MATH 361: NUMBER THEORY — TWELFTH LECTURELetω = ζ3= e2πi/3=−1 +√−32.The subjects of this lecture are the arithmetic of the ringD = Z[ω],and the cubic reciprocity law.1. Unique FactorizationThe ring D is Euclidean with norm functionN : D −→ Z≥0, Nz = zz, N (a + bω) = a2− ab + b2.Hence D is a PID and consequently a UFD. That is, every nonzero z ∈ D takesthe formz = ugYi=1πeii, u ∈ D×, each πiirreducible, each ei∈ Z+,and if also z = ˜uQ˜gi=1˜π˜eiithen ˜g = g and after indexing we may take each ˜πi= uiπiwith ui∈ D×(that is, ˜πiand πiare associate) and ˜ei= ei.2. UnitsOnly 0Dhas norm 0. For any nonzero u ∈ D, we have the equivalenceu ∈ D×⇐⇒ Nu = 1.( ⇐= is immediate since if uu = 1 then u has inverse u; =⇒ is also immediatesince if u ∈ D×then Nu = (Nu−1)−1is both a positive integer and a reciprocalpositive integer, so it is 1.) The equivalence shows thatD×= {±1, ±ω, ±ω2} = hζ6i.Structurally, D×∼=Z/6Z, with generators ζ6= −ω2and ζ−16= −ω.3. Irreducibles (Nonzero Primes)Let π ∈ D be irreducible. Then π | ππ = N π ∈ Z>1. Thus (since π is prime)π | p for at least one rational prime p. If also π | q for a different rational prime qthen consequently π | 1, a false statement. So in factπ | p for a unique rational prime p.(The result just displayed can also be established by working with ideals, as follows.Since πD is a prime ideal of D, πD ∩ Z is a prime ideal of Z, nonzero since itcontains ππ. Thus πD ∩ Z = pZ for some rational prime p. Since pZ ⊂ πD, alsopD ⊂ πD, showing that π | p. If also π | q for a different rational prime q thenconsequently qD ⊂ πD and thus 1 ∈ πD, leading to the false statement that π is aunit.)12 MATH 361: NUMBER THEORY — TWELFTH LECTURENote: If π | q then π | q = q, showing that q = p. Thus Nπ = ππ is a powerof p. Let the letter f denote the relevant power. That is, define f by the formulaNπ = pf.4. Factorization of Rational PrimesSince each irreducible π is a factor of a unique rational prime p, the questionnow is how rational primes factor in D. The factorization of any rational prime pisp = ugYi=1πeii, u ∈ D×, N πi= pfifor each i.It follows thatp2= Np = N ugYi=1πeii!= 1 ·gYi=1(pfi)ei=gYi=1peifi= pPgi=1eifi.Therefore, the positive integers ei, fi, and g satisfy the relationgXi=1eifi= 2.There are three possibilities.• p splits: g = 2, e1= e2= 1, f1= f2= 1. Here we have p = uπ1π2whereNπ1= Nπ2= p, so that in factp = ππ and π, π are nonassociate.• p is inert: g = 1, e = 1, f = 2. Here we have p = uπ where Nπ = p2, sothatp is irreducible in D.• p ramifies: g = 1, e = 2, f = 1. Here we have p = uπ2where Nπ = p, sothatp = ππ and π, π are associate.The next question is: Which rational primes p split, which are inert, and whichramify?• The prime p = 3 ramifies. Specifically,3 = −ω2λ2where λ = 1 − ω .This was a homework problem. To see where the factorization comes from,set X = 1 in the relation X2+ X + 1 = (X − ω2)(X − ω) to get3 = (1 − ω2)(1 − ω) = (1 + ω)(1 − ω)2= −ω2λ2.• The prime p = 3 is the only prime that ramifies. If p ramifies then p = ππwith π/π ∈ D×. After replacing π by ωπ or ω2π if necessary, we mayassume that π/π = ±1, and hence π2= ±p. Let π = a + bω (with b 6= 0),so that π2= (a2−b2) + (2ab − b2)ω. Since π2= ±p and b 6= 0, necessarilyb = 2a. Thus N π = a2− ab + b2equals 3a2. Thus p = 3, and furthermoreπ = ±(1 + 2ω) = ±ω(ω2+ 2) = ±ω(1 − ω) = ±ωλ.MATH 361: NUMBER THEORY — TWELFTH LECTURE 3• If p = 1 mod 3 then p splits. Indeed, the character groupdF×pcontains anelement χ of order 3. Note that χ(F×p) ⊂ D×. Let π = J(χ, χ) ∈ D. Bythe table of Jacobi sum values, N π = p. So p is not inert. Nor does itramify, so the remaining possibility is that it splits.• If p = 2 mod 3 then p is inert. We show this by contraposition. If p is notinert then p = Nπ for some π, i.e., p = a2− ab + b2for some a and b. So4p = (2a −b)2+ 3b2for some a and b, so that p is a square modulo 3. Thusp 6= 2 mod 3. Note:From now on we use the symbol q to denote a 2 mod 3 prime.In sum, we have shown that the Legendre symbol (·/3) describes factorization in D,p splits ⇐⇒ (p/3) = 1,p is inert ⇐⇒ (p/3) = −1,p ramifies ⇐⇒ (p/3) = 0.5. Canonical Representative of Each Associate ClassesEach irreducible in D is one of six associates. We now specify one associate fromeach class of six.As before, we specify λ = 1 − ω among the irreducibles that divide 3.For any rational prime p 6= 3, each divisor π of p has a so-called primaryassociate, meaning the associate π0such thatπ0= 2 mod 3.That is,π0= a + bω, a = 2 mod 3, b = 0 mod 3.Indeed, if π divides a rational prime q = 2 mod 3 then its primary associate issimply q. Otherwise, π divides a rational prime p = 1 mod 3 and N π = p. Thefirst part of the proof of 9.3.5 in Ireland and Rosen shows that π has a primaryassociate. (It may help to modify the wording after (a)–(f) to say, we may assume(after replacing a + bω by an associate if necessary) that 3 - a. . . ) In both cases,the second part of the proof of Ireland and Rosen 9.3.5 shows that the primaryassociate is unique.Given a rational prime p = 1 mod 3, to find a primary prime π lying over p,proceed as follows. We know that π = a + bω where a = 2 mod 3 and b = 0 mod 3,and we want to find a and b. Since p = N π = a2− ab + b2, it follows that4p = (2a − b)2+ 3b2. The procedure is to find A and B such that 4p = A2+ 27B2and A = 1 mod 3 (note that A and B must have the same parity; also note that wehave two choices for B, leading to the two primary primes π and π lying over p) andthen set b = 3B (which has the same parity as B, hence the same parity as A, andindependently of the parity we have A + b = 4 mod 6), and finally a = (A + b)/2,which is 2 mod 3. Note that this is exactly the procedure described in Gauss’sTheorem about the solution-count of the equation x3+ y3= 1 modulo p.For example, let …