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Let's say you have an equation $f(x)=0$ where $f:\mathbb R \to \mathbb R$ is composed of $$+,-,\times,\div,\sqrt{},n \in \mathbb Z$$ If $f$ has only finitely many solutions, are all the solutions algebraic?

I've tried (structural) induction. The cases $n$, $\div$ and $\times$ are easy. $+$ and $-$ are not.

I've been trying to reduce it to a polynomial equation. Problem is that doing $\sqrt{A}=B \implies A=B^2$ makes the RHS more complicated. If the RHS contains sums of square roots then this doesn't make progress. (Maybe I could do more work here?)

I don't see how to use the fact the algebraic numbers are closed under the above operations.

I've been trying to think of a counterexample.

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    $\begingroup$ Finitely many solutions where? In $\mathbb{C}$? In $\mathbb{R}$? (Also, what do you mean by $\sqrt{}$ exactly?) $\endgroup$ – Eric Wofsey Jan 27 '18 at 21:23
  • $\begingroup$ @EricWofsey Does it make a big difference? By $\sqrt{}$, I mean the function that maps $x$ to $\sqrt{x}^+$ (positive square root)? $\endgroup$ – man and laptop Jan 27 '18 at 21:24
  • $\begingroup$ @EricWofsey Because of the discussion on the square root, I'll say $\mathbb R$ $\endgroup$ – man and laptop Jan 27 '18 at 21:27
  • $\begingroup$ You can remove all square roots by replacing the single equation by a system of polynomial equations in several variables. $\endgroup$ – Mariano Suárez-Álvarez Jan 27 '18 at 21:30
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    $\begingroup$ Are the formulas also finitely many operations? Because if not I think you can generate $e$. $\endgroup$ – Joffan Jan 27 '18 at 23:37
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Any radical expression exists in a tower of radical extensions of $\Bbb Q(x)$. For instance, consider

$$ T(x)=\sqrt{\sqrt{x+1}-\sqrt{x-1}}+\sqrt{x}-3.$$

Then $T$ exists in the top of the tower of radical extensions

$$ \begin{array}{c} \Bbb Q\left(\sqrt{x+1},\sqrt{x},\sqrt{x-1},\sqrt{\sqrt{x+1}-\sqrt{x-1}}\right) \\ | \\ \Bbb Q(\sqrt{x+1},\sqrt{x},\sqrt{x-1}) \\ | \\ \vdots \\ | \\ \Bbb Q(x) \end{array} $$

where $\Bbb Q(x)$ is the field of rational functions in $x$. Thus, $T$ is algebraic over $\Bbb Q(x)$, so it has a minimal polynomial, which in this case must have degree $\le 16$ since that's the degree of the extension up top. So $T$ is a root of an irreducible polynomial

$$ c_{16}(x)T^{16}+\cdots+c_1(x)T+c_0(x) \tag{$\ast$} $$

where all the $c_i$s are in $\Bbb Q[x]$ (by clearing denominators if necessary; the polynomial in $T$ needn't be monic), and $c_0(x)$ is not the zero polynomial in $x$ since the above polynomial in $T$ is irreducible. If $x$ is a root of the radical equation $T(x)=0$, then plugging it into $(\ast)$ yields $c_0(x)=0$, making $x$ the root of a polynomial equation - algebraic.

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  • $\begingroup$ The ingredients of this answer are: 1) You start off with the field $\mathbb Q(x)$ 2) You form a polynomial ring over it $\mathbb Q(x)[u]$ 3) You quotient by an irreducible polynomial in $u$ to get e.g. $\mathbb Q[x][u]/(u^2-x)$ 4) You repeat the preceding two steps, each time doubling the dimensionality of the resulting field relative to $\mathbb Q(x)$ 5) You express $T(x)$ (as in your answer) in the resulting field 6) You interpret multiplication by $T(x)$ to be a linear operator on a vector space of $\mathbb Q(x)$... $\endgroup$ – man and laptop Feb 4 '18 at 20:53
  • $\begingroup$ ... 7) You apply the Cayley-Hamilton theorem to get out a minimal polynomial $P$ of $T(x)$ with coefficients in $\mathbb Q(x)$; this polynomial satisfies $P(T(x))=0$ 8) You consider a $k$ for which $T(k)=0$ 9) You substitute $k$ into both sides of the equation $P(T(x)) = 0$ to get $P(T(k))=0 \implies c_0(k)=0$. Hence $k$ is algebraic $\endgroup$ – man and laptop Feb 4 '18 at 20:58
  • $\begingroup$ It's a good answer. I think I understood it. Thank you. (I don't know why I wrote out any of the above) $\endgroup$ – man and laptop Feb 4 '18 at 20:59
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The set of algebraic numbers is algebraically closed. This seems to be the result you are referring to. Here's a link: https://en.m.wikipedia.org/wiki/Algebraic_number#Properties

Algebraic closure, together with the fact that $\sqrt a$ is algebraic for $a\in \mathbb Z$ (consider $x^2-a$), should suffice to show that combinations of doing the four arithmetic operations plus $x\rightarrow \sqrt x$ to rational numbers results in an algebraic number...

It's not clear that you can replace $Q$ with $\bar {\mathbb Q}$,the set of algebraic numbers, here...

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  • $\begingroup$ I guess I should add something about $n$-th roots... $\endgroup$ – Chris Custer Jan 27 '18 at 21:33
  • $\begingroup$ @jkabrg So, $\sqrt3- {\sqrt 2}$... etc?... $\endgroup$ – Chris Custer Jan 27 '18 at 21:45
  • $\begingroup$ Ok. I'll give it a shot... $\endgroup$ – Chris Custer Jan 27 '18 at 22:27
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    $\begingroup$ (you seem to be thinking that the problem is to show that any number constructed using the field operations and square roots from integers is algebraic, but it is not that) $\endgroup$ – Mariano Suárez-Álvarez Jan 27 '18 at 22:41
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    $\begingroup$ I agree with Mariano that it seems you misunderstood the question. $\endgroup$ – anon Jan 28 '18 at 1:18

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