Let $R$ be a ring with a unique minimal prime ideal $\mathfrak{p}$. Let us assume that every zero divisor in $R$ is nilpotent, so that $R \to R_{\mathfrak{p}}$ is injective. If $R_{\mathfrak{p}}$ is an Artin ring, is it necessarily the case that $\mathfrak{p}$ is finitely generated?
One can show that $\mathfrak{p}$ is at least an associated prime, that is, it is the annihilator of an element, which is not necessarily the case for the nilradical in an arbitrary ring.
Let $A=\mathbb{Q}[x_n,y_n:n\in\mathbb{N}]$ be a polynomial ring in infinitely many variables, and define the ideals $\mathfrak{q}=\langle y_n:n\in\mathbb{N}\rangle$ and $\mathfrak{a}=\mathfrak{q}^2+\langle x_ny_n-x_0y_0:n\in\mathbb{N}\rangle$ of $A$. Also, define $R=A\big/\mathfrak{a}$; by the lemma below, $\mathfrak{a}$ is a primary ideal of $A$, so every zero-divisor of $R$ is nilpotent, as desired. Also, note that $\mathfrak{q}=\sqrt{\mathfrak{a}}$, so that $\mathfrak{p}:=\mathfrak{q}\big/\mathfrak{a}$ is the unique minimal prime of $R$. For convenience, denote the images of $x_n$ and $y_n$ in $R$ as $a_n$ and $b_n$; note that $\mathfrak{p}$ is generated by the $b_n$.
We claim that $R_\mathfrak{p}$ is Artinian; indeed, since $\mathfrak{p}$ is a minimal prime of $R$, the ring $R_\mathfrak{p}$ is automatically local of dimension $0$, with unique prime ideal $\mathfrak{p}_\mathfrak{p}$. Thus, to show $R_\mathfrak{p}$ is Artinian, it suffices to show that $R_\mathfrak{p}$ is Noetherian, and, to show that $R_\mathfrak{p}$ is Noetherian, it suffices to show that $\mathfrak{p}_\mathfrak{p}$ is finitely generated. In fact, $\mathfrak{p}_\mathfrak{p}$ is principal, and generated by $b_0\big/1$. To see this, first note that $\mathfrak{p}_\mathfrak{p}$ is certainly generated the $b_n\big/1$, since $\mathfrak{p}$ is generated by the $b_n$; but also, since $x_n\notin\mathfrak{q}$ for any $n\in\mathbb{N}$, we have $a_n\notin\mathfrak{p}$ for any $n\in\mathbb{N}$, and thus $$\frac{b_n}{1}=\frac{a_nb_n}{a_n}=\frac{a_0b_0}{a_n}\in R_\mathfrak{p}\frac{b_0}{1}$$ for every $n\in\mathbb{N}$, as desired. However, $\mathfrak{p}<R$ is not finitely generated: defining $\mathfrak{b}=\mathfrak{q}^2+\langle x_ny_n:n\in\mathbb{N}\rangle<A$, note that $\mathfrak{q}$ is the union of the ascending chain of ideals $$\mathfrak{b}\leqslant\mathfrak{b}+\langle y_0\rangle\leqslant\mathfrak{b}+\langle y_0,y_1\rangle\dots.$$ This chain is strictly ascending, since the only monomial terms of degree $1$ that can appear in an element of $\mathfrak{b}+\langle y_0,\dots,y_n\rangle$ are $y_0,\dots,y_n$, and hence $y_{n+1}\notin\mathfrak{b}+\langle y_0,\dots,y_n\rangle$ for each $n\in\mathbb{N}$. On the other hand, $\mathfrak{b}>\mathfrak{a}$, so this chain descends to a strictly ascending chain in $R$ whose union is $\mathfrak{p}$. In particular, $\mathfrak{p}$ is not finitely generated, so $R$ gives the desired counterexample.
Lemma: $\mathfrak{a}$ is a primary ideal of $A$.
Proof: Let $B$ denote the subring $\mathbb{Q}[x_n:n\in\mathbb{N}]\subset A$, and note that any element $f\in A$ may be written uniquely in the form $f_0+f_1$ for some $f_0\in B$ and $f_1\in\mathfrak{q}$. Furthermore, for any $g\in\mathfrak{q}$, there exist some $n\in\mathbb{N}$, $p_i\in B$, and $h\in\mathfrak{q}^2$, such that $$g=p_0y_0+\dots+p_my_m+h.$$ Now, suppose $fg\in \mathfrak{a}$ for some $f,g\in A$. Then $fg\in\mathfrak{q}$, so assume without loss of generality that $g\in\mathfrak{q}$, and decompose $f$ and $g$ as indicated above. Also, suppose that $f\notin\mathfrak{q}$; we want to show $g\in\mathfrak{a}$. Note that, since $f\notin\mathfrak{q}$, also $f_0\notin\mathfrak{q}$, and that $f_1g\in\mathfrak{q}^2<\mathfrak{a}$, so in fact $f_0g\in\mathfrak{a}$ and we may assume that $f=f_0$, ie that $f\in B$. Furthermore, since $h\in\mathfrak{q}^2$, it suffices to show that $p_0y_0+\dots+p_my_m\in\mathfrak{a}$, and we may assume that $h=0$. Thus we have $$\mathfrak{a}\ni fg=fp_0y_0+\dots+fp_my_m;$$ let $h\in\mathfrak{q}^2$ and $q_i\in A$ be such that $$fg=(x_1y_1-x_0y_0)q_1+\dots+(x_ny_n-x_0y_0)q_n+h.$$ As above, decompose each $q_i$ as $q_{i0}+q_{i1}$, where $q_{i0}\in B$ and $q_{i1}\in \mathfrak{q}$. Then $(x_iy_i-x_0y_0)q_{i1}\in\mathfrak{q}^2$ for each $i$, so we may replace each $q_{i}$ with $q_{i0}$, and replace $h$ with $h+\sum_{i=1}^n(x_iy_i-x_0y_0)q_{i1}$, and thus assume that $q_i\in B$ for each $i$. But now we must have $h=0$; indeed, since $f\in B$, and by construction of $g$ and $q_i$, neither the product $fg$ nor the product $(x_iy_i-x_0y_0)q_i$ has monomial terms divisible by a monomial generator of $\mathfrak{q}^2$ for any $i$. On the other hand, every monomial term of $h$ is divisible by a monomial generator of $\mathfrak{q}^2$, so indeed $h=0$.
Thus, we can rewrite the equations above as \begin{align} fp_my_m+\dots+fp_0y_0&=fg \\ &=(x_1y_1-x_0y_0)q_1+\dots+(x_ny_n-x_0y_0)q_n \\ &=x_nq_ny_n+\dots+x_1q_1y_1-\left[q_1+\dots+q_n\right]x_0y_0 \end{align} where $f\in B$ and $p_i,q_j\in B$ for each $i,j$. By adding appropriate zero terms, assume for convenience that $m=n$. Now, since all of the coefficients of the $y_i$ in this equation lie in $B$, this forces \begin{align} fp_n&= x_nq_n \\ \vdots \\ fp_1&= x_1q_1 \\ fp_0&=-x_0(q_1+\dots+q_n). \end{align} So, $f$ divides the right hand sides in each of these equations. Now, recall that $A$ is a UFD; we can use this to show that in fact $f$ divides $q_i$ for each $0<i\leqslant n$. Indeed, suppose not. Then $f$ must be divisible by $x_i$; say $f=x_i^kf'$, where $k\geqslant 1$ is the highest power of $x_i$ dividing $f$. Then $x_i^k$ divides $q_j$ for all $j\neq i$. In particular, rearranging the last equation, we get that \begin{align} x_0q_i&=-fp_0+x_0\sum_{1<j\neq i\leqslant n}q_j \\ &=-x_i^kf'p_0+x_0\sum_{1<j\neq i\leqslant n}q_j \end{align} is divisible by $x^k_i$, so $x^k_i$ divides $q_i$. Now, since $f=x^k_if'$ divides $x_iq_i$, we know $x_i^{k-1}f'$ divides $q_i$. But $x_i^k$ also divides $q_i$ by the above, and $x_i$ does not divide $f'$ by construction, so in fact $x_i^{k}f'$ divides $q_i$, ie $f$ divides $q_i$, as claimed. So indeed $f$ divides $q_i$ for each $i\leqslant n$, and now we have \begin{align}g&=p_ny_n+\dots+p_0y_0 \\ &= \frac{q_n}{f}x_ny_n+\dots\frac{q_1}{f}x_1y_1-\frac{q_1+\dots+q_n}{f}x_0y_0 \\ &= \frac{q_n}{f}(x_ny_n-x_0y_0)+\dots+\frac{q_1}{f}(x_1y_1-x_0y_0)\in\mathfrak{a}, \end{align} as desired. $\blacksquare$
