I was asked to find a minimal polynomial of $$\alpha = \frac{3\sqrt{5} - 2\sqrt{7} + \sqrt{35}}{1 - \sqrt{5} + \sqrt{7}}$$ over Q.

I'm not able to find it without the help of WolframAlpha, which says that the minimal polynomial of $\alpha$ is $$19x^4 - 156x^3 - 280x^2 + 2312x + 3596.$$ (Truely it is - $\alpha$ is a root of the above polynomial and the above polynomial is also irreducible over Q.)

Can anyone help me with this?

Thank you!


5 Answers 5


To begin, clear denominators: $$(1 - \sqrt{5} + \sqrt{7}) \alpha = 3 \sqrt{5} - 2 \sqrt{7} + \sqrt{35}$$ We need to make the coefficient of $\alpha$ rational, so use a difference-of-squares trick to get rid of the $\sqrt{7}$ on the LHS (i.e. multiply both sides by $1 - \sqrt{5} - \sqrt{7}$), $$((1 - \sqrt{5})^2 - 7) \alpha = (3 \sqrt{5} - 2 \sqrt{7} + \sqrt{35})(1 - \sqrt{5} - \sqrt{7})$$ and after expanding and collecting like terms: $$(1 + 2 \sqrt{5}) \alpha = 1 + 4 \sqrt{5} + 7 \sqrt{7}$$ Now do the same again to deal with the $\sqrt{5}$ on the LHS: $$19 \alpha = 39 - 2 \sqrt{5} - 7 \sqrt{7} + 14 \sqrt{35}$$ Next, we have to deal with the irrational numbers on the RHS. First, we deal with $\sqrt{5}$ (and $\sqrt{35}$): move all the other terms over to the LHS, and square the resulting equation, $$(19 \alpha - 39 + 7 \sqrt{7})^2 = (-2 + 14 \sqrt{7})^2 \cdot 5$$ which expands to this

$$361 \alpha^2 - 1482 \alpha + 266 \sqrt{7} \alpha + 1864 -546 \sqrt{7} = 6880 - 280 \sqrt{7}$$

To finish off, we deal with $\sqrt{7}$: put all multiples of $\sqrt{7}$ on the RHS and all others on the LHS, and then square the resulting equation: $$(361 \alpha^2 - 1482 \alpha - 5016)^2 = (- 266 \alpha + 266)^2 \cdot 7$$ Note that $19$ divides all the coefficients, so we can cancel that common factor: $$(19 \alpha^2 - 78 \alpha - 264)^2 = (-14 \alpha + 14)^2 \cdot 7$$ Finally, we obtain, $$361 \alpha^4 - 2964 \alpha^3 - 3984 \alpha^2 + 41184 \alpha + 69696 = 1372 \alpha^2 - 2744 \alpha + 1372$$ which simplifies to the desired equation: $$19 \alpha^4 - 156 \alpha^3 - 280 \alpha^2 + 2312 \alpha + 3596 = 0$$


$$\alpha-\alpha\sqrt{5}+\alpha\sqrt{7}=3\sqrt{5}-2\sqrt{7}+\sqrt{35},$$ $$\alpha-\sqrt{35}=(\alpha+3)\sqrt{5}-(\alpha+2)\sqrt{7},$$ $$(\alpha-\sqrt{35})^2=[(\alpha+3)\sqrt{5}-(\alpha+2)\sqrt{7}]^2,$$ $$\alpha^2+35-2\alpha\sqrt{35}=5(\alpha+3)^2+7(\alpha+2)^2-2\sqrt{35}(\alpha+2)(\alpha+3),$$ $$2\sqrt{35}(\alpha^2+4\alpha+6)=11\alpha^2+58\alpha+38,$$ $$[2\sqrt{35}(\alpha^2+4\alpha+6)]^2=(11\alpha^2+58\alpha+38)^2.$$ From the last equality, we get that minimal polynomial is $19x^4-156x^3-280x^2+2312x+3596$.

  • 1
    $\begingroup$ Thank you. Is this approach you wrote ad hoc or is there some general and easy (means that there doesn't have to be a whole paper written about it) algorithm for finding minimal polynomials? $\endgroup$
    – pizet
    Jun 16, 2013 at 21:32
  • 2
    $\begingroup$ That is a some kind of trick - to get rid of $\sqrt{}$, but it can be applied in similar problems. $\endgroup$
    – alans
    Jun 16, 2013 at 21:44
  • $\begingroup$ Impressive nonetheless $\endgroup$ Apr 8, 2017 at 1:21
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    $\begingroup$ It only proves that $\alpha$ is a zero of this polynomial. How can we prove that this polynomial is minimal? $\endgroup$ Aug 16, 2017 at 4:03
  • 2
    $\begingroup$ That's not the minimal polynomial, you need to divide by 19 as the minimal polynomial is monic by definition. Also you need to prove it's irreducible over $\mathbb{Q}$, using the rational roots theorem it should be easy. $\endgroup$
    – RJ Acuña
    Feb 14, 2021 at 5:51

A general purpose method is that the equation

$$ \sum_{k=0}^n c_k \alpha^k = 0 $$

is a linear equation in the unknowns $c_k$, and thus this can be solved with linear algebra.

Since the number itself is a rational linear combination of the four linearly independent numbers $1, \sqrt{5}, \sqrt{7}, \sqrt{35}$, we get "$4$ equations in $n+1$ unknowns", so $n=4$ will guarantee a solution exists.

If desired, you can avoid computing the quotient by using the fact $$ \sum_{k=0}^n c_k \left(\frac{\mu}{\nu}\right)^k = 0 \quad \Longleftrightarrow\quad \sum_{k=0}^n c_k \mu^k \nu^{n-k} = 0 $$


I know nothing about Galois Theory, and haven't really thought about this one, and am replying quickly (and am not in shape) so I might be saying stupid things, but I guess we could use the fact that $\alpha \in \mathbb{Q}(\sqrt{5},\sqrt{7})$, and then define $x \mapsto \bar{x}$ as sending -for example- $\sqrt{7}$ to $-\sqrt{7}$, and $x \mapsto x^{*}$ sending $\sqrt{5}$ to $-\sqrt{5}$, then generate a -monic- polynomial over $\mathbb{Q}$ by taking: $$p(X) := (X-\alpha)(X-\bar{\alpha})(X-a^{*})(X-\bar{\alpha}^{*}) $$ then if we want the coefficients of the Polynomial to be integers, we define $$D := 1 - \sqrt{5} + \sqrt{7} $$ i.e. the denominator of $\alpha$, then multiply $p(X)$ by $\pm D \bar{D} D^{*} \bar{D}^{*}$ (you should obviously get $19$) to get a polynomial whose coefficients are all integers (which should be the one you're looking for [at least I made a quick try on Python and it seems to work...]) (Those are just the main ideas to get a polynomial in $\mathbb{Z}[X]$ or a monic polynomial in $\mathbb{Q}[X]$ that sends some $\alpha$ to $0$ -by using what I guess is called the "Galois Group"- but I guess a priori, this obviously doesn't guarantee you will "always" get "THE Minimal polynomial" of $\alpha$; this would rather in general give you a multiple of it [I guess... as I said I know nothing about Galois Theory and haven't really thought about the question; therefore, I guess there should probably be much more efficient methods than this...])


(Part answer).Start by rationalizing the denominator.Let $x=1-\sqrt 5+\sqrt 7.$ Then $$1/x=(1-\sqrt 5-\sqrt 7)/(x(1-\sqrt 5-\sqrt 7))=(1-\sqrt 5-\sqrt 7)/(19-2\sqrt 5).$$ Let $y=19-2\sqrt 5.$ Then $$1/y=(19+2\sqrt 5)/(y(19+2\sqrt 5))=(19+2\sqrt 5)/341.$$ So $$1/x=(1-\sqrt 5-\sqrt 7)(19+2\sqrt 5)/341.$$ Inserting this into $\alpha$ and expanding it, we obtain $$\alpha =A+B\sqrt 5+C\sqrt 7+D\sqrt {35}$$ where $A,B,C, D$ are rational.

If you wish, send me a comment so I can be reminded to finish this.


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