# Is there a finite set of primes such that if K over Q is completely split at all those primes then it is Q?

Is there a finite set of primes, $S \subset Spec(\mathbb{Z})$ such that if $K$ over $\mathbb{Q}$ is a number field such that every $p \in S$ is completely split in $K$ then $K=\mathbb{Q}$? If so, what do we know about such $S$'s?

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I don't think such S exists. If you can find an odd prime q such that all the primes in S are congruent to 1 mod q then every prime in S is totally split in q-cyclotomic field. So, my example boils down to: Given a finite set S of primes is it always possible to find a prime q as above? maybe some analytic number theorist knows the answer of this of the top of his/ger head. – Guillermo Mantilla Jan 8 '11 at 22:49
However there will be an infinite, density zero set of primes with this property. Because you can just enumerate the non-trivial extensions of the rationals as $K_1$, $K_2,\ldots$ and then for each $n$ choose a prime $p>10^{10^n}$ which isn't completely split in $K_n$ and add it to $S$. The ridiculous growth rate I assumed forces the set to be very sparse. – Kevin Buzzard Jan 8 '11 at 23:04
@Guillermo: no. Take {2, 3}. – Qiaochu Yuan Jan 8 '11 at 23:59

This is not even true for quadratic extensions. Given primes $p_1, p_2, ... p_n$ find a prime $q \equiv 1 \bmod 4p_1 ... p_n$, which exists by Dirichlet's theorem, and consider $K = \mathbb{Q}(\sqrt{q})$.

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No. In fact, given a finite set of primes $S=\lbrace p_1,p_2,\ldots,p_n\rbrace$, we can find a finite extension $K$ of $\mathbb{Q}$ in which elements of $S$ decompose in any way that we like. That is, for any numbers $r_i\ge1$ and $e_{i1},e_{i2},\ldots,e_{ir_i}\ge1$, we can find an extension $K$ in which $p_i$ decomposes as $$p_i\mathcal{O}\_K=\mathfrak{p}\_{i1}^{e\_{i1}}\mathfrak{p}\_{i2}^{e\_{i2}}\cdots\mathfrak{p}\_{ir_i}^{e\_{ir_i}}$$ for all $i=1,\ldots,n$, with $\mathfrak{p}\_{ij}$ being distinct primes of $\mathcal{O}\_K$. Furthermore, if $f_{ij}$ are positive integers with $d=\sum_{j=1}^{r_i}e_{ij}f_{ij}$ independent of $i$ then this can always be done such that $[K\colon\mathbb{Q}]=d$ and $[\mathcal{O}\_K/\mathfrak{p}\_{ij}\colon\mathbb{Z}/p_i\mathbb{Z}]=f_{ij}$. For $p_i$ to split completely, this just means that we use $e_{ij}=f_{ij}=1$ and $r_i=d$.

Here's a construction I came up with: Let $k_{ij}$ be an extension of the $p_i$-adic numbers $\mathbb{Q}\_{p_i}$ in which $p_i$ has ramification index $e_{ij}$ and such that the extension of residue fields is of degree $f_i$. We can do this by adding an $e_{ij}$'th root of $p_i$ to the splitting field of $X^{p_i^{f_i}}-X$ over $\mathbb{Q}_{p_i}$. Letting $R_{ij}$ be the elements of $k_{ij}$ which are integral over $\mathbb{Z}\_{p_i}$, the primitive element theorem implies that $k_{ij}=\mathbb{Q}_{p_i}(x_{ij})$ for some $x_{ij}\in R_{ij}$. Let $g_{ij}\in\mathbb{Z}\_{p_i}[X]$ be the minimal (monic) polynomial of $x_{ij}$. This will have degree $[k_{ij}\colon \mathbb{Q}\_{p_i}]=e_{ij}f_{ij}$. Also, $g_{ij}$ will be irreducible in $\mathbb{Z}/p_i^r\mathbb{Z}$ for sufficiently large $r$. Set $g_i=\prod_{j}g_{ij}$ which is monic of degree $d$.

Now, using the chinese remainder theorem, there will be a monic degree $d$ polynomial $g\in\mathbb{Z}[X]$ such that $g=g_i$ mod $p_i^r$ for each $i$, and $f=X^d+p$ mod $p^2$, for some prime $p\not\in S$. Eisenstein's criterion implies that $g$ is irreducible in $\mathbb{Q}[X]$, and Hensel's lemma implies that in $\mathbb{Q}\_{p_i}[X]$, $g$ splits into a product of irreducible terms equal to $g_{ij}$ mod $p_i^r$. Taking $K=\mathbb{Q}(x)$ for a root $x$ of $g$ gives the claimed extension.

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This is a great answer emphasizing many number-theory results. Thanks! – Abhishek Parab Aug 31 '14 at 2:21

No. If d is a square modulo p for all p in S, then all p in S split in $\mathbb{Q}(\sqrt d)$.

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