This naive question came as the last problem in my homework. The author asked me to use linear relations of the discriminant like $\operatorname{disc}(ra_{1},a_{2},...,a_{n})=r^{2}\operatorname{disc}(a_{1},...,a_{n})$, $\operatorname{disc}(a_{1}+\beta,a_{2},...,a_{n})=\operatorname{disc}(a_{1},...,a_{n})$ for $\beta$ a linear combination of $a_{i}$. I could not see how to make use of the hint or how to solve the problem in an easier way than taking the norm and evaluate, which is impractical since $46\times 23$ terms are involved.
Is there any trivial reason for $2$ is irreducible in $\mathbb{Z}[\omega],\omega=e^{\frac{2\pi i}{23}}$?
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0Do you mean irreducible as an element in $\mathbf Z[\omega]$? – 2012-03-08
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0Remember that for a general number field $K$ with integers $\mathcal O$ and $\alpha \in \mathcal O$ we have $|N^K_{\mathbf Q}(\alpha)| = (\mathcal O : (\alpha))$. This last number is something you could get from relating the discriminants of $\mathcal O$ and $(\alpha)$. What's a good basis for the latter? – 2012-03-08
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1You may find something useful in the links in my answer to http://math.stackexchange.com/questions/85520/fermats-last-theorem-and-kummers-objection – 2012-03-08
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0@DylanMoreland: I could prove it is irreducible in $A\cap \mathbb{Q}(\sqrt{-23})$, but I could not prove it is irreducible in $A\cap \mathbb{Q}(\omega)$. – 2012-03-08
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0@DylanMoreland: I think it is a naive approach, but supoose $(2)\in (\alpha)$, and write $\alpha=\sum^{22}_{i=0} a_{i}\omega^{i}$. Then by your hint we have $|N(\alpha)|=(\mathbb{Z}[\omega]/\alpha)$. But this is not very clear - oh, do you mean using the integral basis of $O$ including $\alpha$? – 2012-03-08
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0@ChangweiZhou It seemed like a way to use your hint. Now I'm not so sure. I'll try to think about it later. – 2012-03-08
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0@DylanMoreland: I should ask my professor about this on class. – 2012-03-08
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0@GerryMyerson: The links you gave in that page is only tangentially related to this problem. But the factorization you give is very surprising to me. – 2012-03-10
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0@ChangweiZhou I don't think it's trivial to prove that $2$ is irreducible in the given number ring. – 2012-12-26