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Is $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$?

If $\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\mid a,b,c,d\in\mathbf{Q}\}$ and $\mathbf{Q}(\sqrt{6})= \{a+b\sqrt{6} | a,b \in \mathbf{Q}\}$

Assume an element in $\mathbf{Q}(\sqrt{6})$ , then obviously it is also in $\mathbf{Q}(\sqrt{2},\sqrt{3})$.

Assume an element in $\mathbf{Q}(\sqrt{2},\sqrt{3})$ , because we can write: $\sqrt{6} = \sqrt{3} \cdot \sqrt{2} $ and $\sqrt{6} = \sqrt{2} \cdot \sqrt{3}$ the sum of $b\sqrt{3} + c\sqrt{2}$ is equal to a fraction of $\sqrt{6}$.

So $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$.

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    $\begingroup$ Hmmm, $4\sqrt{3}\neq \sqrt{6}$. $\endgroup$ – Álvaro Lozano-Robledo Dec 22 '11 at 13:58
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    $\begingroup$ Alvaro's answer is good. Basically what you showed is that $\sqrt{6} \in \mathbb{Q}(\sqrt{2}, \sqrt{3})$, which translates to $\mathbb{Q}(\sqrt{6}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$. To get equality, you would need to show $\sqrt{2}, \sqrt{3} \in \mathbb{Q}(\sqrt{6})$, i.e. $\mathbb{Q}(\sqrt{6}) \supseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$. That would give you equality. But, you did not show that and Alvaro's answer shows it is impossible. $\endgroup$ – Graphth Dec 22 '11 at 14:16
  • $\begingroup$ Related: math.stackexchange.com/questions/93453 $\endgroup$ – Watson Nov 26 '18 at 10:04
  • $\begingroup$ Possible duplicate of Is $\mathbb{Q}(\sqrt{2}, \sqrt{5})=\mathbb{Q}(\sqrt{10})$ in field extension? $\endgroup$ – Empty Nov 26 '18 at 15:52
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Suppose $\sqrt{2}\in \mathbb{Q}(\sqrt{6})$. Then there are rational numbers $a,b\in\mathbb{Q}$ such that $a+b\sqrt{6}=\sqrt{2}$. If $a=0$ but $b\neq 0$, then $\sqrt{3}=1/b\in\mathbb{Q}$. That's impossible, as we know that $\sqrt{3}$ is irrational. Similarly, if $b=0$ and $a\neq 0$, then $\sqrt{2}=a$, and again we reach a contradiction. Otherwise, assume $ab\neq 0$. Thus, $$2 = a^2+6b^2+2ab\sqrt{6}.$$ In particular, $$\sqrt{6} = \frac{2-a^2-6b^2}{2ab},$$ so $\sqrt{6}\in \mathbb{Q}$. This is absurd, so we have reached a contradiction. Our original assumption $\sqrt{2}\in \mathbb{Q}(\sqrt{6})$ was false, and $\mathbb{Q}(\sqrt{2})\not\subseteq \mathbb{Q}(\sqrt{6})$. In particular, $\mathbb{Q}(\sqrt{2},\sqrt{3})\neq \mathbb{Q}(\sqrt{6})$.

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Hint $\ $ It is an immediate consequence of the following more general lemma, which is the basis of a general result on linear independence of square roots due to Besicovitch (see below).

Lemma $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\:b}\ $ all are not in $\rm\:K\:$ and $\rm\: 2 \ne 0\:$ in $\rm\:K$

Proof $\ \ $ Let $\rm\ L = K(\sqrt{b}).\,$ $\rm\, [L:K] = 2\:$ via $\rm\:\sqrt{b} \not\in K,\,$ so it is suffices to prove $\rm\: [L(\sqrt{a}):L] = 2.\:$ It fails only if $\rm\:\sqrt{a} \in L = K(\sqrt{b}).\, $ Then $\rm\ \sqrt{a}\ =\ r + s\ \sqrt{b}\ $ for $\rm\ r,s\in K.\:$ But that is impossible since squaring yields $\ \rm\color{#c00}{(1)}:\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\ \sqrt{b}\:,\: $ contra hypotheses, as follows

$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \ \sqrt{b}\ \in\ K\ \ $ by solving $\,\color{#c00}{(1)}\,$ for $\rm\sqrt{b}\:,\:$ using $\rm\:2 \ne 0$

$\rm\qquad\qquad s = 0\ \ \Rightarrow\ \ \ \sqrt{a}\ \in\ K\ \ $ via $\rm\ \sqrt{a}\ =\ r \in K$

$\rm\qquad\qquad r = 0\ \ \Rightarrow\ \ \sqrt{a\:b}\in K\ \ $ via $\rm\ \sqrt{a}\ =\ s\ \sqrt{b},\: $ times $\rm\:\sqrt{b}\quad\,$ QED


Remark $\ $ Using the above as the inductive step one easily proves the following

Theorem $\ $ Let $\rm\:Q\:$ be a field with $2 \ne 0\:,\:$ and $\rm\ L = Q(S)\ $ be an extension of $\rm\:Q\:$ generated by $\rm\: n\:$ square roots $\rm\ S = \{ \sqrt{a}, \sqrt{b},\ldots \}$ of elts $\rm\ a,\:b,\:\ldots \in Q\:.\:$ If every nonempty subset of $\rm\:S\:$ has product not in $\rm\:Q\:$ then each successive adjunction $\rm\ Q(\sqrt{a}),\ Q(\sqrt{a},\:\sqrt{b}),\:\ldots$ doubles the degree over $\rm\:Q\:,\:$ so, in total, $\rm\: [L:Q] \ =\ 2^n.\:$ Hence the $\rm\:2^n\:$ subproducts of the product of $\rm\:S\:$ comprise a basis of $\rm\:L\:$ over $\rm\:Q.$

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