Chapter 15: Problem 6
Let \(R\) be a ring, M a module, \(L\) a left ideal. Show that \(L M\) is a submodule of \(M\). Assume that both \(L\) and \(M\) are simple. Show that \(L M=M\) or \(L M=\\{0\\}\).
Short Answer
Expert verified
In summary, we have shown that the product of a left ideal \(L\) and a module \(M\), denoted by \(LM\), is a submodule of \(M\). Furthermore, if both \(L\) and \(M\) are simple, then \(LM\) is either equal to \(M\) or the zero ideal \(\{0\}\).
Step by step solution
01
Recall the definition of a submodule
First, we need to recall the definition of a submodule. Let M be an R-module, and let N be a non-empty subset of M. N is called a submodule of M if for all m, n in N and r in R, the following conditions hold:
1. m + n is in N.
2. rm is in N.
02
Show LM is a submodule of M
Now, let's prove that LM is a submodule of M. Consider any l_1, l_2 in L, and m_1, m_2 in M. Then we have:
1. (l_1m_1) + (l_2m_2) is in LM, since for any element x in L and y in M, x(y+z) = xy + xz, and the sum of any two elements in L is still in L, hence (l_1m_1) + (l_2m_2) is an element in L(M_1 + M_2).
2. For any r in R, r(l_1m_1) = (rl_1)m_1, and by the definition of a left ideal, rl_1 is in L, and as M is an R-module, m_1 is in M. So, r(l_1m_1) is in LM.
Since LM satisfies both conditions for being a submodule, we conclude that LM is a submodule of M.
03
Recall the definition of a simple module and simple left ideal
Before we move to the second part of the problem, let's recall the definition of a simple module and simple left ideal.
A module M is called simple if it has no nontrivial submodules, i.e., the only submodules of M are M itself and the zero submodule {0}.
A left ideal L is called simple if it has no nontrivial left ideals, i.e., the only left ideals of L are L itself and the zero left ideal {0}.
04
Show LM = M or LM = {0}
Now we assume that both L and M are simple. Let's consider the submodule LM. If LM = {0}, then the statement is already proven. If LM ≠{0}, then we must show that LM = M.
Since M is simple and LM is a nontrivial submodule of M (LM ≠{0}), M must not have any other nontrivial submodule. Thus, LM = M.
Hence, we have proven that if both L and M are simple, then either LM = M or LM = {0}.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Submodules
To demystify the concept of a submodule, we can compare it to a familiar mathematical structure: a subset. Think of a submodule as a special type of subset that comes from the universe of module theory. A submodule satisfies two crucial conditions: it's closed under addition and closed under scalar multiplication.
In a more formal sense, suppose we have a module, let's call it \( M \), over a ring \( R \). Now, if we pick a non-empty subset (say \( N \)) of \( M \), this \( N \) is a submodule if for any elements \( m, n \in N \) and for any scalar \( r \in R \), the following statements hold true: First, when you add any two elements from \( N \), the result stays in \( N \). This reflects closure under addition. Second, when you multiply any element from \( N \) by a scalar from \( R \), the product is still in \( N \). This is closure under scalar multiplication.
A submodule is, therefore, like a self-contained world within the universe of \( M \): While it's part of the greater world, it operates by its own rules, but still in harmony with the structure defined by \( M \). Understanding this concept helps us appreciate the stability and structure in the study of algebraic systems such as rings and modules.
In a more formal sense, suppose we have a module, let's call it \( M \), over a ring \( R \). Now, if we pick a non-empty subset (say \( N \)) of \( M \), this \( N \) is a submodule if for any elements \( m, n \in N \) and for any scalar \( r \in R \), the following statements hold true: First, when you add any two elements from \( N \), the result stays in \( N \). This reflects closure under addition. Second, when you multiply any element from \( N \) by a scalar from \( R \), the product is still in \( N \). This is closure under scalar multiplication.
A submodule is, therefore, like a self-contained world within the universe of \( M \): While it's part of the greater world, it operates by its own rules, but still in harmony with the structure defined by \( M \). Understanding this concept helps us appreciate the stability and structure in the study of algebraic systems such as rings and modules.
Simple Modules and Simplicity
A simple module has a minimalistic beauty. It's a module that can't be broken down into smaller, non-trivial pieces, much like prime numbers in the realm of integers. In the language of modules, a simple module \( M \) does not have any submodules, except for the trivial ones: the module itself, and the zero module (just containing the zero element).
A module \( M \) declared as 'simple' means it's indivisible in terms of module theory; there are no little chunks (submodules) you can take out of it except for \( M \) and \{0\}. This concept is significant because it echoes the idea of atoms in chemistry – indivisible entities that form the building blocks of a compound. In a ring-theoretic sense, simple modules are the building blocks of module theory. They offer a foundation from which more complex module structures can be understood.
If you are trying to visualize this, imagine a piece of paper (\( M \)) that cannot be torn into smaller pieces, without either keeping the whole paper intact or ending up with nothing, not even a speck. Simple modules have a pure, unbreakable algebraic structure, and thus play a crucial role in the structure theory of modules.
A module \( M \) declared as 'simple' means it's indivisible in terms of module theory; there are no little chunks (submodules) you can take out of it except for \( M \) and \{0\}. This concept is significant because it echoes the idea of atoms in chemistry – indivisible entities that form the building blocks of a compound. In a ring-theoretic sense, simple modules are the building blocks of module theory. They offer a foundation from which more complex module structures can be understood.
If you are trying to visualize this, imagine a piece of paper (\( M \)) that cannot be torn into smaller pieces, without either keeping the whole paper intact or ending up with nothing, not even a speck. Simple modules have a pure, unbreakable algebraic structure, and thus play a crucial role in the structure theory of modules.
Left Ideals and Module Actions
Imagine you have a set where you're allowed to multiply its elements by elements from a ring on the left side – this set would be called a 'left ideal'. Precisely, a left ideal \( L \) in a ring \( R \) is a subset of \( R \) that absorbs the multiplication by the elements of \( R \) from the left, meaning for any \( r \in R \) and \( l \in L \), the product \( rl \) stays in \( L \). To connect it to our earlier topic, just like a submodule must respect module operations, a left ideal must respect the ring operations.
Within the context of our exercise, when we talk about \( L M \) where \( L \) is a left ideal and \( M \) is a module, we are considering the set of all possible products between elements of \( L \) and \( M \). This action effectively links the concept of a left ideal with the module, resulting in a new structure that is examined for its properties, such as being a submodule.
Within the context of our exercise, when we talk about \( L M \) where \( L \) is a left ideal and \( M \) is a module, we are considering the set of all possible products between elements of \( L \) and \( M \). This action effectively links the concept of a left ideal with the module, resulting in a new structure that is examined for its properties, such as being a submodule.