Gaussian polynomials (Q2785942)

Let \(A\) be a commutative ring, and let \(x\) be an indeterminate over \(A\). For a polynomial \(f\in A[x]\), denote by \(c(f)\) -- the so-called content of \(f\) -- the ideal of \(A\) generated by the coefficients of \(f\). The content ideal of \(f\) satisfies certain multiplicative properties:NEWLINENEWLINENEWLINEi. \(c(fg) \subset c(f)c(g)\) for any \(g\in A[x]\).NEWLINENEWLINENEWLINEii. \(c(f)^n c(fg)= c(f)^n c(f)c(g)\) for any \(g\in A[x]\) with \(\deg g=n\).NEWLINENEWLINENEWLINEContainment i. becomes equality in certain cases. A well known result states that \(c(fg)= c(f)c(g)\) for all \(g\in A[x]\) when \(c(f)\) is an invertible ideal. More generally this equality holds when \(c(f)\) is a locally principal ideal.NEWLINENEWLINENEWLINEDefinition: A polynomial \(f\in A[x]\) is called a Gaussian polynomial if \(c(fg)= c(f)c(g)\) for any \(g\in A[x]\). NEWLINENEWLINENEWLINEGiven the results mentioned above it is natural to ask the following questions: Let \(A\) be a commutative ring and let \(f\in A[x]\) be a Gaussian polynomial. Is \(c(f)\) an invertible ideal of \(A\)? -- Let \(A\) be a commutative ring and let \(f\in A[x]\) be a Gaussian polynomial. Is \(c(f)\) a locally principal ideal of \(A\)?NEWLINENEWLINENEWLINEConjecture: Let \(A\) be a domain and let \(f\in A[x]\) be a Gaussian polynomial. Then \(c(f)\) is an invertible ideal of \(A\).NEWLINENEWLINENEWLINEThis article surveys the work done about and around this conjecture. Its center, sections 3 and 4, is a reformulation, to the extent that is possible, of earlier results (1995) [cf. \textit{S. Glaz} and \textit{W. V. Vasconcelos}, J. Algebra 202, No. 1, 1-9 (1998)]. Section 2 presents a history of the results related to this conjecture, while section 5 consists of a brief exposition of work by \textit{W. Heinzer} and \textit{C. Huneke} [Proc. Am. Math. Soc. 125, No. 3, 739-745 (1997; Zbl 0860.13005)] extending the authors' (1995) results.NEWLINENEWLINEFor the entire collection see [Zbl 0855.00015].