Timeline for Principal maximal ideals in Z[x]/(F)
Current License: CC BY-SA 3.0
17 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Mar 26, 2013 at 21:04 | vote | accept | Martin Brandenburg | ||
| Mar 20, 2013 at 4:34 | answer | added | user30180 | timeline score: 7 | |
| Mar 19, 2013 at 1:12 | comment | added | Venkataramana | @ayanta thanks for this. I think you may want to put this up as an answer since it answers the question completely (in the negative). | |
| Mar 19, 2013 at 0:49 | comment | added | user30180 | @Aakumadula: I misremembered the bijection between the sets of maximal ideals $P$ of $O_K$ with residue char. not dividing $[O_K:A]$ and maximal ideals $P'$ of $A$ with residue char. not dividing $[O_K:A]$ ($P \mapsto P \cap A$, $P' \mapsto P'O_K$), as we see by localizing at rational primes. Better idea: for $N := [O_K:A]$ we have $1 + NO_K \subset A$, so it suffices to find infinitely many maximal ideals of $O_K$ that are principal and admit a generator congruent to 1 modulo $N$. Now we apply "Dirichlet's proof" of Chebotarev, with characters of generalized ideal class groups. | |
| Mar 18, 2013 at 17:55 | comment | added | Martin Brandenburg | @Peter: Yes, this polynomial is better. I have checked for all primes $p \leq 101$ and all factors $f$ that $(p,f) \neq (f)$. Is it possible that $(p,f) \neq (f)$, but $(p,f)$ is principal? | |
| Mar 18, 2013 at 16:53 | comment | added | Peter Mueller | $x^4-4x^2+9$ seems to be a promising candidate. | |
| Mar 18, 2013 at 16:40 | comment | added | Peter Mueller | I think $F=x^4-10x^2+1$ doesn't work either, for $f=x+2$ is a divisor for $F$ mod $23$, but $23=(x^3 - 2x^2 - 6x + 12)f-F$. | |
| Mar 18, 2013 at 15:25 | comment | added | Venkataramana | @ayanta, I first thought you were right, but I am now confused. If $P\subset O_K$ is a maximal ideal whose residue char does not divide $[O_K:A]$, how can it lie in $A$? If $P$ lies in $A$, it means that $[O_K:P]$ is in fact a multiple of $[O_K:A]$. @martin brandenburg: $Z[x]/(F)$ is definitely an order (since its ${\mathbb Q}$ span is the number field and this is a subring of $O_K$. | |
| Mar 18, 2013 at 15:17 | comment | added | Martin Brandenburg | @ayanta: Thank you. Is this already an answer in the general case? If yes, can you post it as an answer and include more details for those (like me) who are not experienced with algebraic number theory? I'm not sure if $\mathbb{Z}[x]/(F)$ is an order. | |
| Mar 18, 2013 at 15:15 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Mar 18, 2013 at 14:52 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Mar 18, 2013 at 13:08 | comment | added | Peter Mueller | Why the downvote - looks like an attractive question to me! (Now compensated by an upvote) | |
| Mar 18, 2013 at 13:02 | comment | added | user30180 | @Aakumadula: To complete your thought, for any number field $K$ and order $A$ in the ring of integers $O_K$, the finiteness of the index of $A$ in $O_K$ as an abelian group implies that maximal ideals of $O_K$ whose residue characteristic does not divide $[O_K:A]$ lie entirely inside $A$. Thus, principal maximal ideals of $O_K$ whose residue characteristic does not divide $[O_K:A]$ do the job (i.e., they are also principal maximal ideals of $A$). | |
| Mar 18, 2013 at 12:03 | comment | added | Venkataramana | It seems to me that Cebotarev density will imply that there are infinitely many maximal ideals which are principal (this will certainly be true if $A={\mathbb Z}[x]/(F)$ is the ring of integers in the number field ${\mathbb Q}[x]/(F)$). The issue may be to prove Cebotarev for these orders $A$. | |
| Mar 18, 2013 at 11:23 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Mar 18, 2013 at 10:58 | history | asked | Martin Brandenburg | CC BY-SA 3.0 |