Proof. We argue by induction on k. The case $k = 1$ is trivial and the inductive step is reduced to the case $k = 2$ via the decomposition $R = (R_1 \times \cdots \times R_{k - 1}) \times R_k$ . Thus, it is enough to prove the statement for $k = 2$ . We have $R = R_1 \times R_2$ . If condition $(Rg)$ holds in R for any ideals, then obviously it holds in $R_1$ and $R_2$ . Now, suppose that condition $(Rg)$ holds for ideals in each $R_i$ . Let $I_1, I_2, \ldots , I_m$ be ideals in R and let $a_1, a_2, \ldots , a_m \in R$ . We want to prove that $| \bigcup _{j = 1}^{m}(a_j + I_j) | \ge | \bigcup _{j = 1}^{m} I_j |$ . Denote by $\varphi _i : R \to R_i$ the canonical projection. We have

$$ \begin{align*} \bigcup_{j = 1}^{m} (a_j + I_j) = \bigcup_{j = 1}^{m} ( \varphi_1(a_j + I_j) \times \varphi_2(a_j + I_j) ). \end{align*} $$

Let $x \in R_1$ . Let $s(x) := \{ j : x \in \varphi _1(a_j + I_j) \}$ . Then,

$$ \begin{align*} \bigg| \bigcup_{j = 1}^{m} (a_j + I_j) \bigg| & = \bigg| \bigcup_{x \in R_1} \{x \} \times \bigcup_{j \in s(x)} \varphi_2(a_j + I_j) \bigg| = \sum_{x \in R_1} \bigg| \bigcup_{j \in s(x)} \varphi_2(a_j + I_j) \bigg| \\ & \ge \sum_{x \in R_1} \bigg| \bigcup_{j \in s(x)} \varphi_2(I_j) \bigg| = \bigg| \bigcup_{x \in R_1} \{x \} \times \bigcup_{j \in s(x)} \varphi_2(I_j) \bigg| = \bigg| \bigcup_{j = 1}^{m}(b_j + I_j) \bigg|, \end{align*} $$

where $b_j = (\varphi _1(a_j), 0) \in R_1 \times R_2$ . Repeating the argument, we find that $| \bigcup _{j = 1}^{m}(b_j + I_j) | \ge | \bigcup _{j = 1}^{m} I_j |$ .

Hence, any finite set of ideals in R satisfies condition $(Rg_1')$ and, hence, satisfies condition $(Rg)$ .

Proof of Theorem 1.9

Let R be a finite ring. It can be uniquely (up to isomorphism) decomposed into a finite direct product of local rings [Reference Atiyah and Macdonald1, Theorem 8.7]. Thus, Lemma 2.1 implies that it is enough to prove Theorem 1.9 for finite local rings.

Let R be a finite local ring. We want to show that any finite set of ideals in R satisfies condition $(Rg)$ if and only if ideals of R are totally ordered. Let us prove this statement by induction on the number of elements of R.

Suppose that the ideals of R are totally ordered. After relabelling the ideals so that $H_1 \subset \cdots \subset H_r$ , the right-hand side of $(Rg_1')$ equals $|H_r|$ , while the left-hand side is at least $|a_r + H_r| = |H_r|$ . We see that condition $(Rg_1')$ and, hence, condition $(Rg)$ are satisfied.

Suppose that the ideals of R are not totally ordered. First, suppose that R has a unique minimal ideal $J_0$ . Then, any nonzero ideal of R contains $J_0$ . Hence, the ideals of $R/J_0$ are not totally ordered and, by the induction hypothesis, there exist ideals $I_1, \ldots , I_r$ in $R / J_0$ that do not satisfy condition $(Rg)$ . Let $\varphi : R \to R/J_0$ be the canonical projection. It is easy to see that $\varphi ^{-1}(I_1), \ldots , \varphi ^{-1}(I_r)$ do not satisfy condition $(Rg)$ .

Finally, suppose that R has more than one minimal ideal. Let $\mathfrak {m}$ be the unique maximal ideal of the local ring R. Let $J = \text {Ann}(\mathfrak {m})$ be the annihilator of $\mathfrak {m}$ . The ideal J is a vector space over $R / \mathfrak {m}$ and its one-dimensional subspaces are the minimal ideals. Let $(a)$ be a minimal ideal in R (it is clear that minimal ideals are principal). Suppose that $\mathfrak {m} (a) \ne (0)$ . Then, $\mathfrak {m} (a) = (a)$ since $\mathfrak {m} (a) \subset (a)$ and $(a)$ is minimal. By Nakayama’s lemma [Reference Atiyah and Macdonald1, Proposition 2.6], $(a) = 0$ , which leads to a contradiction. Hence, $(a) \subset J$ . This shows that minimal ideals are precisely the one-dimensional vector subspaces of J. By our hypothesis, there exist at least two minimal ideals and, hence, J has dimension at least $2$ over $R/\mathfrak {m}$ . Let H be a two-dimensional subspace in J. Let $I_1, I_2, I_3$ be distinct one-dimensional subspaces of H. They do not satisfy condition $(Rg)$ by Example 1.4.

Proof. ( $\Rightarrow $ ) Suppose that R is a Dedekind domain and let I be a nonzero ideal. Then, $I = \mathfrak {p}_1^{\alpha _1} \cdots \mathfrak {p}_k^{\alpha _k}$ for some prime ideals $\mathfrak {p}_1, \ldots , \mathfrak {p}_k$ , and this representation is unique [Reference Atiyah and Macdonald1, Corollary 9.4]. By the Chinese remainder theorem [Reference Atiyah and Macdonald1, Proposition 1.10],

$$ \begin{align*} R / I \cong R / \mathfrak{p}_1^{\alpha_1} \oplus \cdots \oplus R / \mathfrak{p}_k^{\alpha_k}. \end{align*} $$

Thus, it is enough to show that for any k and any prime ideal $\mathfrak {p}$ , the ring $R / \mathfrak {p}^k$ is local with linearly ordered ideals. The ideals of $R / \mathfrak {p}^k$ are in a one-to-one order-preserving correspondence with ideals of R that contain $\mathfrak {p}^k$ [Reference Atiyah and Macdonald1, Proposition 1.1]. Hence, they are of the form $\mathfrak {p}^{\alpha }$ and, hence, $R / \mathfrak {p}^k$ is a local ring with linearly ordered ideals.

( $\Leftarrow $ ) We need to show that R is integrally closed, assuming that for any nonzero ideal I, the ring $R/I$ is a finite direct product of local rings with linearly ordered ideals. By [Reference Atiyah and Macdonald1, Proposition 5.13], R is integrally closed if and only if the localisation $R_{\mathfrak {p}}$ is integrally closed for each nonzero prime ideal $\mathfrak {p}$ in R. Let us prove that the ideals of $R_{\mathfrak {p}}$ are totally ordered. This is equivalent to $R_{\mathfrak {p}}$ being a valuation ring [Reference Atiyah and Macdonald1, Ch. 5, Exercise 28], which implies that $R_{\mathfrak {p}}$ is integrally closed [Reference Atiyah and Macdonald1, Proposition 5.18].

Let I and J be ideals in $R_{\mathfrak {p}}$ . We want to show that $I \subset J$ or $J \subset I$ . We can suppose that $I, J$ are nonzero ideals, otherwise, the statement is trivial. The ring $R_{\mathfrak {p}}$ is one-dimensional [Reference Atiyah and Macdonald1, Proposition 3.11(iv)] and, hence, $\mathfrak {p} R_{\mathfrak {p}}$ is its unique nonzero prime ideal. The radical of I equals the intersection of prime ideals that contain I [Reference Atiyah and Macdonald1, Proposition 1.14] and, hence, equals $\mathfrak {p}R_{\mathfrak {p}}$ . In a Noetherian ring, any ideal contains a power of its radical [Reference Atiyah and Macdonald1, Proposition 7.14]. Hence, there is an integer k such that $\mathfrak {p}^k R_{\mathfrak {p}} \subset I \cap J$ . Let K be a proper ideal of R that contains $\mathfrak {p}^k$ . Then, K is contained in $\mathfrak {p}$ . Indeed, otherwise, there exists $a \in K \setminus \mathfrak {p}$ . Since R is one-dimensional, every nonzero prime ideal of R is maximal. In particular, $\mathfrak {p}$ is maximal. Because $a \notin \mathfrak {p}$ , the ideal $(a)+\mathfrak {p}$ strictly contains $\mathfrak {p}$ . Hence, $(a) + \mathfrak {p} = (1)$ . Raising to the kth power, we obtain $(1) = (a) + \mathfrak {p}^{k} \subset K$ , which is a contradiction. This shows that the ideals of $R_{\mathfrak {p}} / \mathfrak {p}^k R_{\mathfrak {p}}$ are in one-to-one order-preserving correspondence with the ideals of $R / \mathfrak {p}^k$ , which are totally ordered by assumption. Hence, $I \subset J$ or $J \subset I$ .

Proof of Theorem 1.6

It is easy to see that, in an integral domain, condition $(Rg)$ is satisfied for any finite set of ideals if and only if condition $(Rg)$ is satisfied for each quotient ring $R/I$ modulo a nonzero ideal I. Any order is a one-dimensional Noetherian domain [Reference Neukirch3, Ch. I, Proposition 12.2]. Therefore, Theorem 1.6 follows from Theorem 1.9 and Lemma 3.1.