= Solution
{c}
Let $e_j$ be the sequence with $1$ in coordinate $j$ and zero elsewhere. In the <direct product of rings> $R=k^{\mathbb N}$, the <ideals>
$$
(e_1)\subsetneq(e_1,e_2)\subsetneq(e_1,e_2,e_3)\subsetneq\cdots
$$
form a strict ascending chain: every element of $(e_1,\ldots,e_n)$ vanishes after coordinate $n$, whereas $e_{n+1}$ does not. Hence \b[$R$ is not a Noetherian ring].
Nevertheless, each $a=(a_j)\in R$ has a coordinatewise generalized inverse $b=(b_j)$, defined by $b_j=a_j^{-1}$ when $a_j\ne0$ and $b_j=0$ otherwise. It satisfies
$$
a=a^2b.
$$
This is the defining property of a <Von Neumann regular ring>. Fix any <prime ideal> $P$, and take $a\in P$. Since $ab\in P$, $1-ab\notin P$; but
$$
(1-ab)a=0.
$$
Consequently $a/1=0$ in the <localization at a prime ideal> $R_P$. All elements of $PR_P$ therefore vanish. By the <local ring> description in part (a), $PR_P$ is the unique <maximal ideal>, so $R_P$ is a nonzero <field>. A <field> has only the <ideals> zero and itself, and thus is <Noetherian>.
We have proved the stronger conclusion
$$
\boxed{R_P\text{ is a field for every prime ideal }P\subset R.}
$$
The argument applies to every <prime ideal>, without assuming that it comes from a coordinate projection.
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