[back] [prev] [next] [index] [root]

<p>&nbsp;<p>
<hr noshade size="5">
<h2><a name="EltAutomorphism"></a>
EltAutomorphism
</h2>

Applies an automorphism to an algebraic number.

<p><p>

<h3>
Syntax:
</h3>
<pre>L := EltAutomorphism (a);
c := EltAutomorphism (a,i);</pre>

<p>

<table>
<tr>
<td>list</td>
<td><pre>  L  </pre></td>
<td></td>
</tr>
<tr>
<td>algebraic element</td>
<td><pre>  c  </pre></td>
<td></td>
</tr>
<tr>
<td>algebraic element</td>
<td><pre>  a  </pre></td>
<td></td>
</tr>
<tr>
<td>integer</td>
<td><pre>  i  </pre></td>
<td></td>
</tr>
</table>

<p>


See also:&nbsp;&nbsp;<a href="kashref.html#OrderAutomorphisms">OrderAutomorphisms</a>

<p>

<h3>
Description:
</h3>
Let F be a normal number field with
Q-automorphisms \sigma_1, &hellip; ,\sigma_n.
Let a be an algebraic number in F.\bigskip

<tt> EltAutomorphism(a)</tt>
returns the list <tt> L</tt> = { \sigma_1(a), &hellip; ,\sigma_n(a)}.
Each entry of <tt> L</tt> is an algebraic number whose
underlying order is the same order as a is defined over.\medskip

<tt> EltAutomorphism(a,i)</tt>
returns the algebraic number \sigma_i(a). Note that \sigma_1
is always the identity mapping of F. \bigskip

Before calling <tt> EltAutomorphism</tt>, the automorphisms
\sigma_1, &hellip; ,\sigma_n of F must be computed by the
<tt> OrderAutomorphisms</tt> routine.

<p><p>

<p>
<hr>
<p>
<h3>
Example:
</h3>
Consider the normal field Q(\sqrt[6]{-108}) : 
<pre>

kash> o := Order(Z,6,-108);
Generating polynomial: x^6 + 108

kash> OrderAutomorphisms(o);;
kash> a := Elt(o,[1,2,3,4,5,6]);
[1, 2, 3, 4, 5, 6]
kash> A := EltAutomorphism(a);
[ [1, 2, 3, 4, 5, 6], [12, -552, 630, 48, -28, -39] / 12, 
  [12, 528, -666, 48, -32, -33] / 12, [1, -2, 3, -4, 5, -6], 
  [12, 552, 630, -48, -28, 39] / 12, [12, -528, -666, -48, -32, 33] / 12 ]
kash> A[1]+A[2]+A[3]+A[4]+A[5]+A[6];
6
kash> EltTrace(a);
6
kash> EltAutomorphism(a,1);
> [1, 2, 3, 4, 5, 6]

</pre>

<p>


<hr>

<p>
<a href="../kashref.html"><- back</a>[back] [prev] [next] [index] [root]