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.

  • $\begingroup$ 1. is easy (as far as I understand you don't limit the coefficients to be integer there). Take $x-\sqrt{5}-\sqrt{7}$. $\endgroup$ – Fabian May 12 '11 at 20:36
  • $\begingroup$ Is the given root one of two or several roots, or is the root the only root (or perhaps a double root?)? $\endgroup$ – Namaste May 12 '11 at 20:41
  • $\begingroup$ @Amy there might be more than one. The only information was that the polynomial coefficients needs to be integers. $\endgroup$ – Maayan May 12 '11 at 21:16
  • $\begingroup$ Yes, it is clear that for integer coefficients, there must be more than the given root, as adeptly illustrated below. $\endgroup$ – Namaste May 12 '11 at 21:33

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
    $\begingroup$ @Arturo the level of math in your response is a bit more than I understand, but thanks for the explanation. I will read up on Galois Theory. $\endgroup$ – Maayan May 12 '11 at 21:25
  • $\begingroup$ @Maayan: Sorry about that; but you gave no indication of what level course it was where you received this assignment. This is a perfectly reasonable assignment for someone who is just learning about automorphisms of fields (on the way to the Fundamental Theorem of Galois Theory); of course, it can also be assigned at much lower level. $\endgroup$ – Arturo Magidin May 13 '11 at 3:20
  • $\begingroup$ @Arturo: Just a quick question on the notation, as I'm not a mathematician... Does $\mathbb{Q}(\alpha,\beta)$ mean the field of rational numbers extended to include the irrational numbers $\alpha$ and $\beta$? $\endgroup$ – user4423 May 13 '11 at 6:05
  • $\begingroup$ @yoda: Yes; it means the smallest subfield of $\mathbb{C}$ that contains $\mathbb{Q}$, $\alpha$, and $\beta$. Notation doesn't require $\alpha$ and $\beta$ to be irrationals, but if, say, $\alpha$ is rational, then $\mathbb{Q}(\alpha,\beta) = \mathbb{Q}(\beta)$. $\endgroup$ – Arturo Magidin May 13 '11 at 16:54
  • $\begingroup$ @Arturo: Thanks. You say that the notation doesn't require $\alpha$ and $\beta$ to be irrationals, but don't they have to be irrational, as any $\alpha \notin \mathbb{Q}$ implies $\alpha\in\mathbb{J}$? $\endgroup$ – user4423 May 13 '11 at 18:24
  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

  • $\begingroup$ Hi; I think that this method may also be made to address this question: math.stackexchange.com/questions/118361/effective-equality If you feel that's the case, would you mind saying something there? (I am barely starting to understand LLL myself so I don't feel confident this is what I was looking for.) $\endgroup$ – Bruce George May 14 '12 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.