Corrigendum: on graded pseudo-valuation domains

Abstract

I correct a lemma and two theorems of “On graded pseudo-valuation domains, Comm. Algebra 50 (2022), 247–254”.

Keywords:

• Graded integral domain
• UMT-domain
• valuation domain
• pseudo-valuation domain
• monoid domain

2020 Mathematics Subject Classification:

• 13A02
• 13A15
• 13F05

This article refers to:

On graded pseudo-valuation domains

My paper contains 3 sections including introduction. The subject of Section 2 is the same as subject of [1] with similar results.

Correction 1.

Part (3) of Lemma 1 should be:

3. If $\alpha \in \Gamma$ is not a unit, then $R_{\alpha }={(R_H)}_{0}x$ for every $0\ne x\in R_{\alpha }$

Proof.

See [1, Lemma 3.2(4)]. □

Then Theorem 6 and subsequent corollaries should be stated as:

Theorem 6.

Let $R=\underset{\alpha \in \Gamma }{\oplus }{R}_{\alpha }$ be a graded integral domain, K be the quotient field of R₀, $R_0\ne K$ and assume that $\Gamma \cap -\Gamma =\left\{0\right\}.$ Then R is a gr-PVD if and only if the following conditions hold:

1. R₀ is a PVD.

2. Γ is a valuation monoid.

3. $R_{\alpha }=Kx$ for every $0\ne x\in R_{\alpha }$ whenever $\alpha \in \Gamma \setminus \left\{0\right\}.$

Proof.

See [1, Theorem 3.7].□

Corollary 7.

Let $A\subseteq B$ be an extension of integral domains, A is not a field, Γ a torsionless grading monoid with $\Gamma \cap -\Gamma =\left\{0\right\},{\Gamma }^{*}:=\Gamma \setminus \left\{0\right\},$ and $R=A+B[X;{\Gamma }^{*}]=\{f\in B[X;\Gamma ]|f(0)\in A\}.$ Then R is a gr-PVD if and only if A is a PVD, B is the quotient field of A, and Γ is a valuation monoid.

Corollary 8.

Let D be a non-field integral domain with quotient field K and $R=D+K[X;{\mathbb{Q}}_{+}^{*}].$ Then R is a gr-PVD if and only if D is a PVD.

In Section 3, there is only one correction.

Correction 2.

The statement before part (8) of Theorem 12 should be:

If in addition $\Gamma \cap -\Gamma =\left\{0\right\}$ and R₀ is not a field, then the above statements are equivalent to

(8) Each homogeneous overring of R is a gr-PVD.

Proof.

…Now assume that $\Gamma \cap -\Gamma =\left\{0\right\}$ and R₀ is not a field.

$(8)\Rightarrow (3)$ Without loss of generality we can assume that $R=\overline{R}.$ Assume that $R⊊(M:M)$ and choose a homogeneous element $v\in (M:M)\setminus R.$ We assert that $\lambda :=\mathrm{deg}(v)=0.$ Let $m_0:=M\cap R_0$ be the maximal ideal of R₀. As R₀ is not a field, choose $0\ne r\in m_0.$ Since $vM\subseteq M,$ we have $\lambda =\mathrm{deg}(vr)\in \Gamma .$ Moreover, by Part 3 of Lemma 1, we have $R_{\alpha }={(R_H)}_{0}x={(M:M)}_{\alpha }$ for every $0\ne x\in R_{\alpha }$ whenever $\alpha \in \Gamma$ is not a unit. Hence $\lambda \in \Gamma \cap -\Gamma =\left\{0\right\},$ as asserted. The rest of the proof is correct.□

Acknowledgement

I would like to thank Najib Mahdou and Abdelkbir Riffi for pointing me the errors and sending a copy of their paper.