Problem 9

Question

Prove the first isomorphism theorem of algebras: A homomorphism $\sigma: A \rightarrow B$ of $F$-algebras induces an isomorphism $\bar{\sigma}: A / \operatorname{ker}(\sigma) \approx \operatorname{im}(\sigma)$ defined by $\bar{\sigma}(a \operatorname{ker}(\sigma))=\sigma a$.

Step-by-Step Solution

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Answer

To prove the first isomorphism theorem, we showed that the function $\bar{\sigma}: A /\operatorname{ker}(\sigma) \rightarrow \operatorname{im}(\sigma)$ defined by $\bar{\sigma}(a \operatorname{ker}(\sigma))=\sigma a$ is an isomorphism by establishing that $\bar{\sigma}$ is well-defined, a homomorphism, injective, and surjective. Thus, $\bar{\sigma}: A / \operatorname{ker}(\sigma) \approx \operatorname{im}(\sigma)$ is an isomorphism, proving the first isomorphism theorem of algebras.

1Step 1: Show $\bar{\sigma}$ is well-defined

To show that $\bar{\sigma}$ is well-defined, let $a\operatorname{ker}(\sigma) = b\operatorname{ker}(\sigma)$. This means that $a^{-1}b \in \operatorname{ker}(\sigma)$. Since $a^{-1}b$ is in the kernel of $\sigma$, we have: \[\sigma(a^{-1}b) = 1_B\] Now, we need to show that $\sigma a = \sigma b$. Using the properties of a homomorphism: \[\sigma a = \sigma a \cdot 1_B = \sigma a \cdot \sigma(a^{-1}b) = \sigma(aa^{-1}b) = \sigma b\] Thus, if $a\operatorname{ker}(\sigma) = b\operatorname{ker}(\sigma)$, then $\sigma a = \sigma b$, proving that $\bar{\sigma}$ is well-defined.

2Step 2: Show $\bar{\sigma}$ is a homomorphism

To show that $\bar{\sigma}$ is a homomorphism, we need to show that it preserves the $F$-algebra operations. Let $a\operatorname{ker}(\sigma), b\operatorname{ker}(\sigma) \in A/\operatorname{ker}(\sigma)$. Then, we have: \[\bar{\sigma}((a\operatorname{ker}(\sigma)) \cdot (b\operatorname{ker}(\sigma))) = \bar{\sigma}(ab\operatorname{ker}(\sigma)) = \sigma (ab) = \sigma a \cdot \sigma b = \bar{\sigma}(a\operatorname{ker}(\sigma)) \cdot \bar{\sigma}(b\operatorname{ker}(\sigma))\] This demonstrates that $\bar{\sigma}$ preserves the algebra operation and is an $F$-algebra homomorphism.

3Step 3: Show $\bar{\sigma}$ is injective

To show that $\bar{\sigma}$ is injective, let $\bar{\sigma}(a\operatorname{ker}(\sigma)) = \bar{\sigma}(b\operatorname{ker}(\sigma))$. This means that: \[\sigma a = \sigma b\] Hence, $\sigma(a^{-1}b) = \sigma a^{-1} \sigma b = 1_B$. Since $a^{-1}b \in \operatorname{ker}(\sigma)$, we have: \[a\operatorname{ker}(\sigma) = b\operatorname{ker}(\sigma)\] This shows that $\bar{\sigma}$ is injective.

4Step 4: Show $\bar{\sigma}$ is surjective

To show that $\bar{\sigma}$ is surjective, let $b \in \operatorname{im}(\sigma)$. This means that there exists an $a \in A$ such that $\sigma a = b$. We then have: \[\bar{\sigma}(a\operatorname{ker}(\sigma)) = \sigma a = b\] Thus, for any $b \in \operatorname{im}(\sigma)$, there exists an element in $A/\operatorname{ker}(\sigma)$ that maps to $b$ under $\bar{\sigma}$. This shows that $\bar{\sigma}$ is surjective. Since $\bar{\sigma}$ is well-defined, a homomorphism, injective, and surjective, we can conclude that $\bar{\sigma}$ is an isomorphism, proving the first isomorphism theorem of algebras.

Key Concepts

F-algebra homomorphismkernel and imageisomorphism proof

F-algebra homomorphism

An $F$-algebra homomorphism is a special type of function between two $F$-algebras. This concept is foundational in understanding how algebraic structures can relate to one another. An $F$-algebra itself is an algebraic structure equipped with operations like addition, multiplication, and scalar multiplication by elements of a field $F$.

To classify a function as an $F$-algebra homomorphism, it must preserve these operations between algebras. Specifically, for a function $\sigma: A \rightarrow B$ to be a homomorphism:

$\sigma(a + b) = \sigma(a) + \sigma(b)$ for addition
$\sigma(ab) = \sigma(a)\sigma(b)$ for multiplication
$\sigma(ca) = c\sigma(a)$ for scalar multiplication

These requirements ensure that the algebraic structure remains consistent through the mapping.

In the context of the first isomorphism theorem, this homomorphism $\sigma$ from $A$ to $B$ becomes the central focus. We use it to explore deeper aspects like the kernel and image, which helps to establish an isomorphism between two related algebraic structures.

kernel and image

The kernel and image are pivotal concepts when analyzing the behavior of $F$-algebra homomorphisms. They help reveal a lot about the structure of the homomorphism $\sigma: A \rightarrow B$.

Kernel: The kernel of a homomorphism $\sigma$, denoted as $\operatorname{ker}(\sigma)$, consists of all elements in the domain $A$ that map to the zero element in $B$. Formally, $\operatorname{ker}(\sigma) = \{a \in A \mid \sigma(a) = 0\} $. The kernel measures how the homomorphism fails to be injective, and it always forms an ideal within the algebra.
Image: The image of a homomorphism, represented as $\operatorname{im}(\sigma)$, is the set of all elements in $B$ that have a pre-image in $A$. Put simply, $\operatorname{im}(\sigma) = \{\sigma(a) \mid a \in A\}$. The image reflects the part of $B$ that can be mapped from $A$, indicating how $\sigma$ translates elements across the algebras.

Understanding these concepts is key to leveraging the first isomorphism theorem, which involves quotienting $A$ by the kernel to effectively "strip away" the non-trivial part that collapses under $\sigma$, leaving a structure isomorphic to the image.

isomorphism proof

The isomorphism proof tied to the first isomorphism theorem is a fascinating exploration of mapping and equivalence between algebraic structures. Here, we are tasked with demonstrating that $\bar{\sigma}: A / \operatorname{ker}(\sigma) \approx \operatorname{im}(\sigma)$ is an actual isomorphism.

This proof involves several key components:

Well-defined Mapping: To start, $\bar{\sigma}$ must be well-defined. This ensures that each element in the quotient $A / \operatorname{ker}(\sigma)$ maps neatly without ambiguity to $\operatorname{im}(\sigma)$. For each representative in the quotient class, applying $\sigma$ yields the same result.
Homomorphism Properties: Next, $\bar{\sigma}$ must preserve the algebra structure. This means demonstrating $\bar{\sigma}(a \cdot b) = \bar{\sigma}(a) \cdot \bar{\sigma}(b)$ and similar rules for addition and scalar multiplication, confirming it as an $F$-algebra homomorphism.
Being Injective: Proving injectivity requires showing that if $\bar{\sigma}(x) = \bar{\sigma}(y)$, then $x$ must equal $y$. This step ensures a one-to-one correspondence without overlaps in the mapping.
Being Surjective: Finally, surjectivity is shown by demonstrating that every element in $\operatorname{im}(\sigma)$ has a pre-image in $A / \operatorname{ker}(\sigma)$.

Once these points are established, the completeness of $\bar{\sigma}$ as an isomorphism is clear, showcasing the intrinsic symmetry and similarity in structure between the quotient and the image, thus confirming the theorem.

Problem 8

Problem 10

Other exercises in this chapter

Problem 6

Show that $\mathbb{Z}_{6}$ is not an algebra over any field.

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Problem 8

Let $G=\left\\{1=a_{0}, \ldots, a_{n}\right\\}$ be a finite group. For $x \in F[G]$ of the form $x=r_{1} a_{1}+\cdots+r_{n} a_{n}$ let \(T(x)=r_{1}+\cdots

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Problem 10

Prove that the quaternion field is an $F$-algebra and a field. Hint: For $$ x=r_{0}+r_{1} i+r_{2} j+r_{3} k \neq 0 $$ $\left(r_{0}=r_{0} 1\right)$ consider

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Problem 11

Describe the left regular representation of the quaternions using the ordered basis $\mathcal{B}=(1, i, j, k)$.

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