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The complex number $z = e^{2\pi i/5}$ is a fifth root of the unity: $z^5 = 1$. Find the minimal polynomial of $z$ over $\mathbb{Q}$.

I tried to solve this by converting the z into the term of $a + bi$, which is $\cos\left(\frac{2\pi}{5}\right) + i\sin\left(\frac{2\pi}{5}\right)$. However I found it doesn't work. So I assume the minimal polynomial is just $x^5-1$. I am not sure about it. Could you please help me to solve this question?

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    $\begingroup$ $x-1$ is a factor of $x^5-1$, so the latter is not the minimum poly, but - hint - the min poly is another factor of $x^5 -1$... to be a min poly, it will have to be irreducible, and have $z$ as a root - you will have to prove both. $\endgroup$
    – peter a g
    Mar 23, 2016 at 2:50

2 Answers 2

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$0=z^5-1=(z-1)(z^4+z^3+z^2+z+1)$ and $z\ne 1$ so $f(z)=0$ where, for $x\in C,$ $$f(x)=x^4+x^3+x^2+x+1.$$ $$\text {Let }\quad g(x)= f(x+1)=x^4+5 x^3+10 x^2+10 x+5.$$ By Eisenstein's Criterion, $g(x)$ is irreducible over $Z$, and by a theorem of Gauss, $g(x)$ is therefore irreducible over $Q.$ So $g(x-1)=f (x)$ is also irreducible over $Q.$ So $f$ is the minimal polynomial of $z$ over $Q.$

Theorem (Gauss). If a polynomial with integer co-efficients is irreducible over $Z$ then it is irreducible over $Q.$

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  • $\begingroup$ On the first line, you wrote $z \not = 0$; don't you mean $z\not =1 $? $\endgroup$
    – peter a g
    Mar 23, 2016 at 11:26
  • $\begingroup$ @peterag. Yes thanks. Yet another in my big fine collection of missed typos. I fixed it. $\endgroup$
    – DanielWainfleet
    Mar 23, 2016 at 19:00
  • $\begingroup$ My very first post contained an incredibly embarrassing one - something like $ \ln a^b = a \ln b$. $\endgroup$
    – peter a g
    Mar 23, 2016 at 19:09
  • $\begingroup$ How does $f(x)$ such that $f(a)=0$ being irreducible imply that $f(x)$ is the minimal polynomial for which it is zero at $a$? $\endgroup$
    – Benji Altman
    Jun 27, 2018 at 0:47
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    $\begingroup$ @BenjiAltman. Let $h(x)$ be the minimal polynomial of $a$ over $\Bbb Q.$ We have deg $(h)>0.$ There exist polynomials $j,k$ with $ f(x)=h(x)j(x)+k(x),$ by the long-division algorithm, with deg ($k)<$deg ($g$). (All co-efficients being rational.) So if $f(a)=h(a)=0$ then $k(a)=0.$ By the minimality of deg ($h$), therefore $k(x)=0$ for all $x.$ So $f(x)=h(x)j(x)$ for all $x.$ But $ f$ is irreducible over $\Bbb Q$ so $ j$ is a constant. $\endgroup$
    – DanielWainfleet
    Jun 27, 2018 at 10:28
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As others have noted, $x=e^{2\pi i/5}$ is a root of $P(x)=x^4+x^3+x^2+x+1$. To show that $P$ is irreducible, work mod $2$, in which the irreducible polynomials of degree $1$ and $2$ are $x$, $x+1$, and $x^2+x+1$. Since $P(0)\equiv P(1)\equiv1$ mod $2$, $P$ has no linear factors mod $2$, and

$$(x^2+1+1)^2\equiv x^4+x^2+1\not\equiv P(x)\mod2$$

Hence $P$ is irreducible.

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