Analogies with Polynomials - Lecture Notes | MATH 3240, Papers of Number Theory

Download Analogies with Polynomials - Lecture Notes | MATH 3240 and more Papers Number Theory in PDF only on Docsity! ANALOGIES WITH POLYNOMIALS KEITH CONRAD 1. The Basic Analogies Similarities between Z and F [T ] are an important theme in number theory. The following table collects some analogous concepts in Z and in F [T ]. Z F [T ] Similarity ±1 nonzero constants these are the units prime irreducible have only trivial factors |n| deg f role in division theorem positive monic (lead. coeff = 1) standard unit multiple A polynomial is called monic when it has leading coefficient 1, such as T 2 + 7T + 3 but not 2T 2 + 5T − 1. Every nonzero integer has exactly one positive unit multiple, while every nonzero polynomial in F [T ] has exactly one monic unit multiple: just multiply through the polynomial by the inverse of the leading coefficient. Example 1.1. In Q[T ], the monic unit multiple of 2T 2 + 5T − 1 is 12(2T 2 + 5T − 1) = T 2 + 52T − 1 2 . In F7, 2 · 4 = 1, so the monic unit multiple of 2T 2 + 5T − 1 in F7[T ] is 4(2T 2 + 5T − 1) = T 2 + 6T + 3. Positive integers are closed under multiplication and monic polynomials are closed under multiplication. Positive integers are also closed under addition but monic polynomials are not generally closed under addition. This is an important difference! The standard notation for a prime number in Z is p. The standard notation for an irreducible polynomial in F [T ] is π = π(T ). (This has nothing to do with 3.14159 . . . , of course.) By definition, a prime in Z is a number which is not ±1 and its only factors are ±1 and ± itself. Similarly, a polynomial in F [T ] is called irreducible when it is nonconstant (that is, is not a unit) and its only factors are nonzero constants and nonzero constant multiples of itself. Here are some analogous results in Z and F [T ]: (1) In Z, |mn| = |m||n|. In F [T ], deg fg = deg f + deg g. (2) The units in Z have absolute value 1 (which is the smallest absolute value possible for nonzero integers) and the units in F [T ] have degree 0 (the smallest degree possible for nonzero polynomials). (3) In Z if a|b then |a| ≤ |b|. In F [T ], if f |g then deg f ≤ deg g. (4) If a|b and b|a in Z then a = ±b, while if f |g and g|f in F [T ] then f = cg for some nonzero constant c. (5) Every integer other than 0 and ±1 is a product of primes (allowing negative primes!), while every polynomial in F [T ] other than a constant is a product of irreducible polynomials. The most important similarity between Z and F [T ] is the division theorem in both settings. We state them without proof, using similar wording. 1 2 KEITH CONRAD Theorem 1.2. For a, b ∈ Z with b 6= 0, there are unique q and r in Z such that a = bq+ r with 0 ≤ r < |b|. Theorem 1.3. For f, g ∈ F [T ] with g 6= 0, there are unique q and r in F [T ] such that f = gq + r with r = 0 or deg r < deg g. The greatest common divisor of two integers is the common divisor largest in size (so always positive). In F [T ], the greatest common divisor of two polynomials is the common monic polynomial factor with the largest degree. Examples will be worked out in the next section. Two integers are called relatively prime when their only common factors are ±1. Simi- larly, two polynomials in F [T ] are called relatively prime when their only common factors are nonzero constants. In both Z and F [T ], relative primality means the only common factors are units. Euclid’s algorithm is the standard method to compute greatest common divisors in Z (so, in particular, to determine relative primality) while a variant of Euclid’s algorithm in F [T ] will perform the same role for polynomials. The standard chain of reasoning div. thm. Euclid Bezout if p|ab then p|a or p|b unique factn in Z carries over to F [T ] nearly verbatim, with only minor changes needed in most proofs: div. thm. Euclid Bezout if π|fg then π|f or π|g unique factn. There is one important difference between Z and F [T ]. Division in Z involves remainders ≥ 0, so if two integers are relatively prime Euclid’s algorithm will always have last nonzero remainder 1. But this is false with polynomials: the last nonzero remainder in Euclid’s algorithm for polynomials might be a nonzero constant other than 1, so writing an F [T ]- linear combination of relatively prime polynomials as 1 can involve some additional scaling which we don’t have to do in Z. Example 1.4. In R[T ], let f(T ) = T 2 + 1 and g(T ) = T − 1. Certainly f(T ) and g(T ) are relatively prime: they have no common factor in R[T ] other than nonzero constants. When we carry out Euclid’s algorithm on these two polynomials we find T 2 + 1 = (T − 1)(T + 1) + 2 T − 1 = 2 ( 1 2 T − 1 2 ) + 0, so the last nonzero remainder is 2. This is a nonzero constant in R[T ] but it is not 1. By convention we normalize the gcd of two polynomials to be monic, so the gcd of T 2 + 1 and T − 1 is called 1, not 2. 2. Euclid and Bezout: examples Bezout’s identity in Z says for a and b in Z that we can write ax+ by = (a, b) for some integers x and y. Values for x and y can be found by using back-substitution into Euclid’s algorithm for a and b. Similarly, Bezout’s identity for F [T ] says for f(T ) and g(T ) in F [T ] that f(T )u(T ) + g(T )v(T ) = (f, g), for some u(T ) and v(T ) in F [T ]. Here too the polynomials u(T ) and v(T ) can be found using back-susbtitution into Euclid’s algorithm for f(T ) and g(T ). ANALOGIES WITH POLYNOMIALS 5 Using back-substitution in F3[T ], 1 = g − (T + 2)(T 2 + x+ 2) = g − (f − g(T + 1))(T 2 + T + 2) = f · (2T 2 + 2T + 1) + g · (1 + (T + 1)(T 2 + T + 2)) = f · (2T 2 + 2T + 1) + g · (T 3 + 2T 2). In F5[T ], 4T + 1 = g − (3T 2 + 2T )(2T + 2) = g − (f − g(T + 1)(2T + 2)) = f · (3T + 3) + g · (1 + (T + 1)(2T + 2)) = f · (3T + 3) + g · (2T 2 + 4T + 3). The gcd we found in Euclid’s algorithm, 4T + 1, is not monic. To write the monic gcd of f and g as an F5[T ]-linear combination of f and g we simply multiply through the equations by −1 = 4: T + 4 = f · (2T + 2) + g · (3T 2 + T + 2). In F7[T ], 4 = (3T 2 + 5)− (T + 3)(3T + 5) = (3T 2 + 5)− (g − (3T 2 + 5)(5T ))(3T + 5) = (3T 2 + 5)(1 + 5T (3T + 5)) + g(4T + 2) = (3T 2 + 5)(T 2 + 4T + 1) + g(4T + 2) = (f − g(T + 1))(T 2 + 4T + 1) + g(4T + 2) = f · (T 2 + 4T + 1) + g · (6T 3 + 2T 2 + 6T + 1). Multiplying through by 4−1 = 2, 1 = f · (2T 2 + T + 2) + g · (5T 3 + 4T 2 + 5T + 2). We can also do back-substitution in Q[T ]. The calculations would be quite tedious to do by hand on account of the large fractions arising in Euclid’s algorithm. The result will express 1395256 as a Q[T ]-linear combination of f(T ) and g(T ), and then we have to multiply through by the reciprocal 2561395 to write 1 as a Q[T ]-linear combination of f(T ) and g(T ). Omitting the intermediate details, the final result is 1 = f · ( 16 155 T 2 + 1 155 T − 61 155 ) + g · ( − 16 155 T 3 − 17 155 T 2 + 12 155 T − 54 155 ) . Remark 2.4. The common denominator 155 appearing in this equation factors as 5 · 31. This is related to the special roles of 5 and 31 in Remark 2.2! 3. Solving simultaneous congruences: an example In Z, if we want to solve the pair of congruence conditions x ≡ 2 mod 5, x ≡ 11 mod 19, we lift the first congruence to Z in the form x = 2 + 5y for some y ∈ Z and substitute that into the second congruence and solve for y: 2 + 5y ≡ 11 mod 19⇒ 5y ≡ 9 mod 19⇒ y ≡ 17 mod 19. 6 KEITH CONRAD Thus y = 17 + 19z for some integer z, so x = 2 + 5(17 + 19z) = 87 + 95z, so x ≡ 87 mod 95. Conversely, if x ≡ 87 mod 95 then x ≡ 2 mod 5 and x ≡ 11 mod 19 since 87 fits both conditions and the modulus 95 is divisible by 5 and 19. This is a special instance of the Chinese remainder theorem in Z. We can solve polynomial congruences in the same way. Consider in F5[T ] the two con- gruence conditions f(T ) ≡ 3T mod T 2 + 1, f(T ) ≡ 2T 2 + 1 mod T 3. Here the unknown we are looking for is f(T ), not T : T is just a variable for the polynomials. We want an f(T ) in F5[T ] which fits both congruence conditions. Lift the first congruence into F5[T ] by writing it as (3.1) f(T ) = 3T + (T 2 + 1)g(T ) for some g(T ) ∈ F5[T ]. Substitute this into the second congruence: 3T + (T 2 + 1)g(T ) ≡ 2T 2 + 1 mod T 3. Subtracting 3T from both sides (note −3T = 2T in F5[T ]), (3.2) (T 2 + 1)g(T ) ≡ 2T 2 + 2T + 1 mod T 3. We now need to invert T 2 + 1 mod T 3. This will be done with Euclid: in F5[T ], T 3 = (T 2 + 1)T + 4T, T 2 + 1 = 4T (4T ) + 1, so 1 = T 2 + 1− 4T (4T ) = T 2 + 1− 4T (T 3 − (T 2 + 1)T ) = (T 2 + 1)(4T 2 + 1) + T 3(−4T ), so (T 2 + 1)(4T 2 + 1) ≡ 1 mod T 3. Therefore in F5[T ], the inverse of T 2 + 1 mod T 3 is 4T 2 + 1, so multiplying both sides of (3.2) by 4T 2 + 1 gives g(T ) ≡ (4T 2 + 1)(2T 2 + 2T + 1) mod T 3 ≡ 3T 4 + 3T 3 + T 2 + 2T + 1 mod T 3 ≡ T 2 + 2T + 1 mod T 3. Therefore g(T ) = T 2 + 2T + 1 + T 3h(T ) for some h(T ) ∈ F5[T ], and substituting this formula for g(T ) into (3.1) shows any f(T ) fitting the two original congruence conditions has the form f(T ) = 3T + (T 2 + 1)(T 2 + 2T + 1 + T 3h(T )) = T 4 + 2T 3 + 2T 2 + 1 + (T 2 + 1)T 3h(T ), so f(T ) ≡ T 4 + 2T 3 + 2T 2 + 1 mod (T 2 + 1)T 3. As a check that T 4 + 2T 3 + 2T 2 + 1 fits the original two congruence conditions, in F5[T ] (T 4 + 2T 3 + 2T 2 + 1)− 3T = (T 2 + 1)(T + 1) and (T 4 + 2T 3 + 2T 2 + 1)− (2T 2 + 1) = T 3(T + 2). ANALOGIES WITH POLYNOMIALS 7 Therefore T 4 + 2T 3 + 2T 2 + 1 works, and more generally any polynomial congruent to T 4 + 2T 3 + 2T 2 + 1 mod (T 2 + 1)T 3 works. This is the complete set of solutions to both congruences.