Mapping on Quadratic Integers over 3 to Conjugate is Automorphism

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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

that is, all numbers of the form $a + b \sqrt 3$ where $a$ and $b$ are integers.


Then the mapping $\phi: \Z \sqbrk {\sqrt 3} \to \Z \sqbrk {\sqrt 3}$ defined as:

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

is a ring automorphism.


Proof

We have Quadratic Integers over 2 form Subdomain of Reals.

First we note that:

$\forall x \in \Z \sqbrk {\sqrt 3}: \map \phi x \in \Z \sqbrk {\sqrt 3}$


Proof of Bijection

Let $\map \phi {a + b \sqrt 3} = \map \phi {a' + b' \sqrt 3}$.

Then:

$a - b \sqrt 3 = a' - b' \sqrt 3$

and so:

$a + b \sqrt 3 = a' + b' \sqrt 3$

So $\phi$ is injective.


Now let $y = c + d \sqrt 3 \in \Z \sqbrk {\sqrt 3}$.

We have that:

\(\ds \map \phi {c + \paren {-d} \sqrt 3}\) \(=\) \(\ds c - \paren {-d} \sqrt 3\)
\(\ds \) \(=\) \(\ds c + d \sqrt 3\)


Hence:

$\forall y \in \Z \sqbrk {\sqrt 3}: \exists x \in \Z \sqbrk {\sqrt 3}: \map \phi x = y$

and so $\phi$ is surjective.

So $\phi$ is a bijection.


Proof of Morphism

Now consider $x = a + b \sqrt 3, y = c + d \sqrt 3$.

\(\ds \map \phi x + \map \phi y\) \(=\) \(\ds \map \phi {a + b \sqrt 3} + \map \phi {c + d \sqrt 3}\)
\(\ds \) \(=\) \(\ds \paren {a - b \sqrt 3} + \paren {c - d \sqrt 3}\) Definition of $\phi$
\(\ds \) \(=\) \(\ds \paren {a + c} - \paren {b + d} \sqrt 3\)
\(\ds \) \(=\) \(\ds \map \phi {\paren {a + c} + \paren {b + d} \sqrt 3}\) Definition of $\phi$
\(\ds \) \(=\) \(\ds \map \phi {x + y}\)


Then:

\(\ds \map \phi x \map \phi y\) \(=\) \(\ds \map \phi {a + b \sqrt 3} \map \phi {c + d \sqrt 3}\)
\(\ds \) \(=\) \(\ds \paren {a - b \sqrt 3} \paren {c - d \sqrt 3}\) Definition of $\phi$
\(\ds \) \(=\) \(\ds \paren {a c + 2 b d} - \paren {b c + a d} \sqrt 3\)
\(\ds \) \(=\) \(\ds \map \phi {\paren {a c + 2 b d} + \paren {b c + a d} \sqrt 3}\) Definition of $\phi$
\(\ds \) \(=\) \(\ds \map \phi {x y}\)


So $\phi$ is a ring homomorphism which is also a bijection, whose image equals its domain.

Hence the result by definition of ring automorphism.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Mapping_on_Quadratic_Integers_over_3_to_Conjugate_is_Automorphism&oldid=544111"