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@ -306,10 +306,14 @@
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$\QQ$. In particolare vale che:
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\[ \mu_\alpha(x) = \prod_{i=0}^1 \prod_{j=0}^1 (x + (-1)^i \sqrt{2} + (-1)^j \sqrt{3}) = x^4 - 10x^2 + 1. \]
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In modo analogo si ottengono i polinomi minimi
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di $\sqrt{2} + \sqrt{3}$ su $\QQ(\sqrt2)$,
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$\QQ(\sqrt3)$ e $\QQ(\sqrt6)$, rispettivamente
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$x^2-2\sqrt{2}x-1 = (x-\sqrt{2})^2 - 3$,
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$x^2-2\sqrt{3}x-1 = (x-\sqrt{3})^2 - 2$ e
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$x^2-(\sqrt{2} + \sqrt{3})^2 = x^2 - 2\sqrt{6} - 5$.
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Tutte le informazioni sono infine raccolte nel seguente
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diagramma di estensioni: %TODO: terminare esempio
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diagramma di estensioni:
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\[\begin{tikzcd}[column sep=2.25em]
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&& {\overbrace{\mathbb{Q}(\sqrt{2}, \sqrt{3})}^{\mathbb{Q}(\sqrt{2} + \sqrt3)}} \\
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\\
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@ -326,9 +330,25 @@
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\arrow["{x^2-3}", curve={height=-12pt}, no head, from=3-5, to=5-3]
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\arrow["{x^2-2}"', curve={height=12pt}, no head, from=3-1, to=5-3]
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\arrow["{x^2-6}"'{pos=0.3}, shift left, curve={height=12pt}, no head, from=3-3, to=5-3]
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\arrow["{x^2+2\sqrt3\,x+1}", curve={height=-18pt}, no head, from=1-3, to=3-5]
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\arrow["{x^2+2\sqrt2\,x-1}"', curve={height=18pt}, no head, from=1-3, to=3-1]
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\arrow["{x^2-2\sqrt3\,x+1}", curve={height=-18pt}, no head, from=1-3, to=3-5]
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\arrow["{x^2-2\sqrt2\,x-1}"', curve={height=18pt}, no head, from=1-3, to=3-1]
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\arrow["{\small x^2-2\sqrt6-5}"'{pos=0.8}, curve={height=12pt}, no head, from=1-3, to=3-3]
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\end{tikzcd}\]
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Tramite la corrispondenza di Galois abbiamo fatto
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corrispondere questo diagramma al seguente diagramma
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di gruppi:
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\[\begin{tikzcd}
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& {\{ \text{Id}_L \equiv \varphi_{00} \}} \\
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\\
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{\{ \varphi_{00}, \varphi_{01} \}} & {\{ \varphi_{00}, \varphi_{11} \}} & {\{\varphi_{00}, \varphi_{10}\}} \\
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\\
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& {\{\varphi_{01}, \varphi_{10}, \varphi_{01}, \varphi_{11}\}}
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\arrow[no head, from=1-2, to=3-2]
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\arrow[no head, from=1-2, to=3-1]
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\arrow[no head, from=1-2, to=3-3]
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\arrow[no head, from=3-2, to=5-2]
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\arrow[no head, from=3-3, to=5-2]
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\arrow[no head, from=3-1, to=5-2]
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\end{tikzcd}\]
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\end{example}
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\end{document}
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