Cyclotomic integers: Multiplicities
In today's blog, I continue to review results from Harold Edward's Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory. I am continuing down the same chapter that I started in an earlier blog. If you would like to start at the beginning of cyclotomic integer properties, start here.
Definition 1: divisibility by a prime divisor
A cyclotomic integer g(α) is said to be divisible μ times by the prime divisor of p corresponding to u1, u2, ..., ue if g(α)ψ(η)μ is divisible by pμ, where ψ(η) is the product of the ep-e factors j - ηi (i=1,2,.....,e; j=0,1,...,p-1; j ≠ ui) that are not divisible by this prime divisor. It is said to be divisible exactly μ times by the prime divisor if it is divisible μ times but not divisible μ+1 times.
Proposition 1:
The notion of "divisible one time" coincides with the notion of "divisible" as defined earlier (see Definition 2, here). The notion of "divisible exactly zero times" coincides with "not divisible". The integer p is divisible exactly once; the integer 1 is not divisible at all. If g1(α) and g2(α) are both divisible μ times, then so is g1(α) + g2(α). If g1(α) is divisible exactly μ times and g2(α) is divisible exactly ν times then g1(α)g2(α) is divisible exactly μ + ν times. Finally, if g(α) ≠ 0 then there is a unique integer μ ≥ 0 such that g(α) is divisible exactly μ times.
Proof:
(1) g(α) is divisible by a prime divisor if and only if g(α)ψ(η) ≡ 0 (mod p) [See definition 2, here]
(2) This shows that g(α) is "divisible one time" if and only if g(α) is "divisible one time" [since g(α) is divisible "one time" if and only if g(α)ψ(η) ≡ 0 (mod p), see definition above]
(3) Since we know that ψ(η) not ≡ 0 (mod p) [See here if you need to review the construction of ψ(η)], we know that ψ(η)ψ(η) not ≡ 0 (mod p) and we know that pψ(η)ψ(η) not ≡ 0 (mod p2)
(4) Step #3 combined with step #2 shows that p is divisible exactly once.
(5) Because ψ(η) not ≡ 0 (mod p), we know that (1)ψ(η) not ≡ 0 (mod p) or in other words, the integer 1 is not divisible at all.
(6) The statement that g1(α) + g2(α) is divisible μ times whenever g1(α), g2(α) are, follows immediately from the definition.
(7) From the definition above, if g1(α) is divisible exactly μ times then g1(α)ψ(η)μ+1 is divisible by pμ but not by pμ+1
(8) Therefore, φ1(α) = p-μg1(η)ψ(η)μ is a cyclotomic integer.
(9) Assume that φ1(α) is divisible by one of the prime divisors.
(10) Since ψ(η) is divisible by the remaining e-1 prime divisors of p [since each remaining divisor is characterized by the formula ψ(η)σkψ(η) ≡ 0 (mod p), see Theorem 4 here for review], it follows that φ1(α)ψ(η) is divisible by all e prime divisors of p.
(11) Hence, it would follow that p divides φ1(α)ψ(η) [Note: this follows from Theorem 4 here]
(12) But then g1(α) is divisible μ + 1 times by the prime divisor in question which contradicts the given so we can reject our assumption in step #9.
(13) Therefore φ1(α) is not divisible by the prime divisor in question.
(14) Similarly, if g2(α) is divisible exactly ν times, then φ2 = p-νg2(α)ψ(η)ν is a cyclotomic integer that is not divisible by the prime divisor in question.
(15) Therefore g1(α)g2(α)ψ(η)μ+ν is divisible by pμ + ν and the quotient is a product φ1(α)φ2(α) of two cyclotomic integers neither of which is divisible by the prime divisor in question.
(16) Because this prime divisor is prime, φ1(α)φ2(α) is not divisible by the prime divisor in question.
(17) Since ψ(η) is not divisible by this prime divisor [ψ(η)ψ(η) not ≡ 0 (mod p)] it follows that φ1(α)φ2(α)ψ(η) = g1(α)g2(α)ψ(η)μ+ν+1p-μ-ν is not divisible by the prime divisor of p and a fortiori that is not divisible by p. [Again, using Theoreom 4 from here]
(18) Therefore g1(α)g2(α)ψ(η)μ+ν+1 is not divisible by pμ + ν + 1 and g1(α)g2(α) is divisible with multiplicity exactly μ + ν as was to be shown.
(19) Let g(α) be a cyclotomic integer such that g(α) ≠ 0.
(20) If there is an integer k greater than 0 such that g(α) is not divisible k times by the prime divisor of p in question, then directly from the definition, there must be at least one integer μ such that 0 ≤ μ is less than k such that g(α) is divisible exactly μ times.
(21) As was seen above, this implies that g(α)ψ(η)μ/pμ is not divisible at all.
(22) Therefore, for any j ≥ 0, g(α)ψ(η)μ+j/pμ is not divisible at all.
(23) Consequently, g(α) is not divisible with multiplicity μ + j for j greater than 0.
(24) In particular, g(α) is not divisible with multiplicity m whenever m is greater than k.
(25) To prove the last statement of the proposition, it will suffice to prove that there is an integer k such that g(α) is not divisible k times, because then the exact multiplicity μ with which g(α) is divisible is uniquely determined as the largest integer μ such that g(α) is not divisible μ times.
(26) To this end, note that if g(α) is divisible k times, so is Ng(α).
(27) Let Ng(α) = pnq where n ≥ 0 and q is not divisible by p.
(28) Then, Ng(α) is divisible exactly n times and cannot therefore be divisible k times whenever k is greater than n and the proof is complete.
QED
Definition 2: Multiplicity of a prime divisor
A prime divisor in the arithmetic of cyclotomic integers (for a fixed prime λ greater than 2) is either (a) one of the e prime divisors of some prime p ≠ λ that have been defined above or (b) the prime divisor α - 1 of λ. The prime divisors of the first type are described by giving a prime p ≠ λ and the integers u1, u2, ..., ue. Given a cyclotomic integer g(α) ≠ 0 and a prime divisor there is a multiplicity μ ≥ 0 with which g(α) is divisible by the prime divisor. For prime divisors of the first type this multiplicity is defined to be the exact multiplicity with which g(α) is divisible by the prime divisor in the sense just defined above. For the single prime divisor of the second type it is defined to be the number of times that α - 1 divides g(α)
Proposition 2:
Congruence mod α - 1 is reflexive, symmetric, transitive, and consistent with addition and multiplication. Moreover, α - 1 is prime; that is, g1(α)g2(α) ≡ 0 (mod α - 1) implies that either g1(α) ≡ 0 or g2(α) ≡ 0 (mod α - 1). An integer is divisible by α - 1 if and only if it is divisible by λ. The multiplicity with which g(α) ≠ 0 is divisible by α - 1 is well defined; that is, there is a μ ≥ 0 such that (α - 1)μ divides g(α) but (α - 1)μ+1 does not. If α - 1 divides both g1(α) and g2(α) with multiplicity at least μ, then it divides g1(α) + g2(α) with multiplicity at least μ. If α - 1 divides g1(α) with multiplicity exactly μ and g2(α) with multiplicity exactly ν then it divides g1(α)g2(α) with multiplicity exactly μ + ν. Finally, λ divides g(α) if and only if α - 1 divides g(α) with multiplicity at least λ - 1.
Proof:
(1) α - 1 divides g(α) if and only if g(1) ≡ 0 (mod λ) since:
(a) N(α - 1) = λ [See Lemma 6 here for details.]
(b) Since α - 1 divides g(α) implies N(α-1) divides Ng(α), we know that this is only the case if Ng(α) is divisible by λ which means that if g(α) is an integer, that g(α) must be divisible by λ.
(c) Assume α - 1 divides g(α) so that g(α) ≡ 0 (mod α -1 )
(d) Then g(1) ≡ 0 (mod α - 1) since α ≡ 1 (mod α - 1) so any equation with α is congruent to the same equation with 1.
(e) Then step d implies that α - 1 divides g(1).
(f) But then λ must divide g(1) since N(α - 1) divides Ng(1) and the only way that this can happen is if g(1) is divisible by λ
(g) Assume g(1) ≡ 0 (mod λ)
(h) Then g(1) ≡ 0 (mod α - 1) since (α - 1)(α2 - 1)*...*(αλ-1 - 1) = λ
(2) We know that α - 1 is prime since N(α - 1)= λ which is a prime.
(3) All statements but the last follow from the same reasoning as the details behind Theorem 1.
(4) The last statement follows from the fact that λ = (α - 1)(α2 - 1) * ... *(αλ-1 - 1) = (α - 1)λ-1 * unit
QED
Definition 1: divisibility by a prime divisor
A cyclotomic integer g(α) is said to be divisible μ times by the prime divisor of p corresponding to u1, u2, ..., ue if g(α)ψ(η)μ is divisible by pμ, where ψ(η) is the product of the ep-e factors j - ηi (i=1,2,.....,e; j=0,1,...,p-1; j ≠ ui) that are not divisible by this prime divisor. It is said to be divisible exactly μ times by the prime divisor if it is divisible μ times but not divisible μ+1 times.
Proposition 1:
The notion of "divisible one time" coincides with the notion of "divisible" as defined earlier (see Definition 2, here). The notion of "divisible exactly zero times" coincides with "not divisible". The integer p is divisible exactly once; the integer 1 is not divisible at all. If g1(α) and g2(α) are both divisible μ times, then so is g1(α) + g2(α). If g1(α) is divisible exactly μ times and g2(α) is divisible exactly ν times then g1(α)g2(α) is divisible exactly μ + ν times. Finally, if g(α) ≠ 0 then there is a unique integer μ ≥ 0 such that g(α) is divisible exactly μ times.
Proof:
(1) g(α) is divisible by a prime divisor if and only if g(α)ψ(η) ≡ 0 (mod p) [See definition 2, here]
(2) This shows that g(α) is "divisible one time" if and only if g(α) is "divisible one time" [since g(α) is divisible "one time" if and only if g(α)ψ(η) ≡ 0 (mod p), see definition above]
(3) Since we know that ψ(η) not ≡ 0 (mod p) [See here if you need to review the construction of ψ(η)], we know that ψ(η)ψ(η) not ≡ 0 (mod p) and we know that pψ(η)ψ(η) not ≡ 0 (mod p2)
(4) Step #3 combined with step #2 shows that p is divisible exactly once.
(5) Because ψ(η) not ≡ 0 (mod p), we know that (1)ψ(η) not ≡ 0 (mod p) or in other words, the integer 1 is not divisible at all.
(6) The statement that g1(α) + g2(α) is divisible μ times whenever g1(α), g2(α) are, follows immediately from the definition.
(7) From the definition above, if g1(α) is divisible exactly μ times then g1(α)ψ(η)μ+1 is divisible by pμ but not by pμ+1
(8) Therefore, φ1(α) = p-μg1(η)ψ(η)μ is a cyclotomic integer.
(9) Assume that φ1(α) is divisible by one of the prime divisors.
(10) Since ψ(η) is divisible by the remaining e-1 prime divisors of p [since each remaining divisor is characterized by the formula ψ(η)σkψ(η) ≡ 0 (mod p), see Theorem 4 here for review], it follows that φ1(α)ψ(η) is divisible by all e prime divisors of p.
(11) Hence, it would follow that p divides φ1(α)ψ(η) [Note: this follows from Theorem 4 here]
(12) But then g1(α) is divisible μ + 1 times by the prime divisor in question which contradicts the given so we can reject our assumption in step #9.
(13) Therefore φ1(α) is not divisible by the prime divisor in question.
(14) Similarly, if g2(α) is divisible exactly ν times, then φ2 = p-νg2(α)ψ(η)ν is a cyclotomic integer that is not divisible by the prime divisor in question.
(15) Therefore g1(α)g2(α)ψ(η)μ+ν is divisible by pμ + ν and the quotient is a product φ1(α)φ2(α) of two cyclotomic integers neither of which is divisible by the prime divisor in question.
(16) Because this prime divisor is prime, φ1(α)φ2(α) is not divisible by the prime divisor in question.
(17) Since ψ(η) is not divisible by this prime divisor [ψ(η)ψ(η) not ≡ 0 (mod p)] it follows that φ1(α)φ2(α)ψ(η) = g1(α)g2(α)ψ(η)μ+ν+1p-μ-ν is not divisible by the prime divisor of p and a fortiori that is not divisible by p. [Again, using Theoreom 4 from here]
(18) Therefore g1(α)g2(α)ψ(η)μ+ν+1 is not divisible by pμ + ν + 1 and g1(α)g2(α) is divisible with multiplicity exactly μ + ν as was to be shown.
(19) Let g(α) be a cyclotomic integer such that g(α) ≠ 0.
(20) If there is an integer k greater than 0 such that g(α) is not divisible k times by the prime divisor of p in question, then directly from the definition, there must be at least one integer μ such that 0 ≤ μ is less than k such that g(α) is divisible exactly μ times.
(21) As was seen above, this implies that g(α)ψ(η)μ/pμ is not divisible at all.
(22) Therefore, for any j ≥ 0, g(α)ψ(η)μ+j/pμ is not divisible at all.
(23) Consequently, g(α) is not divisible with multiplicity μ + j for j greater than 0.
(24) In particular, g(α) is not divisible with multiplicity m whenever m is greater than k.
(25) To prove the last statement of the proposition, it will suffice to prove that there is an integer k such that g(α) is not divisible k times, because then the exact multiplicity μ with which g(α) is divisible is uniquely determined as the largest integer μ such that g(α) is not divisible μ times.
(26) To this end, note that if g(α) is divisible k times, so is Ng(α).
(27) Let Ng(α) = pnq where n ≥ 0 and q is not divisible by p.
(28) Then, Ng(α) is divisible exactly n times and cannot therefore be divisible k times whenever k is greater than n and the proof is complete.
QED
Definition 2: Multiplicity of a prime divisor
A prime divisor in the arithmetic of cyclotomic integers (for a fixed prime λ greater than 2) is either (a) one of the e prime divisors of some prime p ≠ λ that have been defined above or (b) the prime divisor α - 1 of λ. The prime divisors of the first type are described by giving a prime p ≠ λ and the integers u1, u2, ..., ue. Given a cyclotomic integer g(α) ≠ 0 and a prime divisor there is a multiplicity μ ≥ 0 with which g(α) is divisible by the prime divisor. For prime divisors of the first type this multiplicity is defined to be the exact multiplicity with which g(α) is divisible by the prime divisor in the sense just defined above. For the single prime divisor of the second type it is defined to be the number of times that α - 1 divides g(α)
Proposition 2:
Congruence mod α - 1 is reflexive, symmetric, transitive, and consistent with addition and multiplication. Moreover, α - 1 is prime; that is, g1(α)g2(α) ≡ 0 (mod α - 1) implies that either g1(α) ≡ 0 or g2(α) ≡ 0 (mod α - 1). An integer is divisible by α - 1 if and only if it is divisible by λ. The multiplicity with which g(α) ≠ 0 is divisible by α - 1 is well defined; that is, there is a μ ≥ 0 such that (α - 1)μ divides g(α) but (α - 1)μ+1 does not. If α - 1 divides both g1(α) and g2(α) with multiplicity at least μ, then it divides g1(α) + g2(α) with multiplicity at least μ. If α - 1 divides g1(α) with multiplicity exactly μ and g2(α) with multiplicity exactly ν then it divides g1(α)g2(α) with multiplicity exactly μ + ν. Finally, λ divides g(α) if and only if α - 1 divides g(α) with multiplicity at least λ - 1.
Proof:
(1) α - 1 divides g(α) if and only if g(1) ≡ 0 (mod λ) since:
(a) N(α - 1) = λ [See Lemma 6 here for details.]
(b) Since α - 1 divides g(α) implies N(α-1) divides Ng(α), we know that this is only the case if Ng(α) is divisible by λ which means that if g(α) is an integer, that g(α) must be divisible by λ.
(c) Assume α - 1 divides g(α) so that g(α) ≡ 0 (mod α -1 )
(d) Then g(1) ≡ 0 (mod α - 1) since α ≡ 1 (mod α - 1) so any equation with α is congruent to the same equation with 1.
(e) Then step d implies that α - 1 divides g(1).
(f) But then λ must divide g(1) since N(α - 1) divides Ng(1) and the only way that this can happen is if g(1) is divisible by λ
(g) Assume g(1) ≡ 0 (mod λ)
(h) Then g(1) ≡ 0 (mod α - 1) since (α - 1)(α2 - 1)*...*(αλ-1 - 1) = λ
(2) We know that α - 1 is prime since N(α - 1)= λ which is a prime.
(3) All statements but the last follow from the same reasoning as the details behind Theorem 1.
(4) The last statement follows from the fact that λ = (α - 1)(α2 - 1) * ... *(αλ-1 - 1) = (α - 1)λ-1 * unit
QED
posted by Larry Freeman at 8:52 AM
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