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How can I prove that $\sqrt5$ is not in $\mathbb{Q}(\sqrt7)$ ?

I can only think of trying to write $\sqrt5 = a+b\sqrt7$ (where $a,b$ are in $\mathbb{Q}$), but I can't think of a good reason that shoes such $a$ and $b$ doesn't exist.

Any ideas ?

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Sounds like homework to me. Do you remember the proof of $\sqrt{2}\not\in\mathbb{Q}$? – AD. Mar 28 '12 at 21:39
See this answer – Arturo Magidin Mar 28 '12 at 21:39
@AD. of course I do, but I don't see how this helps... – Belgi Mar 28 '12 at 21:44
@Arturo: is there not a closer "duplicate"? I can't recall the specific values (here they are 5,7), but this seems familiar... – The Chaz 2.0 Mar 28 '12 at 22:23
up vote 11 down vote accepted

If $\sqrt{5}\in\mathbb{Q}(\sqrt{7})$, then there exist rationals $a$ and $b$ such that $(a+b\sqrt{7})^2 = 5$. But $$(a+b\sqrt{7})^2 = a^2 + 7b^2 + 2ab\sqrt{7} = 5$$ implies $ab=0$ (since $1,\sqrt{7}$ are linearly independent over $\mathbb{Q}$, since $7$ is not a square), which implies $5=a^2$ or $5=7b^2$, both of which are impossible with $a$ and $b$ rational.

(In fact, not only are they different, they are not even isomorphic)

More generally, if $p$ and $q$ are distinct squarefree integers different from $1$, then $\mathbb{Q}(\sqrt{p})\neq\mathbb{Q}(\sqrt{q})$, since $$(a+b\sqrt{q})^2 = a^2+qb^2 + 2ab\sqrt{q}=p$$ implies $ab=0$, hence $p=a^2$ or $p=qb^2$, and the fact that $p$ is squarefree yields a contradiction.

See also this previous answer covering the same ground, or Bill Dubuque's answer to the same question which shows that for rationals $d$ and $d'$, $\mathbb{Q}(\sqrt{d})=\mathbb{Q}(\sqrt{d'})$ if and only if $d/d'$ is a (rational) square.

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thank you very much, I will accept once those 10 minutes pass. – Belgi Mar 28 '12 at 21:44
@Belgi Note that the proof requires only that $5,7,35$ are not squares of rationals (for the last $5 = 7b^2\:$ $\!\iff\!$ $35 = (7b)^2$). This method works generally - see my answer. – Bill Dubuque Mar 28 '12 at 21:51
@Arturo It would be helpful to explicitly state that your inference $ab = 0$ follows from $\sqrt{7}\not\in \mathbb Q$. This allows readers to more easily see the connection between your proof and mine. – Bill Dubuque Mar 28 '12 at 22:07

Hint $\ $ Apply the following lemma, noting $\sqrt{5},\sqrt{7},\sqrt{35}$ all $\rm\not\in K = \mathbb Q$

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

Proof $\ \ $ Let $\rm\ L = K(\sqrt{b})\:.\:$ $\rm\: [L:K] = 2\:$ by $\rm\:\sqrt{b} \not\in K,\:$ so it suffices to show $\rm\: [L(\sqrt{a}):L] = 2\:.\:$ This fails only if $\rm\:\sqrt{a} \in L = K(\sqrt{b})$ $\:\Rightarrow\:$ $\rm \sqrt{a}\ =\ r + s\ \sqrt{b}\ $ for $\rm\ r,s\in K,\:$ which is false, because squaring yields $\rm\:(1):\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\ \sqrt{b}\:,\: $ which is contra to hypotheses as follows:

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

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

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

See my post here for generalizations.

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It suffices to find a field containing $\mathbb{Q}(\sqrt 7)$ in which $5$ is not a square. The field of $3$-adic numbers $\mathbb{Q}_3$ is such a field, since $7 \equiv1 \mod 3$ and thus $7$ is a square, but $5\equiv 2 \mod 3$ so $5$ is not a square. (For any odd prime $p$, an integer prime to $p$ is a square in $\mathbb{Q}_p$ if and only if it is a square modulo $p$.)

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a nitpick - in the last sentence you mean "an integer prime to $p$ is a square..." – user8268 Mar 28 '12 at 22:17
Indeed. I've made the change above. – Brett Frankel Mar 28 '12 at 22:19

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