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In 300 BC or so Euclid pointed out that if $S$ is any finite set of prime numbers then the prime factors of $1+\prod S$ are not in $S$, so that $S$ can always be extended to a larger finite set.

Much later (19th century? 20th?) someone (who?) pointed out that if $F$ is a finite field then $1+\prod\limits_{\alpha\in F}(x-\alpha)$ has no zeros in $F$, so we must extend $F$ to a larger finite field if we want all polynomials to have zeros.

What interesting general propositions are these both special cases of?

  • There is actually a very extensive analogy between the ring of integers and the rings of polynomials, particularly polynomials over finite fields. Another simple one is that both rings have unique factorization into prime elements. –  Jul 06 '14 at 23:55
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    If $S$ is any finite set of nonunits of a commutative ring $R$, then $1+\prod S$ is not divisible by any element of $S$. – anon Jul 06 '14 at 23:56
  • @Hurkyl : True: You can apply Euclid's algorithm to find GCDs in both cases. – Michael Hardy Jul 06 '14 at 23:56
  • Caution: Your post seems to imply that $1 + \prod S$ is a prime. This is not necessarily so. – awkward Jul 07 '14 at 00:00
  • @awkward Not mentioning that it isn't generally prime is far from 'seeming to imply it is a prime.' All of this stuff is well-known and needn't be hashed out again here. – anon Jul 07 '14 at 00:01
  • @awkward : I don't think it implies that. It implies that the factors are not first-degree polynomials. Some people phrase Euclid's argument in convoluted ways that make it look as if they're saying $1+\prod S$ is prime, and that's certainly false even when $S$ is assumed to be the smallest $n$ primes for some $n$. ${}\qquad{}$ – Michael Hardy Jul 07 '14 at 00:05
  • @blue : Maybe you should make your comment an answer. The term "non-units" is interesting. Euclid referred to "the unit", and I think he might have considered it to be something other than a number, so that the smallest of all "numbers" ("$\textstyle{\alpha\rho\iota\theta\mu\omicron\iota}$") that exist is $2$. ${}\qquad{}$ – Michael Hardy Jul 07 '14 at 00:07
  • @JyrkiLahtonen : Yours is another comment that could be an answer. – Michael Hardy Jul 08 '14 at 00:49
  • @Michael The method mentioned by blue does not work in general - see my comment to his answer (only now did I stumble upon this question). – Bill Dubuque Jul 23 '14 at 01:04

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If $S$ is any finite set of nonunits of a commutative ring $R$, then $1+\prod S$ is not divisible by any element of $S$. Thus if $R$ is atomic, unless $1+\prod S$ is a unit then the set $S$ fails to contain all irreducibles. (I had originally failed to consider the possibility that $1+\prod S$ could be a unit. As Bill Dubuque notes, we need to amend our reasoning to get around this obstacle. For $\Bbb Z$ and $\Bbb F_q[x]$ one can choose positive naturals and use absolute value or use degrees respectively to verify that $1+\prod S$ is a nonunit.)

Specializing, if $\frak P$ is any set of irreducibles (say, of a certain type, like "degree one polynomials") and $S$ any finite subset of $\frak P$, and $1+\prod S\not\in R^\times$ then either $S$ is proper or not all elements of $R$ are have a divisor in $\frak P$. If $F$ is a finite field then we know set set $\frak P$ of linear polynommials is finite, so this argument shows there are polynomials without linear factors, equivalently without roots. In any ring primes are a certain type of irreducible element, which gives Euclid's conclusion for $\Bbb Z$.

anon
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    The claim in the first paragraph is false. For example, let $,R,$ be a PID with finitely many primes $S,$ (e.g. invert all but finitely many primes in $,\Bbb Z).,$ Then $,S,$ does contain all irreducibles (up to associate), and $,1+\prod S,$ has no prime factor so is a unit. So in general rings Euclid's construction may fail because it produces only units. However, one can avoid this obstacle if the ring has fewer units than elements - see my generalization of Euclids' proof to fewunit rings. – Bill Dubuque Jul 23 '14 at 01:03
  • @BillDubuque Ah, I had failed to consider $1+\prod S$ being a unit. – anon Jul 23 '14 at 01:13
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"Early in the development of the subject it was noticed that $\Bbb{Z}$ has many properties in common with $A=\Bbb{F}[T]$, the ring of polynomials over a finite field. Both rings are principal ideal domains, both have the property that the residue class ring of any non-zero ideal is finite, bothe have infinitely many prime elements, and both rings have finitely many units. Thus, one is led to suspect that many results which hold in $\Bbb{Z}$ have analogues in $A$."

That quote is from the preface of Michael Rosen's book Number Theory in Function Fields, and is a central theme in that book.

The early parts of that book are IMHO very accessible. The analogues of for example the following concepts/results are explained:

  • Euler totient function
  • Wilson's theorem
  • Prime number theorem (the result $\pi(x)\approx \dfrac{\ln x}x$ in $\Bbb{Z}$)
  • Basic arithmetic functions

These rely on the idea that the degree of a polynomial is an adequate measure of its size much like the absolute value of an integer.

Reciprocity laws are also easy to derive in the ring $A$. If $Q$ and $P$ are monic irreducibles of respective degree $m, n$ in $A$, then we have the symmetric function $$ \prod_{\alpha,\beta}(\beta-\alpha)=(-1)^{mn}\prod_{\beta,\alpha}(\alpha-\beta), $$ where $\alpha$ (resp. $\beta$) ranges over zeros of $Q$ (resp. $P$). The reciprocity laws follow from this.

  • The theory of $L$-functions is also easier in $A$ (the analytical problems are easier to handle), and leads to the equidistribution of irreducible monic polynomials into (coprime) residue classes modulo a given polynomial - a perfect analogy with Dirichlet's result on equidistribution of prime numbers into cosets modulo a given integer. Several people have found very explicit results in interesting enough subcases.
  • The study of extension fields of function fields (an analogue of algebraic number theory) is a natural extension. There is an analogue of the Riemann hypothesis - introduced in Rosen's chapter 4, and proven(!) in chapter 8. Algebraic geometric thinking and tools help here, but Rosen's book is IIRC self-contained to this extent.

Other analogues studied in Rosen's book include:

  • Artin's primitive root Conjecture (proven by Bilharz)
  • ABC-conjecture (a Theorem on the function field side)
  • Material leading to class field theory (this is more advanced)

An analogue that I know to have been studied, but was left out from Rosen's book is the game related to twin primes. The analogue here depends on the size of the field $|\Bbb{F}|=q$. Remember that we view low degree polynomials as small, so two polynomials are close to each other, if their difference has low degree.

  • If $q>2$, then a natural analogue of the twin prime conjecture is to ask about the existence of irreducible polynomials differing only at their constant term.
  • If $q=2$, then all irreducible polynomials other than $T$ have constant term equal to $1$. Furthermore, the number of terms of an irreducible has to be odd (otherwise $T-1$ is a factor). Thus the smallest possible difference between two irreducible monic polynomials is $T^2+T$.

I know that at least Stephen D. Cohen (Glasgow/Bristol) has pursued this, but IIRC the main question is still open.


In general number theory in $A$ seems to be easier than in $\Bbb{Z}$. I'm not the right person to explain the reason for that. Viewed in one way (IIRC Milne's lecture notes?) we get that the usual zeta-function is that of a single point, and when dealing with $A$ only need to deal with the extra layer brought about by a single prime (the direct analogue of the zeta-function looks a lot like a single factor in the Euler product version of Riemann zeta), and that is easier.

Jyrki Lahtonen
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