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Let ω be a primitive 7th root of unity in $\Bbb C$ and set $α := ω + ω ^6$ . Determine, with justification, the minimum polynomial of α over Q.

Would one use logs in such a question , or how should one begin? it's the roots of unity that are throwing me off, I know how to these minimal polynomial questions for radicals etc..

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  • $\begingroup$ How could you find the other roots of this minimal polynomial? $\endgroup$
    – Geoffroi
    Apr 5, 2018 at 4:56
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    $\begingroup$ Multiply both sides of $\alpha=\omega+ \omega^6$ by $\omega$ $\endgroup$
    – saulspatz
    Apr 5, 2018 at 4:56
  • $\begingroup$ Alt. hint: $\omega^6 = \overline{\omega}$, so $\alpha = 2 \operatorname{Re}(\omega)$. You'd expect to find a cubic for it in the end. $\endgroup$
    – dxiv
    Apr 5, 2018 at 5:03
  • $\begingroup$ @dxiv what is the form of w ? $\endgroup$
    – excalibirr
    Apr 5, 2018 at 5:22
  • $\begingroup$ @dxiv by that i mean , as an equation involving real and imaginary parts what does it look like ? $\endgroup$
    – excalibirr
    Apr 5, 2018 at 5:28

4 Answers 4

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One has the following $\omega^7=1$ and $\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+1=0$ Now we compute

$$\alpha=\omega+\omega^6$$ $$\alpha^2=\omega^2+\omega^5+2$$ $$\alpha^3=\omega^3+\omega^4+3\alpha$$

Adding the three identities and rearranging one gets

$$\alpha^3+\alpha^2-2\alpha-1=0$$

And $X^3+X^2-2X-1$ is irreducible over $\Bbb{Q}$ because it has no integer roots and therefore no rational roots (it is monic). So we have the minimal polynomial of $\alpha$ over $\Bbb{Q}$

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Let ω be a primitive $7^{th}$ root of unity in $\Bbb C$.

Hint: $\;\omega^7=1, \omega \ne 1\,$, so $\,\omega^6 = \dfrac{1}{\omega}\,$ and $\,\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+1=0\,$. Dividing by $\,\omega^3\,$:

$$ \left(\omega^3 + \dfrac{1}{\omega^3}\right) + \left(\omega^2 + \dfrac{1}{\omega^2}\right) + \left(\omega + \dfrac{1}{\omega}\right) + 1 = 0 \tag{1} $$

and set $α := ω + ω ^6$ . Determine, with justification, the minimum polynomial of α over Q.

$\alpha=\omega+\omega^6 = \omega + \dfrac{1}{\omega}\,$, then:

$$ \alpha^2 = \omega^2+\dfrac{1}{\omega^2}+2 \\ \alpha^3 = \omega^3+\dfrac{1}{\omega^3}+3\left(\omega + \dfrac{1}{\omega}\right) $$

Express $\,(1)\,$ in terms of $\,\alpha\,$ using the above, and you get the cubic equation that $\,\alpha\,$ satisfies. Then show that's the minimal polynomial, indeed.

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OK, here's what I did:

Starting with

$\omega^7 = 1 \ne \omega, \tag 1$

$\alpha = \omega^6 + \omega, \tag 2$

I computed as follows:

$\alpha^2 = \omega^{12} + 2\omega^7 + \omega^2 = \omega^5 + \omega^2 + 2. \tag 3$

where I have used (1) to reduce $\omega^{12}$ to $\omega^5$ and $\omega^7$ to $1$; then,

$\alpha^3 = \omega^{18} + 3\omega^{13} + 3\omega^8 + \omega^3$ $= \omega^4 + 3\omega^6 + 3 \omega + \omega^3 = \omega^6 + \omega^4 + \omega^3 + \omega + 2(\omega^6 + \omega); \tag 4$

$\alpha^3 + \alpha^2 = \omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 2(\omega^6 + \omega) + 2$ $= \omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1 + (1 + 2(\omega^6 + \omega)), \tag 5$

again using (1); also from (1),

$\omega^7 - 1 = 0, \tag 6$

or

$(\omega - 1)(\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1) = 0; \tag 7$

since

$\omega \ne 1, \tag 8$

we have

$\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1 = 0; \tag 9$

now (5) becomes

$\alpha^3 + \alpha^2 = 1 + 2\alpha, \tag{10}$

or

$\alpha^3 + \alpha^2 - 2\alpha - 1 = 0. \tag{11}$

Thus $\alpha$ is a root of the polynomial

$\chi(x) = x^3 + x^2 - 2x - 1 \in \Bbb Q[x]; \tag{12}$

I found this polynomial to be irreducible over $\Bbb Q$ as follows: if $\chi(x)$ is reducible over $\Bbb Q$, it must have a monic factor of degree $1$, since $\deg \chi = 3$; such a factor must be of the form $x - \rho$, where $\rho \in \Bbb Q$; in this event $\rho$ must be a zero of $\chi(x)$, as is well-known; thus we may write

$\rho = \dfrac{p}{q}, \; p, q \in \Bbb Z, \tag{13}$

and we may of course assume that

$\gcd(p, q) = 1; \tag{14}$

thus

$\chi \left (\dfrac{p}{q} \right ) = \chi(\rho) = 0; \tag{15}$

we write out $\chi(p/q)$ using (12); then (15) yields

$\left ( \dfrac{p}{q} \right )^3 + \left ( \dfrac{p}{q} \right )^2 - 2 \left ( \dfrac{p}{q} \right ) - 1 = 0, \tag{16}$

or

$\dfrac{p^3}{q^3} + \dfrac{p^2}{q^2} - 2\dfrac{p}{q} - 1 = 0; \tag{17}$

we multiply (17) by $q^3$:

$p^3 + p^2 q - 2pq^2 - q^3 = 0, \tag{18}$

which may be written

$p(p^2 + pq - 2q^2) = q^3, \tag{19}$

and we thus see that

$p \mid q^3; \tag{20}$

now suppose $p = 1$; then (18) yields

$1 + q - 2q^2 - q^3 = 0, \tag{21}$

or

$q^3 + 2q^2 - q - 1 = 0; \tag{22}$

it is easy to see that this equation has no integer solutions; if we adopt the notation

$\theta(x) = x^3 + 2x^2 - x - 1 = x^3 + 2x^2 - (x + 1) \in \Bbb Q[x], \tag{23}$

then

$\theta(0) = -1; \; \theta(1) = 1; \; \theta(2) = 15; \; \theta(3) = 41, \tag{24}$

and $\theta(m)$ continues to grow with increasing $m$ since the $x^3 + 2x^2$ terms dominate $x + 1$; one can similarly see that

$\theta(-1) = 1; \; \theta(-2) = 1; \; \theta(-3) = -7; \; \theta(-4) = -29; \theta(-5) = -71, \tag{25}$

and $\theta(m)$ continues to decrease with decreasing $m$ since now the cubic term dominates; we thus see that there is no $q \in \Bbb Z$ satisfying (22), therefore we may rule out the possibility that $p = 1$; then there is some prime $r$ with

$r \mid p; \tag{26}$

from (20) we then also have

$r \mid q^3, \tag{27}$

thus since $r$ is prime we conclude

$r \mid q; \tag{28}$

we now have both $r \mid p$ and $r \mid q$, whence

$r \mid \gcd(p, q) \Longrightarrow \gcd(p, q) \ne 1, \tag{29}$

in contradiction to our assumption (14); therefore $\chi(x)$ has no rational zeroes; thus, it is irreducible over $\Bbb Q$ and hence it is minimal for $\alpha$.

Note: Phew! That's pretty long-winded! It certainly would have been substantially shorter if I had remembered Gauss's Lemma, which I only seem to have recalled after reading marwalix's answer. I'm getting the feeling there are a few things it would behoove me to review . . . End of Note.

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It's important to see the easy reason you expect a cubic, since that also gives you a direct way to compute the polynomial. The Galois group for $\omega$ is $\mathbb{Z}/6\Bbb{Z}$, generated by sending $\omega$ to $\omega^{3}$. Applying this once, $\omega^{3} + \omega^{-3}$ is a conjugate of $\omega + \omega^{-1}$. Applying it again yields $\omega^{2} + \omega^{-2}$ since 3*3 = 2 mod 7. Applying a third time gets you back where you started since 3*3*3 = 27 = -1 mod 7. Thus, the minimal polynomial must be $(x - (\omega + \omega^{-1})) + (x - (\omega^{2} + \omega^{-2})) + (x - (\omega^{3} + \omega^{-3}))$, and the rest is simple arithmetic (and remembering the all 7 7th roots of 1 sum to 0, so that the six of them other than 1 sum to -1).

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