots,f_s\in I$, and let $N=\max_i\deg f_i$. For each degree below $N$, the leading coefficients of members of $I$ of at most that degree likewise form a finitely generated ideal; choose finitely many corresponding polynomials. Any $f\in I$ of degree at least $N$ can have its leading term cancelled by a linear combination of monomial multiples of the $f_i$. Repeating lowers its degree below $N$, where the second finite list completes the reduction. These finitely many selected polynomials generate $I$, proving the theorem.
For every $\alpha\in\mathbb C$, the ring $\mathbb Z[\alpha]$ is Noetherian. If $\alpha$ is transcendental, evaluation identifies it with $\mathbb Z[X]$. If $\alpha$ is algebraic, it is a quotient of $\mathbb Z[X]$. The integers are Noetherian, the Hilbert basis theorem handles $\mathbb Z[X]$, and quotients preserve Noetherianity.
For a <unique factorization domain> that is not Noetherian, take
$$
k[x_1,x_2,\ldots].
$$
Every polynomial and every factorization uses only finitely many variables, so existence and uniqueness of factorization reduce to a finite-variable polynomial ring, which is a UFD. But
$$
(x_1)\subsetneq(x_1,x_2)\subsetneq(x_1,x_2,x_3)\subsetneq\cdots
$$
is a strictly ascending chain, so the ring is not Noetherian.
Finally, the ring $C(\mathbb R,\mathbb R)$ is not Noetherian. For $n\geq1$, let
$$
I_n=\{f:f(x)=0\text{ for all }|x|\leq1/n\}.
$$
These are ideals and $I_n\subsetneq I_{n+1}$: the distance function to $[-1/(n+1),1/(n+1)]$ belongs to $I_{n+1}$ but not to $I_n$. This strict ascending chain proves the claim.
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