# On GCD and LCM of elements in integral domain with Krull-dimension 1

Let $R$ be an integral domain with Krull dimension $1$. If $0\ne a \in R$ is such that for every $b \in R$ , the ideal $Ra \cap Rb$ is principal , then is it true that for every $b\in R$, the ideal $Ra+Rb$ is also principal ?

It is clear that if $0\ne a,b$ and $Ra+Rb=Rd$ , then $Ra \cap Rb=R(ab/d)$. So if we know $Ra \cap Rb =Rc$ and want to prove $Ra+Rb$ is principal, then the unique (upto aasociates) possible candidate for the generator is $ab/c$. I'm unable to say anything else.

• In "Prime ideals in GCD domains", Philip B. Sheldon proves with Corollary 3.9 at page 102 that a one-dimensional GCD domain is a Bézout domain. So WLOG, we may assume that $R$ is $not$ a GCD domain. – Luc Guyot Mar 6 '18 at 17:20
• Perhaps worth pointing out: you may assume without loss of generality that $R$ is not a unique factorization domain. Hence a counterexample can be neither of the form $R=K[x]$, nor $R=K[[x]]$, nor $R=A[x]$ with $A$ some UFD. – Peter Heinig Mar 6 '18 at 17:20
• @LucGuyot : If possible, I would like to avoid using the result that dimension 1 and GCD domain implies Bezout domain. But if you have an answer using that , please feel free to write it. – user111524 Mar 7 '18 at 15:42
• I am failing to find any counter-example. I can only show that Dedekind domains won't provide any counter-example. About orders in Dedekind domains or GGCD domains (Generalized GCD domains), I don't now. – Luc Guyot Mar 8 '18 at 18:05
• @LucGuyot: Can you elaborate on why Dedekind domains won't provide any counterexample ? – user111524 Mar 11 '18 at 18:11

This is only a comment about classes of rings in which no counter-examples can be found.

Since a one-dimensional GCD domain is a Bézout domain [1, Corollary 3.9], GCD domains will not provide any counter-example.

We will show that Dedekind domains will not provide any counter-example either. To do so, we will use the following well-known lemma (see e.g., Matsumura's "Commutative Ring Theory") and two subsequent claims.

Lemma. Let $R$ be a commutative ring with identity. Let $\mathfrak{a}, \mathfrak{b}, \mathfrak{c}$ be ideals of $R$. Assume that $\mathfrak{c}$ is co-maximal with both $\mathfrak{a}$ and $\mathfrak{b}$, i.e., $\mathfrak{a} + \mathfrak{c} = \mathfrak{b} + \mathfrak{c} = R$. Then $\mathfrak{c}$ is co-maximal with $\mathfrak{ab}$. Proof. $R = (\mathfrak{a} + \mathfrak{c})(\mathfrak{b} + \mathfrak{c}) \subseteq \mathfrak{ab} + \mathfrak{c} \subseteq R$.

The following claim is a general fact about commutative domains.

Claim 1. Let $R$ be a commutative domain with identity. Let $M(R)$ be the submonoid of $R \setminus \{0\}$ generated by the units of $R$ together with the prime elements $p$ such that $Rp$ is a maximal ideal of $R$. Let $S(R)$ be the subset of $R \setminus \{0\}$ consisting of the non-zero elements $a$ such that $Ra + Rb$ is principal for every $b \in R$. Then $M(R) \subseteq S(R)$.

Proof. Let $a = up_1^{\alpha_1} \cdots p_n^{\alpha_n} \in M(R)$ where $u$ is a unit of $R$ and where the elements $p_i$ are distinct prime elements such that $Rp_i$ is a maximal ideal for every $i$. Let us show by induction on $s = \alpha_1 + \cdots + \alpha_n$ that $a$ belongs to $S(R)$. If $s = 0$, it is immediate. Let us suppose that $s > 0$ and let $b \in R$. We can certainly assume that $b \neq 0$. Let $\beta_i$ be the largest integer such that $p_i^{\beta_i}$ divides both $a$ and $b$ and set $d = p_1^{\beta_1} \cdots p_n^{\beta_n}$. If $d$ is not a unit, then the induction hypothesis applies to $a/d$ and yields a Bézout relation with $b/d$ so that $a \in S(R)$. Otherwise, the element $b$ cannot be divided by any of the $p_i$. Thus $Rb$ is co-maximal with $Rp_i$ for every $i$. By the above lemma, we deduce that $Rb$ is co-maximal with $Ra$, and hence $a \in S(R)$.

Our last claim establishes that no counter-example to OP's condition can be found in a Dedekind domain.

Claim 2. Let $R$ be Dedekind domain and let $a \in R$ such that $Ra \cap Rb$ is principal for every $b \in R$. Then $a \in M(R)$.

Proof. Let $a$ be a non-zero element in $R \setminus M(R)$. By hypothesis, there is a maximal ideal $\mathfrak{m}$ appearing in the decomposition of $Ra$ which is not principal. By the Chinese Remainder Theorem, we can find $b \in R$ such that $b \in \mathfrak{m} \setminus \mathfrak{m}^2$ and $b$ doesn't belong to any of the other maximal ideals containing $a$. In the group of fractional ideals of $R$, we have $Ra \cap Rb = (Rab)\mathfrak{m}^{-1}$ so that $Ra \cap Rb$ cannot be principal.

[1] P. Sheldon, "Prime ideals in GCD domains".

• In your proof of claim 2, since $R$ is a Noetherian domain, I can see by Nakayama that $\mathfrak m \ne \mathfrak m^2$ , but how do you find $b \in \mathfrak m \setminus \mathfrak m^2$ such that $b$ doesn't belong to any other maximal ideal containing $a$ ? – user111524 Mar 20 '18 at 16:24
• Write $Ra = \mathfrak{m}^k \mathfrak{m}_1^{k_1} \cdots \mathfrak{m}_m^{k_m}$ where each $\mathfrak{m}_j$ is a maximal ideal. Pick $b_0 \in \mathfrak{m} \setminus \mathfrak{m}^2$ and $b_i \notin \mathfrak{m}_i$ for every $i \ge 1$. The CRT applied to the co-maximal ideals $\mathfrak{m}^2, \mathfrak{m}_1,\dots, \mathfrak{m}_m$ gives you $b$. – Luc Guyot Mar 20 '18 at 16:45
• Thanks ... one more thing ... could you please elaborate on how you got $Ra \cap Rb = (Rab)\mathfrak{m}^{-1}$ ? – user111524 Mar 20 '18 at 17:41
• Write $Ra = \mathfrak{m}^k \mathfrak{m}_1^{k_1} \cdots \mathfrak{m}_m^{k_m}$, $Rb = \mathfrak{m} \mathfrak{n}_1^{l_1} \cdots \mathfrak{n}_n^{l_n}$ where $\mathfrak{m}_i \neq \mathfrak{n}_j$ for every $i, j$. Now compare $(Ra)(Rb)$ with $Ra \cap Rb$. The decomposition of the latter ideal can be inferred from the following well-know fact: If $I$ and $J$ are ideals of a Dedekind domain, then $\nu_{\mathfrak{p}}(I \cap J) = \max(\nu_{\mathfrak{p}}(I), \nu_{\mathfrak{p}}(J))$ where $\nu_{\mathfrak{p}}(K)$ denotes the exponent of the maximal ideal $\mathfrak{p}$ in the decomposition of $K$. – Luc Guyot Mar 20 '18 at 22:35