Problem 6 - Algebra

Question

If \(a\) is an ideal of \(A\), let \(a_{s}=S^{-1} a\). If \(\varphi_{s}: A \rightarrow S^{-1} A\) is the canonical map, abbreviate \(\varphi_{s}^{-1}\left(a_{s}\right)\) by \(a_{s} \cap A\), even though \(\varphi_{s}\) is not injective. Show that there is a bijection between the prime ideals of \(A\) which do not intersect \(S\) and the prime ideals of \(S^{-1} A_{+}\) given by $$ p \mapsto p_{s} \text { and } p_{5} \mapsto p_{s} \cap A=p \text { . } $$

Step-by-Step Solution

Answer

In this problem, we are given a ring A, an ideal a of A, and a multiplicative set S. We have shown that there is a bijection between the prime ideals of A that do not intersect S and the prime ideals of the localization \(S^{-1}A\). The bijection is given by: \(p \mapsto p_s\) and \(p_s \mapsto p_s \cap A = p\). This was proved by showing that: 1. If \(p\) is a prime ideal in A such that \(p \cap S = \emptyset\), then \(p_s = S^{-1}p\) is a prime ideal in \(S^{-1}A\). 2. If \(p_s\) is a prime ideal in \(S^{-1}A\), then \(p = p_s \cap A\) is a prime ideal in A that does not intersect S. 3. The mappings \(p \mapsto p_s\) and \(p_s \mapsto p_s \cap A\) are both bijections.

Step 1: Show that \(p_s = S^{-1}p\) is a prime ideal in \(S^{-1}A\)

Let \(p\) be a prime ideal in A such that \(p \cap S = \emptyset\). We want to show that \(p_s = S^{-1}p\) is a prime ideal in \(S^{-1}A\). Let \(x, y \in S^{-1}A\) be such that their product \(x \cdot y \in p_s\). By definition of the localization, we have \(x = a/s\) and \(y = b/t\) for some \(a, b \in A\) and \(s, t \in S\). Then, their product \(x \cdot y = (a/s)(b/t) = ab/st \in p_s\), which implies that \((st)^{-1}(ab) \in p\). Since \(st \in S\) and \(p \cap S = \emptyset\), we know that \(ab \in p\), as localization does not affect which elements are in the ideal. This means that either \(a \in p\) or \(b \in p\), or equivalently, either \(x \in p_s\) or \(y \in p_s\). Hence, \(p_s\) is a prime ideal in \(S^{-1}A\).

Step 2: Show that \(p = p_s \cap A\) is a prime ideal in A that does not intersect S

Now, let \(p_s\) be a prime ideal in \(S^{-1}A\), and let's consider \(p = p_s \cap A\). We want to show that \(p\) is a prime ideal in A that does not intersect S. First, let's show that \(p\) does not intersect S. Suppose for contradiction that there exists \(s \in p \cap S\). Then, \(s^{-1} \in S^{-1}A\), and \((s^{-1}s) \in p_s\), which would mean that 1 is in \(p_s\), a contradiction since prime ideals cannot be the entire ring. Therefore, \(p \cap S = \emptyset\). Now, we need to show that \(p\) is a prime ideal in A. Suppose \(a, b \in A\) are such that their product \(ab \in p\). Since \(ab \in p\), it follows that \((ab)/1 \in p_s\). Because \(p_s\) is a prime ideal in \(S^{-1}A\), either \((a/1) \in p_s\) or \((b/1) \in p_s\). If \((a/1) \in p_s\), then \(a \in p\) and vice versa for \(b \in p\). Hence, \(p\) is a prime ideal in A.

Step 3: Prove that the mappings are bijections

We are given two mappings between the prime ideals of A and \(S^{-1}A\): (i) \(p \mapsto p_s\) (ii) \(p_s \mapsto p_s \cap A\) We need to prove that these mappings are bijections. Let's show that both mappings are injective and surjective. Injective: 1. Suppose \(p_1\) and \(p_2\) are prime ideals in A such that \(p_{1_s} = p_{2_s}\). Then, \(p_1 \cap S^{-1}A = p_2 \cap S^{-1}A\), which implies \(p_1 = p_2\). Thus, the first mapping is injective. 2. Suppose \(p_{s_1}\) and \(p_{s_2}\) are prime ideals in \(S^{-1}A\) such that \(p_{s_1} \cap A = p_{s_2} \cap A\). Then, \(p_{s_1} = p_{s_2}\). Thus, the second mapping is injective. Surjective: 1. Let \(p_s\) be a prime ideal in \(S^{-1}A\). We showed in Step 2 that \(p = p_s \cap A\) is a prime ideal in A. Then, by applying the first mapping, we get \(p_s\), meaning the first mapping is surjective. 2. Let \(p\) be a prime ideal in A such that \(p \cap S = \emptyset\). We showed in Step 1 that \(p_s = S^{-1}p\) is a prime ideal in \(S^{-1}A\). Then, by applying the second mapping, we get \(p\), meaning the second mapping is surjective. Since both mappings are injective and surjective, they are bijections. This proves that there is a bijection between the prime ideals of A that do not intersect S and the prime ideals of \(S^{-1}A\), given by the mappings: \(p \mapsto p_s\) and \(p_s \mapsto p_s \cap A = p\).

Key Concepts

Understanding Ring Theory

Ring theory is a fundamental area of algebra that deals with structures called rings. A ring is a set equipped with two binary operations: addition and multiplication, following certain rules.

Rings generalize many number systems, including integers, and provide a framework for more complex algebraic structures. In the context of our exercise, we're focusing on ideals within rings, particularly prime ideals. A prime ideal is an ideal that has properties mimicking those of prime numbers within the integers. Specifically, if a product of two elements is within a prime ideal, at least one of those elements must be in the ideal too. This property is central to determining prime ideals when working with ring localizations, as seen in the exercise.

Exploring Localization of Rings

Localization is a process in ring theory that allows us to invert certain elements systematically. Essentially, given a ring 'A' and a multiplicatively closed subset 'S' of 'A', we create a new ring, often denoted as 'S^{-1}A', where the elements of 'S' become invertible. This construction is similar to how we obtain rational numbers from integers by inverting non-zero integers.

Localization is powerful because it permits focused study on certain elements within a ring while 'ignoring' others. In the prime ideals exercise, the localization process is used to create a correspondence between the prime ideals in the original ring 'A' that do not meet 'S' and the prime ideals in the localized ring 'S^{-1}A'. This is illustrated by showing that certain mappings related to these ideals are both injective and surjective, which brings us to our next concept.

Injective and Surjective Mappings

In the context of functions between sets—and in ring theory, mappings between ideals—injective (one-to-one) and surjective (onto) properties are crucial. A function is injective if each element of the domain maps to distinct elements in the co-domain. It is surjective if every element of the co-domain is an image of at least one element from the domain. A function that is both injective and surjective is called bijective, and it ensures a perfect 'pairing' between the domain and co-domain.

In our exercise, we observe two mappings among prime ideals where the injective property ensures no two different prime ideals map to the same localized prime ideal, and surjective property verifies that every prime ideal in the localized ring comes from a prime ideal in the original ring, thus proving a bijection. Understanding these mappings is critical as it confirms a tight correspondence between the structures of the original ring and its localization.