reduced noetherian total fractions and normal components

Given: The objects and hypotheses in the statement. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

total ring of fractions: For a nonzero commutative ring $R$, let $S$ be the set of its nonzerodivisors, meaning elements whose multiplication maps on $R$ are injective. Its total ring of fractions is $Q(R)=S^{-1}R$. The set $S$ is multiplicative since composites of injective multiplication maps are injective. The natural map $R\to Q(R)$ is injective: $a\mathrm{/}1=0$ implies $sa=0$ for some $s\in S$, hence $a=0$. Set $Q(0)=0$. For a domain this recovers the fraction field; for a ring with zero divisors it need not be a field.

normal noetherian ring: A commutative Noetherian ring $R$ is normal if every prime localization $R_{\mathfrak{p}}$ is an integrally closed domain. This is a local condition and does not require $R$ itself to be a domain. The zero ring satisfies it vacuously. For a domain, integrally closed means that every element of its fraction field integral over it belongs to it.

A Noetherian ring has finitely many minimal prime ideals: Let $R$ be a Noetherian commutative ring. Then $R$ has only finitely many minimal prime ideals. This theorem inherits only the dependent-choice cost already recorded in the cited Noetherian-induction corollary.

A radical ideal in a Noetherian ring is the intersection of its minimal primes: Assume Dependent Choice. Let $R$ be a Noetherian commutative ring and let $I⊴R$ be a radical ideal. Then there exist finitely many prime ideals ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_m$ minimal over $I$ such that $I={\mathfrak{p}}_1\cap \cdots \cap {\mathfrak{p}}_m\mathrm{.}$ When $I=R$, this is the empty intersection.

A domain is integrally closed if and only if its prime localisations are, equivalently if and only if its maximal localisations are: Assume the Axiom of Choice. Let $A$ be a domain. Then the following are equivalent: 1. $A$ is integrally closed. 2. For every prime ideal $\mathfrak{p}$ of $A$, the localisation $A_{\mathfrak{p}}$ is integrally closed. 3. For every maximal ideal $\mathfrak{m}$ of $A$, the localisation $A_{\mathfrak{m}}$ is integrally closed.

Chinese remainder theorem for pairwise comaximal ideals: Let $R$ be a commutative ring and let $I_1,\dots ,I_r$ be pairwise comaximal ideals, where $r\ge 1$. Then the canonical map $R⟶{\prod }_{i=1}^{r}R\mathrm{/}{I}_i,x⟼(x+I_1,\dots ,x+I_r)$ is surjective, its kernel is ${\bigcap }_{i=1}^r{I}_i$, and ${\bigcap }_{i=1}^r{I}_i={\prod }_{i=1}^r{I}_i\mathrm{.}$ Equivalently, $R\mathrm{/}{\prod }_{i=1}^r{I}_i\cong {\prod }_{i=1}^{r}R\mathrm{/}{I}_i\mathrm{.}$

An ideal contained in a finite union of prime ideals lies in one of them: Let $R$ be a commutative ring, let $I⊴R$ be an ideal, and let ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_n$ be prime ideals with $n\ge 1$. If $I\subseteq {\mathfrak{p}}_1\cup \cdots \cup {\mathfrak{p}}_n,$ then $I\subseteq {\mathfrak{p}}_i$ for some $i$.