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I'm looking for a polynomial $f \in \mathbb{Q}$ of degree $3$ that is irreducible over $\mathbb{Q}$, such that the polynomial's splitting field $L/\mathbb{Q}$ has a Galois group $G\text{al}(L/\mathbb{Q}) = G$ with $G$ isomorfic to the cyclic group of order 3.

The question actually asks me to give this polinomial in a seventh root of unity $\zeta$ with $\zeta^7 = 1$. I guess i should be able to write it in $\mathbb{Q}(\zeta)$.

Could anyone help me find this polynomial, or give me a hint?

Thanks!

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  • $\begingroup$ Find the minimal polynomial for $\zeta+(1/\zeta)$. $\endgroup$
    – Paramanand Singh ♦
    Apr 1 at 0:40
  • $\begingroup$ Let's do that $\gamma=\zeta + \frac{1}{\zeta}$ with $\zeta$ being a seventh root of unity, $\zeta^7 = 1$ hence $\zeta^{7-i} = \zeta^{-i}$ for $i \in \mathbb{Z}$. We have, $$\gamma^2 = \zeta^2 + 2 + \zeta^{5}$$ $$\gamma^3= \zeta^3 + 3\zeta + 3\zeta^{6} + \zeta^4 $$ $$\gamma^4= \zeta^4 + 4\zeta^2 + 6 + 4\zeta^5 + \zeta^3$$ Now we see that for $f(x) = x^4 - x^3 - 4x^2 + 3x + 2$ our $\gamma$ is a root. This is no degree 3 polynomial though, would it be wise to invoke the fact that $\gamma$ is a real value because we're adding complex conjugates? $\endgroup$
    – Dorelanië
    Apr 5 at 12:14
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    $\begingroup$ If $\zeta$ is a 7th root of unity and $\zeta\neq 1$ then we have $\sum_{i=0}^6\zeta^i=0$. Divide this by $\zeta^3$ to get $1+\sum_{i=1}^3\zeta^i+\zeta^{-i}=0$ The sum in last equation can be written in terms of $\zeta+(1/\zeta)$ and result will be a degree 3 polynomial $\endgroup$
    – Paramanand Singh ♦
    Apr 6 at 0:47
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    $\begingroup$ By the way the degree 4 polynomial you found is reducible and it has $x=2$ as root corresponding to $\zeta=1$ (real root of $\zeta^7=1$). Factor out the $(x-2)$ from $f(x) $ and you get the desired degree 3 irreducible polynomial. $\endgroup$
    – Paramanand Singh ♦
    Apr 6 at 0:52

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Note that $\mathbb{Q}(\zeta_7)$ is definitely not quite the extension you want, since that gives you the whole splitting field of 7-th cyclotomic polynomial, which has degree $\phi(7) = 6$. However you can find the extension you're looking for as an intermediate extension of $\mathbb{Q}(\zeta_7)/\mathbb{Q}$. A nice way to approach this is to note that in order to get an extension of degree three, all roots must be real, or in other words the extension must be fixed by complex conjugation.

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