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?
Minimal polynomial
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0@Arturo This really helped! Thanks – 2011-03-11
2 Answers
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.
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0@Bean: If you notice the last paragraph before the first "Added", it says "...giving 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. – 2012-01-21
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 $\rm\ q(x) = p(x^2) = x^4 - 8\ x^2 + 25\:.$