Isomorphism (Abstract Algebra)/Examples/Quadratic Integers over 3 with Numbers of Form 2^m 3^n

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Example of Isomorphism

Let $\Z \sqbrk {\sqrt 3}$ denote the set of quadratic integers over $3$:

$\Z \sqbrk {\sqrt 3} = \set {a + b \sqrt 3: a, b \in \Z}$


Let $S$ be the set defined as:

$S := \set {2^m 3^n: m, n \in \Z}$


Let $\struct {\Z \sqbrk {\sqrt 3}, +}$ and $\struct {S, \times}$ be the algebraic structures formed from the above with addition and multiplication respectively.


Then $\struct {\Z \sqbrk {\sqrt 3}, +}$ and $\struct {S, \times}$ are isomorphic.


Proof

Let us define the mapping $\phi: \Z \sqbrk {\sqrt 3} \to S$ as:

$\forall a + b \sqrt 3 \in \Z \sqbrk {\sqrt 3}: \map \phi {a + b \sqrt 3} = 2^a 3^b$


Let $a_1 + b_1 \sqrt 3$ and $a_2 + b_2 \sqrt 3$ be arbitrary elements of $\Z \sqbrk {\sqrt 3}$.

Then we have:

\(\ds \map \phi {\paren {a_1 + b_1 \sqrt 3} + \paren {a_2 + b_2 \sqrt 3} }\) \(=\) \(\ds \map \phi {\paren {a_1 + a_2} + \paren {b_1 + b_2} \sqrt 3}\)
\(\ds \) \(=\) \(\ds 2^{a_1 + a_2} 3^{b_1 + b_2}\) Definition of $\phi$
\(\ds \) \(=\) \(\ds \paren {2^{a_1} 2^{a_2} } \paren {3^{b_1} 3^{b_2} }\) Product of Powers
\(\ds \) \(=\) \(\ds \paren {2^{a_1} 3^{b_1} } \times \paren {2^{a_2} 3^{b_2} }\) Real Multiplication is Commutative and Real Multiplication is Associative
\(\ds \) \(=\) \(\ds \map \phi {a_1 + b_1 \sqrt 3} \times \map \phi {a_2 + b_2 \sqrt 3}\) Definition of $\phi$

This demonstrates that $\phi$ is a homomorphism.


Now we have:

\(\ds \map \phi {a_1 + b_1 \sqrt 3}\) \(=\) \(\ds \map \phi {a_2 + b_2 \sqrt 3}\)
\(\ds \leadsto \ \ \) \(\ds 2^{a_1} 3^{b_1}\) \(=\) \(\ds 2^{a_2} 3^{b_2}\)
\(\ds \leadsto \ \ \) \(\ds 2^{a_1}\) \(=\) \(\ds 2^{a_2}\) as $2$ and $3$ are coprime
\(\, \ds \land \, \) \(\ds 3^{b_1}\) \(=\) \(\ds 3^{b_2}\)
\(\ds \leadsto \ \ \) \(\ds a_1\) \(=\) \(\ds a_2\)
\(\, \ds \land \, \) \(\ds b_1\) \(=\) \(\ds b_2\)
\(\ds \leadsto \ \ \) \(\ds a_1 + b_1 \sqrt 3\) \(=\) \(\ds a_2 + b_2 \sqrt 3\)

demonstrating that $\phi$ is injective.


Then:

\(\ds \forall x \in S: \exists a, b \in \Z: \, \) \(\ds x\) \(=\) \(\ds 2^a 3^b\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \map \phi {a + b \sqrt 3}\)

demonstrating that $\phi$ is surjective.

The result follows by definition of isomorphic.

$\blacksquare$


Sources

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