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Sorry to ask the trivial notion.

If we consider the relative extension of number field L=K(a) ( a in S, a primitive element) over K and S & R are ring of integers of L & K respectively.

Also, Let B be the prime of S so that lying above the prime p in R. My question is the generator for the residue field S/B over R/p (Finite field extension always be simple extension) can be always written R/p[a'] ?,

where a' is a projection of a from S to S/B. Actually, I think this is true for S is monogenic.

Thank you very much!

Greg Martin
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Thomas
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  • Many thanks for your comments. I mean the generator in S/B over R/p since S/B is finite extension over finite field R/p. – Thomas Oct 25 '16 at 03:19
  • There we see that $p = \prod_{j=1}^k P_j^{e_j}$ and with $F_j = \mathcal{O}_L/(P_j), F = \mathcal{O}_K/(p)$ then $[F_j:F] = f_j$. So $F_j = F(\alpha_j)$, but they don't mention $\alpha_j$ having a simple description – reuns Oct 25 '16 at 03:22
  • Good Point. So that's why I am wondering if we can always choose the generator as same as the field under certain assumption. I know if S is monogenic then this holds. – Thomas Oct 25 '16 at 03:26
  • Can you explain why in the case $\mathcal{O}_L = \mathcal{O}_K[a]$ it becomes easier ? – reuns Oct 25 '16 at 03:28
  • Since the coefficient modulo p is the same modulo B. – Thomas Oct 25 '16 at 03:38

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Almost surely not is the short answer. The main problem is that there are usually multiple primes above a given prime in an extension, so the problem is not even well-defined. Consider, for example, $K=\Bbb Q(\sqrt[3]{2})$ considered as an extension of $\Bbb Q$, and here we know the ring of integers is monogenetic, i.e. $\mathcal{O}_K=\Bbb Z[\sqrt[3]{2}]$. However, let $p=5$. Then we know that the minimal polynomial $x^3-2$ factors as $(x+3)(x^2+2x+4)$ Then there are two primes, each of which is generated by $5$ and some root of the minimal polynomial.

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Adam Hughes
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  • Many thanks for the comments. If we consider $\mathcal{O}{K}/(5,x^2+2x+4)$ is finite field of 25 elements and we can view $\mathcal{O}{K}/(5,x^2+2x+4)$ as $\Bbb{Z}{5}(\bar{\sqrt[3]{2}})$ where $\bar{\sqrt[3]{2}}$ is the element in $\mathcal{O}{K}/(5,x^2+2x+4)$. So is the $\mathcal{O}_{K}/(5, x+3)$. That is, we can take the same generator from ring of integer. I am not sure it is correct? Thank you! – Thomas Oct 25 '16 at 21:35
  • @Thomas the symbol $\sqrt[3]{2}$ is somewhat abused in math, remember there are 3 different cube roots of $2$. One of them generates a degree $2$ extension for $\Bbb Z/5$, and the others do not. That is to say: there is already one cube root of 2 in $\Bbb Z/5$, namely $3$ since $3^3=27\equiv 2\mod 5$. if you adjoin $3$, you aren't extending anything, but if you adjoin a different cube root of $2$, it does. That's why your question doesn't make as much sense, the extension depends on which choice of root you make, and there is no canonical choice. – Adam Hughes Oct 25 '16 at 21:50
  • Got it! Thank you. But we know finite extension of finite fields is always simple extension is there any relation between the generator of residue field and the generator of ring of integer if under specific assumption? – Thomas Oct 26 '16 at 00:21
  • @Thomas not really, this phenomenon I've showed is quite general. If you're not dealing with a number field, but instead a Henselian ring, then yes, but that's more of less by definition. – Adam Hughes Oct 26 '16 at 02:57