Nettet30. nov. 2024 · Let R be an integrally closed domain with finite Krull dimension d \ge 1, and let n be a positive integer. If O (R) =n+d, then the following statements hold true: 1. The number of maximal ideals is finite and satisfies the inequalities: \begin {aligned} \log _ {d+1} (n+d)\le Max (R) \le \log _ {2} (n+1). \end {aligned} 2.

On Some Sufficient Conditions for Polynomials to Be Closed

Nettet28. mar. 2024 · In this paper, we study closed polynomials of the polynomial ring in n variables over an integral domain. By using the techniques on \(\mathbb {Z}\)-gradings on the polynomial ring, we give some sufficient conditions for a polynomial f to be a closed polynomial. We also give a correspondence between closed polynomials and … Nettet28. mar. 2024 · Closed 5 years ago. Show that if an integral domain A is integrally closed in its field of fractions K, then so is A [ T] in its ring of fractions, K ( T) := F r a c ( A [ … godflesh album covers

When is R[θ] integrally closed? ScienceGate

NettetTo prove in the case of the ring integrally closed, show the following: Assume $A$ commutative with $1$ and in $A [X]$ we have the equality between monic polynomials $f= g\cdot h$, where $f = X^m + a_1 X^ {m-1} + \cdots + a_m$, $g = X^p + b_1 X^ {p-1} + \cdots + b_p$, $h =X^q + c_1 X^ {q-1} + \cdot + c_m$. NettetAn integral domain is a UFD if and only if it is a GCD domain (i.e., a domain where every two elements have a greatest common divisor) satisfying the ascending chain condition on principal ideals. An integral domain is a Bézout domain if and only if any two elements in it have a gcd that is a linear combination of the two. http://math.stanford.edu/~conrad/210BPage/handouts/math210b-dedekind-domains.pdf booble coco free