# automorphisms

• November 29th 2009, 11:05 AM
Sampras
automorphisms
Suppose $E = \mathbb{Q}(\sqrt{2}, \sqrt{3})$. Does the map defined by $\sigma(a+b \sqrt{2} + c \sqrt{3}+d \sqrt{6}) = a+ b\sqrt{2}-c \sqrt{3}- d \sqrt{6}$ leave something fixed?

First of all, $E = \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is the set $\mathbb{Q}$ with the elements $\sqrt{2}$ and $\sqrt{3}$ added in? It is an autmomorphism of E?
• November 29th 2009, 06:41 PM
aliceinwonderland
Quote:

Originally Posted by Sampras
Suppose $E = \mathbb{Q}(\sqrt{2}, \sqrt{3})$. Does the map defined by $\sigma(a+b \sqrt{2} + c \sqrt{3}+d \sqrt{6}) = a+ b\sqrt{2}-c \sqrt{3}- d \sqrt{6}$ leave something fixed?

First of all, $E = \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is the set $\mathbb{Q}$ with the elements $\sqrt{2}$ and $\sqrt{3}$ added in? It is an autmomorphism of E?

$\sigma$ leaves $\mathbb{Q}(\sqrt{2})$ fixed. If F is a field and if $\alpha$ and $\beta$ is algebraic over F with $deg(\alpha, \beta)=n$, the map $\psi_{\alpha, \beta}:F(\alpha) \rightarrow F(\beta)$ defined by

$\psi_{\alpha, \beta}(c_0 + c_1\alpha+ \cdots +c_{n-1}\alpha^{n-1})= c_0 +c_1\beta + \cdots +c_{n-1}\beta^{n-1}$

is an isomorphism if and only if $\alpha$ and $\beta$ are conjugate over F where $c_i \in F$. In your example $\sigma$ can be thought of as $\psi_{\sqrt{3}, -\sqrt{3}}$ and it is an automorphism in E.

$E = \mathbb{Q}(\sqrt{2}, \sqrt{3})$ can be thought of as an extension field (actually it is an algebraic extension) of a field $\mathbb{Q}$ adjoining to F the elements $\sqrt{2}$ and $\sqrt{3}$ that are not elements of $\mathbb{Q}$.
$E = \mathbb{Q}(\sqrt{2}, \sqrt{3})$ can also be thought of as a splitting field of $\{x^4 -5x^2 + 6\}$.