A Primer on Sizes of Polynomials. and an Important Application

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1 A Primer on Sizes of Polynomials and an Important Application Suppose p is a prime number. By the Fundamental Theorem of Arithmetic unique factorization of integers), every non-zero integer n can be uniquely written in the form n = p k m where p m. This unique power k is called the order of n at p and denoted ord p n). By convention, ord p 0) =. Definition: Suppose r/s Q, where r, s Z. Then ord p r/s) = ord p r) ord p s). Note how this definition really is... a definition. In other words, it doesn t depend on how you write the rational number. For instance, ord p 1/2) really is the same as ord p 2/4) and ord p 17/34). This actually follows from the integer case of the following lemma. Lemma 1: Suppose p is a prime and a, b Q. Then ord p ab) = ord p a) + ord p b). Also, ord p a + b) min{ord p a), ord p b)}, with equality if ord p a) ord p b). Proof: This is obvious if either a or b is zero, so we will assume ab 0. First suppose a, b Z. Write a = p ordpa) m 1 and b = p ordpb) m 2, where p m 1 and p m 2. Then ab = p ord pa)+ord p b) m 1 m 2 where p m 1 m 2, which proves the first statement. Without loss of generality, ord p a) ord p b). We then have a + b = p ord pa) m 1 + p ord pb) ord p a) m 2 ), and clearly p m 1 + p ord pb) ord p a) m 2 ) if ord p b) > ord p a). Thus, the lemma is true for integers a and b. Now suppose a, b Q and write a = r/s, b = u/v for r, s, u, v Z. Then ab = ru/sv) and by what we ve already shown ord p ab) = ord p ru) ord p sv) = ord p r) + ord p u) ord p s) ord p v) = ord p a) + ord p b). 1

2 Further, ) ord p sva + b) = ordp vr + su) min{ord p vr), ord p su)} = min{ord p v) + ord p r), ord p s) + ord p u)} = ord p s) + ord p v) ) min{ord p r) ord p s), ord p u) ord p v)} = ord p sv) ) min{ord p a), ord p b)}, with equality if ord p vr) ord p su). By what we have already shown, this implies that ord p a+b) min{ord p a), ord p b)}, with equality if ord p v) + ord p r) ord p s) + ord p u). Thus, ord p a + b) = min{ord p a), ord p b)} if ord p r) ord p s) ord p u) ord p v), i.e., if ord p a) ord p b). So the behavior of the ord p function is somewhat reminiscent of the degree function. In fact, it behaves just like minus the degree function except the degree is a function on polynomials whereas ord p is a function on Q. Our goal is really to get a size for polynomials with properties similar to those of the degree function. But before we jump to that, a couple remarks are in order. First of all, notice how ord p r/s) will typically be 0; the only time it won t be zero is if p is a factor of either r or s. Second, r/s will be an integer if and only if ord p r/s) 0 for all primes p. Definition: If P = a 0 + a 1 X + + a n X n Q[X], then ord p P ) = min 1 i n {ord pa i )}. In other words, the order at p of a polynomial is defined to be the minimum of the orders at p of the coefficients. This is the same thing as the order of the greatest common divisor of the coefficients when P Z[X], which is the same as the highest power of p that divides all the coefficients. Remarks: Only the zero polynomial has order equal to. For any non-zero polynomial P, ord p P ) = 0 for all but finitely many primes p. Also, P has integer coefficients if and only if ord p P ) 0 for all primes p. The idea here is that the order at p behaves exactly like minus the degree function. Specifically, we have the following. Lemma 2 Gauss Lemma): Suppose p is a prime number and P, Q Q[X]. Then ord p P Q) = ord p P ) + ord p Q) 2

3 and with equality when ord p P ) ord p Q). ord p P + Q) min{ord p P ), ord p Q)}, Proof: Let n be the maximum of the degree of P and the degree of Q. Write P = a 0 + a 1 X + + a n X n, Q = b 0 + b 1 X + + b n X n. Let l be the largest index i where ord p P ) = ord p a i ) and let k be the largest index j where ord p Q) = ord p b j ). We then have ord p a i ) ord p a l ) for all i and ord p a i ) > ord p a l ) for all i > l and similarly for the b j s). By Lemma 1, this implies that ord p a i b j ) = ord p a i ) + ord p b j ) > ord p P ) + ord p Q) if either i > l or j > k, and ord p a i b j ) = ord p a i ) + ord p b j ) ord p P ) + ord p Q) in general. Hence by Lemma 1 m ) ord p a i b m i ord p P ) + ord p Q) i=0 for any m 0. We also have l+k ) ord p a i b l+k) i = ord p P ) + ord p Q), i=0 since either i > l or l + k) i > k except for when i = l and then l + k) i = k). This proves the first statement. Without loss of generality, we may assume ord p P ) ord p Q). Since ord p a k ) ord p P ) ord p Q) = ord p b k ), by Lemma 1 ord p a k + b k ) ord p b k ) = ord p Q), with equality if ord p P ) > ord p Q). Also, ord p a j + b j ) min{ord p a j ), ord p b j )} ord p Q) for all j. This proves the second statement. Definition: Suppose P Q[X] is a non-zero polynomial. Then ord p P ) = 0 for all but finitely many primes p; denote them by p 1,..., p n. The content of P, denoted contp ), is defined to be contp ) = p ord p 1 P ) 1 p ord pn P ) n. 3

4 Remark: It isn t difficult to check that contp ) is the greatest common divisor of the coefficients when P Z[X]. Also, contp ) Z if and only if P Z[X]. Thus, P is primitive the definition is in section 4.4 of the textbook) if and only if contp ) = 1. What we re doing here is really just an elaboration of the book s approach. Lemma 3: If P, Q Q[X] are non-zero polynomials, then contp Q) = contp ) contq). Proof: Just apply Lemma 2 to the finite collection of primes p where either ord p P ) 0 or ord p Q) 0. Note how this certainly takes into account all primes where ord p P Q) 0 by Lemma 2. Definition: Suppose P Q[X] is a non-zero polynomial. Define P = contp ) ) 1 P. Lemma 4: Suppose P Q[X] is a non-zero polynomial. Then ord p P ) = 0 for all primes p, i.e., P Z[X] is a primitive polynomial. Also, if Q is another non-zero polynomial, then P Q) = P Q. Proof: By definition we have ord p contp ) ) = ordp P ) for all primes p. We can view contp ) ) 1 as a polynomial of degree 0. By Lemma 1, the order at p of contp ) ) 1 is the negative of the order at p of the content of P. The first statement thus follows from Lemma 2. The last statement follows directly from Lemma 3 and the definitions. Corollary to all this stuff: Suppose P Q[X] is a non-zero polynomial and write P as a product of irreducible polynomials: P = P 1 P 2 P n. Then P = P 1 P 2 P n. In particular, if P Z[X], then P can be written uniquely as a product of its content and primitive irreducible polynomials: P = contp )P1 P2 Pn. 4

5 Now suppose if, instead of Z and its field of fractions Q, we start with F [X] polynomials with coefficients in a field F ) and its field of fractions F X) rational functions with coefficients in the field F ). Should everything go the same? Well, you ll notice that we only used particular irreducible integers p here, namely primes; by which we mean positive, irreducible integers. So perhaps when one turns to polynomials one can let monic, irreducible polynomials play the role of primes. After that, will everything else go exactly as it did with integers? 5