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 polynimial 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?

  • $\begingroup$ 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. $\endgroup$ – user228113 Aug 27 '17 at 11:26
  • 1
    $\begingroup$ $$\implies a = ({e^{2πi}})^{1/5}\implies a = 1^{1/5}\implies a - 1= 0$$ Since $a-1=0$ implies $a=1$, it should be obvious you made a mistake somewhere along the way... $\endgroup$ – 5xum Aug 27 '17 at 11:26
  • $\begingroup$ It's wrong because $e^{\frac{2\pi i}{5} } \ne 1$ $\endgroup$ – rtybase Aug 27 '17 at 11:27
  • $\begingroup$ $a^5=1$, correct. But it does not follow that $a=1$. $\endgroup$ – GEdgar Aug 27 '17 at 11:41


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?


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