Decomposition of primes in extensions

LemmaIn any ring R, if a1, . . . , an are pairwise coprime ideals, then a1 · · · an = a1 ∩ · · · ∩ an .
Proof Induction on n. If n = 2, a1 + a2 = R means that there
exist x1 ∈ a1 and x2 ∈ a2, such that x1 + x2 = 1. So, if
x ∈ a1 ∩ a2, x = xx1 + xx2 ∈ a1a2.
Assume the result holds for n − 1 pairwise coprime ideals, and,for i = j, let xij ∈ ai , yij ∈ aj be such that xij + yij = 1. So,xj = (1 − yij ) = 1 − yj , with yj ∈ aj .
RemarkNotice that we have proven in particular that in any ring R, if theideals a1, . . . , an are pairwise coprime, then and aj are coprime. It follows that, for any j = 1, . . . , n, we mayfind elements xj ∈ R such that 1 = xj + yj , with xj ∈ yj ∈ aj . Therefore, for any j = 1, . . . , n, we may find elementsxj ∈ R such that xj ≡ 0 mod ai ∀ i = j , and xj ≡ 1 mod aj .
Taking into account the previous lemma, the following theoremdeserves its name.
Theorem
Chinese Remainder Theorem. Let a1, . . . , an be ideals in a
ring R such that ai + aj = R, for i = j. Then, if a =
i . It therefore suffices to show that it is surjective. For this, let yi mod ai ∈ R/ai , i = 1, . . . , n, be given.
Then we may find elements xj ∈ R as before, and then puttingx = x1y1 + · · · + xnyn, we find x ≡ yi mod ai , i = 1, . . . , n.
We assume here that K ⊂ L are number fields of finite degree,and that L = K (θ), with θ ∈ OL. Let p(X ) ∈ OK [X ] be theminimal (monic) polynomial of θ over K .
DefinitionThe conductor of OK [θ] in OL is the biggest ideal F of OL whichis contained in OK [θ]. In other words F = {α ∈ OL | αOL ⊂ OK [θ] } .
Since OL is a finitely generated OK -module, F = 0.
We are now interested in the decomposition of a prime ideal pof OK in OL. We will prove next week that where Pi , for i = 1, . . . , r , is a prime ideal of OL, andei =: e(Pi /p) is a non-negative integer, called the ramificationindex of Pi /p. We set κ(Pi ) = OL/Pi , κ(p) = OK /p, and, definefi = f (Pi /p) := [κ(Pi ) : κ(p)], the residue degree of Pi /p.
We say that p ramifies in L, if ei > 1, for some i. We say that pis totally ramified in L, if p OL = P[L:K], for a prime P of OL. Wesay that p splits completely in L, if e1 = · · · = er = 1. We saythat p is inert in L, if p OL = P is prime, so that the residuedegree f (P/p) = [κ(P) : κ(p)] = [L : K ].
Granting this, we will explicitly determine the coefficients ei andfi , in a rather general situation.
PropositionLet L/K be as before. Let p be a prime ideal of OK which isrelatively prime to the conductor F of OK [θ] in OL, and let be the factorization of the polynomial p(X ) = p(X ) mod p overthe residue class field κ := κ(p) = OK /p. We assume that allpi(X ) = pi(X ) mod p, for pi(X ) ∈ OK [X ], monic. Then Pi = pOL + pi (θ)OL , i = 1, . . . , r , are distinct prime ideal of OL above p. The inertia degree fi ofPi is the degree of pi(X ), and one has Remark
“p relatively prime to F” means pOL + F = OL. This however is
equivalent to “p is relatively prime to F ∩ OK ”. So, there exist
p ∈ p and f ∈ F such that p + f = 1.
Proof In fact pOL + F = OL implies POL + F = OL, for any
prime of OL with P|p. But if p + (F ∩ OK ) = OK , it would follow
F ∩ OK ⊂ p, hence ∃ P, prime of OL, with P|p and P|F.
Contradiction.
The first isomorphism follows from the fact relative primality of pand F. In fact, as F ⊂ O , we have OL = pOL + O . So, thehomomorphism O → OL/pOL is surjective. It has kernelpOL ∩ O . But this equals pO : since (p, F ∩ OK ) = 1, for p, fas before we have pOL ∩ O = (p + f )(pOL ∩ O ) ⊂ pO + pf OL ⊂ The second isomorphism is deduced from the surjectivehomomorphism Its kernel is the ideal generated by p and p(X ), and sinceO = OK [θ] = OK [X ]/(p(X )), we have O /pO ∼ where the r.h.s. is a direct sum of rings. This shows that theprime ideals of the ring R = κ[X ]/(p(X )) are the principalideals (pi) generated by the pi(X ) mod p(X ), for i = 1, . . . , r ,that the degree [R/(pi) : κ] equals the degree of the polynomialpi(X ), and that In view of the isomorphism κ[X ]/(p(X )) ∼ f (X ) → f (θ), the same situation holds in the ring OL = OL/pOL.
Thus the prime ideals Pi of OL correspond to the prime ideals(pi), and they are the principal ideals generated by the pi(θ)mod pOL. Notice that the prime ideal (pi) = piR = piRi/pi)ei Ri ⊕j=i Rj .
The degree [OL/Pi : κ] is the degree of the polynomials pi(X ), the preimage of Pi via the canonical homomorphism Then Pi , for i = 1, . . . , r , varies over the prime ideals of OLabove p, fi = [OL/Pi : κ] is the degree of the polynomial pi(X ).
ei fi = [L : K ] = deg p(X ) = deg p(X ) = dimκ OL/pOL .
Let m be a squarefree integer, and let K = Q( m), R = OK .
We know that R = Z ⊕ Z m, if m ≡ 2, 3 mod 4, with with discriminant dK = m. Let p be a prime in Z. We know that there are only 3 possibilities of decomposition of p in R.
so the conductor of Z[θm] in R is R, and the previous theoremapplies.
where the factors in the middle formula are distinct.
3. If p is odd and does not divide m, then where the factors in the first formula are distinct.
m)2 = (p2, p m, m). This is contained in pR, since p|m. On the other hand, it contains the G.C.D. of p2 and m,which is p; hence, it contains pR.
since m is odd. On the other hand, it contains 1 + m + 2 m − 2 − 2 m = m − 1 ≡ 2 mod 4. So, it containsthe G.C.D. of 4 and m − 1, which is 2. We conclude that The two factors are distinct because 2 does not divide thediscriminant m in this case.
3.cases1,2. The factorization of p in R depends on thefactorization of the polynomial p(X ) = X 2 − m (if m ≡ 2, 3mod 4) or p(X ) = X 2 − X + 1−m (if m ≡ 1 mod 4) in Independently of this, if m ≡ n2 mod p, we have m) = (p2, pn, 2p m, n2 − m) ⊂ pR, and in fact = pR, since (n, p) = 1 implies ∃a such that an ≡ 1 mod p,so that p belongs to the ideal. Since p does not divide thediscriminant 4m, p does not ramify. So we get 3.case1unconditionally.
3.case2. This is clear if m ≡ 2, 3 mod 4 and m is not a squaremod p. If now m ≡ 1 mod 4, observe that the discriminant ofthe polynomial p(X ) = X 2 − X + 1−m in splits completely in R, then p(X ) has both solutions α1, α2 ∈ Fp,hence the discriminant m = (α1 − α2)2 is a square in Fp.
2.case3. This is the only case we left out. Here m ≡ 1 mod 4, mod 2, we get p(X ) = X 2 − X + 1, which is irreducible Let α3 = 2, and let E = Q(α). It can be shown that the ring ofalgebraic integers OE is precisely Z[α]. In fact, the discriminantis ∆(1, α, α2) = −27 × 4. So, we have to check for integers ofthe form with λ0, λ1, λ2 ∈ {0, 1}, or of the form with λ0, λ1, λ2 ∈ {0, 1, 2}.
Since Tr (α) = Tr (α2) = 0, while Tr 1 = 3, we may assume thatλ0 = 0 in both cases. Now, N (α) = 2, N (α2) = 4.
with λ0, λ1, λ2 ∈ {0, 1}, not all 0, and the second with λ0, λ1, λ2 ∈ {0, 1, 2}, not all 0. It is easy to see that there isno solution.
The decomposition of (2) in Z[α] is then and (α) is a prime ideal, with ramification index 3 over 2Z.
For p = 3, we have where p1 has residue class degree 1, and p2 has residue classdegree 2 over F5.
Let p = 7. Then X 3 − 2 is irreducible over F7[X ], since it has nozeros. So, 7Z is inert in E. This means that (7) is a prime idealof OE , so that its residue degree is 3.
Let p = 11. Here we check that 11|(73 − 2) = 341, and that where p1 has residue class degree 1, and p2 has residue classdegree 2 over F11.
CorollaryLet p be a prime number ≡ 1 mod 4. Then there exist integersx , y ∈ Z such that p = x2 + y2.
Proof This means that the ideal p splits in K = Q( −1). In fact
OK = Z[i] is a PID. So, pZ[i] = p1p2 means that
p = unit × (x + iy )(x − iy ). But the units of Z[i] are ±1, ±i, so
p = x 2 + y 2. Now, p splits in K if and only if −1 is a square
mod p. We know that this is the case iff
= (−1) 2 = +1, that is iff p ≡ 1 mod 4.
Let α be a primitive 8-th root of 1 in Fp. The ’elementy = α + α−1 satisfies y 2 = 2 (from α4 = −1, it follows thatα2 + α−2 = 0). We have y p = (α + α−1)p = αp + α−p .
If p ≡ ±1 mod 8, this implies y p = y , so that = 2 2 = y p−1 = 1. If p ≡ ±5 mod 8, then y p = α5 + α−5 = −(α + α−1) = −y . We deduce from this thaty p−1 = −1, which implies the result.
CorollaryLet p be a prime number ≡ 1, 3 mod 8. Then there existintegers x , y ∈ Z such that p = x2 + 2y2.
Proof This means that the ideal p splits in K = Q( −2). In fact
OK = Z[ −2] is a PID. So, pZ[ −2] = p1p2 means that p = unit × (x + i 2y )(x − i 2y ). But the units of Z[ −2] are±1, so p = x2 − 2y 2. Now, p splits in K if and only if −2 is asquare = +1, that is iff either (p ≡ 1 mod 4 and p ≡ ±1 mod 8) or (p ≡ 3 mod 4 and p ≡ 3, 5 mod 8). All inall this says either p ≡ 1 mod 8 or p ≡ 3 mod 8.
Similarly, one can prove, using QRL, that CorollaryLet p be a prime ≡ 1 mod 3. Then there are x , y ∈ Z such thatp = x 2 + 3y 2.
CorollaryLet p be a prime ≡ 1, 7 mod 8. Then there are x , y ∈ Z suchthat p = x 2 − 2y 2.
Proof This means that the ideal p splits in K = Q( 2). In fact
OK = Z[ 2] is a PID. So, pZ[ 2] = p1p2 means that 2)Z, so p = x 2 + 2y 2. Now, p splits in K if and only if 2

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