# Find the irreducible polynomial over $Q$(linear combination of primitive cubic roots)

Hi I'm student who just started the algebra.

There are some question that bothering me.

Let $$\omega = e^\frac{2\pi i }{7}$$

I've already known the $$irr(\omega,Q) = w^6 +w^5 +w^4+w^3+w^2+w+1$$

(Here, $$irr(\omega,Q )$$ means irruducible polynomials over Q whose root is $$\omega$$)

Then the question bothering me that

*Find the $$irr(w+w^2+w^4,Q)$$ (linear combination of Roots of unity)

Let the $$\beta =w+w^2+w^4$$ for convinience

To find $$irr(\beta,Q)$$ , I used the $$irr(\omega,Q) = w^6 +w^5 +w^4+w^3+w^2+w+1$$

Like $$w^2(w^4+w^2+w)+...$$ or divide $$w^3$$

But failed. :(

HOW COULD I SOLVE THAT????????

Plus, Not only the case the $$\omega = e^\frac{2\pi i }{7}$$

If we consider the more generized form like when the $$\omega_1 = e^\frac{2\pi i }{n}$$

Are there any method to find the irruducible polynomials over Q whose roots are linear combination of Roots of unity??? (I guess there aren't any generized method which means It depends on situations.)

If there are any generlized form exists please answer me. :)

e.g.) Find the$$irr( \beta =w_1+w_1^k+w_1^m,Q)$$

## 1 Answer

In this case, note that for $$\beta=\omega+\omega^2+\omega^4$$, $$\beta^2= \underbrace{\omega^2+\omega^4+\omega^8}_\beta+2(\underbrace{\omega^3+\omega^5+\omega^6}_{-1-\beta})=-\beta-2$$, so $$\beta^2+\beta+2=0$$.

One other approach, which can be used more generally, if you know some Galois theory, is to calculate $$Gal(\mathbb Q(\omega)/\mathbb Q)$$. Then you calculate $$\{f(\beta)\mid f\in Gal(\mathbb Q(\omega)/\mathbb Q)\}$$. If this set is $$\{\beta_1,\dots,\beta_k\}$$, then the minimal polynomial of $$\beta$$ over $$\mathbb Q$$ is $$(X-\beta_1)\dots(X-\beta_k)$$.

• I appreciated your support Mr. @SMM. – se-hyuck yang May 17 at 11:33