Proof. Suppose that \(R\) is an \(S\)-Noetherian ring. Then every ideal of \(R\) is \(S\)-finite. In particular, every nil ideal and every nonnil ideal of \(R\) is \(S\)-finite. Hence \(R\) is both Nil-\(S\)-Noetherian and Nonnil-\(S\)-Noetherian. Conversely, assume that \(R\) is both Nil-\(S\)-Noetherian and Nonnil-\(S\)-Noetherian. Let \(I\) be an ideal of \(R\). If \(I \subseteq \mathrm{Nil}(R)\), then \(I\) is \(S\)-finite. If \(I \nsubseteq \mathrm{Nil}(R)\), then \(I\) is \(S\)-finite. Thus, in either case, \(I\) is \(S\)-finite. Therefore, every ideal of \(R\) is \(S\)-finite, and hence \(R\) is \(S\)-Noetherian. ◻

Proof. (1) \(\Rightarrow\) (2). Let \((I_n)_{n\in\wedge}\) be an increasing sequence of nil ideals of \(R\). Define the ideal \(I = \bigcup\limits_{n\in\wedge}I_n\). Let \(\alpha\in I\). Then \(\alpha\in I_{k}\) for some \(k\in\wedge\). Since \(I_{k}\) is nil, there exists \(m\in\mathbb{N}\) such that \(\alpha^{m}=0\), and hence \(I\) is nil. This implies that \(I\) is \(S\)-finite since \(R\) is Nil-\(S\)-Noetherian. Consequently, there exist a finitely generated ideal \(F\subseteq R\) and \(s\in S\) such that \(sI\subseteq F\subseteq I\). Since \(F\) is finitely generated, there is a \(k\in\wedge\) satisfying \(F\subseteq I_{k}\). Then we have \(sI\subseteq F\subseteq I_{k}\), from which it follows that \(sI_n \subseteq I_{k}\) for each \(n\in\wedge\).

(2) \(\Rightarrow\) (3). Let \(\mathcal{D}\) be an \(S\)-saturated set of nil ideals of \(R\). Given any chain \(\{I_n\}_{n \in\wedge} \subseteq \mathcal{D}\), we claim that \(I = \bigcup\limits_{n\in\wedge} I_n\) belongs to \(\mathcal{D}\), which will establish that \(I\) as an upper bound for the chain. Since each \(I_n\) is a nil ideal of \(R\), so \(I\) is a nil ideal of \(R\). Also by (2), there exist \(k \in \wedge\) and \(s \in S\) such that \(s I_n \subseteq I_k\) for every \(n \in \wedge\). Consequently, we obtain \(sI = s \left(\bigcup\limits_{n\in\wedge} I_n \right) \subseteq I_k\). Since \(\mathcal{D}\) is \(S\)-saturated, it follows that \(I \in \mathcal{D}\), as required. Applying Zorn’s lemma, we conclude that \(\mathcal{D}\) has a maximal element.

(3) \(\Rightarrow\) (4). Let \(\mathcal{D}\) be a nonempty set of nil ideals of \(R\). Consider the family \(\mathcal{D}^{S}\) of all nil ideals \(L \subseteq R\) such that there exist some \(s\in S\) and \(L_0 \in \mathcal{D}\) with \(sL \subseteq L_0\). Clearly, \(\mathcal{D}\subseteq \mathcal{D}^{S}\), so \(\mathcal{D}^{S}\neq\emptyset\). It is straightforward to see that \(\mathcal{D}^{S}\) is \(S\)-saturated. Thus, by (3) \(\mathcal{D}^{S}\) has a maximal element \(K\in\mathcal{D}^{S}\). Fix \(s \in S\) and \(Q\in \mathcal{D}\) such that \(sK\subseteq Q\). Now, we claim that \(Q\) is an \(S\)-maximal element of \(\mathcal{D}\); specifically, given \(L\in \mathcal{D}\) with \(Q\subseteq L\), we will show that \(sL \subseteq Q\). Note that \(K + L\) satisfies \(s(K + L)= sK + sL \subseteq Q + L \subseteq L\), so that \(K + L \in \mathcal{D}^{S}\). Also, \(K+L\) is a nil ideal of \(R\). Therefore maximality of \(K\) implies \(K = K + L\), so that \(L \subseteq K\). But then \(sL \subseteq sK \subseteq Q\), as desired.

(4) \(\Rightarrow\) (1). Let \(I\) be a nil ideal of \(R\). Let \(\mathcal{D}\) denote the family of finitely generated nil ideal \(J\) of \(R\) such that \(J\subseteq I\), which is nonempty as \(0\in\mathcal{D}\). Then \(\mathcal{D}\) has an \(S\)-maximal element \(K\in\mathcal{D}\). Fixing \(x\in I\), take a finitely generated ideal of the form \(Q= K + xR\). Since \(K\subseteq I\) and \(x\in I\), so \(Q\subseteq I\). Consequently, \(Q\in\mathcal{D}\) such that \(K\subseteq Q\). This implies that there exists \(s\in S\) such that \(sQ\subseteq K\); in particular, \(sx\in K\). This verifies \(sI\subseteq K\subseteq I\), so that \(I\) is \(S\)-finite. It follows that \(R\) is Nil-\(S\)-Noetherian. ◻

Proof. Let \(R\) be a uniformly-\(S\)-reduced ring, and \(I\) be a nil ideal of \(R\). Let \(\alpha\in I\). Then there exists \(n\in\mathbb{N}\) such that \(\alpha^{n}=0\). Since \(R\) is uniformly-\(S\)-reduced ring, it follows that there exists fixed \(t\in S\) such that \(t\alpha=0\). This implies that \(tI=(0)\) i.e., \(I\) is \(S\)-finite. Therefore \(R\) is a Nil-\(S\)-Noetherian ring. ◻

Proof. Let \(K\) be a nil ideal of \(R/I\). Then \(K = J/I\) for some ideal \(J\) of \(R\) containing \(I\). If \(K\cap \overline{S}\neq\emptyset\), then \(K\) is \(\overline{S}\)-finite, by [3]. Now, let \(K\cap \overline{S}=\emptyset\) and \(a+I\in K\), where \(a\in J\setminus I\). Then there exists \(n\in\mathbb{N}\) such that \((a+I)^{n}=I\) since \(K\) is nil. It follows that \(a^{n}\in I\). Since \(I\) is nil, there exist \(k\in\mathbb{N}\) such that \(a^{nk}=0\). Put \(m=nk\in\mathbb{N}\). Thus \(a^{m}=0\), and hence \(J\) is a nil ideal. Since \(R\) is Nil-\(S\)-Noetherian, there exist \(s\in S\) and \(x_{1},\ldots, x_{n}\in J\) such that \(sJ\subseteq \langle x_1,\ldots, x_{n} \rangle\subseteq J\). Let \(x\in J\). Then we can find \(a_1,\ldots, a_n\in R\) such that \(sx=a_1x_1+\cdots+a_nx_n\). It follows that \((s+I)(x+I)=(a_1+I)(x_1+I)+\cdots+(a_n+I)(x_n+I)\), where \(s+I\in\overline{S}\) and \(a_{1}+I,\ldots, a_{n}+I\in R/I\). This implies that \((s+I)(J/I)\subseteq \langle x_1+I,\ldots, x_n+I \rangle\subseteq J/I\), i.e., \(K\) is \(\overline{S}\)-finite. Hence \(R/I\) is an \(\overline{S}\)-Noetherian ring. ◻

Proof. Suppose \(R\) is Nil-\(S\)-Noetherian and let \(I_{i}\) be a nil ideal of \(R_{i}\). Then the corresponding nil ideal of \(R\) is \(E_{i}=\{0\}\times \cdots \times I_{i}\times \cdots \times \{0\}\). The projection of \(E_{i}\) onto the \(i\)-th coordinate is precisely \(I_{i}\). Since \(R\) is Nil-\(S\)-Noetherian, \(E_{i}\) must be \(S\)-finite. Hence, there exist \(s\in S\) and a finitely generated ideal \(J\subseteq R\) such that \(sE_{i}\subseteq J \subseteq E_{i}\). Applying the projection map \(\pi_{i}:R\to R_{i}\), we obtain \(\pi_{i}(s)I_{i}=\pi_{i}(sE_{i}) \subseteq \pi_{i}(J)\subseteq I_{i}\), where \(\pi_{i}(s)\in S_{i}=\pi_{i}(S)\) and \(\pi_{i}(J)\) is a finitely generated ideal of \(R_{i}\). Consequently, \(I_{i}\) is \(S_{i}\)-finite, and therefore \(R_{i}\) is Nil-\(S_{i}\)-Noetherian.

Conversely, suppose each \(R_{i}\) is Nil-\(S_{i}\)-Noetherian. Let \(I\) be a nil ideal of \(R\). Then \(I = I_{1} \times \cdots \times I_{n}\), where each \(I_{i}\) is a nil ideal of \(R_{i}\). By assumption, each \(I_{i}\) is \(S\)-finite, and hence \(I\) is \(S\)-finite in \(R\). Therefore, \(R\) is Nil-\(S\)-Noetherian. ◻

Proof. Let \(I\) be a nil ideal of \(R\). For \(i\in I\), since \(\varphi\) is a ring homomorphism, we have \(\varphi(i)^n=\varphi(i^n)=\varphi(0)=0\), so \(\varphi(i)\) is nilpotent in \(R'\). Hence every element of \(\varphi(I)\) is nilpotent. Now consider the ideal \(\varphi(I)R'=\Big\{\;\sum_{j}\varphi(i_j)r'_j \;\Big|\; i_j\in I,\; r'_j\in R'\;\Big\}\). This is the ideal generated by the nilpotent set \(\varphi(I)\), and it is well known that an ideal generated by nilpotent elements is a nil ideal. Thus \(\varphi(I)R'\) is a nil ideal of \(R'\). Since \(R'\) is Nil-\(\varphi(S)\)-Noetherian, there exist \(s'\in \varphi(S)\) and a finitely generated ideal \(J'\subseteq R'\) such that \(s'\varphi(I)R'\subseteq J' \subseteq \varphi(I)R'\). Choose \(s\in S\) with \(\varphi(s)=s'\). Applying \(\rho\) to the chain gives \(\rho(s'\varphi(I)R') \subseteq \rho(J') \subseteq \rho(\varphi(I)R')\). We claim that \(\rho(s'\varphi(I)R')=sI\). Indeed, if \(x'\in s'\varphi(I)R'\), then \(x'=s'\sum_j\varphi(i_j)r'_j\), so \(\rho(x')=\sum_j \rho(s')\,\rho(\varphi(i_j))\,\rho(r'_j) = \sum_j s i_j \rho(r'_j)\in sI\). Hence \(\rho(s'\varphi(I)R')\subseteq sI\). Conversely, every element of \(sI\) has the form \(\sum_k (s i_k) r_k\) with \(i_k\in I,\; r_k\in R\), and \(\sum_k (s i_k) r_k = \rho\!\left(\sum_k \varphi(s)\varphi(i_k)\varphi(r_k)\right)\in \rho(s'\varphi(I)R')\). Thus \(sI\subseteq \rho(s'\varphi(I)R')\), proving the equality. Finally, note that \(\rho(\varphi(I)R')\subseteq I\). Therefore, \(sI=\rho(s'\varphi(I)R') \subseteq \rho(J') \subseteq \rho(\varphi(I)R')\subseteq I\). Since \(J'\) is finitely generated ideal in \(R'\), its image \(\rho(J')\) is finitely generated in ideal in \(R\). Thus there exists \(s\in S\) with \(sI\subseteq \rho(J')\subseteq I\), showing that \(I\) is \(S\)-finite. Hence \(R\) is Nil-\(S\)-Noetherian. ◻

Proof. \((1) \Rightarrow (2)\): The result is immediate from the definition 1.

\((2) \Rightarrow (1)\): Suppose every prime ideal \(P \subseteq Nil(R)\) with \(P \cap S = \emptyset\) is \(S\)-finite. Let \(I\) be a nil ideal of \(R\). On the contrary, assume that \(I\) is not \(S\)-finite. Consider the collection \(\mathcal{F} = \{J \subseteq Nil(R) \mid J~is~ ~not~\)S\(-finite \}\). Clearly \(\mathcal{F}\) is nonempty since \(I \in \mathcal{F}\). Ordered by inclusion, every chain in \(\mathcal{F}\) has an upper bound given by its union, which is again a nil, non-\(S\)-finite ideal. By Zorn’s lemma, there exists a maximal element \(P\in \mathcal{F}\). We claim that \(P\) is a prime ideal. Clearly, \(P\cap S=\emptyset\) otherwise \(P\) is automatic \(S\)-finite, by [3]. Suppose there exist \(a, b \in R \setminus P\) such that \(ab \in P\). Then \(P \subsetneq (P+Ra)\cap Nil(R)\). By maximality of \(P\), the nil ideal \((P+Ra)\cap Nil(R)\) must be \(S\)-finite. Hence there exist \(s \in S\), \(\alpha_{1},\ldots, \alpha_{n} \in (P+Ra)\cap Nil(R)\) such that \(s\big((P+Ra)\cap Nil(R)\big) \subseteq \langle \alpha_{1},\ldots,\alpha_{n} \rangle \subseteq (P+Ra)\cap Nil(R)\). Now consider the ideal \(Q=(P:a)=\{x\in R\mid ax\in P\}\). Evidently, \(b \in Q \setminus P\), so \(P \subsetneq Q\). Define \(Q' = Q \cap Nil(R)\), which is a nil ideal strictly containing \(P\). By maximality of \(P\), \(Q'\) is \(S\)-finite. Thus there exist \(t\in S\) and \(\beta_{1},\ldots,\beta_{k}\in Q'\) such that \(tQ' \subseteq \langle \beta_{1},\ldots,\beta_{k}\rangle \subseteq Q'\). Let \(x \in P\). Since \(sx\in s\big((P+Ra)\cap Nil(R)\big)\), we can write \(sx = u_{1}\alpha_{1} + \cdots + u_{n}\alpha_{n}\) with \(u_{i}\in R\). Each \(\alpha_{i}\) has the form \(p_{i} + ax_{i}\) with \(p_{i}\in P\), \(x_{i}\in R\) and \(ax_{i}\in Nil(R)\). Hence \(sx = u_{1}p_{1} + \cdots + u_{n}p_{n} + a(u_{1}x_{1} + \cdots + u_{n}x_{n})\). Thus \(a(u_{1}x_{1}+\cdots+u_{n}x_{n}) \in P\subseteq Nil(R)\), which implies \(u_{1}x_{1}+\cdots+u_{n}x_{n} \in (P:a)=Q\). Therefore \(u_{1}ax_{1}+\cdots+u_{n}ax_{n}\in Q\cap Nil(R)=Q'\). Since \(Q'\) is \(S\)-finite, there exist \(w_{1},\ldots,w_{k}\in R\) such that \(t(au_{1}x_{1}+\cdots+au_{n}x_{n}) = w_{1}\beta_{1} + \cdots + w_{k}\beta_{k}\). Now, we have \(stx = t(u_{1}p_{1}+\cdots+u_{n}p_{n}) + w_{1}\beta_{1}+\cdots+w_{k}\beta_{k}\). This shows that \(uP \subseteq \langle p_{1},\ldots,p_{n},\beta_{1},\ldots, \beta_{k} \rangle \subseteq P\), where \(u=st\in S\). Hence \(P\) is \(S\)-finite, contradicting the choice of \(P \in \mathcal{F}\). Therefore no such \(P\) exists, and so every nil ideal is \(S\)-finite. Thus \(R\) is Nil-\(S\)-Noetherian. ◻

Proof. Let \(\alpha\in s(I\cap J)\). Then \(\alpha=s\beta\), where \(\beta\in I\cap J\). This implies that \(s\beta\in sI\) and \(s\beta\in sJ\), \(\alpha=s\beta\in sI\cap sJ\). Thus \(s(I\cap J)\subseteq sI\cap sJ\). On the other hand, let \(\alpha\in sI\cap sJ\). Then \(\alpha=s\beta_1=s\beta_2\) for some \(\beta_1\in I\), \(\beta_2\in J\). Consequently, \(s(\beta_1-\beta_2)=0\), \(\beta_1=\beta_2\) since \(S\cap Z(R)=\emptyset\). Therefore \(\alpha\in s(I\cap J)\), and hence \(s(I\cap J)=sI\cap sJ\). ◻

Proof. Since \(rad(I)\) is \(S\)-finite, \(t(rad(I))\subseteq J\subseteq rad(I)\) for some \(t\in S\) and a finitely generated ideal \(J=\langle x_1,x_2,\dots,x_l \rangle\) of \(R\). Suppose \(n_i\in \mathbb{N}\) be such that \(x_i^{n_i}\in I\). Take \(m=\sum_{i=1}^l (n_i-1)+1\). Then \(J^m\subseteq I\). Consequently, \(s(rad(I))^m\subseteq J^m\subseteq I\), where \(s=t^m\). ◻

Proof. Since \(R\) is \(S\)-Noetherian, it follows from [9] that the zero ideal \((0)\) admits a minimal \(S\)-primary decomposition \((0) = \bigcap_{i=1}^{n} Q_{i}\), where each \(Q_i\) is a \(P_i\)-\(S\)-primary ideal and \(P_{i} = rad(Q_{i})\) is an \(S\)-prime ideal. Consequently, \(Nil(R)= rad(0)=\bigcap\limits_{i=1}^{n} P_{i}\). By Lemma 17, for each \(i=1,\ldots,n\), there exist an integer \(k_i \geq 1\) and an element \(s_i \in S\) such that \(s_{i} P_i^{\,k_i}\subseteq Q_i\). Let \(k = \max\{k_1, \dots, k_n\}\) and \(s = \prod_{i=1}^{n} s_{i}\). Then, by Lemma 15, \(s \,(Nil(R))^{k} = s \,\Big(\bigcap\limits_{i=1}^{n} P_{i}\Big)^{k}\subseteq\; s \Big(\bigcap\limits_{i=1}^{n} P_{i}^{\,k_{i}}\Big) = \bigcap\limits_{i=1}^{n} sP^{k_{i}}_{i}\). Consequently, \(s (Nil(R))^{k}= sP^{k_{1}}_{1}\cap \cdots \;\cap\;sP^{k_{n}}_{n}\subseteq s_1P^{k_{1}}_{1}\cap \cdots \;\cap\; s_nP^{k_{n}}_{n}\subseteq Q_{1}\cap\cdots\cap Q_{n} = (0)\), by Lemma 17. Hence the nilradical \(Nil(R)\) is \(S\)-nilpotent. Since \(I \subseteq Nil(R)\), it follows that \(sI^k \;\subseteq\; s(Nil(R))^k \;=\; (0)\), \(sI^{k}=0\). Since \(S\cap Z(R)=\emptyset\), so \(I^{k}=0\). Thus \(I\) is a nilpotent ideal of \(R\). ◻

Proof. Let \(I\) be a nil ideal of \(R\). By the Nil-\(S\)-Noetherian hypothesis, \(I\) is \(S\)-finite; hence there exist \(s\in S\) and a finitely generated ideal \(J\subseteq R\) with \(sI \subseteq J\subseteq I\). Write \(J=\langle a_1,\dots,a_m \rangle\), where \(a_{i}\in R\) for all \(i=1,\ldots, m\). Since \(J\subseteq I\) and \(I\) is nil, each generator \(a_i\) is nilpotent. Thus for each \(i\) there exists \(n_i\ge1\) with \(a_i^{n_i}=0\). Set \(N=\sum\limits_{i=1}^m (n_i-1)+ 1\). Thus \(J^N = 0\). Now observe that \((sI)^N = s^N I^N \subseteq J^N = 0\), so \(s^N I^N = 0\). Setting \(t= s^N \in S\) and \(n= N\) yields \(tI^n = 0\), which shows that \(I\) is \(S\)-nilpotent. ◻

Proof. Let \(P\) be a nil prime ideal of \(R[x]\) such that \(P\cap S\neq\emptyset\). For each \(n\ge0\), define \(P_n=\{\,a\in R\mid \exists~ f(x)\in P \text{ with } \deg f=n \text{ and leading coefficient } a\,\}\cup\{0\}\). Since a polynomial is nilpotent if and only if each of its coefficients is nilpotent, it follows that each \(P_n\) is a nil ideal of \(R\). Suppose \(r\) is the smallest integer such that \(P_r\neq0\). Then \(P_r\subseteq P_{r+1}\subseteq\cdots\) is an ascending chain of ideals of \(R\). Also \(P\cap S\neq\emptyset\) and \(0\notin S\), so \(P_i\cap S=\emptyset\) since \(P_i\subseteq Nil(R)\) and \(Nil(R)\cap S=\emptyset\). Let \(I=\bigcup_{n\ge r}P_n\). Then \(I\) is a nil ideal of \(R\) as each \(P_n\) is nil. Since \(R\) is nil-\(S\)-Noetherian, \(I\) is \(S\)-finite. Hence there exists \(t\in S\) such that \(tI\subseteq\langle b_1,b_2,\ldots,b_m\rangle\) for some \(b_i\in R\). Take \(f_i\in P\) such that \(f_i(x)=b_i x^{d_i}+\text{lower terms}\) for \(i=1,2,\ldots,m\). Put \(d=\max\{d_i\}_{i=1}^{m}\). Since each \(P_k\) is a nil ideal of \(R\) and \(R\) is nil-\(S\)-Noetherian, there exists \(t_k\in S\) such that \(t_kP_k\subseteq\langle a_{k1},a_{k2},\ldots,a_{kn_k}\rangle\subseteq P_k\). For each \(a_{ki}\), take \(f_{ki}\in P\) such that \(f_{ki}(x)=a_{ki}x^k\) for all \(k\in\{r,r+1,\ldots,d\}\). Let \(f=\sum_{i=0}^{\nu}p_ix^i\in P\). If \(\nu>d\), then \(p_\nu\in I\). This implies that \(tp_\nu=r_1b_1+r_2b_2+\cdots+r_mb_m\) for some \(r_i\in R\). Hence \(tf-tp_\nu x^\nu=tf-\sum_{i=1}^{m}r_ib_ix^\nu=tf-\sum_{i=1}^{m}r_if_i x^{\nu-d_i}\in P\) is a polynomial of degree at most \(\nu-1\). By repeating this process, we obtain a polynomial \(h=\sum_{i=1}^{l}q_ix^i\) with \(l\le d\). Since \(q_l\in P_l\), we have \(t_lq_l\in\langle a_{l1},a_{l2},\ldots,a_{ln_l}\rangle\subseteq P_l\). Write \(t_lq_l=c_1a_{l1}+c_2a_{l2}+\cdots+c_{n_l}a_{ln_l}\) for some \(c_{i} \in R\). Then \(t_lh-t_lq_lx^l=t_lh-\sum_{i=1}^{n_l}c_ia_{li}x^l=t_lh-\sum_{i=1}^{n_l}c_if_{li}\) is a polynomial of degree at most \(l-1\). Repeating this process, we obtain \(t^{\nu-l}\prod_{k=r}^{l}t_k f=g_1+g_2\) for some \(g_1\in\langle f_1,\ldots,f_m,f_{r1},\ldots,f_{rn_r},f_{(r+1)1},\ldots,f_{(r+1)n_{r+1}},\ldots,f_{d1},\ldots,f_{dn_d}\rangle\) and \(g_2\in P\) with degree at most \(r-1\). But \(r\) was chosen to be the least integer such that \(P_r\neq0\), hence \(g_2=0\). Since \(S\) is anti-Archimedean, \((\bigcap_{n\ge1}t^nR)\cap S\neq\emptyset\). Let \(w\in(\bigcap_{n\ge1}t^nR)\cap S\). Then \(w\prod_{k=r}^{l}t_k f\in\langle f_1,\ldots,f_m,f_{r1},\ldots,f_{rn_r},f_{(r+1)1},\ldots,f_{(r+1)n_{r+1}},\ldots,f_{d1},\ldots,f_{dn_d}\rangle\). Put \(s=w\prod_{k=r}^{l}t_k\). Then \(sP\subseteq\langle f_1,\ldots,f_m,f_{r1},\ldots,f_{rn_r},f_{(r+1)1},\ldots,f_{(r+1)n_{r+1}},\ldots,f_{d1},\ldots,f_{dn_d}\rangle\subseteq P\). Therefore \(P\) is \(S\)-finite. Hence by Theorem 13, \(R[x]\) is a Nil-\(S\)-Noetherian ring. ◻

Proof. Let \(P\) be a nil prime ideal of \(R[[x]]\) such that \(P\cap S\neq\emptyset\). Let \(P(0) = \{f(0) \mid f(x) \in P\}\). Let \(f= \sum_{i=0}^{\infty}a_{i}x^{i} \in P\). Then \(f\) is nilpotent as \(P\) is nil ideal. This implies that each coefficient \(a_{n}\) is nilpotent, in particular, its constant term \(a_{0} = f(0)\) is nilpotent. Consequently, \(P(0)\) is a nil ideal of \(R\). This implies that \(P(0)\) is \(S\)-finite since \(R\) is Nil-\(S\)-Noetherian. Thus, there exists \(s \in S\) and a finite set of polynomials \(\{p_1(x), p_2(x), \dots, p_n(x)\} \subseteq P\) such that \(sP(0) \subseteq \langle p_1(0), p_2(0), \dots, p_n(0) \rangle \subseteq P(0)\). If \(x \in P\), then we can write \(P = P(0) + xR[[x]]\). It follows that \(sP \subseteq \langle p_1(0), p_2(0), \dots, p_n(0), x \rangle \subseteq P\). Hence \(P\) is \(S\)-finite. Now suppose \(x \notin P\). Let \(f \in P\). Then \(f(0) \in P(0)\). Write \(sf(0) = \sum_{i=1}^{n} r_{0i} p_i(0) \quad \text{for some } r_{0i} \in R\). This implies that \(sf - sf(0) = sf- \sum_{i=1}^{n} r_{0i} p_i(0) = xf_1\) for some \(f_1 \in R[[x]]\). Since \(xf_1 \in P\) and \(x \notin P\) (as \(P\) is a prime ideal), it must be that \(f_1 \in P\). By repeating this iterative process, we can similarly write \(sf_1 - \sum_{i=1}^{n} r_{1i} p_i(0) = xf_2\), where \(f_2 \in P\). Continuing this construction, we obtain the representation \(f = \sum_{i=1}^{n} p_i(0) \left( \sum_{j=0}^{\infty} \frac{r_{ji}}{s^{j+1}} x^j \right)\). Note that the coefficients \(\frac{r_{ji}}{s^{j+1}}\) belong to the localization \(S^{-1}R\). Since \(S\) is given to be an anti-Archimedean set, we have \(\left( \bigcap_{i=1}^{\infty} s^i R \right) \cap S \neq \emptyset\). Let \(t \in \left( \bigcap_{i=1}^{\infty} s^i R \right) \cap S\). Then for each \(j\), \(t = s^j c_j\) for some \(c_j \in R\). Multiplying the expression for \(f\) by \(t\), we get \(tf = \sum_{i=1}^{n} p_i(0) \left( \sum_{j=0}^{\infty} c_{j+1} r_{ji} x^j \right) \in \langle p_1(0), \dots, p_n(0) \rangle\). This implies that \(tP \subseteq \langle p_1(0), p_2(0), \dots, p_n(0) \rangle \subseteq P\). Thus \(P\) is \(S\)-finite. Hence by Theorem 13, \(R[[x]]\) is a Nil-\(S\)-Noetherian ring. ◻

Proof. Suppose \(Q\) is an \(S\)-irreducible nil ideal of \(R\). Let \(a,b\in R\) be such that \(ab\in Q\) and \(sb\notin Q\) for all \(s\in S\). Our aim is to show that there exists \(t\in S\) such that \(ta\in rad(Q)\). Consider \(A_n=\{x\in R| a^{n}x\in Q \}\) for \(n\in\mathbb{N}\). Let \(y\in A_{n}\). Then \(a^{n}y\in Q\). Since \(Q\) is nil, there exists \(k\in\mathbb{N}\) such that \((a^{n}y)^{k}=0\). Write \(y^{k}=y^{k} a^{nk-nk}=(a^{n}y)^{k} a^{-nk}=0\). This implies that each \(A_{n}\) is a nil ideal for all \(n\in\mathbb{N}\), and \(A_1\subseteq A_2\subseteq A_3\subseteq \cdots\) is an increasing chain of nil ideals of \(R\). Since \(R\) is a Nil-\(S\)-Noetherian, by Theorem 5, this chain is \(S\)-stationary, i.e., there exist \(k\in \mathbb{N}\) and \(s\in S\) such that \(sA_n\subseteq A_k\) for all \(n\geq k\). Consider the two ideals \(I=(a^{k}) + Q\) and \(K=(b) + Q\) of \(R\). Then \(Q\subseteq I\cap K\). For the reverse containment, let \(u\in I\cap K\). Write \(u=a^{k}v+q\) for some \(v\in R\) and \(q\in Q\). Since \(ab\in Q\), \(aK\subseteq Q\); whence \(au\in Q\). Now, \(a^{k+1}v=a(a^{k}v)=a(u-q)\in Q\). This implies that \(v\in A_{k+1}\), and so \(sv\in sA_{k+1}\subseteq A_k\). Consequently, \(a^{k}sv\in Q\) which implies that \(a^{k}sv +sq=su\in Q\). Thus we have \(s(I\cap K)\subseteq Q\subseteq I\cap K\). This implies that there exists \(s'\in S\) such that either \(ss'I\subseteq Q\) or \(ss'K\subseteq Q\) since \(Q\) is \(S\)-irreducible. If \(ss'K\subseteq Q\), then \(ss'b\in Q\) which is not possible. Therefore \(ss'I\subseteq Q\) which implies that \(ss'a^{k}\in Q\). Put \(t=ss'\in S\). Then \((ta)^{k} \in Q\) and hence \(ta\in rad(Q)\), as desired. ◻

Proof. Let \(E\) be the collection of nil ideals of \(R\) which are disjoint with \(S\) and can not be written as a finite intersection of \(S\)-primary ideals. We wish to show \(E=\emptyset\). On the contrary suppose \(E\neq\emptyset\). Since \(R\) is Nil-\(S\)-Noetherian ring, by Theorem 5, there exists an \(S\)-maximal element in \(E\), say \(I\). Evidently, \(I\) is not an \(S\)-primary ideal, by Lemma 24, \(I\) is not an \(S\)-irreducible ideal, and so \(I\) is not an irreducible ideal. This implies that \(I=K\cap L\) for some ideals \(K\) and \(L\) of \(R\) with \(I\neq K\) and \(I\neq L\). As \(I\) is not \(S\)-irreducible, and so \(sK\nsubseteq I\) and \(sL\nsubseteq I\) for all \(s\in S\). Now, we claim that \(K\), \(L\notin E\). For this, if \(K\) (respectively, \(L\)) belongs to \(E\), then since \(I\) is an \(S\)-maximal element of \(E\) and \(I\subset K\) (respectively, \(I\subset L\)), there exists \(s'\) (respectively, \(s''\)) from \(S\) such that \(s'K\subseteq I\) (respectively, \(s''L\subseteq I\)). This is not possible, as \(I\) is not \(S\)-irreducible. Therefore \(K, L\notin E\). Also, if \(K\cap S\neq\emptyset\) \((respectively, L\cap S\neq\emptyset)\), then there exist \(s_{1}\in K\cap S\) \((respectively, s_{2}\in L\cap S)\). This implies that \(s's_{1}\in s'K\subseteq I\) \((respectively, s''s_{2}\in s''L\subseteq I)\), which is a contradiction because \(I\) disjoint with \(S\). Thus \(K\) and \(L\) are also disjoint with \(S\). This implies that \(K\) and \(L\) can be written as a finite intersection of \(S\)-primary ideals. Consequently, \(I\) can also be written as a finite intersection of \(S\)-primary ideals since \(I=K\cap L\), a contradiction as \(I\in E\). Thus \(E=\emptyset\), i.e., every proper nil ideal of \(R\) disjoint with \(S\) can be written as a finite intersection of \(S\)-primary ideals. ◻