Commit c2b12809 authored by velleto's avatar velleto
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Accepted heldf's MR.

parents d017b45a a779f9c8
......@@ -59,7 +59,7 @@ We can easily prove that $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}$ is 4
\sqrt{3}
& = x + y\sqrt{2} \\
\implies 3
& = x^2 + 2y^2 + \sqrt{2}\underset{= 0}{\underbrace{xy}}
& = x^2 + 2y^2 + 2\sqrt{2}\underset{= 0}{\underbrace{xy}}
\end{align*}
We know that $xy = 0$ as otherwise the RHS contains irrationals whilst the LHS does not. If, however $xy = 0$, then either $x = 0 \implies y = \sqrt{\frac{3}{2}} \notin \mathbb{Q}$ or $y = 0 \implies x = \sqrt{3} \notin \mathbb{Q}$. Both contradictions. We can do the same for $\mathbb{Q}(\sqrt{2}) \not\subseteq \mathbb{Q}(\sqrt{3})$. Hence we can state that, $\mathbb{Q}(\sqrt{3}, \sqrt{2})$ is an extension over $\mathbb{Q}(\sqrt{2}$, which in turn is an extension over $\mathbb{Q}$, to which we can apply the dimension formula to arrive at our result. $\blacksquare$
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