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How would one find the minimal polynomial of $e^{2πi/5}$ over $\mathbb Q$?

I have tried this:

$$\text {Let } a = e^{2πi/5}$$ $$\implies a = ({e^{2πi}})^{1/5}\implies a = 1^{1/5}\implies a - 1= 0$$

Since the polynomial $a - 1$ is monic and of least degree this therefore must be the minimal polynomial of $e^{2πi/5}$ over $\mathbb Q$?

I can't see why it's wrong, although it's different form related questions on this forum. Can someone show me what's wrong?

Bernard
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mildog8
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  • The identity $a^{bc}=(a^b)^c$ fails much more often for complex numbers than it does for real numbers. However, it holds true when $c$ is an integer. –  Aug 27 '17 at 11:26
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    $$\implies a = ({e^{2πi}})^{1/5}\implies a = 1^{1/5}\implies a - 1= 0$$

    Since $a-1=0$ implies $a=1$, it should be obvious you made a mistake somewhere along the way...

     – 5xum Aug 27 '17 at 11:26
  • It's wrong because $e^{\frac{2\pi i}{5} } \ne 1$ – rtybase Aug 27 '17 at 11:27
  • $a^5=1$, correct. But it does not follow that $a=1$. – GEdgar Aug 27 '17 at 11:41

1 Answers1

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Hint:

It is not, because $\mathrm e^{\tfrac{2\pi i}5}\ne 1$. Hence the minimal polynomial is a divisor of $$\frac{X^5-1}{X-1}=X^4+X^3+X^2+X+1.$$ Can you prove this polynomial is irreducible?

Bernard
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