How to find a polynomial from a given root?

I was asked to find a polynomial with integer coefficients from a given root/solution.
Lets say for example that the root is: $\sqrt{5} + \sqrt{7}$.

  1. How do I go about finding a polynomial that has this number as a root?
  2. Is there a specific way of finding a polynomial with integer coefficients?

Any help would be appreciated. Thanks.

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One can start from the equation $x=\sqrt5+\sqrt7$ and try to get rid of the square roots one at a time. For example, $x-\sqrt5=\sqrt7$, squaring yields $(x-\sqrt5)^2=7$, developing the square yields $x^2-2=2x\sqrt5$, and squaring again yields $(x^2-2)^2=20x^2$, that is, $x^4-24x^2+4=0$.

The simplest polynomial that has $r$ as a root is just $x-r$. But of course, that is not what you are dealing with here.

This is really a question about finding the minimal polynomial of $\alpha=\sqrt{5}+\sqrt{7}$ over $\mathbb{Q}$; such polynomials exist for any algebraic number, by definition of algebraic.

Galois Theory provides all the necessary tools to solve this problem. Basically:

  1. Consider the field $\mathbb{Q}(\sqrt{5},\sqrt{7})$; this field contains the number $\alpha=\sqrt{5}+\sqrt{7}$, so we can work there. Every element of this field can be written uniquely as
    $$a + b\sqrt{5} + c\sqrt{7} + d\sqrt{35}$$
    for some $a,b,c,d\in\mathbb{Q}$.

  2. The field $\mathbb{Q}(\sqrt{5},\sqrt{7})$ has four automorphisms (functions $f\colon\mathbb{Q}(\sqrt{5}+\sqrt{7})\to\mathbb{Q}(\sqrt{5},\sqrt{7})$ that are additive, $f(a+b)=f(a)+f(b)$, multiplicative, $f(ab)=f(a)f(b)$, and invertible), and these automorphisms fix elements of $\mathbb{Q}$ (that is, $f(q)=q$ for all $q\in\mathbb{Q}$).

    These automorphism are the following:
    $$\begin{align*}
    \sigma_1\colon a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35} &\longmapsto a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35} &\qquad&\text{(the identity)};\\
    \sigma_2\colon a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35} &\longmapsto a-b\sqrt{5}+c\sqrt{7}-d\sqrt{35}\\
    \sigma_3\colon a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35} &\longmapsto a+b\sqrt{5}-c\sqrt{7}-d\sqrt{35}\\
    \sigma_4\colon a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35} &\longmapsto a-b\sqrt{5}-c\sqrt{7}+d\sqrt{35} &&\text{(equal to }\sigma_3\circ\sigma_2\text{)}
    \end{align*}$$
    The maps are induced by the conjugation maps $\sqrt{5}\mapsto-\sqrt{5}$ and $\sqrt{7}\mapsto-\sqrt{7}$.

    An important feature is that an element of $\mathbb{Q}(\sqrt{5},\sqrt{7})$ lies in $\mathbb{Q}$ if and only if it is fixed by all four maps.

  3. Suppose $p(x)$ is a polynomial with coefficients in $\mathbb{Q}$, and that $a\in\mathbb{Q}(\sqrt{5},\sqrt{7})$. If
    $$p(x) = a_nx^n + \cdots + a_0$$
    then
    $$\sigma_i(p(a)) = \sigma_i(a_na^n+\cdots+a_0) = a_n\sigma_i(a)^n+\cdots+a_0 = p(\sigma_i(a)).$$
    In particular, if $p(a)\in\mathbb{Q}$, then $p(a)=\sigma_i(p(a)) = p(\sigma_i(a))$ for $i=1,2,3,4$.

  4. Now suppose you find a polynomial $p(x)$ with coefficients in $\mathbb{Q}$ that has $\sqrt{5}+\sqrt{7}$ as a root. Then $p(\sqrt{5}+\sqrt{7})=0$, so by 3 above, it must also be true that $p(\sigma_i(\sqrt{5}+\sqrt{7}))=0$ for $i=1,2,3,4$. That means that $p(x)$ must also have $\sqrt{5}-\sqrt{7}$, $-\sqrt{5}+\sqrt{7}$, and $-\sqrt{5}-\sqrt{7}$ as roots. By unique factorization, we conclude that $p(x)$ must be divisible by
    $$\left(x – (\sqrt{5}+\sqrt{7})\right)\left(x – (-\sqrt{5}+\sqrt{7})\right)\left(x – (\sqrt{5}-\sqrt{7})\right)\left(x – (-\sqrt{5}-\sqrt{7})\right).$$
    That is, any polynomial with coefficients in $\mathbb{Q}$ that has $\sqrt{5}+\sqrt{7}$ as a root must be a multiple of this product.

    But if you multiply out this product, you will discover that this polynomial already has coefficients in $\mathbb{Q}$.

Once you have a polynomial with coefficients in $\mathbb{Q}$ that has $\sqrt{5}+\sqrt{7}$ as a root, you can get a polynomial with coefficients in $\mathbb{Z}$ by simply clearing denominators.

  1. $x-\sqrt 5 – \sqrt 7$ is a polynomial, so you should specify the kind of coefficients you are seeking.

  2. In the case of square roots there exists the low-level technique of “squaring the roots away”.

    $$x=\sqrt 5 + \sqrt 7$$
    $$x^2=5+7+2\sqrt {35}$$
    $$(x^2-12)^2=140$$
    $$x^4-24x^2+4=0$$

  3. If you know that you should expect the conjugates to be roots, you can multiply four linear factors.

    $$(x-\sqrt 5 – \sqrt 7)(x-\sqrt 5 + \sqrt 7)(x+\sqrt 5 – \sqrt 7)(x+\sqrt 5 + \sqrt 7).$$

  4. In general, there are methods for constructing a polynomial for the sum of two numbers whose polynomials are known, for example one can use the resultant.
    http://en.wikipedia.org/wiki/Resultant

Let me just remark that there are algorithms for doing this even when the number is not given in terms of radicals but only as a decimal expansion (but is suspected to be algebraic). Namely, one uses the LLL algorithm to try to construct an integral linear combination of $1,\alpha,\alpha^2,\alpha^3,\ldots$ that is zero up to a rounding error. If no such linear combination is found then one can conclude that the number is either transcendental, or the minimal polynomial must have enormous coefficients or enormous degree.

See http://en.wikipedia.org/wiki/LLL_algorithm#Applications