Problem 13 - Algebra

Question

Let \(A\) be any commutative ring, and \(M, N\) modules over \(A\). If \(M\) is finitely presented, and \(S\) is a multiplicative subset of \(A\), show that $$ S^{-1} \operatorname{Hom}_{A}(M, N) \approx \operatorname{Hom}_{s-1_{A}}\left(S^{-1} M, S^{-1} N\right) $$ This is usually applied when \(A\) is Noetherian and \(M\) finitely generated, in which case \(M\) is also finitely presented since the module of relations is a submodule of a finitely generated free module.

Step-by-Step Solution

Answer

To prove the desired isomorphism, we construct a map \(\Phi: S^{-1} \operatorname{Hom}_{A}(M, N) \rightarrow \operatorname{Hom}_{S^{-1}A}(S^{-1}M, S^{-1}N)\) defined by \(\Phi(u)(\frac{m}{s}) = \frac{u(m)}{s}\) for \(u \in \operatorname{Hom}_{A}(M, N)\) and \(\frac{m}{s} \in S^{-1}M\). We first show that this map is well-defined by verifying that it assigns an element in \(S^{-1}N\) to each element of \(S^{-1}M\), and that it is a homomorphism in \(\operatorname{Hom}_{S^{-1}A}(S^{-1}M, S^{-1}N)\). Then, we establish that \(\Phi\) is both injective and surjective, making it an isomorphism. This result is often applied in the context of Noetherian rings where the module \(M\) is finitely generated, as it allows for the application of the isomorphism in various areas such as homological algebra and algebraic geometry.

Step 1: Notation and definitions

Let \(A\) be a commutative ring, \(M, N\) be modules over \(A\), and \(S\) be a multiplicative subset of \(A\). The functor \(\operatorname{Hom}(M, -)\) takes an \(A\)-module to an abelian group, in this case, \(\operatorname{Hom}_{A}(M, N)\) which denotes the set of all \(A\)-module homomorphisms from \(M\) to \(N\). \(S^{-1}A\) is the localization of the ring \(A\) with respect to \(S\), and \(S^{-1}M, S^{-1}N\) denote the localization of the modules \(M, N\) respectively. \(\operatorname{Hom}_{S^{-1}A}(S^{-1}M, S^{-1}N)\) refers to the \(S^{-1}A\)-module homomorphisms between these localized modules.

Step 2: Constructing the isomorphism

Let \(M\) be a finitely presented module with generators \(m_1, ..., m_n\) and relations \(r_1, ..., r_k\). Note that generating elements and relations can be described by maps: $$ \xi: A^n \rightarrow M, \quad\text{and}\quad \rho: A^k \rightarrow A^n. $$ Now, consider the map \(\Phi: S^{-1} \operatorname{Hom}_{A}(M, N) \rightarrow \operatorname{Hom}_{S^{-1}A}(S^{-1}M, S^{-1}N)\) defined by $$ \Phi(u)(\frac{m}{s}) = \frac{u(m)}{s} \quad \text{for} \quad u \in \operatorname{Hom}_{A}(M, N), ~ \frac{m}{s} \in S^{-1}M. $$ We need to show that \(\Phi\) is well-defined, and that it is an isomorphism.

Step 3: Proving well-definedness and isomorphism

First, we check whether \(\Phi\) is well-defined by making sure that \(\Phi(u)\) assigns an element of \(S^{-1}N\) to each element of \(S^{-1}M\). Consider \(\frac{m}{s} \in S^{-1}M\): $$ \Phi(u)(\frac{m}{s}) = \frac{u(m)}{s} \in S^{-1}N. $$ Since \(u(m) \in N\) and \(s \in S\), the fraction is indeed an element of \(S^{-1}N\). Next, we verify that \(\Phi(u)\) is indeed a homomorphism in \(\operatorname{Hom}_{S^{-1}A}(S^{-1}M, S^{-1}N)\): $$ \Phi(u)(c\frac{m}{s} + d\frac{m'}{s'}) = \Phi(u)(\frac{cm + dm'}{ss'}) = \frac{u(cm + dm')}{ss'} = \frac{cu(m) + du(m')}{ss'}. $$ Now we prove that the map \(\Phi\) is injective. If \(\Phi(u) = \Phi(v)\), then for any \(\frac{m}{s} \in S^{-1}M\): $$ \frac{u(m)}{s} = \Phi(u)(\frac{m}{s}) = \Phi(v)(\frac{m}{s}) = \frac{v(m)}{s}. $$ Since the fraction is unique, we have \(u(m) = v(m)\) for all \(m \in M\). Thus, \(u = v\), and \(\Phi\) is injective. To prove surjectivity, consider any \(\psi \in \operatorname{Hom}_{S^{-1}A}(S^{-1}M, S^{-1}N)\). We can define \(u: M \rightarrow N\) by \(u(m) = \psi(\frac{m}{1})\). Then, for any \(\frac{m}{s} \in S^{-1}M\): $$ \Phi(u)(\frac{m}{s}) = \frac{u(m)}{s} = \frac{\psi(\frac{m}{1})}{s} = \psi(\frac{m}{s}), $$ showing that \(\Phi(u) = \psi\), and hence \(\Phi\) is surjective. Since \(\Phi\) is both injective and surjective, it is an isomorphism.

Step 4: Applications in Noetherian rings

The result of this exercise is typically applied in the context where the ring \(A\) is Noetherian and the module \(M\) is finitely generated. In that case, the module \(M\) is also finitely presented since the module of relations is a submodule of a finitely generated free module, allowing us to apply the constructed isomorphism in this exercise. This isomorphism is particularly important in the study of homological algebra and algebraic geometry.

Key Concepts

Finitely Presented Modules

In the study of modules over a ring, a module is termed "finitely presented" if it can be described using a finite set of generators and a finite set of relations. These modules can be seen as a more specific class of finitely generated modules, where not only finitely many generators exist, but also the relations among them can be expressed finitely.
One can think of finitely presented modules as akin to algebraic systems that are completely determined by a finite blueprint, comprising generators and a comprehensive yet concise set of relations.

• Consider the following: For a module \( M \), being finitely presented entails there is a surjective homomorphism from a free module \( A^n \) (with finitely many generators) to \( M \), and the kernel of this map (representing the relations between the generators) is itself finitely generated.
• This property ensures that operations and manipulations involving such modules can be effectively controlled and analyzed.

In practical terms, finitely presented modules find extensive use in computational settings, making them pivotal in algebraic geometry and other areas where explicit constructions are essential.

Localization of Rings and Modules

Localization is a method used to simplify a ring or module by focusing on certain elements deemed "invertible". In the context of a ring \( A \), and a subset \( S \) composed of non-zero elements forming a multiplicative system, we construct \( S^{-1}A \) - the localization of \( A \) at \( S \). This process effectively adds inverses for elements of \( S \) to the ring.
It's a way of "zooming in" on a particular prime ideal or a subset, hence making algebraic structures more flexible and way easier to analyze in certain respects.

• The effect on modules, like with \( M, N \) as in the exercise, is to produce \( S^{-1}M \) and \( S^{-1}N \), which are the modules with local properties extended to the localized ring.
• Through localization, we can make localized modules have nicer properties; for example, they often behave better with respect to certain "bad" elements in the rest of the ring.

This concept is crucial especially in algebraic geometry and number theory, as it allows one to work with localized structures that often possess properties more amenable to analysis.

Isomorphism in Module Homomorphisms

An isomorphism in the context of module homomorphisms is a homomorphism that is bijective - both injective and surjective, meaning it sets up a perfect "one-to-one correspondence" between two modules. In simple terms, an isomorphism implies two modules are structurally the same, even if they are superficially distinct.
In our exercise, we encounter an isomorphism between localized homomorphisms, showcasing how modules and their transformations can maintain form through localization.

• The isomorphism \( \Phi: S^{-1} \operatorname{Hom}_{A}(M, N) \rightarrow \operatorname{Hom}_{S^{-1}A}(S^{-1}M, S^{-1}N) \) proves this by mapping homomorphisms locally, preserving the structure.
• This idea is central in algebra, as it helps in understanding when two modules can be considered "essentially similar" or can be transformed into one another via the given homomorphism.

Understanding and demonstrating these isomorphisms provide deep insights into algebraic structures, making this concept very powerful in advanced mathematical and algebraic studies.