0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.