# In $\mathbb{Z}[\zeta_3]$ is the product of conjugates equal to the conjugate of the product?

For $\alpha, \beta \in \mathbb{Z}[\zeta_3]$ , is it true that $(\alpha \cdot \beta)'=\alpha'\cdot \beta'$?

Also,

For $\alpha, \beta \in \mathbb{Z}[\zeta_3]$ , is it true that $(\alpha + \beta)'=\alpha'+ \beta'$?

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as I have said, this depends on your definition of conjugate. – Qiaochu Yuan Mar 1 '11 at 22:42
What definition do I need for this to work? – Jason Smith Mar 1 '11 at 22:44
I would like to say things like: Taking conjugates of $z-\zeta y=\zeta'\alpha^3$, we have $z-\zeta' y=\zeta (\alpha')^3$. Thanks a lot. – Jason Smith Mar 1 '11 at 22:51
for imaginary quadratic fields the simplest definition is complex conjugation. But my point is that all of the definitions should be equivalent, so the only thing to prove is their equivalence (or prove the standard properties from any given definition), so you should just pick one. – Qiaochu Yuan Mar 2 '11 at 0:24
Thanks. I get it now...it took a bit to sink in. – Jason Smith Mar 2 '11 at 18:04

If $F$ is a field of characteristic different from $2$, and $K$ is an extension with $[K:F]=2$, then the conjugate of $\alpha\in K$ is the image of $\alpha$ under the unique nontidentity element $\sigma\in\mathrm{Gal}(K/F)$ (this agrees with the usual "complex conjugation", since complex conjugation is the unique nonidentity element of $\mathrm{Gal}(\mathbb{C}/\mathbb{F})$). Since $\sigma$ is a field homomorphism, you automatically get both $$\sigma(\alpha+\beta) = \sigma(\alpha)+\sigma(\beta)\qquad\text{and}\qquad \sigma(\alpha\beta) = \sigma(\alpha)\sigma(\beta).$$

In particular, this holds for the unique nonidentity automorphism of $\mathbb{Q}(\zeta_3)$ over $\mathbb{Q}$ (which, as it happens, is complex conjugation). In particular, it holds for elements of $\mathbb{Z}[\zeta_3]$ (which, as it happens, is closed under this automorphism) , so that the restriction of conjugation is in fact also an automorphism of the ring $\mathbb{Z}[\zeta_3]$.

Of course, you can try defining different kinds of "conjugate". Instead of applying the automorphism of $\mathbb{Q}(\zeta_3)$, you could try defining "conjugate" by $a+b\zeta_3\mapsto a-b\zeta_3$ with $a,b\in\mathbb{Z}$. If you do that, then it's not a ring homomorphism, because $$(a+b\zeta_3)(x+y\zeta_3) = ax+\zeta_3^2by + (ay+bx)\zeta_3 = ax-by + (ay+bx-by)\zeta_3$$ but $$(a-b\zeta_3)(x-y\zeta_3) = ax+\zeta_3^2by - (ay+bx)\zeta_3 = ax-by - (ay+bx+by)\zeta_3$$ so the "conjugate" of the product is not the product of the "conjugates". That's why one uses the field automorphism.

Added. I focused on extensions of degree $2$ to be able to refer to the conjugate. One can of course consider extensions of larger degree or in charactersitic $2$.

In the general case, the conjugates (plural in general) of $\alpha$ in $K$ are the images of $\alpha$ under automorphisms that lie in $\mathrm{Aut}(K/F)$. Under any particular automorphism $\sigma$ it is still true that $\sigma(\alpha\beta) = \sigma(\alpha)\sigma(\beta)$, but if you consider different automorphisms (different notions of "conjugate") of course things go astray. One can also consider the norm, $N_{K/F}$, which maps from $K$ to $F$ by $N_{K/F}(\alpha) = \prod_{\sigma\in G} \sigma(\alpha)$, where $G$ is the Galois group of the Galois closure of $K$ over $F$. The norm is multiplicative: $N_{K/F}(\alpha\beta) = N_{K/F}(\alpha)N_{K/F}(\beta)$; and a very important map. But I doubt this is what you were refering to.

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If conjugation means mapping $\zeta$ to $\zeta^2=1/\zeta$, then it coincides with complex conjugation, which is a ring homomorphism.

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