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Does there exist a prime $p$ for each natural number $n$ such that all $n$th degree or less polynomials in $\mathbb{Z}_p$ with coefficients in $[-n,n] \cap \mathbb{Z}$ split?

Motivation:

If you have a finite field and want to add roots of various polynomials, an easy way to do this is to construct an extension field that gives you those new roots; however, if you instead demand that the field be of the form $\mathbb{Z}_p$ for some prime $p,$ then this no longer works. Instead, you must find a new prime that has roots for the additional polynomials.

If you know of a proof of the statement above or a link to a resource discussing this problem or related problems, please let me know. Thanks.

based on the discussion so far it appears that this is not possible for all n, so I therefore would like to know what the largest such n is, Thanks

Regarding the choice of p1, it is easy to show this will never work, consider the polynomial of degree 2 x^2-k, this will only split if k is a square, however in Zp for any prime p there are (p-1)/2 non squares and all k must be considered because [-n,n] is equivalent to [0,p-1]. Therefore we must consider p greater than n if we hope to find such a p. Like I said above, based on the conversation so far it seems like this is not possible for all n so please try to find an n for which this can't be done.

Mathew
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  • I am not sure what is meant by "add[ing] roots to various polynomials." Every finite field is order $p^k$ for some prime $p$ and some positive integer $k.$ Generally, if you adjoin a root $\alpha$ of an irreducible polynomial of degree $d$ to the finite field $F_q,$ you obtain a field of order $q^d.$ (Here, $q$ is prime-power.) – Dylan C. Beck Aug 17 '18 at 03:52
  • Furthermore, there are irreducible polynomials of every degree in $\mathbb{Z}_p$ for every prime $p,$ so given a prime $p$ and a positive integer $n,$ there exists an irreducible polynomial of degree $n$ in $\mathbb{Z}_p,$ and therefore, this polynomial cannot split in $\mathbb{Z}_p.$ – Dylan C. Beck Aug 17 '18 at 03:54
  • sorry, I meant adding roots of various polynomials, I just fixed that – Mathew Aug 17 '18 at 04:12
  • I realize that all finite fields have irreducible polynomials, what I'm asking is that all polynomials of a particular kind be split-able, that is, polynomials of with degree n or less with coefficients in the interval [-n,n] – Mathew Aug 17 '18 at 04:14
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    Do you mean for your degree $n$ and the $n$ in your set ${-n, -n+1, \dots, n-1, n }$ to coincide? – Dylan C. Beck Aug 17 '18 at 04:16
  • Be aware that the only possible coefficients of a polynomial over $\mathbb{Z}_p$ are $0, 1, \dots, p - 1.$ So, your set ${-n, -n+1, \dots, n-1, n }$ reduces to ${0, 1, \dots, p-1 }$ for every positive integer $n \geq p.$ – Dylan C. Beck Aug 17 '18 at 04:19
  • "Do you mean for your degree n and the n in your set {−n,−n+1,…,n−1,n} to coincide?" - yes – Mathew Aug 17 '18 at 04:30
  • "Be aware that the only possible coefficients of a polynomial over ℤp are 0,1,…,p−1. So, your set {−n,−n+1,…,n−1,n} reduces to {0,1,…,p−1} for every positive integer n≥p." I am first picking n then asking you to find me a p, so you just need to pick a p that is very large relative to n and this won't be a problem – Mathew Aug 17 '18 at 04:31
  • Let us continue this discussion in chat. – Dylan C. Beck Aug 17 '18 at 04:34
  • I fully agree with your assesment of the case p<n however I don't understand this part: "if you pick p>n prime, then you have that (p,n)=1, and you encounter the same problem again" – Mathew Aug 17 '18 at 04:59
  • @Dylan_Carlo_Beck I'm not sure I follow your reasoning with cyclotomic polynomials. If $p\equiv1\pmod n$, then the $n$th cyclotomic polynomial splits modulo $p$. – Jyrki Lahtonen Aug 17 '18 at 05:13
  • First of all, I had a major typo in that comment. But also, I had initially misunderstood the aim of the problem. I hope I have not caused too much confusion or that my comments have not been helpful. – Dylan C. Beck Aug 17 '18 at 05:22

1 Answers1

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This follows from Chebotarev's density theorem.

Let $f_n(x)$ be the product of all those (finitely many for a fixed $n$) polynomials, viewed as a polynomial with integer coefficients. Let $L_n$ be the splitting field of $f_n(x)$ over the rationals. If $G_n=Gal(L_n/\Bbb{Q})$, then Chebotarev's theorem implies that the density of primes $p$ that split totally in $L_n$ is $1/|G_n|$. The total splitting of a prime $p$ implies that $f_n(x)$ splits into linear factors modulo $p$.

So there will exist infinitely many primes with the prescribed property, but finding one may take a while.

Jyrki Lahtonen
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  • There are known error terms to Chebotarev's density formula, but I don't remember any. Those would tell how high you need to go to find a suitable prime $p$. – Jyrki Lahtonen Aug 17 '18 at 05:05
  • I am in a discussion with @ Dylan_Carlo_Beck right now, I'm not sure how you would join us if you want to but if you know how and would like to please do – Mathew Aug 17 '18 at 05:17
  • I don't completely understand your answer, I concerned that your looking at spliting fields over the rationals, I am concerned only with fields of the form Zp for some prime p which is of course not characteristic 0 so the rationals shouldn't come into this as far as I can tell, but I'm probably missing your point – Mathew Aug 17 '18 at 05:18
  • @mathew You need a few pieces from algebraic number theory to see the connection. A prime ideal $p$ splits completely in the field $\Bbb{Q}(\alpha)$, if the minimal polynomial of $\alpha$ splits completely modulo $p$. We cannot arrange $f_n(x)$ to be the minimal polynomial of anything, but that doesn't matter, because the splitting of a prime ideal only depends on the field. – Jyrki Lahtonen Aug 17 '18 at 05:23
  • ok thanks for your answer, I will need to examine it a bit more to fully understand – Mathew Aug 17 '18 at 05:31
  • Chebotarev is relatively deep stuff. Getting there is may be year's worth of algebraic number theory. That assuming you already know Galois theory, modules over PID's and a few other things. – Jyrki Lahtonen Aug 17 '18 at 05:34