On Additive Polynomials

Transcription

1 On Additive Polynomials Zachary Scherr 1 Preliminaries The purpose of this article is to explore the following question: Question 1. Let K be a field. For which f(z) K[z] does f(a+b) = f(a)+f(b) for all a, b K? It may not be too surprising that when K is a field of characteristic 0, the only such polynomials are f(z) = cz for some c K. In characteristic p, however, the answer is not so simple. The polynomial f(z) = z p, for example, satisfies f(a + b) = (a + b) p = a p + b p = f(a) + f(b). To understand what additive polynomials look like, we first prove the following useful proposition: Proposition 2. Let K be a field, and suppose f(z) K[z] with deg f < K, satisfies f(a + b) = f(a) + f(b) for all a, b K. Then f(x + y) = f(x) + f(y) identically, in K[x, y]. Proof. For b K, consider the polynomial g(x) = f(x+b) f(x) f(b) K[x]. f is additive, so a K, g(a) = f(a + b) f(a) f(b) = 0. Since deg g < deg f < K, and g(x) vanishes on all of K, we must have that g(x) = 0 in K[x]. Now, consider the polynomial h(x, y) = f(x+y) f(x) f(y) K[x, y]. We may view h(x, y) as a polynomial in one variable in K(x)[y], and for all b K, h(x, b) = f(x + b) f(x) f(b) = 0 identically. As a polynomial in y, h(x, y) has degree strictly smaller than K, and since it vanishes on all of K, we get that h(x, y) = 0 in K(x)[y]. Since h(x, y) actually lies in K[x, y], we have that h(x, y) = f(x + y) f(x) f(y) = 0 identically in K[x, y], which proves the result. The restriction that deg f < K may seem unnatural, but it is only in place to make the proposition true over finite fields. Over F p for example, the polynomial f(z) = (z p z)(z + 1) = z p+1 + z p z 2 z 1

2 has the property that for all a F p, f(a) = 0 so f(z) is trivially additive. It is not the case, however, that f(x + y) = f(x) + f(y) in F p [x, y]. In particular, if one expands f(x + y), then there will be a term (p + 1)xy p, which does not show up in f(x) + f(y). Of course for a finite field K of size q, if f(z) K[z], then the polynomial g(z) = f(z) (mod z q z) satisfies deg g < q, and a K, g(a) = f(a). If f(z) is additive, then g(z) is additive as well, and proposition?? allows us to conclude that g(x + y) = g(x) + g(y) identically. Thus the assumption that deg f < K in proposition?? isn t too restrictive. 2 Structure Theorem Knowing that f(x + y) = f(x) + f(y) for an additive polynomial allows us to completely classify which polynomials f(z) K[z] are additive. Since there are distinctions between characteristic 0 and characteristic p fields, and infinite versus finite fields, I will give the structure of additive polynomials in three separate theorems. Theorem 3. Let K be a field of characteristic 0, then a polynomial f(z) K[z] is additive if and only if f(z) = cz for some c K Proof. The fact that polynomials of the form cz are additive is trivial, so we just prove the converse. Let f(z) K[z] be an additive polynomial. Necessarily, K = since the characteristic of K is 0, so proposition?? implies that f(x + y) = f(x) + f(y) identically. Write with a n 0. Consider f(z) = a i z i f(x + y) = a i (x + y) i. The degree n terms of f(x + y) come from a n (x + y) n, and are n a n ( ) n x j y n j. j We know that f(x + y) = f(x) + f(y), and f(x) + f(y) contains no terms of the form x i y j, i, j 0, so we can immediately conclude that n 1. Thus f(z) = cz + b for some c, b K, and 0 = f(x + y) f(x) f(y) = (c(x + y) + b) (cx + b) (cy + b) = b Therefore b = 0, and indeed we have f(z) = cz for some c K. 2

3 The characteristic p case is a little bit more delicate, and is handled by the following proposition Proposition 4. Let K be a field of characteristic p. The polynomial f(z) K[z] satisfies f(x + y) = f(x) + f(y) identically, if and only if f(z) = a i z pi for some a i K. Proof. The fact that f(z) = a i z pi satisfies f(x + y) = f(x) + f(y) follows from the fact that (x + y) p = x p + y p over a field of characteristic p, so we need only prove the converse. To do this cleanly we proceed by induction on the degree of f(z). In the case that deg f = 0, i.e., f is a constant, if we write f(z) = c for some c K, then the statement that f(x+y) = f(x)+f(y) implies that c = 2c, which can only occur when c = 0. Thus let deg f = n, and assume that the proposition is true for all degrees less than n. Write f(z) = b i z i where b n 0. Write n = p k l where p l, and consider f(x + y) = a i (x + y) i. Each (x + y) i contributes monomials of total degree i, so the terms of degree n in f(x + y) all come from a n (x + y) n. Now a n (x + y) n = a n (x + y) pk l = a n (x pk + y pk ) l l ( ) l = a n x pkj y pk (l j) j so that if l > 1, necessarily f(x + y) contains a monomial of the form x i y j, whereas f(x) + f(y) cannot. Thus n = p k, and the polynomial satisfies: g(z) = f(z) a n z pk g(x + y) = f(x + y) a n (x + y) pk = f(x) + f(y) a n x pk a n y pk = g(x) + g(y). 3

4 deg g < deg f = n, so by the induction hypothesis we get that so that and the proposition is proved. g(z) = m b j z pj f(z) = g(z) + a n z pk = m b j z pj + a n z pk With proposition?? proved, we are now in a position to give a structure theorem for additive polynomials over fields of characteristic p. Theorem 5. Let K be an infinite field of characteristic p > 0. A polynomial f(z) K[z] is additive if and only if f(z) = a i z pi Proof. Proposition?? tells us that f(z) is additive if and only if f(x + y) = f(x) + f(y) in K[x, y]. This combined with proposition?? gives the result. Theorem 6. Let K be a finite field of size q = p k. A polynomial f(z) K[z] k 1 is additive if and only if f(z) = a i z pi + (z q z)g(z) where g(z) is some polynomial in K[z]. Proof. In accordance with the remarks following proposition??, a polynomial f(z) is additive if and only if h(z) = f(z) (mod z q z) is additive. Since deg h < q = K, we can conclude from proposition?? and proposition?? that k 1 h(z) = b i z pi. The theorem follows since f(z) = h(z) + (z q z)g(z) for some polynomial g(z) K[z]. 3 An Application Let K be a field of characteristic p, and let M K be a finite nonempty additive subgroup of K. Consider the minimal monic polynomial vanishing on M, i.e. f(z) = (z α). α M Using our work from above, we can show that f(z) is an additive polynomial: Proposition 7. Defining f(z) as above, we have f(z) = a i K k a i z pi for some 4

5 Proof. We mimic our technique of proof from proposition??. Let deg f = n = M, and for b M, consider the polynomial g(x) = f(x + b) f(x) f(b) K[x]. It is evident that deg g < deg f = n, and for all a M, a + b M so g(a) = f(a + b) f(a) f(b) = = 0. Thus g(x) has at least n roots, and is therefore identically 0. We can then apply the exact same argument as in proposition?? to conclude that f(x + y) = f(x) + f(y) identically, which in turn implies by proposition?? that f(z) takes the required form. It is interesting to note that the polynomial f(z), which is defined to be the minimal monic polynomial vanishing on all of M, turns out to be additive on all of K. This observation provokes the following natural question: Question 8. Let M K be a finite additive subgroup of the field K, a field of characteristic p. If f(z) is the minimal monic polynomial for M, is there any way to see that f(z) is additive on all of K without explicitly proving that f(x + y) = f(x) + f(y) in K[x, y]? At the time of writing this article I am not aware of a satisfactory answer for question??. This doesn t mean that there shouldn t be a good reason, and the enthusiastic reader is encouraged to think about this. The polynomial f(z) has one more rather interesting property, in that its derivative is easy to describe. A quick calculation shows that ( k f (z) = = a i z pi ) k p i a i z pi 1 = a 0 since the characteristic of K is p. Note that a 0 is the coefficient of the linear term in f(z), so up to sign, a 0 is the sum of the roots of f taken n 1 at a time where n = M = deg f. M is an additive subgroup of k, so 0 M which implies that f (z) = a 0 = ± α 4 Final Remarks α M\{0} The theory of additive polynomials is elementary, but is also rather interesting. And while the statements and the proofs above are rather elementary, the theories of Carlitz modules and Drinfeld modules, which play a role in local class field theory, have their roots in some of the ideas presented in this paper. 5