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I am wondering if there exist classification of polynomials that are irreducible in $ \mathbb{Z}$ but reducible $\pmod p$ for all primes $p$.

I am aware that $\Phi_n$ has this property if $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is not cyclic. Are there any other nice families?

In fact, even after researching a bit, I have not been able to find any other polynomials that don't fall into the family listed above. Any single example that is not of the above form is also welcomed.

Actually, I have found this paper which might be of interest: http://www.m-hikari.com/imf-password2008/33-36-2008/pearsonIMF33-36-2008.pdf

Sandeep Silwal
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    Not sure whether you would call it a family, but if a polynomial has an abelian but non-cyclic Galois group then it cannot remain irreducible after reduction modulo $p$. This is because over $\Bbb{Z}/p\Bbb{Z}$ the only finite Galois groups are cyclic. See also Qiaochu's explanation. – Jyrki Lahtonen Aug 04 '14 at 06:41
  • How would you like to see this generalized to finite fields? A polynomial reducible over $\Bbb{F}_p$ is surely also reducible over any extension field, so you probably meant something else? But a cool question (+1). – Jyrki Lahtonen Aug 04 '14 at 06:42
  • Sorry, I was not thinking clearly. I have edited the question. The other thread definately had some relevant information but I am not satisfied yet. – Sandeep Silwal Aug 04 '14 at 22:52
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    One could generalize, kind of. Indeed, let $\cal O$ be the ring of integers of some number field, and $\Bbb A$ the ring of all algebraic integers. We can form the image $\overline{\cal O}$ of $\cal O$ in the ring ${\Bbb A}/{\frak P}{\Bbb A}$ for prime ideals $\frak P$ of $\cal O$, and talk about images $\bar{x}$ of elements $x\in\Bbb A$. Then one can speak of $x\in\Bbb A$ for which $\overline{\cal O}[\bar{x}]$ is not a field for any prime ideal ${\frak P}$ of $\cal O$. – anon Aug 05 '14 at 05:27
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    isn't this equivalent to the Galois group of $P$ being transitive (as a subgroup of $S_n$) and not containing an $n$-cycle ? There are plenty of possible subgroups, so glossing over the inverse galois problem, there should be plenty of possible polynomials. – mercio Aug 05 '14 at 11:55
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    Also note that the smallest degree for such polynomials is $4$. Showing that degree $2$ polynomials fail is a straightforward application of local-global principle (a number that is a square modulo all primes is a perfect square, you can do this yourself :P ). To see that degree $3$ polynomials fail, note that its Galois group over $\mathbb{Q}$ would be $S_3$ (since its a transitive subgroup of $S_3$), so there would be "many" primes such that the polynomial would be irreducible in $\mathbb{Z}/p\mathbb{Z}$ (this is like Chebotarev or something). – tc1729 Aug 07 '14 at 20:05
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    @SandeepSilwal Also here is an article that may be of some interest to you: http://web.mit.edu/rsi/www/pdfs/papers/2005/2005-bretth.pdf – tc1729 Aug 07 '14 at 20:12
  • @SiddharthPrasad: Thanks homie. – Sandeep Silwal Aug 07 '14 at 20:35
  • @EulerianAdventurer: Stalking? I'm afraid that is unacceptable behavior on a maths forum. – Sandeep Silwal Sep 15 '14 at 18:48
  • @EulerianAdventurer: Who are you..? – Sandeep Silwal Sep 15 '14 at 23:58

1 Answers1

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As explained in this math.SE answer and stated in the comments by mercio, for an irreducible polynomial of degree $n$ this condition is equivalent to the Galois group failing to contain an $n$-cycle. There are many transitive subgroups of $S_n$ failing to contain an $n$-cycle and hence (up to solving the inverse Galois problem, also as stated in the comments by mercio) many such polynomials, although no such subgroups exist when $n$ is prime (exercise) and "most" polynomials, as it turns out, have Galois group $S_n$.

In particular, in the linked math.SE answer I give a family of irreducible polynomials of degree $4$ with Galois group $V_4$, namely the polynomials

$$f_a(x) = x^2 - 2ax + 1$$

where $a$ is an integer not equal to $1$ or $\frac{b^2 \pm 2}{2}$ for any integer $b$. I think examples of this form were originally written down by Hilbert. The other four possible Galois groups, namely $S_4, A_4, D_4, C_4$, all contain a $4$-cycle, so $V_4$ is the only acceptable Galois group for a quartic.

A method for generating lots of examples, although it takes a bit of work to make it fully explicit, is the following. Let $G$ be non-cyclic. Construct a Galois extension $K/\mathbb{Q}$ with Galois group $G$, and then let $f$ be the minimal polynomial of a primitive element of $K$. The cyclotomic polynomials occur this way when $G = \mathbb{Z}_n^{\times}$ is not cyclic, but $G$ can be, say, any nonabelian group. This construction in particular shows that examples exist with arbitrarily large degrees other than the cyclotomic polynomials (up to solving the inverse Galois problem, any degree $n$ such that there exists a non-cyclic group of order $n$; it is known that these are precisely the numbers $n$ such that $\gcd(n, \varphi(n)) \neq 1$).

It shouldn't be hard to be explicit about this for for $G = S_3$ (as a transitive subgroup of $S_6$); you can take $K$ to be the splitting field of any monic irreducible cubic polynomial whose discriminant is not a square.

For the sake of having an explicit sequence of examples of arbitrarily large degrees other than the cyclotomic polynomials, take the minimal polynomials of the algebraic integers

$$\sqrt{2} + \sqrt{3} + \sqrt{5} + \dots + \sqrt{p_k}$$ where $p_k$ is the $k^{th}$ prime. These are primitive elements of $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \dots \sqrt{p_k})$, which has Galois group $C_2^k$, which is non-cyclic as soon as $k \ge 2$.

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Qiaochu Yuan
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