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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 ?

Belgi
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3 Answers3

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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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Arturo Magidin
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  • thank you very much, I will accept once those 10 minutes pass. – Belgi Mar 28 '12 at 21:44
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    @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
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Hint $ $ Apply the below lemma (using $\,\sqrt{5},\sqrt{7},\sqrt{35}\,$ 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 $\rm\,\color{#0a0}{2 \ne 0}\,$ 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}^{\phantom{|}} =\ r + s\ \sqrt{b}\ $ for $\rm\ r,s\in K,\,$ which is false, because squaring yields $\rm\,\color{#c00}{[\![1]\!]}\!:\ \ a\ =\ r^2 + b\ s^2 + \color{#0a0}2\,r\,s\,\sqrt{b}^{\phantom{i}}\!,\, $ contra our hypotheses, by below:

$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \ \sqrt{b}\ \in\ K\ \ $ by solving $\color{#c00}{[\![1]\!]}$ for $\rm\sqrt{b},\,$ using $\rm\,\color{#0a0}{2 \ne 0}$

$\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 $\rm\,\sqrt{b}.\qquad$ $\bf\small QED$

Generalizations are in this answer, e.g. a proof of 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 $\rm\ a,b,\,\ldots \in Q.\,$ If all nonempty subsets of $\rm S $ have product $\rm\not\in\! Q\,$ then each successive adjunction $\rm\ Q(\sqrt{a}),\ Q(\sqrt{a},\sqrt{b}),\,\ldots$ doubles degree over $\rm Q,\,$ so, in total, $\rm\, [L:Q] = 2^n.\,$ Thus the $\rm 2^n$ subproducts of the product of $\rm\,S\, $ are a basis of $\rm\,L\,$ over $\rm\,Q.$

Bill Dubuque
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  • Very theoretical. I prefer an high school answer. – Bob Dobbs Dec 17 '23 at 19:59
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    @BobDobbs Again, the OP's tags imply knowledge of ring and field theory, so answers are written assuming knowledge of such. We also have many other variants of this question posed at lower level, so there is no need to duplicate such lower-level answers here. – Bill Dubuque Dec 17 '23 at 20:09
  • Simple is better in my opinion But, developing a theory is other thing. I am here for fun not theory. – Bob Dobbs Dec 17 '23 at 20:18
  • @Bob Context is essential. Exercises like this are posed in field theory courses to help illustrate basic field theory ideas. As is often the case - we can eliminate more abstract algebraic notions to obtain more elementary proofs, but doing so obfuscates the key ideas that are essential to learn when mastering field theory. That's why I did not write the above proof in the slightly shorter form by instead squaring $\rm,\sqrt a-r ,=, s\sqrt b, $ (as is common in more elementary contexts) since e.g. doing so slightly obscures the ideas needed for said generalization – Bill Dubuque Dec 17 '23 at 21:01
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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 $p$.)

http://en.wikipedia.org/wiki/P-adic_number

Brett Frankel
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