# 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.)

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

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