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I try to work out what $\Phi_{12}(z)$ is:

By the fundamental theorem of arithmetic: $$\Phi_{12}(z)=(z-\xi^0)(z-\xi^1)(z-\xi^2)(z-\xi^3)(z-\xi^4)(z-\xi^5)(z-\xi^6)\\(z-\xi^7)(z-\xi^8)(z-\xi^9)(z-\xi^{10})(z-\xi^{11})$$ $$\Phi_{12}(z)=(z-\xi^0)(z-\xi^1)(z-\xi^2)(z-\xi^3)(z-\xi^4)(z-\xi^5)\\ (z+\xi^0)(z+\xi^1)(z+\xi^2)(z+\xi^3)(z+\xi^4)(z+\xi^5) $$

$$\Phi_{12}(z)=(z^2-\xi^{0\cdot 2})(z^2-\xi^{1\cdot 2})(z^2-\xi^{2\cdot 2})(z^2-\xi^{3\cdot 2})(z^2-\xi^{4\cdot 2})(z^2-\xi^{5\cdot 2})$$

$$\Phi_{12}(z)=(z^2-\xi^{0})(z^2-\xi^{2})(z^2-\xi^{4})(z^2-\xi^{6})(z^2-\xi^{8})(z^2-\xi^{10})$$

$$\Phi_{12}(z)=(z^2-\xi^{0})(z^2-\xi^{2})(z^2-\xi^{4})(z^2+\xi^{0})(z^2+\xi^{2})(z^2+\xi^{4})$$

$$\Phi_{12}(z)=(z^4-\xi^{0\cdot 2})(z^4-\xi^{2\cdot 2})(z^4-\xi^{4\cdot 2})$$

$$\Phi_{12}(z)=(z^4-\xi^{0})(z^4-\xi^{4})(z^4-\xi^{8})$$

$$\Phi_{12}(z)=(z^4-\xi^{0})(z^8-z^4(\xi^{4}+\xi^{8})+\xi^{4}\xi^{8})$$

$$\Phi_{12}(z)=(z^4-1)(z^8-z^4+1)$$

Why is $\Phi _{12} (z)$ given as $z^4-z^2+1$?

(Or, more precisely, what is wrong with my method (or am I not noticing their equivalence)?).

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up vote 1 down vote accepted

I suppose $\xi$ is a primitive $12^{th}$ root of unity.

The polynomial $\Phi_{12}(x)$ is the product $\prod(z-\xi^k)$ over the primitive $12^{th}$ roots of unity (i.e. for $k=1,5,7,11$).

What you are calculating is, $\displaystyle{\prod_{k=0}^{11}(z-\xi^k)}$, the product over all $12^{th}$ roots of unity which is just the polynomial $z^{12}-1$.

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Thanks. Just to complete the answer for myself: 'An nth root of unity is primitive if it is not a kth root of unity for some smaller k'. –  Alyosha Jan 17 '13 at 18:33
    
That is right @Alyosha. –  P.. Jan 17 '13 at 18:34
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