I have read that for any finite field $\mathbb{F}_q$ there exists a number field $F$, and some prime ideal $P$ of $\mathcal{O}_F$ such that $$\mathbb{F}_q \cong \mathcal{O}_F / P.$$ The ring of integers of $F$ is denoted by $\mathcal{O}_F$. Is this fact very well-known? I have had some Algebraic Number Theory but never encountered this.


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    $\begingroup$ Please note that the standard notation for a quotient structure is $A/B$, not $\dfrac{A}{B}$. $\endgroup$ – Zev Chonoles Feb 16 '12 at 15:57
  • $\begingroup$ The short version: your statement is essentially equivalent to "For any positive natural number $d$ and (rational) prime $p$, there exists a number field with (a ring of integers that has) a degree $d$ prime ideal lying over $p$". It's nearly equivalent to "There exists an irreducible integer polynomial that has an irreducible degree $d$ factor modulo $p$". $\endgroup$ – user14972 Mar 11 '12 at 18:05

There are lots of facts of this kind that are well-known to algebraic number theorists, but may not be in standard textbook accounts. The general theme of which your question is an instance is that of "global realization of a given local situation", namely can a given local situation (e.g. extension of local fields, or something analogous) be obtained from an appropriate global context (e.g. an extension of number fields) by completing.

In your particular case, there are probably many ways to answer the question, but here is one approach, which is fairly flexible.

Start with your finite extension $\mathbb F_q$ over $\mathbb F_p$. Now this can be obtained from an extension of local fields, i.e. we can find an extension $K$ of $\mathbb Q_p$ whose residue field equals $\mathbb F_q$. E.g. if $q = p^f$ then we can take $K$ to be the unramified extension of $\mathbb Q_p$ of degree $f$.

Now if we write $K = \mathbb Q_p(\alpha),$ let $f(x)$ be the minimal polynomial of $\alpha$ over $\mathbb Q_p$. Since $\mathbb Q$ is dense in $\mathbb Q_p$, we may approximate $f(x)$ as closely as we like by an element $g(x) \in \mathbb Q[x]$. If we assume that $g$ is close enough to $f$, then $g$ will also have a root $\beta$ in $K$ which is very close to $\alpha$ (in particular, so that $K = \mathbb Q_p(\beta)$.) (This is a version of Krasner's lemma; in fact it can be deduced pretty directly from Hensel's lemma.)

Now if let $F = \mathbb Q(\beta)$ then $F$ is a number field (since $\beta$ is the root of a polynomial in $\mathbb Q$), and the given embedding $F = \mathbb Q(\beta)\hookrightarrow K = \mathbb Q_p(\beta)$, which as dense image, corresponds to a prime ideal $P$ of $\mathcal O_F$ such that $F_P = K$. In particular, the residue field of $P$ equals the residue field of $K$, i.e. $\mathbb F_q$.

If we do choose $K$ to be unramified over $\mathbb Q_p$ of degree $f$, then we can choose $\alpha = \zeta$, a $(p^f-1)$st root of $1$, which is already algebraic over $\mathbb Q$. Thus we can take $\alpha = \beta = \zeta$, and so we can set $F = \mathbb Q(\zeta)$, the field generated by $(p^f-1)$st roots of unity. (Thus we get essentially the same answer as that indicated by Cam McLeman.)

Note, though, that the above result proves something stronger --- namely not just that any finite extension of $\mathbb F_p$ can be obtained as the residue field of a prime ideal in a number field, but that any finite extension of $\mathbb Q_p$ can be obtained as the completion at a prime ideal of a number field.

Furhermore, these Krasner-type arguments are very flexible and can be generalized in various directions. Suitably strengthened versions play an important role in current research related to Galois representations and automorphic forms.

  • $\begingroup$ Ah, of course, thanks. I was too focused on adjoining only a prime-th root of unity. $\endgroup$ – Cam McLeman Feb 16 '12 at 17:17

Write $q=p^f$ for some prime $p$, and let $\ell$ be any prime dividing $p^f-1$. Let $F=\mathbb{Q}(\zeta_{\ell})$. Then the inertia degree of $p$ in this extension is $f$, and so if $\mathfrak{p}$ denotes a prime of $F$ above $p$, then $\mathbb{Z}[\zeta_{\ell}]/\mathfrak{p}\cong \mathbb{F}_q$.

Edit: This doesn't work for the reasons I mention in the comments, and actually (to my surprise) seems somewhat unfixable. If $q=49=7^2$, then the only primes $\ell$ dividing $p^f-1$ are $\ell=2$ and $\ell=3$, and 7 has order 2 mod neither. So it appears that $\mathbb{F}_{49}$ is not the residue field of any cyclotomic field. (Of course, $\mathbb{F}_{49}$ itself is easily obtained by a quadratic field).

  • $\begingroup$ Oops, wait. We need $f$ to be the smallest $f$ such that $p^f\cong 1\pmod{\ell}$. Maybe if I take the largest $\ell$ dividing $p^f-1$? $\endgroup$ – Cam McLeman Feb 16 '12 at 15:52
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    $\begingroup$ Dear Cam, $\mathbb F_{49}$ is a residue field of $\mathbb Q(\zeta_{48}).$ Regards, $\endgroup$ – Matt E Feb 16 '12 at 16:54

The following more general result was proved (in an even more general form) by Hasse (Zwei Existenztheoreme in der Theorie der algebraischen Zahlkörper; Math. Ann. 1925): Let $k$ be a number field, let $P$ denote a prime ideal in $k$, and assume that $(e_1,f_1)$, . . . , $(e_s,f_s)$ are pairs of natural numbers $\ge 1$ with $\sum e_j f_j = n$. Then there exist infinitely many extensions $K/k$ with degree $n$ such that $P$ decomposes as $$ P O_K = Q_1^{e_1} \cdots Q_s^{e_s}, $$ where each $Q_j$ has relative inertia degree $f_i$.


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