Exercise 5.1.5

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

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ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

We showed in Section 4.1 that $f =x^4-10x^2+1$ is irreducible over $\Q$. Show that $L = \Q(\sqrt{2}+ \sqrt{3})$ is the splitting field of $f$ over $\Q$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Recall the computing of Exercise 4.1.8(b) : 
\begin{align*}
 f&= (x-\sqrt{2} - \sqrt{3})(x+\sqrt{2} -\sqrt{3}) (x-\sqrt{2} + \sqrt{3})(x+\sqrt{2} + \sqrt{3})\
&=[(x-\sqrt{3})^2-2][(x+\sqrt{3})^2-2]\
&= (x^2 -2\sqrt{3} x +1)(x^2 -2\sqrt{3} x +1)\
&= (x^2+1)^2 - (2\sqrt{3}x)^2\
&=x^4 -10x^2 +1
\end{align*}

The splitting field of $f$ over $\Q$ is thus $$K = \Q(\sqrt{2} + \sqrt{3},\sqrt{2} - \sqrt{3},-\sqrt{2} + \sqrt{3},-\sqrt{2} + \sqrt{3}).$$

As  $\sqrt{2} + \sqrt{3},\sqrt{2} - \sqrt{3},-\sqrt{2} + \sqrt{3},-\sqrt{2} + \sqrt{3} \in  \Q(\sqrt{2},\sqrt{3})$, then $$K \subset \Q(\sqrt{2},\sqrt{3}).$$

Moreover, 
\begin{align*}
\sqrt{2} &= \frac{1}{2}\left[(\sqrt{2}+\sqrt{3}) - (-\sqrt{2}+\sqrt{3})ight] \in K,\
 \sqrt{3} &= \frac{1}{2}\left[(\sqrt{2}+\sqrt{3}) - (\sqrt{2}-\sqrt{3}) ight]\in K,
 \end{align*}
thus $$\Q(\sqrt{2},\sqrt{3}) \subset K.$$

So $K = \Q(\sqrt{2},\sqrt{3})$. Moreover, the Example 4.3.9 shows that $\Q(\sqrt{2},\sqrt{3}) = \Q(\sqrt{2}+\sqrt{3})$.

(Or a direct proof is given in  section 4.2, since $\sqrt{2} = \frac{1}{2}(\alpha^3 - 9 \alpha)$, where $\alpha = \sqrt{2}+\sqrt{3}$, and $\sqrt{3} = \alpha - \sqrt{2}$, so $\sqrt{2},\sqrt{3} \in \Q(\sqrt{2}+\sqrt{3})$.)

Conclusion: the splitting field of $x^4-10x^2+1$ over $\Q$ is $\Q(\sqrt{2}+\sqrt{3})$.
\end{proof}

Exercise 5.1.5

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Exercise 5.1.5

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