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Magazine Chapter 4: Problem 4
If \(T\) is a homomorphism of \(M\) into \(N\) let \(K(T)=|x \in M| x T=0\\}\). Prove that \(K(T)\) is a submodule of \(M\) and that \(I(T)=[x T \mid x \in M\\}\) is a submodule of \(N\).
Short Answer
Expert verified
Both \(K(T)\) and \(I(T)\) are submodules of \(M\) and \(N\), respectively.
Step by step solution
01
Understanding the Kernel
The kernel of a homomorphism, denoted as \(K(T)\), is the set of elements in module \(M\) that are mapped to zero in module \(N\). That is, \(K(T) = \{x \in M \mid xT = 0\}\). Our goal is to show that \(K(T)\) forms a submodule of \(M\).
02
Showing the Kernel is Non-empty
To confirm \(K(T)\) is a submodule, we first prove it is non-empty. The zero element \(0_M\) of \(M\), when applied to the homomorphism \(T\), maps to \(0_N\) since \(0_M T = 0_N\) for any homomorphism. Therefore, \(0_M \in K(T)\), indicating \(K(T)\) is non-empty.
03
Closure Under Addition
Next, we show closure under addition. Assume \(a, b \in K(T)\), which means \(aT = 0\) and \(bT = 0\). For the sum \((a + b)T\), we have \((a + b)T = aT + bT = 0 + 0 = 0\). Hence, \(a + b \in K(T)\).
04
Closure Under Scalar Multiplication
For closure under scalar multiplication, take any scalar \(r\) from the ring over which module \(M\) is defined. If \(a \in K(T)\), then \(aT = 0\). So, \((ra)T = r(aT) = r \cdot 0 = 0\). Thus, \(ra \in K(T)\).
05
Understanding the Image
The image of the homomorphism, denoted as \(I(T)\), is the set \(\{xT \mid x \in M\}\). We need to show \(I(T)\) forms a submodule of \(N\).
06
Showing the Image is Non-empty
To show \(I(T)\) is non-empty, we note that the zero element is always in the image because \(0_M T = 0_N\), thus \(0_N \in I(T)\).
07
Closure of Image Under Addition
For closure under addition, assume \(aT, bT \in I(T)\) for some \(a, b \in M\). Then \(aT + bT = (a + b)T\) is also in \(I(T)\). Thus, \(I(T)\) is closed under addition.
08
Closure of Image Under Scalar Multiplication
For closure under scalar multiplication, let \(r\) be a scalar and \(aT \in I(T)\). Then \(r(aT) = (ra)T\), which is in \(I(T)\) because \(ra \in M\). Thus, \(I(T)\) is closed under scalar multiplication.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Kernel of a Homomorphism
The kernel of a homomorphism is an essential concept in understanding how modules map between each other. It's particularly useful when analyzing the internal structure of a module. Given a homomorphism \( T \) from a module \( M \) to a module \( N \), the kernel, denoted as \( K(T) \), comprises all elements \( x \) in \( M \) that are transformed to zero in \( N \) through the homomorphism.
Here's why \( K(T) \) forms a submodule of \( M \):
Here's why \( K(T) \) forms a submodule of \( M \):
- Non-empty set: At the very least, the zero element \( 0_M \) from \( M \), when mapped by \( T \), results in the zero element \( 0_N \) in \( N \). Thus, \( 0_M \) is in \( K(T) \), confirming its non-emptiness.
- Closure under addition: For any elements \( a, b \in K(T) \), since \( aT = 0 \) and \( bT = 0 \), their sum also maps to zero, i.e., \( (a + b)T = 0 \). Thus, \( a + b \in K(T) \).
- Closure under scalar multiplication: For any scalar \( r \) and any element \( a \in K(T) \), the product \( ra \) also maps to zero, as \( (ra)T = r(aT) = r \cdot 0 = 0 \). Therefore, \( ra \in K(T) \).
Image of a Homomorphism
The image of a homomorphism \( T : M \rightarrow N \), indicated as \( I(T) \), is another fundamental concept. It focuses on understanding what subset of \( N \) contains those elements that are actually images of elements from \( M \).
Specifically, the image \( I(T) \) is the collection of all values \( xT \) for every \( x \in M \). Let's see why \( I(T) \) is a submodule of \( N \):
Specifically, the image \( I(T) \) is the collection of all values \( xT \) for every \( x \in M \). Let's see why \( I(T) \) is a submodule of \( N \):
- Non-empty property: Due to the nature of homomorphisms, \( 0_N \) always resides in the image because \( 0_M T = 0_N \).
- Closure under addition: If \( aT \) and \( bT \) are in \( I(T) \), then \( aT + bT = (a + b)T \) also belongs to \( I(T) \), since \( a + b \in M \).
- Closure under scalar multiplication: Given a scalar \( r \) and an element \( aT \) in \( I(T) \), the scalar product \( r(aT) = (ra)T \) is in \( I(T) \), because it originates from an element \( ra \in M \).
Submodule
Submodules are integral in examining the structure within modules, akin to the role subgroups play in group theory. A submodule is essentially a set within a module that is closed under the operations of addition and scalar multiplication defined on the module.
To differentiate a submodule from just any subset, a few conditions must be satisfied:
To differentiate a submodule from just any subset, a few conditions must be satisfied:
- A submodule must be non-empty, often confirmed by the existence of the zero element in the set.
- It must be closed under addition, meaning if you take any two elements within it and add them, the result must also lie within the submodule.
- The submodule must also be closed under scalar multiplication; for any element in the submodule, multiplying by a scalar should yield another element of the submodule.
