Problem 5 - Algebra

Question

Let \(A\) be an arbitrary commutative ring and let \(S\) be a multiplicative subset. Let \(p\) be a prime ideal and let \(q\) be a p-primary ideal. Then p intersects \(S\) if and only if \(q\) intersects \(S\). Furthermore, if a does not intersect \(S\), then \(S^{-1} q\) is \(S^{-1} p\) -primary in \(S^{-1} A\)

Step-by-Step Solution

Answer

The given statement can be proven in two steps. Step 1: Show that $p$ intersects $S$ if and only if $q$ intersects $S$, which can be done by demonstrating two implications: (1) If $p$ intersects $S$, then $q$ intersects $S$ and (2) If $q$ intersects $S$, then $p$ intersects $S$. Step 2: Assume that $a$ does not intersect $S$ and prove that \( S^{-1}q \) is \( S^{-1}p \)-primary in \( S^{-1}A \) by showing that \( S^{-1}q \) is an ideal in \( S^{-1}A \) and that if \( \frac{a}{s} \), \( \frac{b}{t} \) ∈ \( S^{-1}A \), then \( \frac{a}{s} ∈ S^{-1}p \) or \( \frac{b}{t} ∈ S^{-1}p \).

Step 1: Prove p intersects S if and only if q intersects S

To show that p intersects S if and only if q intersects S, we need to show the following two statements: 1. If p intersects S, then q intersects S. 2. If q intersects S, then p intersects S. To show statement 1, assume that p intersects S, i.e., there exists x ∈ p that also belongs to S. Since q is a p-primary ideal, x^n ∈ q for some n ≥ 1. Since x ∈ S and S is a multiplicative set, x^n ∈ S. Therefore, q intersects S. To show statement 2, assume that q intersects S, i.e., there exists x ∈ q that also belongs to S. Since q is a p-primary ideal, there exists y ∈ A such that x^n = yp for some n ≥ 1. Then x^n ∈ p. Since x ∈ S and S is a multiplicative set, x⁽ⁿ⁻¹) ∈ S. Then yy⁽ⁿ⁻¹) ∈ p. Since p is a prime ideal and y ∈ A, y ∈ p or y⁽ⁿ⁻¹) ∈ p. In either case, p intersects S. This completes the proof of this step.

Step 2: Prove that if a does not intersect S, then \( S^{-1}q \) is \( S^{-1}p \)-primary in \( S^{-1}A \)

First, we need to show that \( S^{-1}q \) is an ideal in \( S^{-1}A \). Note that the localization of q with respect to S is given by \( S^{-1}q = \{ \frac{a}{s} \mid a ∈ q, s ∈ S \} \). Since q is an ideal in A and S is a multiplicative set, \( S^{-1}q \) is an ideal in \( S^{-1}A \). Next, we will show that \( S^{-1}q \) is \( S^{-1}p \)-primary. Let \( \frac{a}{s}, \frac{b}{t} \) ∈ \( S^{-1}A \) and \( ( \frac{a}{s} ) ( \frac{b}{t} ) ∈ S^{-1}q \). Then \( \frac{ab}{st} ∈ S^{-1}q \), which implies that there exists u ∈ S such that \( uab ∈ q \). Since q is p-primary, we have one of the following cases: 1. \( ua ∈ p \) 2. \( b^n ∈ q \) for some n ≥ 1 In case 1, since \( ua ∈ p \) and u ∈ S, we have \( \frac{a}{s} ∈ S^{-1}p \). In case 2, since \( b^n ∈ q \), we have \( \frac{b^n}{t^n} ∈ S^{-1}q \). This implies that \( \frac{b}{t} ∈ S^{-1}p \). Hence, \( S^{-1}q \) is \( S^{-1}p \)-primary in \( S^{-1}A \). This completes the proof.

Key Concepts

Commutative Ring

In the heart of algebra lies the concept of a commutative ring, which is an algebraic structure comprising a set equipped with two binary operations: addition and multiplication. Both operations are associative, addition is commutative and has an identity element (0), and every element has an additive inverse. Moreover, multiplication is commutative as well, with an identity element (typically denoted as 1), and it distributes over the addition operation.

When we refer to a commutative ring in the context of ideal theory, we're looking at the way certain subsets of the ring, known as ideals, behave under these operations. This relates directly to the exercise where the ring A and the ideals p and q must adhere to the ring's structure and properties.

Prime and primary ideals, as seen in the exercise, are types of subsets within a commutative ring that have additional properties related to the ring's multiplication operation. Specifically, a prime ideal has the property that if a product of two elements is inside the ideal, at least one of those elements must be in the ideal too.

Multiplicative Set

A multiplicative set is a subset of a commutative ring that is closed under multiplication. This means that for any two elements in the set, their product is also in the set. Moreover, the most crucial aspect of a multiplicative set is that it does not contain the zero element of the ring since that would make every element of the ring a member of the set because of the multiplication with zero.

In our given exercise, S is a multiplicative set within the commutative ring A. The significance of S is demonstrated when exploring the intersections of p and q with S, as well as their localization. The property of S being a multiplicative set is crucial, as it ensures the stable formation of fractions within the localization and preserves certain ideal properties.

Localization of Ideals

The localization of an ideal is a process that allows us to refine and zoom into a specific region of our commutative ring, akin to adjusting a microscope to get a closer look at a particular area. To put it mathematically, given a ring A, a multiplicative set S, and an ideal I, the localization of I at S, denoted as S^{-1}I, is the set of fractions with numerators in I and denominators in S. This construction yields a new ideal in the localized ring S^{-1}A, where the operations are defined in terms of fractions.

In the exercise, when the prime ideal p or the p-primary ideal q doesn’t intersect S, their localizations can help us understand the structure of ideals in a more localized setting. This is critical because it maintains the relationship between p and q within the context of fractions, preserving the property of being primary related to the localized prime S^{-1}p.