<s>[INST] Polynomial having rational coefficients and one root: $\sqrt{2}+\sqrt{3}-\sqrt{5}$ 
Form a polynomial of smallest degree having rational coefficients and one root as $\sqrt{2}+\sqrt{3}-\sqrt{5}$

Idea 1:
I thought that other roots would be just different combination of signs on the surds, ie 


*

*$\sqrt{2}+\sqrt{3}+\sqrt{5}$

*$\sqrt{2}-\sqrt{3}+\sqrt{5}$


so least degree would be $2^3 = 8$.
Polynomial then could be formed using viete's formulas.
Idea 2:
We let $x = \sqrt{2}+\sqrt{3}-\sqrt{5}$. Then rearranging and squaring repeatedly gives us the polynomial.
Questions


*

*This method seems unsatisfactory and is just a thought. Please help me with a proper method.

*Is the polynomial i found unique? or there are more polynomials with rational coefficients with this root ($\sqrt{2}+\sqrt{3}-\sqrt{5}$)?

*Also can we generalise this result: that the least degree of a polynomial whose root is a sum of $n$ distinct surds is $\sum \binom{n}{k} = 2^n$ ?
Edit
As stated by Hagen Von Elitzen, the result in third question is correct only for square roots of numbers which are pairwise coprime. Eg. ($\sqrt{2}, \sqrt{3}, \sqrt{5}$) and not ($\sqrt{2}, \sqrt{5}, \sqrt{10}$)
 [/INST] 
If $f(x)\in \mathbb{Q}[x]$ has $\sqrt{2}+\sqrt{3}-\sqrt{5}$ as a root,
  then it has all eight numbers $\pm \sqrt{2} \pm \sqrt{3} \pm \sqrt{5}$
  as roots. So the smallest degree of your polynomial is $8$.

The reason is that we have some automorphisms over $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$. One is 
$$\sigma(\sqrt{2}) = \sqrt{2} \quad \sigma(\sqrt{3}) = \sqrt{3}  \quad \sigma(\sqrt{5}) = -\sqrt{5}$$ This automorphism also fixes $\mathbb{Q}$
Assume $f(x)\in \mathbb{Q}[x]$ has $\sqrt{2}+\sqrt{3}-\sqrt{5}$ as root, write $f(x) = a_0 + a_1 x + ... + a_n x^n$, $\alpha = \sqrt{2}+\sqrt{3}-\sqrt{5}, \beta = \sqrt{2}+\sqrt{3}+\sqrt{5}$. Then we have $\sigma(\alpha) = \beta$.
We have $a_0 + a_1 \alpha + ... +a_n \alpha^n = 0$, taking $\sigma$ on both sides give
$$\sigma(a_0)+\sigma(a_1)\sigma(\alpha) + ... + \sigma(a_n)\sigma(\alpha)^n = 0 \quad \Rightarrow \quad a_0 + a_1 \beta + ... + a_n \beta^n = 0$$
Hence $f(\alpha) = 0$ implies $f(\beta) = 0$.
Using the same argument, we can show that all remaining 7 numbers are roots. Proving the existence of these automorphisms has to resort to Galois theory.
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