Amalgamated algebra along an ideal defined by S-noetherian spectrum-like-conditions

Abstract

Let $f:A\longrightarrow B$ be a ring homomorphism and J be an ideal of B. The subring $A\bowtie ^{f}J:=\{(a,f(a)+j)/ a\in A\quad \text {et}\quad j\in J\}$ of   $A\times B$ is called the amalgamation of A with B along with J with respect to f. In this paper we investigate a general concept of ring with Noetherian spectrum, called S-Noetherian spectrum property which was introduced by Hamed, on the $A\bowtie ^{f}J$ for a multiplicative subset S of $A\bowtie ^{f}J.$

1 Introduction

A ring R is said to have Noetherian spectrum if R satisfies the ascending chain conditions (ACC) on radical ideals (cf. Ohm and Pendleton 1968; Chain-Pi 2014). This is equivalent to the condition that R satisfies the ACC on prime ideals and each ideal has only finitely many prime ideals minimal over it. Note that every Noetherian ring has Noetherian spectrum. Later, Anderson and Dumitrescu (2002) generalized the notion of Noetherian ring by introducing the concept of S-Noetherian rings. Let R be a commutative ring, S a multiplicative subset of R and I an ideal of R. According to Anderson and Dumitrescu (2002), we say that I is S-finite, if $sI\subseteq J\subseteq I$ for some finitely generated ideal J of R and some $s\in S.$ Also, R is called S-Noetherian ring if each ideal of R is S-finite. Many authors studied several different results which generalize well-known results on Noetherian rings (cf. Anderson and Dumitrescu 2002; Hamed and Hizem 2015, 2016; Lim 2015).

Ahmed (2018) generalized the notion of Noetherian spectrum property by introducing the concept of S-Noetherian spectrum property. Let R be a commutative ring with identity, $S\subseteq R$ a given multiplicative subset of R and I an ideal of R. According to Ahmed (2018), we say that I is radically S-finite, if $sI\subseteq \sqrt{J}\subseteq \sqrt{I}$ for some finitely generated ideal J of R and some $s\in S.$ Also, we say that R satisfied the S-Noetherian spectrum property if each ideal of R is radically S-finite. Note that if S consists of units of R, then R satisfies the S-Noetherian spectrum property if and only if R has Noetherian spectrum.

Recently S-version of many special rings, ideals and modules draw attention. See, for examples (Hamed and Hizem 2015, 2016; Lim 2015; Lim and Oh 2014).

Let A be a commutative ring and M be an A-module. The trivial ring extension of A by M (or the idealization of M over A) is the ring $R =A (+) M$ whose underlying group is $A\times M$ with multiplication given by $(a, b)(c, d)=(ac, ad + bc)$ (for example see Anderson and Winders 2009).

Let $f:A\longrightarrow B$ a ring homomorphism and let J be an ideal of B. The amalgamation of A with B along J with respect to f is the subring of $A\times B$ given by:

$$\begin{aligned} A\bowtie ^{f}J:=\{(a,f(a)+j)/a\in A,j\in J\} \end{aligned}$$

This construction is a natural generalization of duplications and amalgamations (D’ana et al. 2009; D’ana and Fontana 2007a, b; D’ana et al. 2010).

If A is a commutative ring and I be an ideal of A,  the The amalgamated duplication of a ring A along an ideal I is the subring of $A\times A$ given by:

$$\begin{aligned} A\bowtie I:=\{(a,a+i)\vert a\in A,i\in I\}. \end{aligned}$$

In Lim and Oh (2014) J.W. Lim and D.Y. Oh studied the transfer the notion of S-Noetherian rings to amalgamated algebra and characterize $A\bowtie ^{f}J$ to be S-Noetherian ring.

In this paper, we pursue the investigation on the structure of the rings of the form $A(+)M$ and $A\bowtie ^{f}J$, with a particular attention to the S-Noetherian spectrum property introduced and studied in Ahmed (2018). More precisely, in Sect. 2 we give necessary and sufficient condition for the trivial ring extension to be satisfies the $S(+)N$-Noetherian spectrum property, where S is a multiplicative subset of A and N a submodule of the A-module M. In Sect. 3 we give a necessary and sufficient condition for amalgamation to satisfies the $S'$-Noetherian spectrum where $S'=\{(s,f(s)\vert s\in S\}$ and S are respectively multiplicative subset of $A\bowtie ^{f}J$ and A. Our motivation in this work is to provide new classes of rings which satisfies the S-Noetherian spectrum property.

In Sect. 2, we study the S-Noetherian spectrum property under trivial extension

2 Trivial extension defined by S-Noetherian spectrum property

In this section we give the fisrt result establishes necessary and sufficient conditions for a trivial extension to inherit the S-Noetherian spectrum property. For this purpose, let us adopt the following notations:

Let R a commutative ring and M be an R-module. The product $R\times M$ may be endowed with a structure of ring such that the addition is term wise and the multiplication is defined by $(a,x)(a',x')=(aa',ax'+a'x).$ We note $A(+)M$ this ring and we call it trivial extension of R by M or the idealization of M in R. The zero element is (0, 0) and the unity is (1, 0)(cf. Anderson and Winders 2009).

If I is an ideal of A and N is a submodule of M,  by the Theorem (Anderson and Winders 2009, Theorem 3.1) $A(+)N$ is an ideal of $A(+)M$ if and only if $IM\subseteq N.$ Moreover, the prime ideal of $A(+)M$ are the form $P(+)M$ where P is a prime ideal of A. Let S be a multiplicative subset of A and N an submodule of M,  It is clear to check that $S(+)M$ is a multiplicative subset of $A(+)M.$

Ours next result characterizes the radically $S(+N)$-finite ideals of the form $I(+)N$ where I is an ideal of R and N a submodule of M.

Lemma 2.1

Let A be a commutative ring, M be an A-module and S be a multiplicative subset of A. Let I be a proper ideal of A and N be a submodule of M. Then, $I(+)M$ is an radically $S(+)N$-finite ideal of $A(+)M$ if and only if I is an radically S-finite ideal of A.

Proof

For the “only if” part, let I be an ideal of A. Since $I(+)M$ is a radically $(S(+)N$-finite ideal of $A(+)M,$ there exist $(s,n)\in S(+)N$ and a finitely generated ideal L of $A(+)M$ such that $(s,n)(I(+)M)\subseteq \sqrt{L}\subseteq \sqrt{I(+)M}.$ From the Theorem (Anderson and Winders 2009, Theorem 3.2) we have $\sqrt{I(+)M}=\sqrt{I}(+)M,$ and $\sqrt{L}=\sqrt{K}(+)M$ where $K:=\{r\in A; (r,m)\in L\quad \hbox { for some}\ m\in M\}.$ It is clear to check that $sI\subseteq \sqrt{K}\subseteq \sqrt{I}.$ Now we will show that K is finitely generated ideal of A. Indeed check $r\in K,$ then $(r,m)\in L$ for some $m\in M.$ Since L is finitely generated ideal of $A(+)M,$ then $L=\sum _{i=1}^{n}(A(+)M)(a_{i},e_{i}).$ Hence there exist $(\alpha _{i},f_{i}) \in A(+)M$ such that $(r,m)=\sum _{i=1}^{n}(\alpha _{i},f_{i})(a_{i},e_{i})=\sum _{i=1}^{n}(\alpha _{i}a_{i}, \alpha _{i}e_{i}+a_{i}f_{i}),$ which implies that $r=\sum _{i=1}^{n}\alpha _{i}a_{i}.$ Hence K is finitely generated ideal of A. It follows that I is radically S-finite ideal of A.

For the "If" part, Let I ba an ideal of A,  since I is radically S-finite ideal, there exist $s\in S$ and a finitely generated ideal J of A such that $sI\subseteq \sqrt{J}\subseteq \sqrt{I}.$ Then $(s,0)(I(+)M)\subseteq \sqrt{J}(+)M\subseteq \sqrt{I}(+)M.$ Take $K=J(+)JM,$ since J is finitely generated, so K is, on the other hand $\sqrt{K}=\sqrt{I}(+)M.$ Then $(s,n)(I(+)M)\subseteq \sqrt{K}\subseteq \sqrt{I(+)M}.$ This proves that $I(+)M$ is radically $S(+)N$-finite ideal of $A(+)M.$ This completes the proof. $\square$

Our next goal is to give necessary and sufficient conditions for $A(+)M$ to satisfy the $S(+)N$-Noetherian spectrum.

Theorem 2.2

Let A be a commutative ring, M be an A-module, S be a multiplicative subset of A and N be a submodule of M. Then, the following are equivalent:

• (1) A satisfies the S-Noetherian spectrum property.

• (2) $A(+)M$ satisfies the $S(+)N$-Noetherian spectrum property.

Proof

$(1)\Longrightarrow (2).~~$ Suppose that (1) satisfied. Let Q be a prime ideal of $A(+)M,$ by the Theorem (Anderson and Winders 2009, Theorem 3.2) Q has the form $Q=P(+)M$ for some prime ideal P of A. Now, by the Theorem (Ahmed 2018, Theorem 2.2), we have P is radically S-finite and follows from the Lemma 2.1, we conclude that $P(+)M$ is radically $S(+)N$-finite. Hence from the Theorem (Ahmed 2018, Theorem 2.2) $A(+)M$ satisfies the $S(+)N$-Noetherian spectrum property.

$(2)\Longrightarrow (1).~~$ Assume that (2) holds. Let P a prime ideal of A,  then by the Theorem (Anderson and Winders 2009, Theorem 3.2), we get $P(+)M$ is a prime ideal of $A(+)M,$ and so radically $S(+)N$-finite by assumption. Now the result follows from the Lemma 2.1 and the Theorem (Ahmed 2018, Theorem 2.2). $\square$

Let R be a commutative ring and S be a multiplicative subset of R. recall from Hamed and Hizem (2016) that, an increasing sequence $(I_{k})_{k\ge 1}$ of ideals of R is called S-stationary if there exist a positive integer n and $s\in S$ such that for each $k\ge n, sI_{k}\subset I_{n}.$

From the Theorem (Anderson and Winders 2009, Theorem 3.2 (3)), the radicals ideals of $R(+)M$ have the form $I(+)M$ where I is a radical ideal of R. And, if J is an ideal of $R(+)M,$ then $\sqrt{J}=\sqrt{I}(+)M$ where $I=\{r\in R\vert (r,m)\in J\quad \hbox { for some}\ m\in M\}$ is an ideal of R. Now following the Theorem (Ahmed 2018, Theorem 2.1) and the Theorem (Anderson and Winders 2009, Theorem 3.2(3)), we obtain the following theorem.

Theorem 2.3

Let A be a commutative ring, M an A-module, S be an at most countable multiplicative subset of A and N be a submodule of M. Then the following are equivalent:

• (1) A satisfies the S-Noetherian spectrum property.

• (2) Every increasing sequence of radical ideals of A is S-stationary.

• (3) Every increasing sequence of radical ideals of $A(+)M$ is $S(+)N$-stationary.

• (4) $A(+)M$ satisfies the $S(+)N$-Noetherian spectrum property.

Proof

$(1)\Longleftrightarrow (2).~~$ Follows from the Theorem (Ahmed 2018, Theorem 2.1).

$(3)\Longleftrightarrow (4).~~$ Follows from the Theorem (Ahmed 2018, Theorem 2.1).

Now we will show that $(2)\Longleftrightarrow (3).$

$(2)\Longrightarrow (3).~~$ Let $(J_{k})_{k\ge 1}$ be an increasing sequence of radical ideals of $A(+)M.$ By the Theorem (Anderson and Winders 2009, Theorem 3.2 (3)), for each positive integer k,  there exist a radical ideal $I_{k}$ of A such that $J_{k}=I_{k}(+)M.$ It is easy to check that the sequence $(I_{k})_{k\ge 1}$ is increasing sequence of radicals ideals of A which is S-stationary by the Theorem (Ahmed 2018, Theorem 2.1), that is, there exist $s\in S$ and a positive integer n such that $sI_{k}\subseteq I_{n}$ for each positive integer $k\ge n.$ Now, for each $n\in N,$ we have $(s,n)(I_{n}(+)M)\subseteq I_{k}(+)M,$ since for every $(a,f)\in I_{n}(+)M$ we have $(s,n)(a,f)=(sa,sf+an)\in I_{k}(+)M.$ Hence the increasing sequence $(J_{k})$ of radical ideals of $A(+)M$ is $S(+)N$-stationary, which yields that $R(+)N$ satisfies $S(+)M$- the Noetherian spectrum property.

$(3)\Longrightarrow (2).~~$ Assume that (2) holds. Let $(I_{k})_{k\ge 1}$ be an increasing sequence of radical ideals of A. Then $(I_{k}(+)M)_{k\ge 1}$ is an increasing sequence of radical ideals of $A(+)M.$ By assumption there exist $(s,g)\in S(+)N$ and a positive integer n such that $(s,g)(I_{k}(+)M)\subseteq I_{n}(+)M$ for each $k\ge n,$ so $s I_{k}\subseteq I_{n}$ for each $k\ge n.$ Hence the sequence $(I_{k})_{k\ge 1}$ is S-stationary. Moreover, A satisfies the S-Noetherian spectrum property. $\square$

Corollary 2.4

Let A be a commutative ring, M an A-module. Then the following are equivalent:

• (1) A satisfies the Noetherian spectrum.

• (2) $A(+)M$ satisfies the Noetherian spectrum.

Proof

Follows from the Theorem 2.2 with $S=\{1\}$ and $N=\{0\}.$ $\square$

Example 2.5

Let A be an integral domain and M be a finitely generated A-module. Let $S=A\setminus \{0\}$ which is a multiplicative subset of A. Take I be a non-zero ideal of A,  then for $s\in S\cap I,$ we have $sI\subseteq sA\subseteq \sqrt{sA}\subseteq I\subseteq \sqrt{I}.$ Then every non-zero ideal I of A is radically S-finite, which implies by the Theorem 2.2, that $I(+)M$ is radically-$S(+)M$-finite. Now, the ideal $0(+)M$ is finitely generated, and for every $(s,m)\in S(+)M, (s,m)(0(+)M)=0(+)sM\subseteq \sqrt{0(+)M}=0(+)M.$ Hence $0(+)M$ is radically $S(+)M$-finite ideal of $A(+)M.$ By the Theorem (Anderson and Winders 2009, Theorem 3.2), since the prime ideal of $A(+)M$ are the form $P(+)M$ where P a prime ideal of A. Now, the fact that A is an integral domain, gives that P is radically S-finite for each non zero prime ideal of A. Hence $P(+)M$ is radically $S(+)M$-finite ideal of $A(+)M.$ From the above, we get the prime ideal $0(+)M$ is radically $S(+)M$ finite. Hence, every prime ideal of $A(+)M$ is radically $S(+)M$-finite. Moreover $A(+)M$ satisfies the $S(+)M$-spectrum property. Note that $A(+)M$ is not an integral domain since $(0(+)M)^{2}=0.$

Example 2.6

Let $A=\mathbb {Z}(+)\mathbb {Z}$ and $S\subseteq \mathbb {Z}$ any multiplicative set. It is clear that A satisfies the S-Noetherian spectrum. Since the pime ideals are the form $p\mathbb {Z}(+)\mathbb {Z}$ which clearly are radically S-finite. By the Theorem (Ahmed 2018, Theorem 3.1), we get $(\mathbb {Z}(+)\mathbb {Z})[X]$ satisfies the $S(+)\mathbb {Z}$-Noetherian spectrum. Now let $B=\mathbb {Z}(+)\mathbb {Z}[X]$ not satisfies the $S(+)\mathbb {Z}[X]$-Noetherian spectrum, since the ideal $0(+)X\mathbb {Z}[X]$ is not radically $S(+)\mathbb {Z}[X]$-finite for degrees reasons.

Example 2.7

Let $A=\mathbb {Z}(+)\mathbb {Q}$ and $S=\mathbb {Z}\setminus \{0\}.$ We have A satisfies the S-Noetherian spectrum since $\mathbb {Z}$ is an integral domain. But, by the Theorem (Baeck and Lim 2016, Theorem 3.8) A is not S-Noetherian since the $\mathbb {Z}$-module $\mathbb {Q}$ is not S-finite.

3 Amalgamated algebra along an ideal defined by S-Noetherian spectrum property

In this section we establish the necessary and sufficient conditions for a amalgamated algebra along an ideal to inherit the S-Noetherian spectrum property. For this purpose, let us adopt the following definition and notation. Let A and B be commutative rings with identity, let J be an ideal of B and $f:A\longrightarrow B$ be a ring homomorphism. We set

$$\begin{aligned} A\bowtie ^{f}J=\{(a,f(a)+j)\vert a\in A,j\in J\} \end{aligned}$$

$A\bowtie ^{f}J$ is a subring of $A\times B$ called the amalgamation of A with B along J with respect to f. Such construction was introduced and studied by D’anna, Finacchiaro and Fontana (cf. D’ana et al. 2010; D’ana et al. 2009) this construction is in fact a generalization of the amalgamated duplication of ring along an ideal (cf. D’ana and Fontana 2007a, b).

Let S be a multiplicative subset A. Clearly f(S) is a multiplicative subset of $f(A)+J.$ Set $S'=\{(s,f(s))\vert s\in S\}$ which is a multiplicative subset of $A\bowtie ^{f}J.$

Let I be an ideal of A and K be an ideal of $f(A)+J.$ Notice that

$$\begin{aligned} I\bowtie ^{f}J=\{(i,f(i)+j)\vert i\in I,j\in J\} \end{aligned}$$

and

$$\begin{aligned} {\overline{K}}^{f}=\{(a,f(a)+j)\vert a\in A,j\in J,f(a)+j\in K\} \end{aligned}$$

are ideals of $A\bowtie ^{f}J.$

The next result gives a characterization of the property radically $S'$-finite ideal for ideals of the form $I\bowtie ^{f}J$ where I is an ideal of A.

Lemma 3.1

Let $f:A\longrightarrow B$ be a homomorphism rings, J be an finitely generated ideal of $f(A)+J.$ Let I be an ideal of A. Then the following statements are equivalent:

• (i) I is radically S-finite ideal of A.

• (ii) $I\bowtie ^{f}J$ is radically $S'$-finite ideal of $A\bowtie ^{f}J.$

Proof

$(i)\Longrightarrow (ii).\quad$ Assume that (1) holds. Let I be an ideal of A. We can find $s\in S$ and a finitely generated ideal K of A such that $sI\subseteq \sqrt{K}\subseteq \sqrt{I}.$ Clearly $(s,f(s))(I\bowtie ^{f}J)\subseteq \sqrt{K}\bowtie ^{f}J\subseteq \sqrt{I}\bowtie ^{f}J.$ Following the definition of the radical of an ideal, it is easy to show that $\sqrt{K}\bowtie ^{f}J=\sqrt{K\bowtie ^{f}J}$ and $\sqrt{I}\bowtie ^{f}J=\sqrt{I\bowtie ^{f}J}.$ To succeed in proving (2) it suffices to justify that $K\bowtie ^{f}J$ is finitely generated ideal of $A\bowtie ^{f}J.$ To see it, let $(a,f(a)+j)\in K\bowtie ^{f}J.$ Since $K=\sum _{i=1}^{n}Aa_{i}$ and $J=\sum _{i=1}^{n}(f(A)+J)e_{i}$ where $a_{i}\in I$ and $e_{i}\in J$ for each integer i. There exist $\alpha _{i},\beta _{i}\in A$ and $j_{i}\in J$ such that

$$\begin{aligned} (a,f(a)+j)&=\left( \sum _{i=1}^{n}\alpha _{i}a_{i},\sum _{i=1}^{n}f(\alpha _{i})f(a_{i})\right) +\left( 0,\sum _{i=1}^{n}(f(\beta _{i})+j_{i})e_{i}\right) \\&=\sum _{i=1}^{n}(\alpha _{i},f(\alpha _{i})(a_{i},f(a_{i}))+\sum _{i=1}^{n}(0,f(\beta _{i})+j_{i})(0,e_{i})\\&=\sum _{i=1}^{n}(\alpha _{i},f(\alpha _{i})(a_{i},f(a_{i}))+\sum _{i=1}^{n}(\beta _{i},f(\beta _{i})+j_{i})(0,e_{i})\\ \end{aligned}$$

Hence $(a,f(a)+j)\in \sum _{i=1}^{n}(A\bowtie ^{f}J)(a_{i},f(a_{i}))+\sum _{i=1}^{n}(A\bowtie ^{f}J)(0,e_{i}),$ since $(\alpha _{i},f(\alpha _{i}))\in A\bowtie ^{f}J$ and $(\beta _{i}, f(\beta _{i})+j_{i})\in A\bowtie ^{f}J$ for each integer i. Thus $K\bowtie ^{f}J$ is finitely generated ideal of $A\bowtie ^{f}J.$ The proof is therefore complete and leads to the conclusion that $I\bowtie ^{f}J$ is radically $S'$-finite. As desired.

$(ii)\Longrightarrow (i).~~$ Assume that (2) holds and let I be an ideal of A. By assumption there exist $s\in S$ and a finitely generated ideal L of $A\bowtie ^{f}J$ such that $(s,f(s))(I\bowtie ^{f}J)\subseteq \sqrt{L}\subseteq \sqrt{I\bowtie ^{f}J}.$ Set $p_{A}(L)=L'$ where $p_{A}$ is the canonical projection, then $L'$ is a finitely generated ideal of A. Now, by applying $p_{A}$ to the inclusions $(s,f(s))(I\bowtie ^{f}J)\subseteq \sqrt{L}\subseteq \sqrt{I\bowtie ^{f}J}$, we get $sI\subseteq \sqrt{L'}\subseteq \sqrt{I}.$ On the other hand it is easy to show that $\sqrt{L'}=\sqrt{p_{A}(L)}=p_{A}(\sqrt{L}).$ Hence I is radically S-finite ideal of A. $\square$

Our next result characterizes the property radically $S'$-finite ideal for ideals of the form ${\overline{K}}^{f}.$

Lemma 3.2

Let $f:A\longrightarrow B$ be a homomorphism rings. Let $J\nsubseteq K$ two ideals of $f(A)+J.$ Suppose that $f^{-1}(J)$ is finitely generated ideal of A. Then the following are equivalent:

• (i) K is radically f(S)-finite ideal of $f(A)+J.$

• (ii) ${\overline{K}}^{f}$ is radically $S'$-finite ideal of $A\bowtie ^{f}J.$

Proof

$(i)\Longrightarrow (ii).~~$ Assume that K is radically f(S)-finite ideal of $f(A)+J.$ Then there exist $s\in S$ and a finitely generated ideal L of $f(A)+J$ such that $f(s) K\subseteq \sqrt{L}\subseteq \sqrt{K}.$

First case: $J\subseteq L.$ If J is containing in K,  then ${\overline{K}}^{f}=f^{-1}(L)\bowtie ^{f}J.$ Since A satisfies the S-Noetherian spectrum property, then $f^{-1}(L)$ is radically S-finite. Now by the Lemma 3.1 we get ${\overline{K}}^{f}=f^{-1}(L)\bowtie ^{f}J$ is radically $S'$-finite ideal of $A\bowtie ^{f}J.$

Second case: $J\nsubseteq L.$ If L not containing J,  then ${\overline{L}}^{f}$ is not homogeneous ideal of $A\bowtie ^{f}J.$ Set $L'=p_{B}^{-1}(L)$ which is an ideal of $A\bowtie ^{f}J.$ We claim that

$$\begin{aligned} (s,f(s) {\overline{K}}^{f}\subseteq \sqrt{L'}\subseteq \sqrt{{\overline{K}}^{f}}. \end{aligned}$$

Indeed, let $(a,f(a)+j)\in {\overline{K}}^{f},$ then $f(as)+f(s)j\in \sqrt{L},$ which implies that $f(s)^{n}(f(a)+j)^{n}\in L$ for some positive integer n. That is $p_{B}((as)^{n},f(s)^{n}(f(a)+j)^{n})\in L,$ hence $\left( (s,f(s)(a,f(a)+j)\right) ^{n}\in L'.$ This yields that $(s,f(s))(a,f(a)+j)\in \sqrt{L'}.$ Thus $(s,f(s) {\overline{K}}^{f}\subseteq \sqrt{L'}.$ Now, let $y\in A\bowtie ^{f}J$ such that $y\in \sqrt{L'},$ then $y^{n}\in L'$ for some positive integer n, hence $(p_{B}(y))^{n}\in K,$ this implies that $y^{n}\in p_{B}^{-1}(K)\subseteq {\overline{K}}^{f}.$ Thus $y\in \sqrt{{\overline{K}}^{f}},$ as desired. Now we will show that ${\overline{L}}^{f}$ is a finitely generated ideal of $A\bowtie ^{f}J.$ Since $f^{-1}(J)$ is finitely generated of A,  there exist $z_{1},\dots ,z_{m}\in f^{-1}(J)$ such that $f^{-1}(Z)=\sum _{k=1}^{k=m}Az_{k}.$ Now let $(a,f(a)+j)\in {\overline{L}}^{f},$ then $f(a)+j\in p_{B}({\overline{L}}^{f}),$ where $p_{B}: A\bowtie ^{f}J\longrightarrow f(A)+J$ the canonical projection, so there exist $f(\alpha _{1})+k_{1},\dots ,f(\alpha _{n})+k_{n}\in f(A)+J$ and $f(a_{1})+l_{1},\cdot ,f(a_{n})+l_{n}\in f(A)+J$ such that $f(a)+j=\sum _{i=1}^{n}(f(\alpha _{i})+k_{i})(f(a_{i})+l_{i})),$ so $f(a)-\sum _{i=1}^{n}f(\alpha _{i})f(a_{i})\in f^{-1}(J)$ and thus $a-\sum _{i=1}^{n}\alpha _{i}a_{i}=\sum _{k=1}^{m}r'_{k}z_{k}$ for some $r'_{1},\dots ,r'_{k} \in A.$ Hence $a=\sum _{k=1}^{m}r'_{k}z_{k}+\sum _{i=1}^{n}\alpha _{i}a_{i}$ Then,

$$\begin{aligned} (a,f(a)+j)&=\left( \sum _{i=1}^{n}\alpha _{i}a_{i}+\sum _{k=1}^{m}r'_{k}z_{k},\sum _{i=1}^{n}(f(\alpha _{i})+k_{i})(f(a_{i}\right) +l_{i})\\&=\left( \sum _{i=1}^{n}\alpha _{i}a_{i}+\sum _{i=1}^{n}(f(\alpha _{i})+k_{i})(f(a_{i})+l_{i})\right) +\left( \sum _{k=1}^{m}r'_{k}z_{k},0\right) \\&=\sum _{i=1}^{n}(\alpha _{i}a_{i}, (f(\alpha _{i})+k_{i})(f(a_{i})+l_{i}))+\sum _{k=1}^{m}(r'_{k}z_{k},0)\\&=\sum _{i=1}^{n}(\alpha _{i},f(\alpha _{i})+k_{i})(a_{i},f(a_{i})+l_{i})+\left( \sum _{k=1}^{m}(r'{k},f(r'_{k})(z_{k},0)\right. \\&\left. \in (a_{i},f(a_{i})+k_{i}), (z_{i},0), 1\le i\le m,1\le i\le n\right) A\bowtie ^{f}J. \\ \end{aligned}$$

With $\{a_{i},f(a_{i})+k_{i}),(z_{k},0), 1\le i\le n, 1\le k\le m\}\subseteq {\overline{L}}^{f}.$ Consequently ${\overline{L}}^{f}$ is finitely generated ideal of $A\bowtie ^{f}J.$

$(ii)\Longrightarrow (i).~~$ Assume that ${\overline{K}}^{f}$ is radically $S'$-finite ideal of $A\bowtie ^{f}J$ and put $K:=p_{B}({\overline{K}}^{f})$ which is an ideal of $f(A)+J.$ By assumption, there exist $s\in S$ and a finitely generated ideal L of $A\bowtie ^{f}J$ such that $(s,f(s)){\overline{K}}^{f}\subseteq \sqrt{L}\subseteq \sqrt{{\overline{K}}^{f}}.$ The fact that the canonical projection $p_{B}:A\bowtie J \longrightarrow f(A)+J$ is surjective gives that $L'=p_{B}(L)$ is finitely generated ideal of $f(A)+J.$ Now, let $a\in A$ and $j\in J.$ We have $f(a)+j \in \sqrt{L'}\Longleftrightarrow (f(a)+j)^{n}\in L\quad \hbox { for some positive integer}\ n\Longleftrightarrow (a^{n},(f(a)+j)^{n})=(a,f(a)+j)^{n}\in L\Longleftrightarrow (a,f(a)+j))\in \sqrt{L}\Longrightarrow f(a)+j\in p_{B}(\sqrt{L}).$ Since $f(a)+j\in p_{B}(\sqrt{L}),$ then $f(a)+j=p_{B}(y)$ for some $y\in \sqrt{L}\subseteq A\bowtie ^{f}J,$ which implies $(f(a)+j)^{n}=p_{B}(y^{n})\in p_{B}(L)=L'$ for some positive integer n. Hence $f(a)+j\in \sqrt{L'}.$ Consequently we get $p_{B}(\sqrt{L})=\sqrt{L'}.$ Now, let $a\in A$ and $j\in J,$ $(s,f(s)(a,f(a)+j)\in \sqrt{L}$, we deduce that $f(s)(f(a)+j)\in p_{B}(\sqrt{L})=\sqrt{L'}.$ Thus $f(s)K\subseteq \sqrt{L'}.$ On the other hand the inclusion $\sqrt{L'}\subseteq \sqrt{K}$ is clear. Finally $f(s)K\subseteq \sqrt{L'}\subseteq \sqrt{K}.$ Hence K is radically f(S)-finite ideal of $f(A)+J.$ $\square$

Now, using the previous lemmas we will give a characterization for $A\bowtie ^{f}J$ to satisfies the $S'$-Noetherian spectrum property.

Theorem 3.3

Let $f:A\longrightarrow B$ be a homomorphism rings, S be a multiplicative subset of A. Suppose that J and $f^{-1}(J)$ are finitely generated ideal of $f(A)+J$ and A respectively. Then the following statements are equivalent:

• (1) $A\bowtie ^{f}J$ satisfies the $S'$-Noetherian spectrum property.

• (2) A satisfies the S-Noetherian property and the subring $f(A)+J$ satisfies the f(S)-Noetherian spectrum property

Proof

$(1)\Longrightarrow (2).~~$ Assume that (1) holds. To establish (2) it suffices by (Ahmed 2018, Theorem 2.2) to justify that the prime ideals of A are radically S-finite and the prime ideal of $f(A)+J$ are radically f(S)-finite. Indeed, let P a prime ideal of A,  then $P\bowtie ^{f}J$ is a prime ideal of $A\bowtie ^{f}J$ which is radically $S'$-finite by assumption. Now, by Lemma 3.1, we deduce that P is radically S-finite. As desired

By the same argument above, we establish that $f(A)+J$ satisfies the f(S)-Noetherian spectrum property. Indeed, let Q be a prime ideal of $f(A)+J,$ then, by assumption we have ${\overline{Q}}^{f}$ is radically $S'$-finite. So by Lemma 3.2, we get that Q is radically f(S)-finite ideal of $f(A)+J.$ Hence $f(A)+J$ satisfies the f(S)-Noetherian spectrum property. This completes the proof.

$(2)\Longrightarrow (1).~~$ Assume that (2) holds, we will show that every prime ideal ${\mathcal {P}}$ of $A\bowtie ^{f}J$ is radically $S'$-finite ideal. Indeed, let ${\mathcal {P}}$ a prime ideal of $A\bowtie ^{f}J.$ According to the spectrum of $A\bowtie ^{f}J$, there are two cases to be treated.

First case: If ${\mathcal {P}}=P\bowtie ^{f}J$ where P is a prime ideal of A,  by assumption P is radically S-finite, then by the Lemma 3.1, we get ${\mathcal {P}}$ is radically $S'$-finite.

Second case: If ${\mathcal {P}}={\overline{Q}}^{f}$ where Q is a prime ideal of B not containing J. Then by Lemma 3.2 we get ${\overline{Q}}^{f}$ is radically $S'$-finite ideal of $A\bowtie ^{f}J.$

Consequently, every prime ideal of $A\bowtie ^{f}J$ is radically $S'$-finite. So Theorem (Ahmed 2018, Theorem 2.2), we conclude that $A\bowtie ^{f}J$ satisfies the $S'$-Noetherian property. $\square$

Corollary 3.4

Let $f:A\longrightarrow B$ be a homomorphism rings, J an ideal of $f(A)+J.$ If J and $f^{-1}(J)$ are finitely generated ideal of $f(A)+J$ and A respectively. Then the following statements are equivalent:

• (1) $A\bowtie ^{f}J$ satisfies the Noetherian spectrum condition.

• (2) A and $f(A)+J$ satisfies the Noetherian spectrum condition.

Proof

Follows from the Theorem 3.3 with $S=\{1\}.$ $\square$

Corollary 3.5

Let A be a commutative, I be an finitely generated ideal of A and S be a multiplicative subset of A. Then the following are equivalent

• (1) A satisfies the S-Noetherian spectrum property.

• (2) $A\bowtie I$ satisfies the $S\times S$-Noetherian spectrum property.

Proof

Set $B=A$ and $f=id$ and $J=I$ in the Theorem 3.3.$\square$

Corollary 3.6

Let A be a commutative, I be an finitely generated ideal of A. Then the following are equivalent

• (1) A has a Noetherian spectrum property.

• (2) $A\bowtie I$ has a Noetherian spectrum property.

Proof

Follows from the Corollary 3.6 with $S=\{1\}.$ $\square$

Example 3.7

Let A be an integral domain, and $S=A\setminus \{0\}.$ We have A satisfies the S-Noetherian spectrum. Let I be an ideal of A. Let $B=A\bowtie I$ satisfies $S\times S$-Noetherian spectrum property.

Example 3.8

Let $f:A\longrightarrow \longrightarrow B$ be a homomorphism ring, J be a finitely generated ideal of B such that $f^{-1}(J)$ is finitely generated ideal of B. Suppose that $J\subseteq Nil(B).$ Then $A\bowtie ^{f}J$ satisfies the $S'$-Noetherian spectrum if and only if A satisfies the S-Noetherian spectrum.

Proof

Since $J\subseteq Nil(B),$ we get the prime ideals of $A\bowtie ^{f}J$ are the form $P\bowtie ^{f}J$ where P is a prime ideal of A. Now the result follows from the Theorem 3.3.

Example 3.9

Let A be a commutative ring and M be a finitely generated A-module. Let S be a multiplicative subset of A. Put $J=0(+)M.$ Since M is finitely generated A-module, so J is finitely generated ideal of $A(+)M.$ Let

$$\begin{aligned} f:\left\{ \begin{array}{llll} A\longrightarrow B:=A(+)M\\ a\longmapsto (a,0)\\ \end{array} \right. \end{aligned}$$

We have on the one hand $f^{-1}(J)=\{0\}$ and on the other hand $J\subseteq Nil(B)$ since $J^{2}=0.$ So $A\bowtie ^{f}J$ satisfies the $S'=\{s,f(s)\vert s\in S\}$-Noetherian spectrum if and only if A satisfies the S-Noetherian spectrum.

$\square$

Example 3.10

Let K be a field and E a K-vector space. Let $A=K(+)E.$ According the Theorem 2.2, we have A satisfies $S:=\{1\}(+)E$-Noetherian spectrum. Set $I=0(+)E.$ Then by the Corollary 3.5, we get $A\bowtie I$ satisfies the S-Noetherian spectrum.