# A special integral polynomial

Given $n \in \mathbf{N}$,is always possible to construct a monic polynomial in $\mathbf{Z}[x]$ of degree $2n$, whose roots are in $\mathbf{C} \setminus \mathbf{R}$ and whose Galois group over $\mathbf{Q}$ is $S_{2n}$? I have an approximate idea of how to solve the problem for the Galois group (I immagine something related to the Hilbert irreducibility theorem), but I have no idea for the condition on the roots. Furthermore, is it possible to give an explicit example?

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Is reasonable to think that the generic monic integral polynomial will have that form? I do not have a precise meaning here for the word generic, but maybe it might be given a number theoretic (and scheme-theoretic on Spec Z?) sense. –  Roberto Svaldi Mar 13 '10 at 17:43
In some sense a generic polynomial with no real roots should have a Galois group of S_2n, but I don't know a sense in which a generic polynomial doesn't have any real roots. –  Douglas Zare Mar 13 '10 at 17:51
No, the condition of having no real roots is not generic. Rather, it defines a nonempty open set (in the analytic topology) of the space of all degree $2n$ polynomials: e.g. for quadratic polynomials the condition is just $b^2-4ac < 0$. Thus if you endow this space with some reasonable measure, the locus you want will have positive, but not full, measure. In contrast the locus of the set where the Galois group is $S_{2n}$ will have full measure, so morally there's your existence proof. But I didn't immediately see how to make this rigorous, so I did something totally different below. –  Pete L. Clark Mar 13 '10 at 18:02

An easy way to ensure that a polynomial $g$ of degree $m$ over $\mathbf{Z}$ has Galois group $S_m$ is to take primes $p_1$, $p_2$ and $p_3$ with $g$ irreducible modulo $p_1$, a linear times an irreducible modulo $p_2$ and a bunch of distinct linears times an irreducible quadratic modulo $p_3$. Then the Galois group must be doubly transitive and have a transposition, so it's $S_m$.

Now take $m=2n$ and a polynomial $f$ over $\mathbf{Q}$ with no real roots (e.g. $(x^2+1)^n$). Replacing the coefficients of $f$ by close rationals won't create any real roots. So replace the $x^k$ coefficient of $f$ by a sufficient close rational $a_k/b_k$ where $a_k$ and $b_k$ are congruent modulo $p_1 p_2 p_3$ respectively to the $x^k$ coefficient of $g$ and to $1$. Then the new polynomial has rational coefficients, no real roots and Galois group $S_{2n}$. You can easily convert it to one with these properties and integer coefficients should you wish.

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It's easier than that to eliminate real roots. Just add a multiple of $p_1p_2p_3$ greater than the minimum. –  Douglas Zare Mar 14 '10 at 0:40
Thanks Douglas, that's a nice trick. +1 My method (basically weak approximation) extends to obtaining any even number of non-real roots. –  Robin Chapman Mar 14 '10 at 7:52

Yes, it is always possible.

First note that it suffices to construct a totally complex Galois extension $K/\mathbb{Q}$ of degree $2n$ with Galois group $S_{2n}$. By the primitive element theorem, this extension is of the form $\mathbb{Q}[t]/(f(t))$ for some irreducible polynomial $f$, the minimal polynomial of an algebraic number $\alpha \in K$. Then there exists $n \in \mathbb{Z}^+$ such that $n \alpha$ is an algebraic integer: take the minimal polynomial of that algebraic integer: it generates the same field extension.

To construct the desired extension $K$, in turn it suffices to find an irreducible polynomial with $\mathbb{Q}$-coefficients with no real roots and whose Galois group is the largest possible $S_{2n}$. This is possible by a weak approximation / Krasner's Lemma argument. I will just sketch it for now; I can fill in more details if needed. The idea is to find a finite set of primes $p$ and degree $2n$ polynomials $f_p$ such that the Galois group of $f_p$, as a group of permutations on the roots of $f_p$, is of a certain form (e.g. contains a specific transposition). Also let $f_{\infty}$ be any degree $2n$ polynomial over $\mathbb{R}$ without real roots. Then by Krasner's Lemma, there exists a polynomial $f$ which is sufficiently $p$-adically close to each $f_p$ and to $f_{\infty}$ to have the same local behavior: in particular, to factor the same way over $\mathbb{Q}_p$ and over $\mathbb{R}$ and to generate the same local Galois groups. Then, by identifying the local Galois groups with decomposition groups at $p$ (of unramified extensions), if one has enough primes so as to get permutations of every possible cycle type, then the global Galois group of $f$ certainly must be $S_{2n}$. Indeed, to see this we use the following result from lecture notes of Keith Conrad (and Bertrand's postulate!):

Theorem: For $n \geq 2$, a transitive subgroup of $S_n$ which contains a transposition and a $p$-cycle for some prime $p > \frac{n}{2}$ is $S_n$.

The condition at infinity means that $\mathbb{Q}[t]/(f(t))$ is totally complex, hence so is its splitting field. To ensure that $f$ is irreducible, we may apply Krasner's Lemma again and take its coefficients sufficiently close to those of an irreducible degree $2n$ polynomial over $\mathbb{Q}_p$ (for a different $p$ from those used thus far) so as to be irreducible over $\mathbb{Q}_p$, which implies irreducibility over $\mathbb{Q}$.

This can in principle be made explicit, but I might search the literature for a known classical family of polynomials doing what you want before I tried to carry out this construction explicitly.

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@Pete: I'm sure this works but I don't quite understand the argument yet. Your strategy shows that I can find f with no real roots and such that for some finite set of primes p in S, f mod p factors in a given way. The upshot is that you can decree the cycle type of Frob_p for p in a finite set. But you can't control which roots are in which cycle, can you? So aren't you left with the following issue: you have to prove that if G is a transitive subgroup of S_{2n} containing an element of each cycle type, then G=S_{2n}. No doubt this is standard but don't you need it to complete the argument? –  Kevin Buzzard Mar 13 '10 at 18:33
@Kevin: Thanks for the comment. I completed the argument along the lines you suggested. –  Pete L. Clark Mar 13 '10 at 20:04
:-/ Now you mention the result I remember using it about 10 years ago to check that the char poly of T_2 on S_k(SL_2(Z)) had Galois group S_n for all k<=2048. Daft story connected with this: after I checked this I emailed William Stein telling him what I had done, and the next day he emailed me back saying he'd just done k=2050 so now he held the record :-) –  Kevin Buzzard Mar 13 '10 at 21:23
@Kevin: that's funny. By the way, why no upvotes? I don't need the reputation, but the corroboration that my argument is correct and understandable is very welcome. –  Pete L. Clark Mar 13 '10 at 21:43
By the way, since this is an application of the sort of weak approximation + Krasner argument that I have used in my own work and now introduced in my course on local fields, having looked at Keith's paper it's natural to try to make a similar argument work to get the alternating groups A_n as Galois groups over any global field. [Yes, I know this is due to Hilbert.] But it's not immediately clear to me how to do it -- can anyone help? –  Pete L. Clark Mar 13 '10 at 21:45