<s>[INST] Which of the following is an irreducible factor of $x^{12}-1$ over $\mathbb Q$? Which of the following is an irreducible factor of $x^{12}-1$ over $\mathbb Q$?  


*

*$x^8+x^4+1$    

*$x^4+1$   

*$x^4-x^2+1$   

*$x^5-x^4+x^3-x^2+x-1$    


Is the answer $(1)$ correct? I am not sure which one is.
 [/INST] By repeatedly using the simple factorizations:
$$x^{2n}-1=(x^n+1)(x^n-1)$$
and
$$x^{3n}-1=(x^{2n}+x^n+1)(x^n-1)$$
and
$$x^{3n}+1=(x^{2n}-x^n+1)(x^n+1)$$
We get:
$$x^{12}-1=(x^4-x^2+1)(x^2+1)(x^2-x+1)(x+1)(x^2+x+1)(x-1)$$
Ruling out every choice except for number 3:
$$x^4-x^2+1$$
Call this $p(x)$. There are many ways to show that $p(x)$ is irreducible. For example, we can note that it has no real roots, and hence it must be a product of two 2nd degree polynomials, if it is not irreducible. But $p(2)=p(-2)=13$ and $p(3)=p(-3)$ and $p(4)=p(-4)=241$, so it is a prime number for at least 6 different integer values of $x$. However, a product of two second degree polynomials with integer coefficients can only be prime for $4$ integer values of $x$, since one of the factors must be either $1$ or $-1$ for $p(x)$ to be prime. 
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