I am trying to determine the minimal polynomial of $e^{\frac{2 \pi i}{3}}$.
$e^{\frac{2 \pi i}{3}}=x$
$e^{2 \pi i}=x^3$
$1=x^3$
$x^3-1=0$
Now my problem is that I can not prove that $x^3-1$ is irreducible. Does anyone know how to show this?
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3Note that $x=1$ is a root of $x^3-1$ – J. W. Tanner Jun 21 '23 at 04:58
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@J.W.Tanner Oh, right. This means that $\frac{x^3-1}{x-1}=x^2+x+1$ is the minimal polynomial? – Sigi Jun 21 '23 at 05:00
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2$x^2+x+1$ is the minimal polynomial – J. W. Tanner Jun 21 '23 at 05:00
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$x=1$ is a root of $x^3-1$, so in fact $\dfrac{x^3-1}{x-1}=x^2+x+1$ is the minimal polynomial of $e^{\frac{2\pi i}3}$.
J. W. Tanner
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2Are you sure this particular question hasn't been covered earlier? Did you search at all? – Jyrki Lahtonen Jun 21 '23 at 05:10
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@JyrkiLahtonen: I searched and didn't find a duplicate. How about you? – J. W. Tanner Jun 21 '23 at 15:37
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11, 2. Wikipedia. This is just too basic to be deserving an answer even if there is no exact duplicate. I didn't look hard at all. – Jyrki Lahtonen Jun 21 '23 at 20:26