Search Results

The two results you mentioned readily follow from well-known facts in multiplicative ideal theory. I provide references for those facts in the proofs below. Claim 1. Let $K$ be a field and let $\{D_t …

This is not an answer, but only an attempt to retrieve, and present, the early result mentioned by the OP when $n = 2$. Hopefully, this will trigger further input from MO readers. The following is i …

Here is a 7-line proof of your statement. Claim. Let $R$ be a commutative ring with identity $1_R$. Let $A \simeq \mathbb{Z}/d_1\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/d_k\mathbb{Z}$ be a …

Let us not leave this question as unanswered: A ring $R$ has the $X$-property if and only if $R$ is arithmetical. The proof is straightforward.

There is a natural generalization of the aforementioned dimension-based reasoning to modules $M$ over commutative domains or commutative Noetherian reduced rings. Let $M$ be a module over a commutati …

In Chapter 3 of Jonas Jankauskas's dissertation thesis, entitled "Heights of Polynomials", we learn that the inequality $$\min_{Q \in \mathbb{Z}[x] \setminus \{0\}} H(PQ) \le \lfloor M(P)\rfloor$$ whe …

Your lemma easily follows from the Smith Normal Form Theorem, a result you already referred to. The short heuristic argument that you gave can indeed be turned into a short proof. Nothing in the sequ …

No, a free Boolean algebra $R$ on an infinite cardinal $\kappa$ (e.g., if $\kappa = \aleph_0$, $R$ is the Cantor algebra), is a commutative von Neumann regular ring which is not well complemented as a …

Given a ring $R$, let us denote by $L(R)$ the lattice of two-sided ideals of $R$ for which the infimum and supremum are given by $\inf(I, J) = I \cap J$ and $\sup(I, J) = I + J$. Such lattices are co …