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- Jan 26, 2012

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Remember to read the POTW submission guidelines to find out how to submit your answers!

- Thread starter Jameson
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- Thread starter
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- #1

- Jan 26, 2012

- 4,034

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Jun 20, 2014

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I will use two theorems without proof:

1. If $a,b$ are algebraic over $F$, then $F(a,b) = [F(a)](b)$.

2. If $F \leq K \leq E$ are fields, with $\{u_i\}$ a basis for $K$ over $F$ and $\{v_j\}$ a basis for $E$ over $K$, then a basis for $E$ over $F$ is $\{u_iv_j\}$.

Together, these imply $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ is a basis for $\Bbb Q(\sqrt{2},\sqrt{3})$ (this hinges on the fact that $x^2 - 2$ is irreducible over $\Bbb Q$; that is, $\sqrt{2}$ is irrational, and that $x^2 - 3$ is irreducible over $\Bbb Q(\sqrt{2})$, which can be seen by squaring:

If $(a + b\sqrt{2})^2 = 3$, with $a,b \in \Bbb Q$, then: $a^2 + 2b^2 = 3$ and $2ab\sqrt{2} = 0$. The second equation tells us $ab = 0$, which then means either $a^2 = 3$ (but $\sqrt{3} \not\in \Bbb Q$), or $2b^2 = 3$, which is likewise impossible (since: $\dfrac{\sqrt{3}}{\sqrt{2}} \not\in \Bbb Q$).).

Now the function $L: \Bbb Q(\sqrt{2},\sqrt{3}) \to \Bbb Q(\sqrt{2},\sqrt{3})$ given by $\alpha \mapsto (\sqrt{2} + \sqrt{3})\alpha$ is linear, so we can find its matrix relative to the basis above, which is:

$[L] = \begin{bmatrix}0&2&3&0\\1&0&0&3\\1&0&0&2\\0&1&1&0\end{bmatrix}$

The POTW asks us to find the trace and determinant of $L$ (which are invariant under a change of basis). The trace is immediately seen to be $0$, the determinant is a bit harder to compute, but expansion by minors gives us:

$\det(L) = (-1)\begin{vmatrix}2&3&0\\0&0&2\\1&1&0\end{vmatrix} + \begin{vmatrix}2&3&0\\0&0&3\\1&1&0\end{vmatrix}$

$= (-1)(6-4) + (9-6) = -2 + 3 = 1$.

Alternatively, if $G = \text{Gal}(\Bbb Q(\sqrt{2},\sqrt{3})/\Bbb Q)$, then if, for $\alpha \in \Bbb Q(\sqrt{2},\sqrt{3})$ we have:

$\text{Trace}(\alpha) = \sum\limits_{\sigma \in G} \sigma(\alpha)$

and:

$\text{Norm}(\alpha) = \prod\limits_{\sigma \in G} \sigma(\alpha)$

(It should be evident that $\Bbb Q(\sqrt{2},\sqrt{3})$ is a Galois extension, since it is the splitting field of $(x^2 - 2)(x^2 - 3)$),

then since the elements of $G$ are determined by: $\sigma(\sqrt{2}),\sigma(\sqrt{3})$ and we must have (since $\sigma$ fixes $\Bbb Q$)

$\sigma(\sqrt{2})^2 - 2 = 0$

$\sigma(\sqrt{3})^2 - 3 = 0$

the elements of $G$ are seen to be:

$\sigma_1 = \text{id}$

$\sigma_2(\sqrt{2}) = -\sqrt{2}, \sigma_2(\sqrt{3}) = \sqrt{3}$ (this fixes $\Bbb Q(\sqrt{3})$)

$\sigma_3(\sqrt{2}) = \sqrt{2}, \sigma_3(\sqrt{3}) = -\sqrt{3}$ (this fixes $\Bbb Q(\sqrt{2})$)

$\sigma_4(\sqrt{2}) = -\sqrt{2}, \sigma_3(\sqrt{3}) = -\sqrt{3}$ (this fixes $\Bbb Q(\sqrt{6})$).

We have 3 automorphisms of order 2, so this group is isomorphic to $V$, the Klein $4$-group.

The trace is thus computed to be:

$(\sqrt{2} + \sqrt{3}) + (-\sqrt{2} + \sqrt{3}) + (\sqrt{2} - \sqrt{3}) + (-\sqrt{2} - \sqrt{3}) = 0$,

while the norm is seen to be:

$(\sqrt{2} + \sqrt{3})(-\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})(-\sqrt{2} - \sqrt{3})$

$= (-2 + 3)(-2 + 3) = 1$.

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