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Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?

Consider the field extension $\mathbb{Q}(\sqrt2)$. I want to show that $\sqrt5 \notin \mathbb{Q}(\sqrt2)$. If this were not the case, then we could write $\sqrt5 = a + b\sqrt2$ for $a,b\in\mathbb{Q}$. However, I do not see the contradiction here. Is there a better/easier way to prove this?

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marked as duplicate by Arturo Magidin, Sasha, Alex Becker, t.b., Asaf Karagila Apr 4 '12 at 22:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I should square both sides of your equation before concluding you've made no progress. – Mark Bennet Apr 4 '12 at 16:22
See this answer for a proof that for squarefree integers $d$ and $d'$, $\mathbb{Q}(\sqrt{d})\cong \mathbb{Q}(\sqrt{d'})$ implies $d=d'$. – Arturo Magidin Apr 4 '12 at 16:44
up vote 4 down vote accepted

HINT $\ $ Since $\sqrt{2},\sqrt{5},\sqrt{10}\not\in\mathbb Q,\:$ it is an immediate consequence of this

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 $\rm\ L = K(\sqrt{b})\:.\:$ Then $\rm\: [L:K] = 2\:$ via $\rm\:\sqrt{b} \not\in K\:,\:$ therefore it is sufficient to prove $\rm\: [L(\sqrt{a}):L] = 2\:.\:$ But this fails only if $\rm\:\sqrt{a} \in L = K(\sqrt{b})\ $ and then $\rm\ \sqrt{a}\ =\ r + s\ \sqrt{b}\ $ for $\rm\ r,s\in K\:.\:$ That's impossible: squaring yields $\rm(1):\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\ \sqrt{b}\:,\: $ contra hypotheses as follows:

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

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

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

Using the above as the inductive step one easily proves the following result of Besicovic.

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 $\rm\:S\:$ form a basis of $\rm\:L\:$ over $\rm\:Q\:.$

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Squaring both sides of $\sqrt5 = a + b\sqrt2$ will imply that $\sqrt2$ is rational, which it is not.

(Actually, you also need to consider the case $b=0$, which would imply $\sqrt5$ is rational, and the case $a=0$, which would imply $\sqrt{\dfrac52}$ is rational.)

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