How to prove that $ (\zeta_{p^{n+1}}-1)^{p} $ = $ (\zeta_{p^{n}}-1) $ as ideals where $ \zeta_{n} $ is a primitive nth root of unity ?
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1$\begingroup$ False statements tend to be hard to prove. $\endgroup$– Emil JeřábekCommented May 16, 2013 at 15:07
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$\begingroup$ It's not false. Look at Washington's book on cyclotomic fields in the appendix on infinite Galois theory and ramification. $\endgroup$– SumanCommented May 16, 2013 at 15:28
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$\begingroup$ Well they certainly have the same valuation. $\endgroup$– ArijitCommented May 16, 2013 at 15:35
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$\begingroup$ I suspect that Emil made the mistake of looking at $\zeta_{p^0}-1$. $\endgroup$– Tom GoodwillieCommented May 16, 2013 at 15:57
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3$\begingroup$ @Tom: My comment concerned the first version of the question, where it was impossible to guess that the OP is actually talking about ideals. It looked like a plain identity between two numbers (which were actually distinct). $\endgroup$– Emil JeřábekCommented May 16, 2013 at 16:06
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1 Answer
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Suppose $\zeta _{p^{n+1}}=x$ say.Then, for every $\omega $ a $p$-th root of unity, $x\omega-1=x^r-1$ (for some $r$) generates the same ideal as $x-1$ (clearly, it is contained in the ideal generated by $x-1$; by switching $x$ and $x\omega$ we get the other statement).
Taking the product over all the $\omega $ we now get the result you wanted.
answered May 16, 2013 at 15:50