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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 ?

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    Not sure if this will help, but maybe use $a^2-b = (a+\sqrt{b})(a-\sqrt{b}) = x(x-2\sqrt{b})$ for $x = m+n+2\sqrt{mn}$?2012-09-09

4 Answers 4

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

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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, $a^2 -b = m^2 + 2mn + n^2 -4mn = (m-n)^2$.

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    Then also $a+\sqrt{b} \in F$ (follows from my second equation), which we know not to be the case.2012-09-10
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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 $F$.
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    very nice solution!2012-09-09
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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.

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    @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.\ $2012-09-09