Commit 7344c8ce authored by heldf's avatar heldf
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Update algebra-II-lecture-2019-03-08.md

Fix of typo

namely a missing ")"
parent e0b8c53b
......@@ -61,7 +61,7 @@ We can easily prove that $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}$ is 4
\implies 3
& = 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. $\blacksquare$
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. $\blacksquare$
We now want to use this information to think about the autmorphism group of the extension of $\mathbb{Q}(\sqrt{2}, \sqrt{3}) / \mathbb{Q}$. Let's go via the extension of the adjoin of only $\sqrt{2}$:
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