Values of Polynomials over Integral Domains

It is well known that no nonconstant polynomial with integer coefficients can take on only prime values. We isolate the property of the integers that accounts for this, and give several examples of integral domains for which there are polynomials that only take on unit or prime values. Throughout this note, we let R denote an arbitrary UFD (unique factorization domain) that is not a field. Of course, R must be infinite. As is well known, any nonconstant polynomial with integer coefficients cannot take on only prime values. We may ask what property of the integers Z accounts for this. Here is the answer. Theorem 1. Suppose that R has only finitely many units. If f (x) ∈ R[x] is any nonconstant polynomial, then f (a) is composite for some a ∈ R. Proof. Suppose that R has k units, and let f (x) have degree d . Choose any distinct kd + 1 elements of R. Then for one of these elements b, f (b) is not a unit, as otherwise f (x) would have to take the same value more than d times, and so would be constant. Let p be any prime dividing f (b). Choose any distinct (k + 1)d + 1 elements of R congruent to b mod p. Then for one of these elements a, f (a) is neither a unit multiple of p nor is equal to 0, for the same reason, so f (a) is composite. We now want to investigate cases when we do have nonconstant polynomials that take on only prime or unit values. Definition 2. Let f (x) ∈ R[x] be a polynomial. Then f (x) is a p-polynomial if f (r) is prime for every r ∈ R, f (x) is a u-polynomial if f (r) is a unit for every r ∈ R, and f (x) is a up-polynomial if f (r) is a unit or is prime for every r ∈ R. Let P be any set of primes in Z, and let ZP denote the localization of Z at P ; concretely, ZP = {rational numbers m/n with no prime factor of n in P}. Example 3. (a) Let P = {p ≡ 3 (mod 4)} ∪ {2}, and let R = ZP . Then f (x) = x2 + 1 is an up-polynomial. (b) Let P = {p ≡ 3 (mod 4)}, and let R = ZP . Then f (x) = x2 + 1 is a upolynomial. (c) Let P = {p ≡ 5 or 7 (mod 8)} ∪ {2}, and let R = ZP . Then f (x) = (x(x + 1))2 + 2 is a p-polynomial. http://dx.doi.org/10.4169/amer.math.monthly.121.01.073 MSC: Primary 13G05 January 2014] NOTES 73 This content downloaded from 128.8.128.139 on Tue, 11 Feb 2014 10:28:35 AM All use subject to JSTOR Terms and Conditions (Parts (a) and (b) use the fact that −1 is not a quadratic residue of any prime p ≡ 3 (mod 4), and part (c) uses the fact that −2 is not a quadratic residue of any prime p ≡ 5 or 7 (mod 8).) Note that in all parts of this example, f (x) is a monic polynomial and the ring R has infinitely many distinct primes. Remark 4. We observe that if f (x) ∈ R[x] has a root in R, i.e., if f (b) = 0 for some b ∈ R, then f (x) must take on composite values. The reason is much the same as above. Choose any composite element c ∈ R. Then for any element r of R with r ≡ b (mod c), f (r) ≡ 0 (mod c), so for some a ≡ b (mod c), 0 6= f (a) ≡ 0 (mod c) is composite. We also observe that if f (x) ∈ R[x] takes on two distinct (i.e., nonassociated) prime values, then f (x) must take on composite values. For if f (b1) ≡ 0 (mod p1) and f (b2) ≡ 0 (mod p2), then by the Chinese Remainder Theorem there is a b with b ≡ b1 (mod p1) and b ≡ b2 (mod p2). For any such b, f (b) ≡ 0 (mod p1 p2), so for some a ≡ b (mod p1 p2), 0 6= f (a) ≡ 0 (mod p1 p2) is composite. Theorem 1 has a generalization, whose proof we leave to the reader. Theorem 5. Suppose that R has only finitely many units. Let Q be the quotient fieldof R. If f (x) ∈ Q[x] is any nonconstant polynomial such that f (q0) ∈ R for someq0 ∈ Q, then f (q1) ∈ R is composite for some q1 ∈ Q.Our focus in this note has been on properties of polynomials over general integraldomains. But we will remark that polynomials over the (ordinary) integers have anumber of special properties. In [1] it is shown that for any integer M , there is apolynomial of any given degree of a specific form that takes on prime values for at leastM positive integer arguments. In [3] it is shown that, assuming a standard conjecture innumber theory, for any integer M there is a quadratic polynomial of a specific form thattakes on prime values for at least M consecutive positive integer arguments. Passing topolynomials in more than one variable, [2] gives an explicit polynomial of degree 25 in26 variables whose values at integer arguments are either negative integers or (positive)primes, and which takes on all prime values. [2] also shows that any algebraic function,of any number of variables, that takes on integer values at all positive integer argumentsmust be a polynomial. (We have just sketched the main results of these papers and werefer the reader to them for more specific statements.) The proofs of these results usevery different ideas than our proofs above. REFERENCES1. B. Garrison, Polynomials with large numbers of prime values, Amer. Math. Monthly 97 (1990) 316–317,available at http://dx.doi.org/10.2307/2324515.2. J. P. Jones, D. Sato, H. Wada, D. Wiens, Diophantine representation of the set of prime numbers, Amer.Math. Monthly 83 (1976) 449–464, available at http://dx.doi.org/10.2307/2318339.3. R. A. Mollin, Prime-producing quadratics, Amer. Math. Monthly 104 (1997) 529–544, available at http://dx.doi.org/10.2307/2975080. Department of Mathematics, Lehigh University, Bethlehem, PA [email protected] 74c© THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121 This content downloaded from 128.8.128.139 on Tue, 11 Feb 2014 10:28:35 AMAll use subject to JSTOR Terms and Conditions