Strongly Graded Rings Over Hereditary Noetherian Prime Rings

Abstract

Let $R=\bigoplus _{n\in \mathbb {Z}}R_{n}$ be a strongly graded ring of type $\mathbb {Z}$ where $R_{0}$ is a hereditary Noetherian prime ring. In this paper, we investigate and completely describe the structure of projective ideals of R and prove that R is a strongly G-HNP ring.

1 Introduction

Strongly graded rings attracted growing attention in the literature, and several papers have been published with different names such as Clifford systems or generalized crossed products. In this paper, we contribute to the theory of strongly graded rings of type $\mathbb {Z}$ by completely describing the structure of its projective ideals when the base ring $R_{0}$ is a hereditary Noetherian prime ring.

In Sect. 2, we obtain results on maximal $\mathbb {Z}$-invariant ideals of $R_{0}$ and prove that each maximal $\mathbb {Z}$-invariant ideal of $R_{0}$ is either idempotent or invertible. Moreover, we show that the invertible $\mathbb {Z}$-invariant ideals of $R_{0}$ generate an abelian group whose generators are maximal $\mathbb {Z}$-invariant ideals of $R_{0}$. The results in Sect. 2 are used to describe the structure of projective ideals of R in Sects. 3 and 4.

Section 3 is devoted to the structure of projective ideals of R with nonzero contracted ideals of $R_{0}$, and in Sect. 4 we study the projective ideals of R with zero contracted ideals of $R_{0}$. After completing the structure theorems on projective ideals, we conclude that R is a strongly G-HNP ring.

Let R be a prime Goldie ring with quotient ring Q. We refer the readers to [8, 9] for unexplained terminology in the arithmetic theory of orders. Let X and Y be subsets of Q. There are two subsets of Q derived from X and Y, namely $(X:Y)_{l}=\{q\in Q|qY\subset X\}$ and $(X:Y)_{r}=\{q\in Q|Yq\subset X\}$, which are of particular importance in our study. If I is any (fractional) right R-ideal, then we define $I_{v}=(R:(R:I)_{l})_{r}$. Note that we always have $I\subseteq I_{v}$, and if the equality holds, then I is called a rightv-ideal. Similarly, we define $_{v}J=(R:(R:J)_{r})_{l}$ for a left R-ideal J, and if $J= {_{v}J}$, then J is called a left v-ideal. An R-ideal A is called a v-ideal provided $_{v}A=A=A_{v}$. An integral v-ideal is referred to as a v-ideal of R.

We refer the readers to [10] for further information on graded rings which are not mentioned in this paper.

2 The Structure of ${\mathbb {Z}}$-Invariant Ideals of $R_{0}$

Let $R = \oplus _{n \in {\mathbb {Z}}} R_n$ be a strongly graded ring of type ${\mathbb {Z}}$ and $R_0$ be a prime Goldie ring with its quotient ring $Q_0$. Let ${\mathcal {C}}_0$ be the set of all regular elements in $R_0$. Then ${\mathcal {C}}_0$ is a regular Ore set of R, and $Q^g = R{{\mathcal {C}}}_0^{-1}$, the quotient ring of R at ${\mathcal {C}}_0$, is of the form $Q^g = \oplus _{n \in {\mathbb {Z}}} Q_0R_n$ with $Q_0R_n = R_nQ_0$ for all $n \in {\mathbb {Z}}$ and is a principal ideal ring (see [12, Corollary 2.3, p.181]). We denote by Q the quotient ring of R which is the quotient ring of $Q^g$.

Definition 2.1

An $(R_0, \ R_0)$-bisubmodule $A_0$ of $Q_0$ is $\mathbb {Z}$-invariant if $R_nA_0 = A_0R_n$ for all $n \in {\mathbb {Z}}$, which is equivalent to $R_1A_0R_{-1} = A_0$.

Lemma 2.2

Let $A_0$ be an $R_0$-ideal in $Q_0$. Then $A_0R$ is an R-ideal if and only if $A_0$ is ${\mathbb {Z}}$-invariant.

Proof

$\Longrightarrow$ : For any $n \in {\mathbb {Z}}$, $R_nA_0R_{-n} \subseteq R_nA_0R \cap R_0 \subseteq A_0R \cap R_0 = A_0$ and similarly $R_{-n}A_0R_n \subseteq A_0$. Hence, $R_nA_0 = A_0R_n$ for all $n \in {\mathbb {Z}}$.

$\Longleftarrow$ : If $A_0$ is ${\mathbb {Z}}$-invariant, then $R_nA_0R = A_0R_nR \subseteq A_0R$ for all $n \in {\mathbb {Z}}$ and so $RA_0R = A_0R$. Similarly, $RA_0R = RA_0$ and hence $A_0R$ is an R-ideal. $\square$

Lemma 2.3

Let A be an R-ideal in Q and $A_0 = A \cap Q_0$. Then A and $A_0$ are both ${\mathbb {Z}}$-invariant.

Proof

It is clear that A is ${\mathbb {Z}}$-invariant, that is, $R_nA = AR_n$ for all $n \in {\mathbb {Z}}$ (see the proof of Lemma 2.2). It follows that $A_0R_n = (A \cap Q_0)R_n = AR_n \cap Q_0R_n = R_nA \cap R_nQ_0 = R_nA_0$. Hence, $A_0$ is ${\mathbb {Z}}$-invariant. $\square$

Lemma 2.4

Let $A_0$ and $B_0$ be ${\mathbb {Z}}$-invariant $R_0$-ideals in $Q_0$. Then

• (1) So are $(R_0 : A_0)_l, \ (R_0 : A_0)_r, \ O_l(A_0)$ and $O_r(A_0)$.

• (2) $A_0 \cap B_0, \ A_0B_0$ and $A_0 + B_0$ are all ${\mathbb {Z}}$-invariant.

Proof

Easy. $\square$

From now on, unless otherwise stated, we assume that $R_0$ is an HNP ring.

Lemma 2.5

An ideal $N_0$ of $R_0$ is a maximal ${\mathbb {Z}}$-invariant ideal of $R_0$ (ideals maximal among ${\mathbb {Z}}$-invariant ideals of $R_0$) if and only if there is a maximal ideal $M_0$ of $R_0$ containing $N_0$ and $n \ge 1$ such that

$$\begin{aligned} N_0= & {} M_0 \cap R_1M_0R_{-1} \cap \cdots \cap R_nM_0R_{-n} \\= & {} M_0 \cap R_1M_0R_{-1} \cap \cdots \cap R_{n+1}M_0R_{-(n+1)} \ \ \ \ \ (*) \end{aligned}$$

Proof

$\Longrightarrow$ : Let $M_0$ be a maximal ideal containing $N_0$. Since $R_0/N_0$ is Artinian and $N_0$ is ${\mathbb {Z}}$-invariant, the chain

$$\begin{aligned} M_0 \supseteq M_0 \cap R_1M_0R_{-1} \supseteq M_0 \cap R_1M_0R_{-1} \cap R_2M_0R_{-2} \supseteq \cdots \supseteq N_0 \end{aligned}$$

must stabilize and so there is an $n \ge 1$ such that $A_n = A_{n+1}$, where $A_n = M_0 \cap R_1M_0R_{-1} \cap \cdots \cap R_nM_0R_{-n}$. We may assume

$$\begin{aligned} A_0 = M_0 \supset A_1 \supset \cdots \supset A_{n-1} \supset A_n \ \hbox { and } \ A_n = A_{n+1} \ \ \ \ \ (\sharp ) \end{aligned}$$

Since $R_iM_0R_{-i}$ are all maximal ideals, there is an $i \ (0 \le i \le n)$ such that $R_{n+1}M_0R_{-(n+1)} = R_iM_0R_{-i}$. Suppose $i \ge 1$. Then $M_0 = R_{n+1-i}M_0R_{-n-1+i}$ with $l = n + 1 - i \le n$ and so $A_{l-1} = A_l$, a contradiction. Hence, $i = 0$, that is, $R_{n+1}M_0R_{-(n+1)} = M_0$. Thus, $R_1A_nR_{-1} = R_1(M_0 \cap \cdots \cap R_nM_0R_{-n})R_{-1} = A_n$ and so $A_n$ is ${\mathbb {Z}}$-invariant. Hence, $A_n = N_0$.

$\Longleftarrow$ : If $N_0$ satisfies ($*$) and ($\sharp$), then it is easy to see that $N_0$ is $\mathbb Z$-invariant. If $B_0$ is a maximal ${\mathbb {Z}}$-invariant ideal of $R_0$ containing $N_0$, then we may assume that $M_0 \supseteq B_0$. Then it follows that $R_nM_0R_{-n} \supseteq B_0$ for all $n \ge 0$ and so $A_n \supseteq B_0$, that is, $B_0 = N_0$. $\square$

Definition 2.6

A ${\mathbb {Z}}$-invariant ideal $P_0$ is called a $\mathbb {Z}$-prime ideal if $P_0 \supseteq A_0B_0$, where $A_0$ and $B_0$ are ${\mathbb {Z}}$-invariant ideals, then either $P_0 \supseteq A_0$ or $P_0 \supseteq B_0$.

Lemma 2.7

A ${\mathbb {Z}}$-invariant nonzero ideal of $R_0$ is a maximal ${\mathbb {Z}}$-invariant ideal if and only if it is a ${\mathbb {Z}}$-prime ideal.

Proof

It is clear that any maximal ${\mathbb {Z}}$-invariant ideal is a ${\mathbb {Z}}$-prime ideal. Conversely, let $P_0$ be a $\mathbb Z$-prime nonzero ideal, $M_{01}, \ldots , M_{0n}$ be the set of all maximal ideals containing $P_0$ and $A_0 = M_{01} \cap \cdots \cap M_{0n}$. Then $A_0/P_0$ is the nilpotent radical of $R_0/P_0$ since $R_0/P_0$ is an Artinian ring. So $P_0 \supseteq A_0^k$ for some $k \ge 1$. Let $N_{0i}$ be the maximal ${\mathbb {Z}}$-invariant ideal induced by $M_{0i}$ (Lemma 2.5). Then $P_0 \supseteq A_0^k \supseteq (N_{01} \cap \cdots \cap N_{0n})^k \supseteq (N_{01} \cdots N_{0n})^k$ and $P_0 \supseteq N_{0i}$ for some i by definition. Hence, $P_0 = N_{0i}$, a maximal ${\mathbb {Z}}$-invariant ideal. $\square$

Lemma 2.8

Each maximal ${\mathbb {Z}}$-invariant ideal of $R_0$ is either idempotent or invertible.

Proof

Let $N_0$ be a maximal ${\mathbb {Z}}$-invariant ideal of $R_0$. If $(R_0 : N_0)_lN_0 = R_0 = N_0(R_0 : N_0)_r$, then $N_0$ is invertible. If $N_0 \subseteq (R_0 : N_0)_lN_0 \subset R_0$, then, by Lemma 2.4, $N_0 = (R_0 : N_0)_lN_0$. Hence,

$$\begin{aligned} N_0^2 = (R_0 : N_0)_lN_0 \cdot (R_0 : N_0)_lN_0 = (R_0 : N_0)_lO_l(N_0)N_0 = (R_0 : N_0)_lN_0 = N_0, \end{aligned}$$

and so $N_0$ is idempotent. $\square$

Lemma 2.9

Let $X_0$ be an invertible ${\mathbb {Z}}$-invariant ideal of $R_0$. Then there is a one-to-one correspondence between idempotent ${\mathbb {Z}}$-invariant ideals $A_0$ such that $X_0 \subseteq A_0 \subseteq R_0$ and ${\mathbb {Z}}$-invariant overrings $S_0$ with $X_0^{-1} \supseteq S_0 \supseteq R_0$.

Proof

The proof is the same as the one in [4, Theorem 1.6]. $\square$

Definition 2.10

A finite set of distinct idempotent maximal ${\mathbb {Z}}$-invariant ideals $M_{01}, \ldots , M_{0n}$ is called a ${\mathbb {Z}}$-cycle if $O_r(M_{01}) = O_l(M_{02}), \ldots , O_r(M_{0n}) = O_l(M_{01})$.

Note that we also consider an invertible maximal $\mathbb Z$-invariant ideal $P_0$ to be a ${\mathbb {Z}}$-cycle since $O_r(P_0) = O_l(P_0)$.

Lemma 2.11

• (1) If $M_{01}, \ldots , M_{0n}$ is a ${\mathbb {Z}}$-cycle, then $P_0 = M_{01} \cap \cdots \cap M_{0n}$ is an invertible ideal and is ${\mathbb {Z}}$-invariant.

• (2) Let $X_0$ be an invertible ${\mathbb {Z}}$-invariant ideal and $M_{01}$ be an idempotent maximal ${\mathbb {Z}}$-invariant ideal containing $X_0$. Then there are idempotent maximal $\mathbb Z$-invariant ideals $M_{0i}$ containing $X_0$$(1 \le i \le n)$ such that $M_{01}, \ldots , M_{0n}$ is a ${\mathbb {Z}}$-cycle.

Proof

• (1) For any $k \ (1 \le k \le n)$, put $S_k = O_l(M_{0k}) = O_r(M_{0(k-1)})$ (if $k = 1$, then $k - 1 = n)$ and $A = M_{01} \cdots M_{0(k-2)}M_{0(k+1)} \cdots M_{0n}$. Since $M_{0k}S_k = S_k$ and $M_{0(k-1)}S_k = M_{0(k-1)}$, we have

  $$\begin{aligned} R_0 \supseteq M_{0(k-1)}S_k \supseteq P_0S_k \supseteq AM_{0(k-1)}M_{0k}S_k = AM_{0(k-1)}, \end{aligned}$$

  and so $R_0 \supseteq P_0(R_0 : P_0)_r \supseteq P_0S_k \supseteq AM_{0(k-1)}$. It follows from Lemma 2.7 that $P_0(R_0 : P_0)_r \not \subseteq M_{0k}$ for all $k \ (1 \le k \le n)$. Hence $P_0(R_0 : P_0)_r = R_0$. Similarly $(R_0 : P_0)_lP_0 = R_0$ and hence $P_0$ is invertible.

• (2) Put $S_0 = O_r(M_{01})$, which is a minimal $\mathbb Z$-invariant overring by Lemma 2.9. Then $M_{02} = (R_0 : S_0)_r$ is an idempotent maximal ${\mathbb {Z}}$-invariant ideal containing $X_0$ by left version of Lemma 2.9 and so

  $$\begin{aligned} O_l(M_{02}) = (R_0 : M_{02})_l = (R_0 : (R_0 : S_0)_r)_l = S_0 = O_r(M_{01}). \end{aligned}$$

  Continuing this method, the chain

  $$\begin{aligned} M_{01} \supseteq M_{01} \cap M_{02} \supseteq M_{01} \cap M_{02} \cap M_{03} \supseteq \cdots \supseteq X_0 \end{aligned}$$

  must stabilize. So there is an $n \ge 1$ such that $M_{01} \cap \cdots \cap M_{0n} = M_{01} \cap \cdots \cap M_{0(n+1)}$. We may assume that n is minimal for this property and then $M_{0(n+1)} = M_{0i}$ for some $i \ (1 \le i \le n)$. If $i > 1$, then $O_r(M_{0n}) = O_l(M_{0(n+1)}) = O_l(M_{0i}) = O_r(M_{0(i-1)})$ and so $M_{0n} = M_{0(i-1)}$ by Lemma 2.9 which contradicts the choice of n. Hence, $i = 1$, that is, $M_{01}, \ldots , M_{0n}$ is a ${\mathbb {Z}}$-cycle. $\square$

The following lemma follows from Lemmas 2.7 and 2.11

Lemma 2.12

An ideal $P_0$ of $R_0$ is a maximal invertible $\mathbb Z$-invariant ideal (ideal maximal among invertible $\mathbb Z$-invariant ideals) if and only if $P_0 = M_{01} \cap \cdots \cap M_{0n}$, where $M_{01}, \ldots , M_{0n}$ is a ${\mathbb {Z}}$-cycle.

Further, combining the proofs of [4, Propositions 2.1 and 2.8] with Lemma 2.4, we have the following proposition.

Proposition 2.13

The invertible ${\mathbb {Z}}$-invariant ideals of $R_0$ generate an abelian group whose generators are maximal invertible $\mathbb Z$-invariant ideals of $R_0$.

Lemma 2.14

Let $M_0 = M_{01} \cap \cdots \cap M_{0n}$ and $N_0 = N_{01} \cap \cdots \cap N_{0l}$ be maximal idempotent ${\mathbb {Z}}$-invariant ideals (ideals maximal among idempotent ${\mathbb {Z}}$-invariant ideals), where $M_{0i}$ and $N_{0j}$ are idempotent maximal ideals of $R_0$ (Lemma 2.5). Then

• (1) $M_0 \cap N_0$ is idempotent if and only if $O_r(M_0) \not = O_l(N_0)$ and $O_r(N_0) \not = O_l(M_0)$. In this case $M_0N_0 = M_0 \cap N_0 = N_0M_0$.

• (2) If $O_r(M_0) = O_l(N_0)$ and $O_r(N_0) \not = O_l(M_0)$, then $M_0 \cap N_0 = M_0N_0$.

Proof

Note that $O_r(M_{0i}) \not = O_l(M_{0j})$ and $O_r(N_{0i}) \not = O_l(N_{0j})$ for all $i, \ j$ by [11, Corrollary 5.5].

• (1) Suppose first $M_0 \cap N_0$ is idempotent. If $R_0' = O_r(M_0) = O_l(N_0)$, then $S_0 = O_r(M_{01})$ is an $R_0$-ideal. $R_0 \supset (R_0 : S_0)_r \supseteq (R_0 : O_l(N_0))_r = N_0$ and so there is an $N_{0i} \supseteq (R_0 : S_0)_r$. It follows that $R_0' \supseteq (R_0 : (R_0 : S_0)_r)_l = S_0 \supset (R_0 : N_{0i})_l = O_l(N_{0i})$ and $O_r(M_{01}) = O_l(N_{0i})$ by [11, Corollary 4.5]. Thus, $M_0 \cap N_0$ is not idempotent by [11, Corollary 5.5], a contradiction. Hence, $O_r(M_0) \not = O_l(N_0)$. Similarly, $O_l(M_0) \not = O_r(N_0)$.

  Conversely, suppose that $O_r(M_0) \not = O_l(N_0)$ and $O_l(M_0) \not = O_r(N_0)$ and assume that $M_0 \cap N_0$ is not idempotent. Then by [11, Corollary 5.5], there are $M_{0i}$ and $N_{0j}$ such that $O_r(M_{0i}) = O_l(N_{0j})$. We may assume that $O_r(M_{01}) = O_l(N_{01})$. Then

  $$\begin{aligned} O_r(M_{0i}) = O_r(R_iM_{01}R_{-i}) = R_iO_r(M_{01})R_{-i} = R_iO_l(N_{01})R_{-i} = O_l(N_{0i}). \end{aligned}$$

  In particular, we have $n = l$ and also, by [5, Lemma 3.2],

  $$\begin{aligned} O_r(M_0)= & {} O_r(M_{01} \cdots M_{0n}) = O_r(M_{01}) \cdots O_r(M_{0n}) = O_l(N_{01}) \cdots O_l(N_{0n}) \\= & {} O_l(N_{01} \cdots N_{0n}) = O_l(N_0), \end{aligned}$$

  since $M_0 = M_{01} \cdots M_{0n}$ and $N_0 = N_{01} \cdots N_{0n}$. This is a contradiction. Hence, $M_0 \cap N_0$ is idempotent.

• (2) Put $R_0' = O_r(M_0) = O_l(N_0)$ and $I_0 = M_0 \cap N_0$. Since $I_0$ is not invertible, either $R_0 \supset I_0(R_0 : I_0)_r$ or $R_0 \supset (R_0 : I_0)_lI_0$. In the first case,

  $$\begin{aligned} R_0 \supset I_0(R_0 : I_0)_r \supseteq I_0R_0' \supseteq M_0N_0R_0' = M_0R_0' = M_0, \end{aligned}$$

  and so $I_0(R_0 : I_0)_r = M_0$ since $I_0(R_0 : I_0)_r$ is $\mathbb Z$-invariant. Thus,

  $$\begin{aligned} I_0 = I_0O_r(I_0) = I_0(R_0 : I_0)_rI_0 = M_0I_0 \subseteq M_0N_0. \end{aligned}$$

  In case $R_0 \supset (R_0 : I_0)_lI_0$, the symmetry argument shows $I_0 = M_0N_0$. $\square$

The following corollary follows from [11, Corollary 5.4] and the proof of Lemma 2.14.

Corollary 2.15

Let $M_{01}, \ldots , M_{0r}$ be maximal idempotent $\mathbb Z$-invariant ideals of $R_0$. Then $M_{01} \cap \cdots \cap M_{0r}$ is idempotent if and only if $O_r(M_{0i}) \not = O_l(M_{0j})$ for all $i, \ j \ (1 \le i, \ j \le r)$.

Let $M_{01}$ and $M_{02}$ be maximal idempotent $\mathbb Z$-invariant ideals of $R_0$ such that $R_0' = O_r(M_{01}) = O_l(M_{02})$ and $O_r(M_{02}) \not = O_l(M_{01})$. Put $R' = \oplus _{n \in {\mathbb {Z}}} R_n' \ (R_n' = R_nR_0' = R_0'R_n)$. Then $R'$ is a strongly graded ring of type ${\mathbb {Z}}$ since $R_0'$ is ${\mathbb {Z}}$-invariant by Lemma 2.4.

Note that an $R_0'$-ideal $A_0'$ is ${\mathbb {Z}}$-invariant as $R_0$-ideals if and only if it is ${\mathbb {Z}}$-invariant as $R_0'$-ideals, that is, $R_n'A_0' = A_0'R_n'$ for all $n \in \mathbb Z$. Further, $M_{01}R_0' = M_{01}, \ R_0'M_{01} = R_0', \ R_0'M_{02} = M_{02}$ and $M_{02}R_0' = R_0'$, which are used in the proof of the following proposition and Sect. 3.

Proposition 2.16

Let $M_{01}, M_{02}$ be maximal idempotent ${\mathbb {Z}}$-invariant ideals of $R_0$ such that $R_0' = O_r(M_{01}) = O_l(M_{02})$. Then $\{ M_0R_0' \mid M_0 : \hbox { maximal } {\mathbb {Z}} \hbox {-invariant ideals with} M_0 \not = M_{0i} \ (i = 1, \ 2) \} \cup \{ M_{02}M_{01} \}$ is the set of maximal ${\mathbb {Z}}$-invariant ideals of $R_0'$.

Proof

Let $M_0$ be a maximal ${\mathbb {Z}}$-invariant ideal of $R_0$ with $M_0 \not = M_{0i} \ (i = 1, \ 2)$. If $R_0' = M_0R_0'$, then write $1 = \sum x_iq_i$, where $x_i \in M_0$ and $q_i \in R_0'$ and $M_{02} = (\sum x_iq_i)M_{02} \subseteq M_0$ which implies $M_{02} = M_0$, a contradiction. Thus, $R_0' \supset M_0R_0'$. We claim $M_0M_{01} = M_0 \cap M_{01}$ and $M_{02}M_0 = M_0 \cap M_{02}$. If $M_0$ is invertible, then $M_0M_{01} = M_0 \cap M_{01}$ as in [4, Proposition 2.8]. If $M_0$ is idempotent, then $O_r(M_{01}) \not = O_l(M_0)$, otherwise $O_l(M_0) = O_l(M_{02})$ and so $M_0 = (R_0 : O_l(M_0))_r = (R_0 : O_l(M_{02}))_r = M_{02}$, a contradiction. Hence, by Lemma 2.14 (1) and (2), $M_0M_{01} = M_0 \cap M_{01}$ follows. Symmetric argument shows $M_0 \cap M_{02} = M_{02}M_0$.

Since $R_0'$ is $R_0$-flat, we have

$$\begin{aligned} R_0'M_0M_{01} = R_0'(M_0 \cap M_{01}) = R_0'M_0 \cap R_0'M_{01} = R_0'M_0. \end{aligned}$$

Since $R_0'M_0M_{01}$ is an ideal of $R_0'$, we have $R_0'M_0 \supseteq M_0R_0'$. Similarly, $M_{02}M_0R_0' = M_0R_0'$ and $M_{02}M_0R_0'$ is an ideal of $R_0'$. So $M_0R_0' \supseteq R_0'M_0$. Hence, $M_0R_0' = R_0'M_0$ and is ${\mathbb {Z}}$-invariant.

Let $N_0'$ be a maximal ${\mathbb {Z}}$-invariant ideal of $R_0'$ containing $M_0R_0'$. Then $N_0 = N_0' \cap R_0 \supseteq M_0$ and so $N_0 = M_0$. By [2, Lemma 3.2], we have $N_0' = N_0R_0' = M_0R_0'$. Hence, $M_0R_0'$ is a maximal ${\mathbb {Z}}$-invariant ideal of $R_0'$.

It is easy to see that $M_{02}M_{01}$ is a ${\mathbb {Z}}$-invariant ideal of $R_0'$. Let $N_0'$ be a maximal ${\mathbb {Z}}$-invariant ideal of $R_0'$ containing $M_{02}M_{01}$ and $N_0 = N_0' \cap R_0$, ${\mathbb {Z}}$-invariant. It is clear that any ${\mathbb {Z}}$-invariant ideal containing $N_0$ is either $M_{01}$ or $M_{02}$. If $M_{01} \supseteq N_0$, then $M_{01} = M_{01}R_0' \supseteq N_0R_0' = N_0'$ and $M_{02}M_{01} \supseteq M_{02}N_0' = N_0'$. If $M_{02} \supseteq N_0$, then $M_{02} = R_0'M_{02} \supseteq R_0'N_0 = N_0'$ and $M_{02}M_{01} \supseteq N_0'M_{01} = N_0'$. Hence, $M_{02}M_{01}$ is a maximal ${\mathbb {Z}}$-invariant ideal of $R_0'$.

Finally, let $N_0'$ be a maximal ${\mathbb {Z}}$-invariant ideal of $R_0'$ and $N_0 = N_0' \cap R_0$. Let $M_0$ be a maximal $\mathbb Z$-invariant ideal with $M_0 \supseteq N_0$. If $M_0 \not = M_{0i} \ (i = 1, \ 2)$, then $M_0' = M_0R_0' \supseteq N_0R_0' = N_0'$ and so $M_0' = N_0'$ since $M_0'$ is a maximal ${\mathbb {Z}}$-invariant ideal of $R_0'$. In case either $M_0 = M_{01}$ or $M_0 = M_{02}$, it follows as before that $N_0' = M_{02}M_{01}$. This completes the proof. $\square$

Remark

Ideals of $R_{0}$ are not necessarily ${\mathbb {Z}}$-invariant as it is seen in the following examples:

Let $M_{01}, \ldots , M_{0n}$ be idempotent maximal ideals of $R_{0}$ which is a cycle ($n\ge 2$) and $X=M_{01} \cap \cdots \cap M_{0n}$ which is a maximal invertible ideal of $R_{0}$ by [4, Proposition 2.5]. Then $R=\bigoplus _{n\in {\mathbb {Z}}}X^{n}$ is a strongly graded ring of type ${\mathbb {Z}}$ and all $M_{0i}$ are not $\mathbb Z$-invariant since $X^{-1}M_{0i}X=M_{0(i+1)}$ ($i+1=1$ if $i=n$).

3 Projective Ideals of R with Nonzero Contracted Ideals of $R_0$

In this section, we describe all projective ideals A of R with $A_0 = A \cap R_0 \not = (0)$. For an ideal A of R and $n \in {\mathbb {Z}}$, define

$$\begin{aligned} L_n(A) = \{ a_n \in R_n \mid \exists \ a = a_n + (\hbox {lower-degree terms}) \in A \} \cup \{ 0 \}. \end{aligned}$$

It is clear that $L_n(A)$ is an $(R_0, R_0)$-bimodule of $R_n$ and is ${\mathbb {Z}}$-invariant since A is ${\mathbb {Z}}$-invariant. In particular, $L_n(A)R_{-n}$ is a ${\mathbb {Z}}$-invariant ideal of $R_0$.

Lemma 3.1

Let A be an ideal of R. Then

• (1) There is an $n \ge 1$ such that $L_n(A)R_k = L_{n+k}(A)$ for all $k \ge 0$. In particular, if $L_n(A) = R_n$ for some $n \ge 1$, then $L_{n+k}(A) = R_{n+k}$ for all $k \ge 0$.

• (2) If $L_n(A) = R_n$ for some $n \in {\mathbb {Z}}$ and $A_0 = A \cap R_0 \not = (0)$, then $(R : A)_l = R$ and $A_v = R$.

Proof

• (1) It is clear that $L_m(A)R_1 \subseteq L_{m+1}(A)$ for any $m \in {\mathbb {Z}}$ and so

  $$\begin{aligned} L_m(A)R_{-m} = L_m(A)R_1R_{-(m+1)} \subseteq L_{m+1}(A)R_{-(m+1)} \subseteq \cdots \subseteq R_0. \end{aligned}$$

  Thus, there is an $n > 1$ such that $L_n(A)R_{-n} = L_{n+1}(A)R_{-(n+1)} = L_{n+k}(A)R_{-(n+k)}$ for all $k \ge 0$, that is, $L_n(A)R_k = L_{n+k}(A)$. If $L_n(A) = R_n$ for some $n \ge 1$, then $L_{n+k}(A) = R_{n+k}$ for all $k \ge 0$.

• (2) Since $A_0 \not = (0)$, $AQ^g = Q^g$ and so, for any $q \in (R : A)_l$, $q \in qQ^g = qAQ^g \subseteq Q^g$. Write $q = q_{l_1} + \cdots + q_{l_k}$, where $q_{l_i} \in Q_0R_{l_i}$ and $l_1> l_2> \cdots > l_k$. For any $r_n \in R_n$, there is an $a = r_n + (\hbox {lower-degree terms}) \in A$ and so $R \supseteq qA \ni qa = q_{l_1}r_n + (\hbox {lower-degree terms})$, which implies $q_{l_1}r_n \subseteq R_{l_1+n}$ and so $q_{l_1}R_n \subseteq R_{l_1+n}$. Hence, $q_{l_1} \in R_{l_1}$. Since $R \ni (q - q_{l_1})a = (q_{l_2} + \cdots + q_{l_k})a$, it similarly follows $q_{l_2} \in R_{l_2}$. Continuing this process, we have $q \in R$. Hence, $(R : A)_l = R$ and so $A_v = R$ follows. $\square$

We set $V_r(R) = \{ A : \hbox { ideal of } R \mid A = A_v \}$ and $V_{(r, m)}(R)$, the set of all maximal members in $V_r(R)$. Since $\hbox {gl.} \dim R \le 2$ ([10, Lemma II, 8.3]), $V_r(R)$ is the set of all ideals which are right projective by [3, Proposition 5.2].

Proposition 3.2

Let $P \in V_r(R)$ with $P_0 = P \cap R_0 \not = (0)$. Then the following conditions are equivalent.

• (1) $P \in V_{(r, m)}(R)$.

• (2) P is a prime ideal.

• (3) $P = P_0R$ and $P_0$ is a maximal ${\mathbb {Z}}$-invariant ideal of $R_0$.

Proof

$(1) \Longrightarrow (2)$ : This follows from [1, Lemma 2.1].

$(2) \Longrightarrow (3)$ : Put $P' = P_0R$. It is an ideal of R by Lemmas 2.2 and 2.3 . We prove that $P'$ is a prime ideal. We assume, on the contrary, that $P'$ is not a prime ideal of R. Then there are ideals A and B such that $P' \supseteq AB$ with $P'\subset A$ and $P' \subset B$. Further, we may assume that $A = (P' : B)_l \cap R$. Let $a \in A \backslash P'$. Then we may assume that $a = a_l + \cdots + a_0$, where $a_i \in R_i \ (0 \le i \le l)$ with $a_l \not \in P_0R_l$ and also l is minimal for this property. It is clear that $R_{-l}a_l \not \subseteq P_0$. We claim $R_{-l}a_l \subseteq A \cap R_0 = A_0$. Since $a^{-1} \cdot P' = \{ b \in R \mid ab \in P' \} \supseteq B$, if we prove that $P' \supseteq R_{-l}a_l(a^{-1} \cdot P')$, then $P' \supseteq R_{-l}a_lB$ and $R_{-l}a_l \subseteq (P' : B)_l \cap R_0 = A_0$, showing the claim. So we assume, on the contrary, that $P' \not \supseteq R_{-l}a_l(a^{-1} \cdot P')$. Then there is a $b \in a^{-1} \cdot P'$ such that $P' \not \supseteq R_{-l}a_lb$. As before, we may assume $b = b_m + \cdots + b_0, b_i \in R_i \ (0 \le i \le m)$ and m is minimal for this property. Then $P' \ni ab = a_lb_m + (\hbox {lower-degree terms})$, $a_lb_m \in P_0R_{l+m}$ and $a_lb_mR_{-m} \subseteq P_0R_l$. Thus, $A \supseteq (a - a_l)b_mR_{-m}$ and so, by the choice of a, $P' \supseteq (a - a_l)b_mR_{-m}$ and $P' \supseteq ab_mR_{-m}$, equivalently, $P' \ni ab_m$. Thus, $P' \ni a(b - b_m)$, that is, $b - b_m \in a^{-1} \cdot P'$ with $\deg (b - b_m) < m$. By the choice of b, $R_{-l}a_l(b - b_m) \subseteq P'$. Again, $a_lb_m \in P_0R_{l+m}$ implies $R_{-l}a_lb_m \subseteq P_0R_m \subseteq P'$. Hence, $R_{-l}a_lb \subseteq P'$, a contradiction, which proves the claim, that is, $A_0 \supseteq R_{-l}a_l$ and $P_0 \not \supseteq R_{-l}a_l$. Hence, $A_0 \supset P_0$ and $B_0 \supset P_0$ by symmetric argument. Since $P_0 = P' \cap R_0 \supseteq A_0B_0$, either $P_0 \supseteq A_0$ or $P_0 \supseteq B_0$. (Note $P_0$ is a ${\mathbb {Z}}$-prime ideal since P is prime.) This is a contradiction. Hence, $P'$ is a prime ideal. Note that $P_0$ is a maximal ${\mathbb {Z}}$-invariant ideal by Lemma 2.7.

To prove $P = P'$, assume that $P \supset P'$. Then there is an $n \ge 1$ such that $R_n \supseteq L_n(P) \supset P_0R_n$ and $R_0 \supseteq L_n(P)R_{-n} \supset P_0$. Since $L_n(P)R_{-n}$ is a ${\mathbb {Z}}$-invariant ideal of $R_0$, it follows that $L_n(P)R_{-n} = R_0$ and so $P_v = R$ by Lemma 3.1, a contradiction. Hence, $P = P_0R$ and $P_0$ is a maximal ${\mathbb {Z}}$-invariant ideal.

(3) $\Longrightarrow$ (1) : Let $M \in V_{(r,m)}(R)$ containing P. Then $M = M_0R$ for some maximal $\mathbb Z$-invariant ideal $M_0$ of $R_0$ and so $M_0 = P_0$. Hence, $P = M$. $\square$

Lemma 3.3

Let $I_0$ be a right $R_0$-ideal, and let $J_0$ be a left $R_0$-ideal. Then

• (1) $(R : I_0R)_l = R(R_0 : I_0)_l$ and $(R : RJ_0)_r = (R_0 : J_0)_rR$.

• (2) $(I_0R)_v = (I_0)_vR$ and ${}_v(RJ_0) = R{}_v(J_0)$.

Proof

• (1) It is clear that $(R : I_0R)_l \supseteq R(R_0 : I_0)_l$. Let $q \in (R : I_0R)_l$, that is, $qI_0R \subseteq R$ and $q \in qQ^g = qI_0Q^g \subseteq Q^g$ follows. Write $q = q_{n_1} + \cdots + q_{n_k}$, where $q_{n_i} \in Q_0R_{n_i}$. $R \supseteq qI_0$ implies $q_{n_i}I_0 \subseteq R_{n_i}$ and $R_{-n_i}q_{n_i}I_0 \subseteq R_0$. Thus, $R_{-n_i}q_{n_i} \in (R_0 : I_0)_l$, $q_{n_i} \in R(R_0 : I_0)_l$ and hence $q \in R(R_0 : I_0)_l$. Similarly, $(R : RJ_0)_r = (R_0 : J_0)_rR$ follows.

• (2) By (1),

  $$\begin{aligned} (I_0R)_v= & {} (R : (R : I_0R)_l)_r = (R : R(R_0 : I_0)_l)_r \\= & {} (R_0 : (R_0 : I_0)_l)_rR = (I_0)_vR. \end{aligned}$$

  Similarly, ${}_v(RJ_0) = R {}_v(J_0)$. $\square$

Proposition 3.4

An ideal P of R with $P_0 = P \cap R_0 \not = (0)$ is a maximal invertible ideal if and only if $P = P_0R$ and $P_0$ is a maximal invertible ${\mathbb {Z}}$-invariant ideal of $R_0$ (ideals maximal among invertible ${\mathbb {Z}}$-invariant ideals).

Proof

Suppose P is a maximal invertible ideal of R with $P_0 = P \cap R_0 \not = (0)$. Thus, $P \in V_r(R)$ since P is invertible. By Proposition 3.2, there is an $M \in V_{(r,m)}(R)$ with $M \supseteq P$ and $M = M_0R$ such that $M_0$ is a maximal $\mathbb Z$-invariant ideal of $R_0$. By Lemma 2.8, either $M_0$ is invertible or idempotent. If $M_0$ is invertible, then $M = P$ and so $P = P_0R$ and $P_0$ is a maximal ${\mathbb {Z}}$-invariant ideal of $R_0$ which is invertible. In case $M_0$ is idempotent. Put $M_0 = M_{01}$ and $M_1 = M_{01}R \supseteq P$. Then

$$\begin{aligned} P^{-1} \supseteq (R : M_{01}R)_r = (R_0 : M_0)_rR = O_r(M_{01})R \end{aligned}$$

by Lemma 3.3, $R = P^{-1}P \supseteq O_r(M_{01})P_0$ and $R_0 \supseteq O_r(M_{01})P_0$. Thus, $O_r(M_{01})$ is an $R_0$-ideal and is ${\mathbb {Z}}$-invariant by Lemma 2.4. Since

$$\begin{aligned} R \supseteq (R : O_r(M_{01})R)_r = (R_0 : O_r(M_{01}))_rR \supseteq (R : P^{-1})_r = P, \end{aligned}$$

we have $R_0 \supseteq (R_0 : O_r(M_{01}))_r \supseteq P_0$. Let $M_{02}$ be a maximal ${\mathbb {Z}}$-invariant ideal containing $(R_0 : O_r(M_{01}))_r$. It follows that

$$\begin{aligned} M_{02}R \supseteq (R_0 : O_r(M_{01}))_rR \supseteq P. \end{aligned}$$

If $M_{02}$ is invertible, then $M_{02}R = P$ and so $M_{02} = M_{01}$ is invertible, a contradiction. Hence, $M_{02}$ is idempotent. Then

$$\begin{aligned} O_l(M_{02}) = (R_0 : M_{02})_l \subseteq (R_0 : (R_0 : O_r(M_{01}))_r)_l = {}_v(O_r(M_{01})) = O_r(M_{01}), \end{aligned}$$

since $O_r(M_{01})$ is projective. So $O_r(M_{01}) = O_l(M_{02})$ since $O_r(M_{01})$ is a minimal ${\mathbb {Z}}$-invariant overring of $R_0$. Continuing this process as Lemma 2.11 (2), we have a ${\mathbb {Z}}$-cycle $M_{01}, \ldots , M_{0n}$ such that $M_{0i}R \supseteq P$ for all $i \ (1 \le i \le n)$. By Lemma 2.12, $P_0 = M_{01} \cap \cdots \cap M_{0n}$ is a maximal invertible ${\mathbb {Z}}$-invariant ideal of $R_0$ and $P_0R \supseteq P$. Hence, $P = P_0R$.

Conversely, suppose that $P = P_0R$ and $P_0$ is a maximal invertible ${\mathbb {Z}}$-invariant ideal of $R_0$. Then it is now clear that P is a maximal invertible ideal of R. $\square$

Lemma 3.5

Let P be a proper prime ideal of R with $P_0 = P \cap R_0 \not = (0)$. Then the following conditions are equivalent.

• (1) $L_n(P) = R_n$ for some $n \ge 1$.

• (2) $P_v = R$.

• (3) $P \not = P_v$.

Proof

(1) $\Longrightarrow$ (2) : This is just Lemma 3.1.

(2) $\Longrightarrow$ (3) : Trivial.

(3) $\Longrightarrow$ (1) : $P \not = P_v$ implies $P \supset P_0R$. So there is an $n \ge 1$ such that $L_n(P) \supset P_0R_n$ and $R_0 \supseteq L_n(P)R_{-n} \supset P_0$. Since $P_0$ is a ${\mathbb {Z}}$-prime ideal, $P_0$ is a maximal ${\mathbb {Z}}$-invariant ideal by Lemma 2.7. Hence, $R_0 = L_n(P)R_{-n}$ and $L_n(P) = R_n$ follows. $\square$

Lemma 3.6

Let $A_i$ be ideals of R such that $L_n(A_i) = R_n$ for some $n \ge 1 \ (1 \le i \le k)$ and $B_j$ be ${\mathbb {Z}}$-invariant ideals of $R_0 \ (0 \le j \le k)$. Then

• (1) $L_{kn}(A_kA_{k-1} \cdots A_1) = R_{kn}$.

• (2) $L_{kn}(A_kB_k A_{k-1}B_{k-1} \cdots A_1B_1) \supseteq B_kB_{k-1} \cdots B_1R_{kn}$.

• (3) $L_{kn}(B_kA_k \cdots B_1A_1B_0) \supseteq B_kB_{k-1} \cdots B_0R_{kn}$.

Proof

• (1) For any $a_n, \ b_n \in R_n$, there are $a \in A_2, \ b \in A_1$ such that

  $$\begin{aligned} a= & {} a_n + (\hbox {lower-degree terms}) \in A_2\\ b= & {} b_n + (\hbox {lower-degree terms}) \in A_1\\ A_2A_1 \ni ab= & {} a_nb_n + (\hbox {lower-degree terms}). \end{aligned}$$

  Thus, $a_nb_n \in L_{2n}(A_2A_1)$, $R_{2n} \subseteq L_{2n}(A_2A_1)$, and hence, $R_{2n} = L_{2n}(A_2A_1)$. Inductively, we may assume that $R_{(k-1)n} = L_{(k-1)n}(A_{k-1} \cdots A_1)$. Let $a_n \in R_n$ and $b_{(k-1)n} \in R_{(k-1)n}$. Then there are $a \in A_k$ and $b \in A_{k-1} \cdots A_1$ such that $a = a_n + (\hbox {lower-degree terms}) \in A_k$ and $b = b_{(k-1)n} + (\hbox {lower-degree terms)} \in A_{k-1} \cdots A_1$. Then $A_kA_{k-1} \cdots A_1 \ni ab = a_nb_{(k-1)n} + (\hbox {lower-degree terms})$ and $a_nb_{(k-1)n} \in L_{kn}(A_kA_{k-1} \cdots A_1)$. Thus, $R_{kn} = R_n \cdot R_{(k-1)n} \subseteq L_{kn}(A_kA_{k-1} \cdots A_1)$, and hence, $L_{kn}(A_kA_{k-1} \cdots A_1) = R_{kn}$.

• (2) $L_n(A_kB_k) \supseteq B_kR_n$ as in (1) since $B_k$ is ${\mathbb {Z}}$-invariant. Inductively, we may assume that $B_{k-1} \cdots B_1R_{(k-1)n} \subseteq L_{(k-1)n}(A_{k-1}B_{k-1} \cdots A_1B_1)$. Let $d_n \in B_kR_n$ and $c_{(k-1)n} \in B_{k-1} \cdots B_1R_{(k-1)n}$. Then there are $d \in A_kB_k$ and $c \in A_{k-1}B_{k-1} \cdots A_1B_1$ such that $d = d_n + (\hbox {lower-degree terms})$ and $c = c_{(k-1)n} + (\hbox {lower-degree terms})$. Thus, $d_nc_{(k-1)n} \in L_{kn}(A_kB_k \cdots A_1B_1)$, and hence,

  $$\begin{aligned} L_{kn}(A_kB_k \cdots A_1B_1) \supseteq B_kR_n \cdot B_{k-1} \cdots B_1R_{(k-1)n} = B_k \cdots B_1 \cdot R_{kn}. \end{aligned}$$

• (3) This is proved in a similar way as in (2). $\square$

Lemma 3.7

Let $A \in V_r(R)$ with $A_0 = A \cap R_0 \not = (0)$, and let $M_i = M_{0i}R \in V_{(r,m)}(R)$ containing $A \ (1 \le i \le r)$, where $M_{0i}$ are maximal ${\mathbb {Z}}$-invariant ideals of $R_0$ (Proposition 3.2). Suppose there are no invertible ideals containing A. Then

• (1) $M_{0i}$ are all idempotent.

• (2) There is an $n \ge 0$ such that $N_0^sR_n \subseteq L_n(A)$ for some $s \ge 1$, where $N_0 = M_{01} \cap \cdots \cap M_{0r}$, is ${\mathbb {Z}}$-invariant.

Proof

• (1) follows from Lemma 2.8 and the assumption.

• (2) Since R is Noetherian, there are prime ideals $P_j \ (1 \le j \le l)$ such that

  $$\begin{aligned} P_1 \cdots P_l \subseteq A \subseteq P_j \ \ \ \ \ \ (*) \end{aligned}$$

  We may assume $M_i \in \{ P_1, \ldots , P_l \}$ for all $i \ (1 \le i \le r)$. If $P_j = (P_j)_v$, then $P_j = M_i$ for some $i \ (1 \le i \le r)$ by Proposition 3.2. Let k be the number of $P_j$ with $P_j \not = (P_j)_v$. If $k = 0$, then $N_0^l \subseteq A$ and so $N_0^l \subseteq L_0(A)$. In case $k \not = 0$, there is an $m \ge 1$ such that $L_m(P_j) = R_m$ for all $P_j$ with $P_j \not = (P_j)_v$ by Lemmas 3.1 and 3.5 . Further, if there is a product of $P_j$ with $P_j \not = (P_j)_v$ in $(*)$, then we take a bigger n than m by Lemma 3.6 (1). Now we replace $P_i$ by $N_0$ in $(*)$ if $P_i = (P_i)_v$. Then $N_0^{l-k}R_{tn} \subseteq L_{tn}(A)$ for some $t \ (1 \le t \le k)$ by Lemma 3.6 (1) or (2). $\square$

Lemma 3.8

Under the same notation and assumptions in Lemma 3.7, if $N_0$ is idempotent, then $A = N_0R$.

Proof

By Lemma 3.7, $N_0R_n \subseteq L_n(A)$ for some $n \ge 0$. Let $q \in (R : A)_l$. Then $q \in Q^g$ since $AQ^g = Q^g$ and write $q = q_{l_1} + \cdots + q_{l_k}$, where $q_{l_i} \in Q_0R_{l_i}$ and $l_1> \cdots > l_k$. For any $a_n \in L_n(A)$, there is $a = a_n + (\hbox {lower-degree terms}) \in A$ and $R \ni qa = q_{l_1}a_n + (\hbox {lower-degree terms})$. Thus, $q_{l_1}a_n \in R_{l_1+n}$ and $q_{l_1}L_n(A) \subseteq R_{l_1+n}$. Thus, $q_{l_1}N_0R_n \subseteq q_{l_1}L_n(A) \subseteq R_{l_1+n}$, which implies $q_{l_1}N_0 \subseteq R_{l_1} \subseteq R$. It follows that $q_{l_1} \in (R : N_0R)_l \subseteq (R : A)_l$ and $(R : A)_l \ni q -q_{l_1} = q_{l_2} + \cdots + q_{l_k}$. So similarly $q_{l_2} \in (R : N_0R)_l$ and continuing this process, we have $q \in (R : N_0R)_l$, that is, $(R : N_0R)_l = (R : A)_l$. Hence, $A = A_v = (R : (R : A)_l)_r = (R : (R : N_0R)_l)_r = (N_0R)_v = N_0R$. $\square$

Let $M_{01}$ and $M_{02}$ be maximal idempotent $\mathbb Z$-invariant ideals of $R_0$ such that $R_0' = O_r(M_{01}) = O_l(M_{02})$ and put $R' = \oplus _{n \in {\mathbb {Z}}} R_n'$, where $R_n' = R_0'R_n = R_nR_0'$ as in Sect. 2. Then $R'$ is a strongly graded ring of type ${\mathbb {Z}}$.

Lemma 3.9

Let $M_{01}$ and $M_{02}$ be maximal idempotent $\mathbb Z$-invariant ideals of $R_0$ and put $R' = \oplus _{n \in {\mathbb {Z}}} R_n'$, where $R_n' = R_0'R_n = R_nR_0'$. Then

• (1) $R' = (R : M_{02}R)_l$ and is a projective R-ideal.

• (2) $(R : R')_l = RM_{01} = M_{01}R$ and is a projective ideal.

Proof

• (1) $(R : M_{02}R)_l = R(R_0 : M_{02})_l = RR' = R'$ by Lemma 3.3. $R'(R : R')_l = R'R(R_0 : M_{02})_l = R'M_{01} = R' \ni 1$ and so $R'$ is right R-projective. The symmetric argument shows that $R'$ is left R-projective.

• (2) $(R : R')_l = (R : R_0'R)_l = R(R_0 : R_0')_l = RM_{01} = M_{01}R$. It follows that

  $$\begin{aligned} M_{01}R(R : M_{01}R)_l= & {} M_{01}R(R_0 : M_{01})_l = RM_{01}(R_0 : M_{01})_l \\= & {} R \cdot O_l(M_{01}) = O_l(M_{01}R). \end{aligned}$$

  Hence, $M_{01}R$ is right projective. Similarly, $M_{01}R$ is left projective. $\square$

Lemma 3.10

Under the same notation and assumptions as in Lemma 3.9, let $A \in V_r(R)$ with $A_0 = A \cap R_0 \not = (0)$ and $M_{01}R, \ldots , M_{0r}R$ be the members in $V_{(r,m)}(R)$ containing A, where $M_{0i}$ are maximal ${\mathbb {Z}}$-invariant ideals of $R_0$ (Proposition 3.2). If $M' \in V_{(r,m)}(R')$ such that $M' \supseteq A$, then $M' \in \{ M_{02}M_{01}R', M_{03}R', \ldots , M_{0r}R' \}$.

Proof

By Proposition 3.2, $M' = M_0'R'$ for some maximal $\mathbb Z$-invariant ideal $M_0'$ of $R_0'$ and $M_0'$ is either $M_{02}M_{01}$ or $M_0' = M_0R_0'$ for some maximal $\mathbb Z$-invariant ideal $M_0$ of $R_0$ with $M_0 \not = M_{0i} \ (i = 1, \ 2)$ by Proposition 2.16. If $M_0' = M_{02}M_{01}$, then $M' = M_{02}M_{01}R'$. If $M_0' = M_0R'$ with $M_0 \not = M_{0i} \ (i = 1, \ 2)$, then $A \subseteq M' \cap R = M_0R' \cap R = M_0R$ and so $M_0 = M_{0i}$ for some $i \ (3 \le i \le r)$, that is, $M' = M_{0i}R'$. $\square$

Proposition 3.11

Let $A \in V_r(R)$ with $A_0 = A \cap R_0 \not = (0)$. Suppose there are no invertible ideals containing A. Then $A = A_0R$.

Proof

Let $M_{0i}R$ be the full members in $V_{(r,m)}(R)$ containing $A \ (1 \le i \le r)$ and $N_0 = M_{01} \cap \cdots \cap M_{0r}$. If $N_0$ is idempotent (including the case of $r = 1$), then $A = N_0R$ by Lemma 3.8. Thus, we may assume that $r \ge 2$ and that the assertion is true for all ideals B of a strongly graded ring R of type ${\mathbb {Z}}$ with $R_0$, an HNP ring such that $B \in V_r(R)$, $B_0 = B \cap R_0 \not = (0)$, there are no invertible ideals containing B and B contained in at most $r - 1$ ideals in $V_{(r,m)}(R)$. Furthermore, we may assume, without loss of generality, that $R_0' = O_r(M_{01}) = O_l(M_{02})$ and $O_r(M_{02}) \not = O_l(M_{01})$ (Lemma 2.12 and Corollary 2.15). Put $R' = \oplus _{n \in {\mathbb {Z}}} R_n' \ (R_n' = R_0'R_n = R_nR_0')$, a strongly graded ring of type ${\mathbb {Z}}$, and denote the v-operation in $R'$ by $v'$. Note that $M_{01}R_0' = M_{01}, \ R_0'M_{01} = R_0', \ R_0'M_{02} = M_{02}$ and $M_{02}R_0' = R_0'$. Put $A_1 = M_{01}AM_{02}, \ A_2 = M_{01}AM_{01}, \ A_3 = M_{02}AM_{02}$ and $A_4 = M_{02}AM_{01}$. Note that $A_i$ are all ideals of R and that $(A_i)_v$ satisfy the assumption of Lemma 3.10. We prove that there are ${\mathbb {Z}}$-invariant ideals $C_{0i} \ (1 \le i \le 4)$ of $R_0$ satisfying the following properties:

• (i) $(A_1)_v = C_{01}R$,

• (ii) $(A_2)_v = C_{02}R$,

• (iii) $(A_3)_v = C_{03}R$ and

• (iv) $(A_4)_v = C_{04}R$.

The proof of (i): Let $A_1' = R'A_1R' = R'AR'$. By Lemma 3.10 and induction hypothesis, $(A_1')_{v'} = B_{01}'R'$ for some ${\mathbb {Z}}$-invariant ideal $B_{01}'$ of $R_0'$. By [2, Lemma 3.2] and [1, Lemma 2.1], we have

$$\begin{aligned} (A_1)_vR'= & {} (A_1R')_{v'} = (M_{01}R'AR')_{v'} = \big (M_{01}R'(R'AR')_{v'}\big )_{v'} = (M_{01}B_{01}'R')_{v'} \\= & {} (M_{01}B_{01}'R)_vR' = M_{01}B_{01}'R', \end{aligned}$$

since $M_{01}B_{01}'R$ is right R-projective. Taking v-operation and using [1, Lemma 2.1 (2)] and Lemma 3.9,

$$\begin{aligned} (A_1)_v= & {} (M_{01}AM_{02}R)_v = (A_1M_{02}R)_v = \big ((A_1)_vM_{02}R\big )_v = \big ((A_1)_vR'M_{02}R\big )_v \\= & {} (M_{01}B_{01}'R'M_{02}R)_v = (M_{01}B_{01}'M_{02}R)_v = M_{01}B_{01}'M_{02}R = C_{01}R, \end{aligned}$$

since $M_{01}B_{01}'M_{02}R$ is right R-projective, where $C_{01} = M_{01}B_{01}'M_{02}$ is a ${\mathbb {Z}}$-invariant ideal of $R_0$.

The proof of (ii): $A_2$ is an $(R, \ R')$-ideal in R. As in (i),

$$\begin{aligned} (R'A_2)_vR' = (R'A_2R')_{v'} = (R'A_2)_{v'} = B_{02}'R' \end{aligned}$$

for some ${\mathbb {Z}}$-invariant ideal $B_{02}'$ of $R_0'$, and so

$$\begin{aligned} M_{01}B_{02}'R = (M_{01}B_{02}'R')_v = \big ((M_{01}(R'A_2)_vR'\big )_v = (M_{01}R'A_2R')_v = (A_2)_v. \end{aligned}$$

Thus, $(A_2)_v = C_{02}R$, where $C_{02} = M_{01}B_{02}'$ and is a ${\mathbb {Z}}$-invariant ideal of $R_0$.

The proof of (iii): $A_3$ is an $(R', \ R)$-ideal. As in (i),

$$\begin{aligned} (A_3)_v = (A_3R')_{v'} = B_{03}'R' \end{aligned}$$

for some ${\mathbb {Z}}$-invariant ideal $B_{03}'$ of $R_0'$ and

$$\begin{aligned} B_{03}'M_{02}R= & {} (B_{03}'M_{02}R)_v = (B_{03}'R'M_{02}R)_v = \big ((A_3)_vR'M_{02}R\big )_v = \big ((A_3)_vM_{02}R\big )_v \\= & {} (A_3M_{02}R)_v = (A_3)_v. \end{aligned}$$

Thus, $(A_3)_v = C_{03}R$, where $C_{03} = B_{03}'M_{02}$ and is a ${\mathbb {Z}}$-invariant ideal of $R_0$.

The proof of (iv): $A_4$ is an ideal of $R'$ and as in (i),

$$\begin{aligned} (A_4)_v = (A_4R')_{v'} = B_{04}'R' = B_{04}'R \end{aligned}$$

for some ${\mathbb {Z}}$-invariant ideal $B_{04}'$ of $R_0'$ and so

$$\begin{aligned} M_{01}B_{04}'R= & {} \big (M_{01}(A_4)_vR'\big )_v = \big (M_{01}\big ((A_4)_vR'\big )_v\big )_v = \big (M_{01}(A_4R')_v\big )_v \\= & {} (M_{01}A_4R')_v = (A_4)_v. \end{aligned}$$

Hence, $(A_4)_v = C_{04}R$, where $C_{04} = M_{01}B_{04}'$ and is a ${\mathbb {Z}}$-invariant ideal of $R_0$.

Now, by using (i) $\sim$ (iv), we have

$$\begin{aligned} (C_{01} + C_{03})R= & {} \big ((C_{01} + C_{03})R\big )_v = (C_{01}R + C_{03}R)_v = \big ((A_1)_v + (A_3)_v\big )_v \\= & {} (A_1 + A_3)_v = \big ((M_{01}A + M_{02}A)M_{02}\big )_v = \big ((M_{01} + M_{02})AM_{02}\big )_v \\= & {} (AM_{02})_v, \end{aligned}$$

since $M_{01} + M_{02} = R_0$. Similarly, $(C_{02} + C_{04})R = (AM_{01})_v$ follows. Hence,

$$\begin{aligned} (C_{01} + C_{02} + C_{03} + C_{04})R= & {} \big ((AM_{02})_v + (AM_{01})_v\big )_v \\= & {} (AM_{02} + AM_{01})_v = A_v = A, \end{aligned}$$

which completes the proof. $\square$

Theorem 3.12

Let A be an ideal of R with $A_0 = A \cap R_0 \not = (0)$. Then $A \in V_r(R)$ if and only if $A = A_0R$. In particular, $A = A_0R$ is a projective ideal of R, that is, left and right R-projective.

Proof

It is clear that $A = A_0R \in V_r(R)$. Let $A \in V_r(R)$, and we assume, on the contrary, that $A \supset A_0R$. We may assume that A is maximal for this property. If there is an invertible ideal P with $P \supseteq A$ and we may assume that P is a maximal invertible ideal of R, then $P = P_0R$ by Proposition 3.4, where $P_0 = P \cap R_0$. $R = PP^{-1} \supseteq AP^{-1} \supseteq A$ and $(AP^{-1})_v = A_vP^{-1} = AP^{-1}$ by [1, Lemma 2.1], that is, $AP^{-1} \in V_r(R)$. If $AP^{-1} = A$, then $A = AP^n$ for all $n \ge 1$ and $A \subseteq \cap AP^n \subseteq \cap P^n = (0)$, a contradiction. Thus, $AP^{-1} \supset A$ and so $AP^{-1} = B_0R$ by the choice of A, where $B_0 = AP^{-1} \cap R_0$. Hence, $A = PB_0R = P_0B_0R$, a contradiction. If there are no invertible ideals containing A, then, by Proposition 3.11, $A = A_0R$, a contradiction. Hence, $A = A_0R$ for any $A \in V_r(R)$ with $A_0 = A \cap R_0 \not = (0)$. Since $A_0$ is a projective ideal of $R_0$, $A = A_0R$ is a projective ideal of R. $\square$

Corollary 3.13

There is a one-to-one correspondence between the set of $\mathbb Z$-invariant ideals of $R_0$ and $\{ A \in V_r(R) \mid A_0 = A \cap R_0 \not = (0) \}$, which is given by

$$\begin{aligned} A_0 \rightarrow A = A_0R \hbox { and } A \rightarrow A_0 = A \cap R_0, \end{aligned}$$

where $A_0$ is a ${\mathbb {Z}}$-invariant ideal of $R_0$ and $A \in \{ A \in V_r(R) \mid A_0 = A \cap R_0 \not = (0) \}$. Furthermore,

• (i) A is invertible if and only if so is $A_0$.

• (ii) A is (eventually) idempotent if and only if so is $A_0$.

4 Projective Ideals with Zero Contracted Ideals of $R_0$

We denote by $\hbox {Spec}_0(R) = \{ P : \hbox {prime ideal} \mid P \cap R_0 = (0) \}$.

Lemma 4.1

• (1) There is a one-to-one correspondence between $\hbox {Spec}_0(R)$ and $\hbox {Spec}(Q^g)$, which is given by

  $$\begin{aligned} P \rightarrow PQ^g, \quad P' \rightarrow P = P' \cap R, \end{aligned}$$

  where $P \in \hbox {Spec}_0(R)$ and $P' \in \hbox {Spec}(Q^g)$.

• (2) Each $P \in \hbox {Spec}_0(R)$ is invertible and $P \in V_{(r,m)}(R)$.

Proof

• (1) is due to [12, Proposition 2.7, p.184].

• (2) Let $P \in \hbox {Spec}_0(R)$. Then, by [2, Lemma 3.2], $PQ^g = (PQ^g)_{v'} = (P_vQ^g)_{v'} = P_vQ^g$ since $Q^g$ is a principal ideal ring, where $v'$ is the v-operation in $Q^g$. Hence, $P = P_v$ and similarly $P = {}_vP$. So P is a projective ideal.

  We claim $P(R : P)_r \in V_r(R)$ with $P(R : P)_r \cap R_0 \not = (0)$. It is clear that $P(R : P)_r \cap R_0 \not = (0)$ since $Q^gP(R : P)_r = Q^gP(Q^g : Q^gP)_r = Q^g$ by [2, Lemma 3.2]. Since $(R : P)_rP = O_r(P)$, it is easy to see that $(R : P(R : P)_r)_l = O_l(P)$. Let $x \in (R : O_l(P))_r$. Then $P(R : P)_lx = O_l(P)x \subseteq R$, $(R : P)_lx \subseteq (R : P)_r$ and

  $$\begin{aligned} x \in O_l(P)x = P(R : P)_lx \subseteq P(R : P)_r. \end{aligned}$$

  Thus, $(R : O_l(P))_r \subseteq P(R : P)_r$. So

  $$\begin{aligned} (P(R : P)_r)_v = (R : (R : P(R : P)_r)_l)_r = (R : O_l(P))_r \subseteq P(R : P)_r \end{aligned}$$

  and hence $(P(R : P)_r)_v = P(R : P)_r$ as claimed. If $R \supset P(R : P)_r$, then there is an $M = M_0R$ containing $P(R : P)_r$, where $M_0$ is a maximal ${\mathbb {Z}}$-invariant ideal of $R_0$ (Proposition 3.2). Since Krull dimension ${{\mathcal {K}}}(R)$ of R$\le 2$ by [10, Lemma II.5.23], we have $2 \ge {\mathcal K}(R)> {{\mathcal {K}}}(R/P) > {{\mathcal {K}}}(R/M_0R)$ by [9, Proposition 6.3.11]. Hence, ${{\mathcal {K}}}(R/M_0R) = 0$. $R/M_0R \cong \oplus _{n \in {\mathbb {Z}}} R_n/M_0R_n$, which is a strongly graded ring of type ${\mathbb {Z}}$ such that the degree zero part is $R_0/M_0$. Since $R_0/M_0$ is a semi-simple Artinian ring by Lemma 2.5, it follows that $R/M_0R \cong (R_0/M_0)[X, X^{-1}, \sigma ]$ for some automorphism $\sigma$ of $R_0/M_0$ by [6, Corollary 1.2] and so ${{\mathcal {K}}}(R/M_0R) \ge 1$. The contradiction shows that $R = P(R : P)_r$ and similarly $(R : P)_lP = R$, that is, P is invertible. If $P \not \in V_{(r,m)}(R)$, then there is an $M \in V_{(r,m)}(R)$ with $M \supset P$ and $M_0 = M \cap R_0 \not = (0)$ by (1). This is impossible from the discussions above. Hence, $P \in V_{(r,m)}(R)$. $\square$

Proposition 4.2

$\{ P = P_0R \mid P_0 \hbox { is a maximal invertible}\ {\mathbb {Z}} \hbox {-invariant ideal of } R_0 \} \cup \hbox {Spec}_0(R)$ is the set of all maximal invertible ideals of R and the invertible ideals of R generates an abelian group.

Proof

The first statement follows from Propositions 3.4 and 4.1, and the second statement is obtained in the similar way as the proofs of [4, Propositions 2.1 and 2.8]. $\square$

Theorem 4.3

Let A be an ideal of R with $A = A_v$. Then $A = XB_0R$, where X is an invertible ideal and $B_0$ is an eventually idempotent ${\mathbb {Z}}$-invariant ideal of $R_0$ and A is a projective ideal.

Proof

We may assume that A is not invertible. Let X be minimal among invertible ideals containing A (such X exists, see [2, Lemma 2.9]). Then there are no invertible ideals containing $X^{-1}A$ and $X^{-1}A \in V_r(R)$ by [1, Lemma 2.1]. If $X^{-1}A \cap R_0 = (0)$, then there is a $P \in \hbox {Spec}_0(R)$ with $P \supseteq X^{-1}A$, a contradiction, since P is invertible. Thus, $X^{-1}A \cap R_0 = B_0 \not = (0)$ and $X^{-1}A = B_0R$ for some eventually idempotent ${\mathbb {Z}}$-invariant ideal of $R_0$ by Corollary 3.13. Hence, $A = XB_0R$ and A is a projective ideal. $\square$

The following remark is obtained in the symmetric arguments.

Remark

Let A be an ideal of R with $A = {}_vA$. Then $A = XB_0R$, where X is an invertible ideal and $B_0$ is an eventually idempotent ${\mathbb {Z}}$-invariant ideal of $R_0$ and A is a projective ideal.

In [2], we defined a concept of G-HNP rings which is, in some sense, polynomial-type generalization of HNP rings ([1] and [7]).

Definition 4.4

A prime Goldie ring S is called a G-HNP ring if

• (1) Any ideal A of S with $A = A_v$ or $A = {}_vA$ is a projective ideal.

• (2) S is $\tau$-Noetherian (see [8, p.116] for the definition).

A G-HNP ring is called a strongly G-HNP ring if any one-sided v-ideal is projective.

A ring S is an HNP ring if and only if $\hbox {gl} \dim S \le 1$. However, the class of G-HNP rings ranges from rings with global dimension two to rings with infinite global dimension ([2]). We note that Noetherian prime rings with global dimension two are not necessarily G-HNP rings ([1]).

We end this paper with the following corollary, which follows from Theorem 4.3, its remark, [3, Proposition 5.2] and [10, Lemma II.8.3].

Corollary 4.5

R is a strongly G-HNP ring.