Polynomial having rational coefficients and one root: $\sqrt{2}+\sqrt{3}-\sqrt{5}$

Question

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

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

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

3. 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 ($\sqrt{2}, \sqrt{5}, \sqrt{10}$)

Answer 1

Here's a partial answer.

Your idea does indeed work. If you apply it, then you will get the polynomial$$p(x)=x^8-40x^6+353x^4-960x^2+576.$$Asserting that it is a polynomial with the smallest degree within the non-null polynomials of which $\sqrt2+\sqrt3-\sqrt5$ is a root is the same thing as asserting that $p(x)$ is irreducible in $\mathbb{Q}[x]$. Right now, I don't see how to prove it. Now, assume that it is true. Then, if $q(x)\in\mathbb{Q}[x]\setminus\{0\}$ is such that $q\bigl(\sqrt2+\sqrt3-\sqrt5\bigr)=0$, I will prove that the degree of $q(x)$ is greater than or equal to $8$.

Since $p(x)$ and $q(x)$ have a common root, then they are not relatively prime in $\mathbb{Q}[x]$. Therefore, there is a non-constant polynomial $r(x)\in\mathbb{Q}[x]$ that divides both of them. But, since $p(x)$ is irreducible, this means that $r(x)$ can only be of the form $\lambda p(x)$, for some $\lambda\in\mathbb Q$. But then $p(x)$ itself divides $q(x)$ and therefore the degree of $q(x)$ is greater than than or equal to the degree of $p(x)$, which is $8$.

Answer 2

The degree of a linear combination of $n$ square roots may be less than $2^n$, like in the case of $\sqrt{2}+\sqrt{3}+\sqrt{6}$.

Answer 3

There is a systematic method based on linear algebra:

• Let $\alpha = \sqrt{2}+\sqrt{3}-\sqrt{5}$. Note that $\alpha \in \mathbb Q(\sqrt{2},\sqrt{3},\sqrt{5})$.

• Write the matrix $A$ of the linear map $x \mapsto \alpha x$ in the basis $1,\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{6},\sqrt{10},\sqrt{15},\sqrt{30}$.

• Find the minimal polynomial of $A$.

Answer 4

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.