Mapping on Integers a plus b root 2 to Conjugate is Automorphism

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Theorem

Let $\Z \sqbrk {\sqrt 2}$ denote the set:

$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$

that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.


Then the mapping $\phi: \Z \sqbrk {\sqrt 2} \to \Z \sqbrk {\sqrt 2}$ defined as:

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

is a ring automorphism.


Proof

We have Numbers of Type Integer a plus b root 2 form Subdomain of Reals.

First we note that:

$\forall x \in \Z \sqbrk {\sqrt 2}: \map \phi x \in \Z \sqbrk {\sqrt 2}$


Proof of Bijection

Let $\map \phi {a + b \sqrt 2} = \map \phi {a' + b' \sqrt 2}$.

Then:

$a - b \sqrt 2 = a' - b' \sqrt 2$

and so:

$a + b \sqrt 2 = a' + b' \sqrt 2$

So $\phi$ is injective.


Now let $y = c + d \sqrt 2 \in \Z \sqbrk {\sqrt 2}$.

We have that:

\(\displaystyle \map \phi {c + \paren {-d} \sqrt 2}\) \(=\) \(\displaystyle c - \paren {-d} \sqrt 2\)
\(\displaystyle \) \(=\) \(\displaystyle c + d \sqrt 2\)


Hence:

$\forall y \in \Z \sqbrk {\sqrt 2}: \exists x \in \Z \sqbrk {\sqrt 2}: \map \phi x = y$

and so $\phi$ is surjective.

So $\phi$ is a bijection.


Proof of Morphism

Now consider $x = a + b \sqrt 2, y = c + d \sqrt 2$.

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


Then:

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


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

Hence the result by definition of ring automorphism.

$\blacksquare$


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