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Suppose you have a consistent radical function $R(x)$ that can be solved by finding the roots of an nth degree polynomial, assumed to be solvable using radicals. Some of the $n$ roots maybe extraneous. Typically the student is told to verify each possible root by inserting it into $R(x)$ and verifying that $R(x_i) = 0$ for $i= 1,..,n$. This step basically amounts to verifying an identity that can involve multiple roots and powers. And the $x_i$'s can be complex. Can it be proved algebraically that each extraneous root can be identified, i.e., $R(x_i) \neq 0$ for some $i$?

Let me try to be specific. Suppose $R(x) = \sqrt{x+a} -(x+b) = 0$. The solution to this simple radical equation results in a quadratic equation. When two roots result it appears that one of them will be extraneous. The roots can be complex involving irrational values. Verifying that a root of the polynomial is also a root of $R(x)$ requires proving an identity: $R(x_i)=0$. This could be very difficult because the root may involve nested radicals and $R(x)$ can result in more nested radicals. But how does one prove the identity without resorting to steps that were used to generate the polynomial? I hope this helps. My gut feeling is that one cannot be assured of proving the identity.

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Could you please give a concrete example of what you have in mind? – lhf Jun 2 '11 at 2:39
I think this may be a clearer reformulation of the question: given a function $R(x)$ that may involve radicals and the usual arithmetic operations, and a number $\alpha$ expressed in terms of radicals and the usual arithmetic operations, is there an algorithm guaranteed to tell you whether or not $R(\alpha)=0$? – Gerry Myerson Jun 2 '11 at 3:35

$\root n\of{a+bi}$ can always be evaluated by going to polar form. By going back and forth between polar and rectangular form, any expression $R(x)$ can be evaluated where $R$ and $x$ are of the form envisaged in the question. Then you can tell whether or not it's zero. In principal. In practice, I can see where there may be some difficulties with this approach. First of all, you get $n$ different values for $\root n\of{a+bi}$, and I guess you have to test all of them, and keep some consistency in branches chosen. Also, expressions can get unwieldy, unless you go to floating point - and then if you get zero, you don't know if it's really zero or just machine zero (and if you don't get zero, maybe that's round-off error on something that really is zero).

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If R(x) is a polynomial in a field F and $\alpha$ involves only a finite number of nested root extractions you can compute the minimal polynomial of $\alpha$ in F and compare it to R(x). Call the minimal polynomial of $\alpha$ g(x). Since the minimal polynomial is the unique polynomial of smallest degree in F so that g($\alpha$)=0, g(x) is a factor of R(x) iff R($\alpha$)=0.

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Of course, it may not be so easy to compute the minimal polynomial if, say, $\alpha=\sqrt2+\sqrt3+\sqrt5+\sqrt7+\sqrt{11}$. Also, while it isn't exactly clear what OP wants, I assumed that $R$ was not necessarily a polynomial, but itself an expression involving radicals. The problem is that in going from an equation involving radicals to a polynomial equation you can introduce extraneous roots, and that's what OP is worried about (I think). – Gerry Myerson Jun 2 '11 at 11:34
My answer treats the case when R is a polynomial with radical coefficients since the radicals in R can be adjoined to $\mathbb{Q}$. I was under the impression that there was a standard algorithm for the minimal polynomial in any finite field but I haven't turned up a reference to one.But if R is a polynomial in $\mathbb{Q}$ the minimal polynomial can be computed by using one of the bounds relating the size of the roots of polynomial to its coefficients and then enumerating all polynomials of coefficients less than the bound and all degrees starting from 2 until the minimal polynomial is found – David Marquis Jun 2 '11 at 13:06
Mainly I was responding to your first comment on the OP. The original question needs clarification. – David Marquis Jun 2 '11 at 13:10

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