Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a cyclotomic field $\mathbb{Q}(\zeta_8)$, and want to know how I can find a minimal polynomial of this element $\zeta_8-2*\zeta_8^3$. Can we generalize this to any number field?

share|cite|improve this question
I don't understand the question. What is the statement that you would like to generalize? It is true that if K is a number field, then each element of K has a minimal polynomial over Q. – Brad Mar 11 '11 at 21:07
over $\mathbb{Q}$? its min poly is linear over $\mathbb{Q}(\zeta_8)$ – yoyo Mar 11 '11 at 21:12
@Brad My aim is to get various elements c of K = Q(\zeta_n) and find their minimal polynomials. I know if c is in K, but c is not in a subfield of K then then c has a minimal polynomial whose degree=\varphi(n). Am really interested with cs which are a linear combination \zeta_n with integer coefficients Thanks – user8129 Mar 11 '11 at 21:45
Use the "determinant method" outlined below, using as a basis $1,\zeta_n,\zeta_n^2,\ldots,\zeta_n^{\varphi(n)-1}$; that will give you a polynomial satisfied by your linear combination. You may need to factor it to find the minimal one in some cases, though. – Arturo Magidin Mar 11 '11 at 21:58
@Arturo This really helped! Thanks – user8129 Mar 11 '11 at 22:17
up vote 6 down vote accepted

If you know the automorphisms of $\mathbb{Q}(\zeta_8)$, then the simplest way is to use them: if $\sigma_1,\ldots,\sigma_4$ are the four automorphisms of $\mathbb{Q}(\zeta_8)$ over $\mathbb{Q}$ (the degree of the $n$th cyclotomic field is $\varphi(n)$, and $\varphi(8) = 4$), then $\zeta_8 - 2\zeta_8^3$ certainly satisfies $$\prod_{i=1}^4\left( x - \sigma_i(\zeta_8 - 2\zeta_8^3)\right) = \prod_{i=1}^4\left(x - \sigma_i(\zeta_8) - 2\sigma_i(\zeta_8)^3\right).$$ So the minimal polynomial will necessarily divide this degree $4$ polynomial; it is not linear, so it is either this polynomial, or is an irreducible quadratic factor (this easily generalizes).

The four automorphisms are the maps $\zeta_8\mapsto\zeta_8$, $\zeta_8\mapsto\zeta_8^3$, $\zeta_8\mapsto\zeta_8^5$, and $\zeta_8\mapsto\zeta_8^7$. You can then compute the polynomial explicitly.

Alternatively, since $\zeta_8$ is an algebraic integer, you can use the "determinant method", outlined in Daniel Marcus's Number Fields (Theorem 2, page 15):

Note that both $\mathbb{Z}[\zeta_8-2\zeta_8]$ is contained in a finitely generated $\mathbb{Z}$-module. Let $a_1,\ldots,a_n$ generate this module.

Now let $\alpha$ be the algebraic integer you want: $\alpha=\zeta_8 - 2\zeta_8^3$; for each $i$, $1\leq i\leq n$, express $\alpha a_i$ as a linear combination of the generators $a_i$. You obtain a matrix equation of the form $$\left(\begin{array}{c} \alpha a_1\\ \alpha a_2\\ \vdots\\ \alpha a_n\end{array}\right) = M\left(\begin{array}{c} a_1\\ a_2\\ \vdots\\ a_n \end{array}\right)$$ where $M$ is an $n\times n$ matrix with coefficients in $\mathbb{Z}$. This means that $\det(\alpha I - M) = 0$, since the equation $(\alpha I - M)\mathbf{x}=\mathbf{0}$ has nontrivial solutions. Expressing this determinant in terms of the $n^2$ coefficients of $\alpha I - M$ you get an expression of the form $\alpha^n + \text{(lower degree terms)} = 0$, giving a monic polynomial with integer coefficients which has $\alpha$ as a root. Factoring it will yield the minimal polynomial. (This last step may be, of course, somewhat difficult in practice).

Added. I gave the above answers because you implied you want to know general methods. In this case, though, we can proceed explicitly. (Of course, since $\zeta_8 = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$, we know that $\zeta_8 - 2\zeta_8^3=\frac{3\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$ has degree $4$ over $\mathbb{Q}$, so it is necessarily the monic polynomial of degree $4$ that you found above; we know it's degree $4$ because if it were degree $2$ it would lie in $\mathbb{Q}(\sqrt{d})$ for some square free integer $d$, and you would need both $\sqrt{2}$ and $\sqrt{-2}$ to be in this quadratic field, which is impossible).

Since $\mathbb{Q}(\zeta_8) = \mathbb{Q}(\sqrt{2},\sqrt{-2}) = \mathbb{Q}(\sqrt{2},i)$, the four automorphisms are the identity, complex conjugation, the map that sends $\sqrt{2}$ to $-\sqrt{2}$ and $i$ to $i$; and the map that sends $\sqrt{2}$ to $-\sqrt{2}$ and $i$ to $-i$. So the minimal polynomial is: \begin{align*} p(x) &= \left( x- \left(\frac{3\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)\right) \left( x- \left(\frac{3\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)\right)\\ &\qquad\qquad \times \left( x- \left(-\frac{3\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)\right) \left( x- \left(-\frac{3\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)\right)\\ &= \left( \left(x-\frac{3\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2\right)\left( \left(x+\frac{3\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2\right)\\ &= \left( x^2 -3\sqrt{2}x + 5\right)(x^2+3\sqrt{2}x+5)\\ &= (x^2+5)^2 - \left(3\sqrt{2}x\right)^2\\ &= x^4 + 10x^2 + 25 -18x^2\\ &= x^4 - 8x^2 + 25. \end{align*} (Unless I made some silly arithmetical error, anyway).

To see the second method in action, we can take $1,\zeta_8, \zeta_8^2,\zeta_8^3$ as our generating set. Then we want to express the products of $\alpha$ with these four elements in terms of these four elements (where $\alpha = \zeta_8 - 2\zeta_8^3$). Noting that $\zeta_8$ is a root of $x^4+1$, we have \begin{align*} \alpha 1 &= \zeta_8 - 2\zeta_8^3\\ \alpha \zeta_8 &= 2 + \zeta_8^2\\ \alpha \zeta_8^2 &= 2\zeta_8 + \zeta_8^3\\ \alpha\zeta_8^3 &= -1+2\zeta_8^2. \end{align*} So the matrix $M$ in question is: $$M = \left(\begin{array}{rrrr} 0 & 1 & 0 & -2\\ 2 & 0 & 1 & 0\\ 0 & 2 & 0 & 1\\ -1 & 0 & 2 & 0 \end{array}\right).$$ Then we compute the determinant of $\alpha I - M$. We have: \begin{align*} \det(\alpha I - M) &= \left|\begin{array}{rrrr} \alpha & -1 & 0 & 2\\ -2 & \alpha & -1 & 0\\ 0 & -2 & \alpha & -1\\ 1 & 0 & -2 & \alpha \end{array}\right|\\ &= \alpha\left|\begin{array}{rrr} \alpha & -1 & 0\\ -2 & \alpha & -1\\ 0 & -2 & \alpha \end{array}\right| + 2 \left|\begin{array}{rrr} -1 & 0 & 2\\ -2 & \alpha & -1\\ 0 & -2 & \alpha \end{array}\right| - \left|\begin{array}{rrr} -1 & 0 & 2\\ \alpha & -1 & 0\\ -2 & \alpha & -1 \end{array}\right|\\ &= \alpha(\alpha^3 - 2\alpha - 2\alpha) +2(-\alpha^2 + 8 + 2) - (-1 + 2\alpha^2 - 4)\\ &= \alpha^4 - 4\alpha^2 - 2\alpha^2 + 20 + 5 - 2\alpha^2\\ &= \alpha^4 - 8\alpha^2 + 25. \end{align*} So $\alpha$ satisfies $x^4 - 8x^2 + 25$, as before.

share|cite|improve this answer
Great this explains it all Thank you very much – user8129 Mar 11 '11 at 21:58
@Mtumodzi: I trust you saw the second the explicit derivation using the second method; I was probably editing it in while you were reading. – Arturo Magidin Mar 11 '11 at 22:00
I'm a bit confused on how this generalizes: if I want to compute the minimal polynomial of, say, $i+\sqrt[4]{2}$, and use the matrix method (and the obvious basis of 8 elts), I get the polynomial $x^8 + 4x^6 + 2x^4 + 28x^2 + 1$. A quick attempt at using Eisenstein with various automorphisms doesn't seem to prove the polynomial is irreducible. Does this method guarantee that it is? In other words, could this just be the characteristic polynomial of the matrix, and not necessarily the minimal polynomial? – JeremyKun Jan 21 '12 at 20:47
@Bean: If you notice the last paragraph before the first "Added", it says " a monic polynomial with integer coefficients which has α as a root. Factoring it will yield the minimal polynomial." So there is no guarantee, in general, that it will be the minimal polynomial (for a simple example, do the process with an integer!). Here, if you look at the obvious automorphisms, I think you'll find that $i+\sqrt[4]{2}$ is a primitive element of $\mathbb{Q}(i,\sqrt[4]{2})$, and so it will have degree $8$, which will guarantee that the polynomial you found is the minimal one. – Arturo Magidin Jan 21 '12 at 21:53

Computing the min polynomial of your number $\rm\:w\:$ is easy since $\rm\ v\: :=\: w^2 $ is a quadratic number.

Namely $\rm\ w = \zeta_8-2\ \zeta_8^3\ \Rightarrow\ w^2 = 4 - 3\ i = v\:,\ $ root of$\rm\ \ p(x)= (x-v)\:(x-v') = x^2 - 8\ x +25\:.$

Hence, since $\rm\: p(w^2) = p(v) = 0\ $ we conclude that $\rm\ q(w) = 0\ $ for $\rm\ q(x) = p(x^2) = x^4 - 8\ x^2 + 25\:.$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.