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Minimal polynomial for $\zeta+\zeta^5$ for a primitive seventh root of unity $\zeta$

I have asked a similar problem Minimal Polynomial of $\zeta+\zeta^{-1}$ and i tried to repeat similar idea

Consider $\alpha=\zeta+\zeta^5$ and consider $\sigma \in Gal(\mathbb{Q}(\zeta)/\mathbb{Q})$ defined as $\sigma(\zeta)=\zeta^2$

so, we consider conjugates of $\alpha$ under $\sigma$

  • $\alpha=\zeta+\zeta^5$
  • $\sigma(\alpha)=\sigma(\zeta+\zeta^5)=\zeta^2+\zeta^3$
  • $\sigma^2(\alpha)=\sigma(\zeta^2+\zeta^3)=\zeta^4+\zeta^6$
  • $\sigma^3(\alpha)=\alpha$

So i would consider $(x-\alpha)(x-\sigma(\alpha))(x-\sigma^2(\alpha))$

This is giving me $x^3+x^2+(2+\zeta+\zeta^2+\zeta^4)x-1$

I do not know how to make coefficient of $x$ as real..

Please help me to fill this gap..

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    $\begingroup$ You only have half the Galois group. The Galois group has 6 elements, but your $\sigma$ has order 3. $\endgroup$ – Gerry Myerson Aug 6 '14 at 7:39
  • $\begingroup$ @GerryMyerson : Oh.. I was thinking i can take any $\sigma$ and consider its distinct conjugates... $\endgroup$ – user87543 Aug 6 '14 at 8:45
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The map $\tau: \zeta \rightarrow \zeta^3$ is a generator of the automorphism group of $Q(\zeta)$, $\zeta^7=1$, $\zeta\ne 1$. Your automorphism $\sigma: \zeta \rightarrow \zeta^2$ is $\tau^2$, so it represents an element of order 3 in the Galois group. So its fixed field is a quadratic extension of $K$ of $Q$.

The coefficients of your cubic polynomial $f$ lie in $K$ not in $Q$. You have computed the minimal polynomial over $K$ instead of $Q$. The field $K$ has one non-trivial automorphism induced by complex conjugation so to obtain the irreducible polynomial over $Q$ we just compute $g=f\bar{f}$ (applying complex conjugation to the coefficients). Complex conjugation is also an automorphism of the field $Q(\zeta)$ also and it is given by $\zeta\rightarrow \zeta^{-1}=\zeta^6$ (it is $\tau^3$). Complex conjugation does not belong to the subgroup generated by your element $\sigma$.

Your cubic is $f=x^3+x^2+\frac{3+\sqrt{-7}}{2}x-1$, $\bar{f}=x^3+x^2+\frac{3-\sqrt{-7}}{2}x-1$ and $g=f\bar{f}=x^6+2x^5+4x^4+x^3+2x^2-3x+1$.

This is the exactly the same as multiplying all $\prod_{i=0}^5 (x-\tau^i(\zeta+\zeta^5))$, since $f=\prod_{i=0}^2 (x-\tau^{2i}(\zeta+\zeta^5))$ and $\bar{f}=\prod_{i=0}^2 (x-\tau^{2i+1}(\zeta+\zeta^5))$

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  • $\begingroup$ This is simple and perfect... How do i make sure this is irreducible?? Reduction can not be into a polynomial of degree $3$ and $3$ and it can not be $5,1$ so it has to be $2,4$ and then should i take $(x-a)(x-b)$ for all $a,b$ in those conjugates and prove this is not a polynomial in $\mathbb{Q}[x]$... $\endgroup$ – user87543 Aug 6 '14 at 19:07
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    $\begingroup$ @PraphullaKoushik I should have said $1, s, \cdots, s^5$ are independent. $\endgroup$ – i. m. soloveichik Aug 6 '14 at 22:00
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    $\begingroup$ And a simpler way of seeing the irreducibility is to observe that $\zeta^5+\zeta$ has six distinct conjugates. As they must also be zeros of $m(x)$, we must have $\deg m(x)=6$. $\endgroup$ – Jyrki Lahtonen Aug 6 '14 at 22:18
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    $\begingroup$ @JyrkiLahtonen That is not correct. For example the conjugates of $\sqrt{2}$ over the rationals are distinct but not linearly independent. $\endgroup$ – i. m. soloveichik Aug 6 '14 at 22:59
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    $\begingroup$ I was just pointing out that the degree of the minimal polynomial is equal to the number of distinct conjugates. Their eventual linear dependence is irrelevant. Also $\sqrt2$ has two distinct conjugates over $\Bbb{Q}$, so its minimal polynomial is quadratic (in accordance with my comment) - quite independently from the fact that its conjugates happen to be linearly dependent. $\endgroup$ – Jyrki Lahtonen Aug 6 '14 at 23:08
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Because $7$ is prime, we know that, $Aut(\mathbb{Q}[\zeta_7]/\mathbb{Q}) \cong \mathbb{Z}_7^\times \cong \mathbb{Z}_6$.

From here, we wish to find a generator for that automorphism group. Note that $\mathbb{Q}[\zeta_7]$ is the splitting field for the irreducible polynomial $f(x) = x^6 + x^5 + \cdots + x + 1$ whose roots are the $6$ nontrivial $7$th roots of unity. Now, it is a theorem that the automorphism group acts transitively on the roots of the polynomial $\iff$ that polynomial is irreducible. Therefore, we know there must exist an automorphism within this group such that $\zeta \mapsto \zeta^n$ for any $n \in \{1, 2, ..., 5\}$. The easiest way I know how to do this is through experimentation.

Once you've found your generator, $\phi$, then your automorphism group is simply $\{id, \phi, \phi^2, ..., \phi^5\}$.

The motivation behind all this work was that you need to let your entire automorphism group act on $\zeta + \zeta_5$ to generate the minimal polynomial. If we denote $\{\alpha_1, \alpha_2, ..., \alpha_k \}$ as the orbit of $\zeta + \zeta^5$ under this action, then the minimal polynomial will be be $m(x) = \displaystyle \prod_{i=1}^k\Big(x-\alpha_i\Big)$.

The moral of the story is that you had the right idea, but the previous sentence is where you went astray.


Edit: It is not necessary to actually find the generator, but it is necessary to let the entire automorphism group act on $\zeta + \zeta^5$. Note that there will be an automorphism that maps $\zeta \mapsto \zeta^k$ for all $k$ by the reasoning above, and each one must (obviously) be unique. Thus, the $6$ of them are:

  • $\zeta \mapsto \zeta$
  • $\zeta \mapsto \zeta^2$
  • $\zeta \mapsto \zeta^3$

etc.

To conclude, if possible, I would recommend you find a copy of Artin and read Chapter 14, Section 4, Proposition 4.4. I think that would get to the crux of where you went wrong initially. You can find a .pdf version of this text here: http://www.drchristiansalas.org.uk/MathsandPhysics/AbstractAlgebra/ArtinAlgebra.pdf

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  • $\begingroup$ Oh.. I was thinking i can take any $σ$ and consider its distinct conjugates.. $\endgroup$ – user87543 Aug 6 '14 at 8:47
  • $\begingroup$ I have got a problem... I could see that $\sigma(\zeta)=\zeta^5$ generates the galois group but the computation is becoming so messy...Could you suggest some way to make it simple.. $\endgroup$ – user87543 Aug 6 '14 at 10:33
  • $\begingroup$ @PraphullaKoushik, indeed. You can mod the exponent out by $7$ each time. That is, $\zeta \mapsto \zeta^5 \mapsto \zeta^{25}$. And at this point, recognize that $\zeta^{25} = \zeta^{4}$. You can continue in that fashion each time you iterate the automorphism. So the next one will be $\zeta^4 \mapsto \zeta^{20} = \zeta^6$. $\endgroup$ – Kaj Hansen Aug 6 '14 at 10:51
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    $\begingroup$ Multiply your cubic by its complex conjugate. Your cubic is $x^3+x^2+ \frac{1}{2}(3+\sqrt{7}I)x-1$ $\endgroup$ – i. m. soloveichik Aug 6 '14 at 13:21
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    $\begingroup$ @Praphulla Your polynomial is irreducible over $Q(\sqrt{-7})$ so multiply by its complex conjugate to get an irreducible polynomial over $Q$ $\endgroup$ – i. m. soloveichik Aug 6 '14 at 17:53
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Credits for this answer are for Jyrki Lahtonen and i.m.soloveichik

I have considered Only $(x-(\zeta+\zeta^5))(x-(\zeta^2+\zeta^3))(x-(\zeta^4+\zeta^6))$ and simplified this to $$x^3+x^2+(2+\zeta+\zeta^2+\zeta^4)x-1$$

Now i would consider other conjugates of $\zeta+\zeta^5$ namely $\zeta^6+\zeta^2;\zeta^5+\zeta^4;\zeta^3+\zeta$

With Jyrki Lahtonen's link Is $\sqrt 7$ the sum of roots of unity? we see that for $S=\zeta+\zeta^2+\zeta^4$ we have $$(2S+1)^2=-7\Rightarrow S=\dfrac{i\sqrt{7}-1}{2}\Rightarrow 2+\zeta+\zeta^2+\zeta^4=2+S=\dfrac{i\sqrt{7}+3}{2}$$

So, My polynomial is $$f(x)=(x^3+x^2+(\dfrac{i\sqrt{7}+3}{2})x+1)$$

As i was missing conjugates I would consider conjugate of the polynomial $f(x)$ and multiply with $f(x)$

Conjugate of $f(x)$ is

$$f(x)=(x^3+x^2+(\dfrac{-i\sqrt{7}+3}{2})x+1)$$

Now i multiply (I did with pen and paper and do not want to write those calculations here) and realized that their product is

$$g(x)=x^6+2x^5+4x^4+x^3+2x^2-3x+1$$

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  • $\begingroup$ You’ve got a couple of typos in your expression for the $X^2$-term of $g$. I got $g(X)=1 - 3X + 2X^2 + X^3 + 4X^4 + 2X^5 + X^6$ $\endgroup$ – Lubin Aug 6 '14 at 19:37
  • $\begingroup$ @Lubin: Edited.. Thank you :) $\endgroup$ – user87543 Aug 6 '14 at 20:02

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