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This is an exercise for the book Abstract Algebra by Dummit and Foote (pg. 530):

Let $F$ be a field of characteristic $\neq2$ . Let $a,b\in F$ with $b$ not a square in $F$. Prove $\sqrt{a+\sqrt{b}}=\sqrt{m}+\sqrt{n}$ for some $m,n\in F$ iff $a^{2}-b$ is a square in $F$.

I am having problem proving this claim, I tried to assume $\sqrt{a+\sqrt{b}}=\sqrt{m}+\sqrt{n}$ and I naturally squared both sides, to try and get $a^{2}$ I squared both sides again and then reduced $2b$ from both sides and rearranged to get $$a^{2}-b=(m+n+2\sqrt{mn})^{2}-2\sqrt{b}(a+\sqrt{b})$$ but I don't see how I can use it.

Can someone please help me prove this claim ?

Fabian
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Belgi
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4 Answers4

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$$\sqrt{a+\sqrt{b}} = \sqrt{m} + \sqrt{n}$$ $$a+\sqrt{b} = m+n+2\sqrt{mn}$$ Thus $a = m+n$ and $b = 4mn$ as $b$ is not a square . Finally, $$a^2 -b = m^2 + 2mn + n^2 -4mn = (m-n)^2$$.

Karolis Juodelė
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$$\sqrt{a+\sqrt b}=\sqrt m + \sqrt n \Rightarrow a+\sqrt b = m+n+2\sqrt{mn}$$ Since $\phi(\alpha+\beta\sqrt b)=\alpha-\beta\sqrt b$ for $\alpha,\beta\in F$ defines an automorphism $\phi\colon F[\sqrt b]\to F[\sqrt b]$ that leaves $F$ fixed, we have that $\phi(\sqrt{mn})= \pm\sqrt{mn}$ because $\phi$ maps the polynomial $X^2-mn$ to itself and can at most interchange its roots. Thus we additionally get $a-\sqrt b=\phi(a+\sqrt b)=\phi(m+n+2\sqrt{mn})$, i.e. $$a-\sqrt b=m+n\pm2\sqrt{mn}.$$ Since $\sqrt b\ne -\sqrt b$ (characteristic $\ne 2$), the left hand sides differ, hence so do the right hand soides, hence "$\pm$" is really "$-$". By adding and subtracting these equations we find that $a=m+n$ and $\sqrt b =2\sqrt{mn}$. Hence $m,n$ are roots of $0=X^2-(m+n)X+mn=X^2-a X+\frac b4$ and can be found as $$\frac{a\pm\sqrt{a^2-b}}2$$ More precisely:

  • If $a^2-b$ is a square, this actually produces $m,n\in F$ with the property that $(\sqrt m +\sqrt n)^2=a+\sqrt b$, i.e. $\sqrt m + \sqrt n$ is a root of $X^2-(a+\sqrt b)$ as desired.
  • If $a^2-b$ is not a square, no solutions for $m,n$ exist in $F$.
Hagen von Eitzen
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  • Can you please explain "It can only map $\sqrt{mn}\mapsto \pm\sqrt{mn}$." ? I don't understand why – Belgi Sep 09 '12 at 20:50
  • $\sqrt{mn}$ is a root of $X^2-mn\in F[X]$. Every automorphism of $F[\sqrt b]$ over $F$ leaves $F$ and hence this polynomial fixed. It can at most interchange the roots, which means flip sign for square roots. – Hagen von Eitzen Sep 09 '12 at 20:53
  • I don't think that every automorphism of $F[\sqrt{b}]$ leaves $F$. for example take $F=\mathbb{Q}(\sqrt{2})$ and $b=3$ then you know that there is an automorphism of $F[\sqrt{b}]$ that takes $\sqrt{2}$ to $-\sqrt{2}$ – Belgi Sep 09 '12 at 20:56
  • Yes. But every automorphism of $F[\sqrt b]$ over $F$. In fact, in this simple case it can be written explicitly as $\phi(\alpha +\beta\sqrt b)=\alpha-\beta\sqrt b$ for $\alpha,\beta\in F$. – Hagen von Eitzen Sep 09 '12 at 21:02
  • sorry I didn't see it's over $F$. How do we get the equality after "Thus we additionally get" ? I know that the LHS it mapped to the RHS but that does not mean their equal... – Belgi Sep 09 '12 at 21:05
  • very nice solution! – Belgi Sep 09 '12 at 21:28
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Let $F$ be a field of charcteristic different from 2. Let $a$ and $b$ be elements of the field $F$ with $b$ not a square in F. Prove that a necessary and sufficient condition for $\sqrt{a+\sqrt{b}}={\sqrt{m}+\sqrt{n}}$ for $m,n\in F$ is that $a^2-b$ is a square in $F.$

Solution. $\Rightarrow:$ Suppose that $a^2-b$ is a square in $F$. Then $\sqrt{a^2-b}\in F.$ Let $$m= \frac{a+\sqrt{a^2-b}}{2}$$ and $$n= \frac{a-\sqrt{a^2-b}}{2}.$$ Then $n,m\in F$ because $\textrm{char}\, F\neq 0.$

$\Leftarrow:$ Now $$m =\frac{a+\sqrt{a^2-b}}{2} = \frac{(a+\sqrt{b})+2\sqrt{a^2-b}+(a-\sqrt{b})}{4} = \left( \frac{\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}}{2} \right)^2,$$ this means that $$\sqrt{m}=\frac{\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}}{2}.$$

Also $$n =\frac{a-\sqrt{a^2-b}}{2} = \frac{(a+\sqrt{b})-2\sqrt{a^2-b}+(a-\sqrt{b})}{4} = \left( \frac{\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}}{2} \right)^2,$$ this means that $$\sqrt{n}=\frac{\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}}{2}.$$

Thus $$\sqrt{m}+\sqrt{n} =\frac{\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}}{2}+\frac{\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}}{2} = \sqrt{a+\sqrt{b}}.$$

Reader
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Hint $\rm\,\ \left[\sqrt{a+\sqrt b}\,=\,\sqrt m + \sqrt n\right]^2\! \Rightarrow\: a+\sqrt b \,=\, m\!+\!n+2\sqrt{mn}\in F[\sqrt{b}].\:$ Thus, taking "norms" $\rm\:a^2 - b\, =\, (m+n)^2 - 4mn = (m-n)^2,\:$ where norm = constant term of minimal polynomial.

Bill Dubuque
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  • How did you calculate those norms ? – Belgi Sep 09 '12 at 22:55
  • @Belgi Compare their monic minimal polynomials $\rm: (x-\alpha)(x-\alpha') = x^2 - (\alpha+\alpha'), x + \alpha\alpha'.$ – Bill Dubuque Sep 09 '12 at 23:09
  • Should I find the minimal polynomial of $a+\sqrt{b}$ to calculate its norm ? – Belgi Sep 09 '12 at 23:12
  • @Belgi The monic minimal polynomial is unique (else subtracting two different ones gives a smaller deg poly, contra min). Thus both sides have the same monic minimal polynomial, so the same constant coefficient - which is the (formal) norm. – Bill Dubuque Sep 09 '12 at 23:17
  • I know that this polynomial is unique, but I don't understand what you calculated to get the quality after the word - " "norms" " – Belgi Sep 09 '12 at 23:19
  • @Belgi One could calculate the entire minimal polynomial for each, but we don't need the linear coeff (= -trace), only the constant coeff (= norm). Thus I took the norm of both sides of the prior equation. The uniqueness of the minimal poly shows that the norm (and trace) of a quadratic irrational are absolute - independent of the choice of discriminant (radicand) representing the quadratic field - you'll get the same result in $\rm:F[\sqrt{d}]:$ and $\rm:F[\sqrt{e^2 d}],\ e,d\in F,\ \sqrt{d}\not\in F.\ $ – Bill Dubuque Sep 09 '12 at 23:41