/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Let \(p\) be a prime of \(R\). P... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 3

Let \(p\) be a prime of \(R\). Prove that \(R / p R\) is a simple \(R\) -module.

Short Answer

Expert verified
Let \(N\) be a submodule of \(R/pR\) and assume that \(N = \phi(M)\) for some ideal \(M\) of \(R\) containing \(pR\). Since \(p\) is prime, it must divide at least one of the factors of any product it divides. If there exists a nontrivial submodule of \(R/pR\), there must be some element outside of \(pR\) that belongs to \(M\). However, this results in a contradiction with the prime property, so the submodule \(N\) of \(R/pR\) must be either the zero submodule or \(R/pR\) itself. Hence, \(R/pR\) is a simple \(R\)-module.

Step by step solution

01

Define a simple module

A simple module is a non-zero module that has no nontrivial submodule (i.e., it contains only itself and the zero submodule). To prove that \(R/pR\) is a simple \(R\)-module, we need to show that any submodule of \(R/pR\) is either the zero submodule or \(R/pR\) itself.
02

Consider an arbitrary submodule of \(R/pR\)

Let \(N\) be a submodule of \(R/pR\). Since \(pR\) is the kernel of the natural map \(\phi: R \rightarrow R/pR\), given by \(\phi(r) = r + pR\), we know that \(N = \phi(M)\) for some ideal \(M\) of \(R\) containing the ideal \(pR\).
03

Use the properties of prime elements

The definition of a prime element states that \(p\) is prime if whenever it divides a product \(ab\) (with \(a, b \in R\)), it must divide at least one of the factors \(a\) or \(b\). Recall that \(M\) is an ideal containing the ideal \(pR\). If \(pR \subseteq M\), and \(M \neq R\), then \(p\) divides some element in \(M\) but is not divisible by any other element in \(M\). Given \(m \in M\), we have an element \(m + pR \in N\), and since \(R\) is a module, we note that for any \(r \in R\), \(r(m + pR) = rm + pR \in N\).
04

Check for non-trivial submodule

If there exists a nontrivial submodule of \(R/pR\), there must be some element outside of \(pR\) that belongs to \(M\). Suppose \(x \in M\) but \(x \notin pR\), then \(x\) is not divisible by \(p\). According to the property mentioned in Step 3, since \(p\) is prime and \(x\) is not divisible by \(p\), the product \((x)(p)\) cannot be divisible by \(p\) either. Therefore, we have a contradiction in our assumption that \(p\) is prime.
05

Conclude that \(R/pR\) is a simple module

Since we found a contradiction in our assumption, we conclude that the submodule \(N\) of \(R/pR\) is either the zero submodule or \(R/pR\) itself. Therefore, the quotient ring \(R/pR\) is a simple \(R\)-module as required.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Prime Element in Algebra
In algebra, a prime element can be thought of as the building block that cannot be broken down into simpler elements in the context of multiplication. Specifically, an element p of a ring R is considered prime if it satisfies two conditions. First, p must not be a unit, meaning it cannot have a multiplicative inverse within the ring. Second, if p divides the product of two elements, ab, then it must divide at least one of those elements, a or b.

This concept is critical to understanding how structures like quotient rings and modules behave when related to a prime element. The step-by-step solution of the exercise seamlessly showed that when dealing with a quotient ring formed by a prime element, any non-zero element that does not contain the prime as a divisor leads to contradictions. This aligns with the key characteristic of primality, which imposes strict divisibility conditions.
Submodule in Module Theory
A module over a ring is an algebraic structure that generalizes important properties found in vector spaces. A submodule, then, is a subset of a module that is itself a module under the same operations. In other words, within a larger module, a submodule maintains the structure and operations that define a module.

For instance, in the provided exercise, a submodule N of the quotient ring R/pR must satisfy module properties. These properties imply closure under addition and the ability to 'absorb' ring elements when multiplied, just like vector spaces do with scalar multiplication. The solution's effectiveness lies in leveraging the fact that if N is a nontrivial submodule, it must include an element not in pR, leading to a contradiction and proving that no such nontrivial submodules exist. Therefore, R/pR itself must be simple.
Quotient Ring
A quotient ring is formed when an ideal of a ring is used to partition the ring into equivalent classes. This mirrors the concept of forming a quotient group in group theory. Specifically, we consider the set of cosets of the ideal, which in our case is pR, where p is a prime element. The operations of addition and multiplication on the quotient ring are defined in a way that respects the structure of the cosets.

Understanding quotient rings is essential when proving that R/pR is a simple module, as the exercise showed. By arguing that any submodule of this quotient ring must either be trivial or the whole of R/pR itself, the exercise concluded that the quotient ring R/pR is indeed simple when p is prime. Such a simple structure is paramount in understanding the fundamental aspects of algebraic structures like rings and modules.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(B\) be another endomorphism of \(V\). Assume that \(A B=B A\) and both \(A, B\) are diagonalizable. Prove that \(A\) and \(B\) are simultaneously diagonalizable, that is \(V\) has a basis consisting of elements which are eigenvectors for both \(A\) and \(B\).

Let \(R\) be a ring, and \(S\) a set consisting of \(n\) elements, say \(s_{1}, \ldots s_{n}\). Let \(F\) be the set of mappings from \(S\) into \(R\). (a) Show that \(F\) is a module. (b) If \(x \in R\), denote by \(x s_{i}\) the function of \(S\) into \(R\) which associates \(x\) to \(s_{i}\) and 0 to \(s_{j}\) for \(j \neq i\). Show that \(F\) is a free module, that \(\left\\{1 s_{1}, \ldots, 1 s_{n}\right\\}\) is a basis for \(F\) over \(R\), and that every element \(v \in F\) has a unique expression of the form \(x_{1} s_{1}+\cdots+x_{n} s_{n}\) with \(x_{i} \in R\).

Let \(A: V \rightarrow V\) be an endomorphism, and \(V\) finite dimensional. Suppose that the characteristic polynomial of \(A\) has the factorization $$ P_{A}(t)=\left(t-\alpha_{1}\right) \cdots\left(t-x_{\infty}\right) $$ where \(x_{1}, \ldots, \alpha_{n}\) are distinct elements of the field \(K\). Show that \(V\) has a basis consisting of eigenvectors for \(A\). For the rest of the exercises, we suppose that \(V \neq\\{0\\}\), and that \(V\) is finite dimensional over the algebraically closed field \(K\). We let \(A: V \rightarrow V\) be an endomorphism.

Let \(K\) be a field, and \(R=K[X]\) the polynomial ring over \(K\). Let \(J\) be the ideal generated by \(X^{2}\). Show that \(R / J\) is a \(K\) -space. What is its dimension?

Assume that \(V\) is a simple \(S\) -space and that \(A B=B A\) for all \(B \in S\). Prove that either \(A\) is invertible or \(A\) is the zero map. Using the fact that \(V\) is finite dimensional and \(K\) algebraically closed, prove that there exists \(\alpha \in K\) such that \(A=x l .\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.