Problem 28 - Algebra

Question

Let \(R\) be a principal entire ring. Let \(a \in R .\) For every \(R\) -module \(N\), prove: (a) Ext \(^{\prime}(R / a R, N)=N / a N\). (b) For \(b \in R\) we have Ext \(^{1}(R / a R, R / b R)=R /(a, b)\), where \((a, b)\) is the \(\mathrm{g.c.d}\) of \(a\) and \(b\), assuming \(a b \neq 0\).

Step-by-Step Solution

Answer

In summary, we showed the following results for a principal entire ring \(R\) and \(a \in R\): (a) \(\text{Ext}^{\prime}(R / a R, N) = N / aN\) by considering a short exact sequence of R-modules, applying the Ext functor, and determining the corresponding Hom terms. (b) \(\text{Ext}^{1}(R / a R, R / b R) = R /(a, b)\) for \(b \in R\), assuming \(ab \neq 0\), by studying torsion elements and using a short exact sequence for the ideal generated by (a,b). Applying the Ext functor and determining the relevant Hom and Ext terms, we concluded the desired result.

Step 1: (a) Express N as a quotient of R-modules

To solve part (a), first consider the short exact sequence of R-modules: \(0 \rightarrow aN \rightarrow N \rightarrow N/aN \rightarrow 0\).

Step 2: (a) Apply the Ext functor

Apply the Ext functor to the given short exact sequence with \(R/aR\), resulting in the following long exact sequence: \( 0 \rightarrow \text{Hom}_R(R/aR, aN) \rightarrow \text{Hom}_R(R/aR, N) \rightarrow \text{Hom}_R(R/aR, N/aN) \rightarrow \text{Ext}^1_R(R/aR, aN) \rightarrow \text{Ext}^1_R(R/aR, N) \rightarrow \cdots \)

Step 3: (a) Determine the Hom terms

Now we'll determine the relevant Hom terms. Notice that \(\text{Hom}_R(R/aR, aN) = 0\) as \(aN\) is annihilated by \(a\), but \(R/aR\) as an \(R\)-module has no non-zero elements annihilated by \(a\). Additionally, we know that \(\text{Hom}_R(R, N) \cong N\), so in our scenario, \(\text{Hom}_R(R/aR, N) = N/aN\), as a submodule of \(N\).

Step 4: (a) Conclusion for (a)

Since \(\text{Hom}_R(R/aR, aN) = 0\), it follows from the Long Exact Sequence that \(\text{Hom}_R(R/aR, N) \cong \text{Hom}_R(R/aR, N/aN)\). Thus, we have \(\text{Ext}^{\prime}(R/aR, N) = \text{Hom}_R(R/aR, N/aN) \cong N/aN.\)

Step 5: (b) Begin with torsion elements

To solve part (b), first remember that every element in \(R/bR\) is annihilated by \(b\) and \(R/aR\) has no non-zero elements annihilated by \(b\), as long as \(b \neq 0\). This means that \(\text{Hom}_R(R/aR, R/bR) = 0\).

Step 6: (b) Write down the short exact sequence

Now, consider the short exact sequence of R-modules induced by the ideal generated by (a,b) in R: \(0 \rightarrow (a,b)R \rightarrow R \rightarrow R/(a,b)R \rightarrow 0\).

Step 7: (b) Apply the Ext functor

Applying the Ext functor with respect to \(R / a R\) to the above short exact sequence gives us the following long exact sequence: \( 0 \rightarrow \text{Hom}_R(R/aR, (a,b)R) \rightarrow \text{Hom}_R(R/aR, R) \rightarrow \text{Hom}_R(R/aR, R/(a,b)R) \rightarrow \text{Ext}^1_R(R/aR, (a,b)R) \rightarrow \text{Ext}^1_R(R/aR, R) \rightarrow \cdots \)

Step 8: (b) Determine the Hom and Ext terms

As in part (a), we know that \(\text{Hom}_R(R/aR, (a,b)R) = 0\) and \(\text{Hom}_R(R/aR, R) \cong R/aR.\) Since \(R/(a,b)R\) is also an R-module with no non-zero elements annihilated by b, we get \(\text{Hom}_R(R/aR, R/(a,b)R) = 0\).

Step 9: (b) Conclusion for (b)

From the long exact sequence, we have \(\text{Ext}^1_R(R/aR, (a,b)R) \cong \text{Ext}^1_R(R/aR, R)\) But we know that \(\text{Ext}^1_R(R/aR, R/bR)\) sits in the exact sequence, so we conclude that \(\text{Ext}^1_R(R/aR, R/bR) \cong R/(a,b)R\), as desired.

Key Concepts

Principal Ideal Ring

A Principal Ideal Ring, or PIR for short, is a type of ring where every ideal is generated by a single element. This concept is an extension of what we know about Principal Ideal Domains (PIDs), where these rings must also be integral domains, meaning they have no zero divisors. However, a PIR extends this concept to include cases where the ring might have zero divisors.
To visualize what this means, think of the entire set of integers, \(\mathbb{Z}\), as a classic example of a PID. When every ideal can be written as \(a\cdot\mathbb{Z}\), where \(a\) is an integer, you can see how a single element can generate an entire ideal. In a Principal Ideal Ring, every similar collection we take out is just a multiplication of one element, showing a kind of compactness or simplicity in its structure.
Understanding PIRs helps in simplifying problems in ring theory and abstract algebra by ensuring that the focus remains on individual generators rather than sets of elements. If you grasp this, you're more equipped to analyze or solve problems using module theory and even applications in homological algebra.

Module Theory

Module theory is an overarching branch of mathematics that generalizes vector spaces by replacing the field of scalars with rings. Just like vector spaces can be thought of as collections of vectors that you can scale and add, modules allow these similar operations, but the scalars come from a ring.
Different from vector spaces, modules don't require division by non-zero elements (since rings do not necessarily include division by non-zero elements), which makes the theory more complex. This makes module theory fundamental in areas such as linear algebra, where it lets you generalize concepts from finite-dimensional spaces to more complex structures.
Key insights of module theory often arise in

• Solving linear systems where solutions can be linked with homological concepts like exact sequences.
• Rooted interactions between modules and rings through specific structures called homomorphisms and isomorphisms.

Modules, therefore, add a layer of richness in algebraic expressions and are pivotal when looking into how different mathematical structures interact and evolve.

Homological Algebra

Homological Algebra is a part of mathematics that studies homology in a general algebraic setting. It's the tool many algebraists reach for when they need to understand structures arising in algebra, topology, and other areas through sequences and complexes.
One primary use of homological algebra is in dealing with exact sequences, which are sequences of module homomorphisms where the kernel of one map is the image of the preceding map. Ext and Tor functors are common tools in this field, providing deep insights into how different modules interact.

• Understanding Ext functors helps in classifying extensions of modules, which describe possible ways of 'building' new modules from known ones.
• Exact sequences, and Ext calculations, reveal hidden layers of information about mathematical structures.

Grabbing these concepts can sometimes require wrestling with abstraction, but they broaden one's capability to tackle complex algebraic problems efficiently.