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Discrete valuation rings Suppose F is a field. A discrete valuation on F is a function v : F ∗ − {0} → Z such that: 1. v is surjective. 2. v(ab) = v(a) + v(b). 3. v(a + b) ≥ min(v(a), v(b)) if a + b 6= 0. Proposition: The set R = {0} ∪ {r ∈ R : v(r) ≥ 0} is a ring, which we will call the valuation ring of v. Def: An integral domain will be called a discrete valuation ring if it is the valuation ring of a discrete valuation of its quotient field. 1 Proof of the Proposition: We have to show that R is closed under subtraction and multiplication, and that 1 ∈ R. Note first that v(1) = v(1 · 1) = v(1) + v(1) forces v(1) = 0, so 1 ∈ R. Then v(−1) + v(−1) = v(−1 · −1) = v(1) = 0 so v(−1) = 0 and −1 ∈ R. If a, b ∈ R, then v(a) ≥ 0 and v(b) ≥ 0. We get v(ab) = v(a) + v(b) ≥ 0 so ab ∈ R. Also, v(a−b) ≥ v(a)+v(−b) = v(a)+v(−1)+v(b) ≥ 0 so a − b ∈ R and we’re done. 2 Theorem: Discrete valuation rings (D.V.R.’s) are Euclidean. Proof: We define N : R → Z≥0 by N (0) = 0 and N (r) = v(r) if 0 6= r ∈ R. To show the Euclidean property, suppose a, b ∈ R and 0 6= b. We have to find q, r ∈ R with a = qb + r and r = 0 or N (r) < N (b). If v(a) ≥ v(b) then v(a/b) = v(a) − v(b) ≥ 0 so q = a/b ∈ R and we can let r = 0. Suppose v(a) < v(b). This case is easy: let q = 0 and r = a. 3 Example: Suppose p is a rational prime. There is a discrete valuation vp : Q∗ → Z defined by letting a vp( ) = ordp(a) − ordp(b) b for all non-zero integers a, b, where ordp(a) is the exponent of the highest power of p which divides a. The valuation ring of vp is just the localization a Z(p) = { ∈ Q : a, b ∈ Z, p 6 |b} b It will be a homework problem on the next homework assignment to check that every discrete valuation of Q has the form vp for some prime p. 4 Example: Suppose F = C(t) is the rational function field in one variable. If α ∈ C, we can define a discrete valuation vα : C(t)∗ → Z by vα(f (x)) = n if s(x) n f (x) = (x − α) · t(x) with s(α) 6= 0 6= t(α). The discrete valuation ring of vα is the localization h(x) C[x](x−α) = { : h(x), g(x) ∈ C[x], g(α) 6= 0} g(x) and all the discrete valuations of C(t) have the form vα for some α ∈ C. 5 Example: If p is a prime, the p-adic integers Zp form the valuation ring of the discrete valuation vp : Q∗p → Z defined in the following way. Let vp(r) = n if rp−n is a unit of Zp. This is equivalent to saying that r can be represented in the form r = pn ∞ X aipi i=0 with ai ∈ {0, . . . , p − 1} for all i and a0 6= 0. 6 Greatest common divisors Def: Suppose R is a commutative ring and that a, b ∈ R. An element d ∈ R is called a G.C.D. (greatest common divisor) for a and b if: 1. d|a in the sense that a = dd0 for some d0 ∈ R, and d|b. 2. If d0 is any other element of R such that d0|a and d0|b then d0|d. Write gcd(a, b) for the set of all G.C.D.’s for a and b. Example: The elements d of R = Z which are G.C.D.’s for a = 2 and b = 3 are ±1. This illustrates the fact if d ∈ gcd(a, b) then du ∈ gcd(a, b) for all units u ∈ R∗. 7 Theorem: Suppose that a, b ∈ R and that the ideal Ra + Rb equals Rd for some d ∈ R. Then d ∈ gcd(a, b). Proof: If d0|a and d0|b then Ra + Rd ⊂ Rd0. So Rd = Ra + Rb ⊂ Rd0 and d = f d0 for some f ∈ R. This means d0|d in R. Warning: The converse is not true. For example, suppose R = Z[x], a = 2 and b = x. We claim that d = 1 is in gcd(a, b), but Ra + Rb 6= Rd. It’s clear that 1|a and 1|b. So to show 1 ∈ gcd(a, b) it will suffice to show that if d0|a and d0|b in R, then d0|1, which is the same as saying d0 ∈ R∗. If d0|2, then d0 must be a constant polynomial, and d ∈ {±1, ±2}. But d0|x implies d0 6∈ {±2}. So d0 = ±1 is a unit in R, and 1 ∈ gcd(2, x). 8 The ideal Ra + Rb = R · 2 + R · x is not equal to R · 1 = R, though, since the constant term of every element of R · 2 + R · x is an even integer. Corollary: If R is a P.I.D, then every pair of elements a, b of R has a G.C.D. If one of a or b is nonzero, then all the elements of gcd(a, b) differ from each other by multiplication by a unit. Proof: Given a, b ∈ R we know Ra + Rb = Rd for some d ∈ R since R is a P.I.D, and d ∈ gcd(a, b) by the previous Theorem. If of a or b is not zero, then d must be non-zero as well. If d0 is some other element of gcd(a, b), then d0|d and d|d0. So d0 = ud and d = vd0 for some u, v ∈ R. Then d = uvd so d(1 − uv) = 0. Since d 6= 0 and R is an integral domain, this forces uv = 1, so u, v ∈ R∗. 9 The Euclidean algorithm Theorem: Suppose R is a Euclidean domain with Euclidean norm N : R → Z≥0. Let a and b be elements of R, and suppose 0 6= b. Construct a sequence of pairs (qi, ri) of elements of R using the Euclidean property of N : a = qb + r1 with r1 = 0 or N (r1) < N (b) b = q1 r1 + r2 with r2 = 0 or N (r2) < N (r1) r1 = q2r2 +r3 with r3 = 0 or N (r3) < N (r2) ··· rk = qk+1rk+1 + 0 The last non-zero remainder rk+1 in this sequence is in gcd(a, b). Note: The sequence terminates because the norms N (ri) form a decreasing sequence of non-negative integers. 10 Proof: We know that R is a P.I.D., so gcd(a, b) is non-empty. An element c of R divides both a and b if and only if it divides both b and r1 = a − qb. So by induction, gcd(a, b) = gcd(b, r1) = gcl(r1, r2) = · · · gcd(rk , rk+1) However, by assumption, rk+1 divides rk , so we find rk+1 ∈ gcd(rk , rk+1) = gcd(a, b) Note: We can write the equations in the Euclidean algorithm in reverse order and use them to find c, d ∈ R such that ca + db = rk+1. For example, rk−1 − qk rk rk−1 − qk (rk−2 − qk−1rk−1) −qk rk−2 + (1 + qk qk−1)rk−1 −qk rk−2 + (1 + qk qk−1)(rk−3 − qk−2rk−2) ··· = ca + db rk+1 = = = = 11 Principal ideal domains We’ve seen that Euclidean domains are P.I.D.’s. At the end of section 8.1 of Dummit and Foote’s book, there is a proof that the ring ! √ 1 + −19 R=Z+Z 2 is a P.I.D. which is not Euclidean with respect to any norm. One way to determine that a ring R is not a P.I.D. is to use the following result: Theorem: Suppose R is a P.I.D.. Then every non-zero prime ideal of R is maximal. Example: The ring C[x, y] = A is not a P.I.D.. This is because the ideal P = Ax is prime, because A/P is isomorphic to the polynomial ring C[y], which is an integral domain. But P is not maximal, since it’s property contained in the maximal ideal Ax + Ay. 12 Proof of the Theorem: Let P be a non-zero prime ideal. Then P = Ra for some 0 6= a ∈ R since R is a P.I.D.. Suppose P is not a maximal ideal. Then P is properly contained in some maximal ideal M . Write M = Rm for some m, again using that R is a P.I.D.. Then P = Ra ⊂ M = Rm implies a = m · m0 for some m0 ∈ R. If we can show m, m0 6∈ P , this will contradict the fact that P is prime and we’ll be done. We have m 6∈ P since P = Ra 6= Rm = M . If m0 ∈ P then m0 = za for some z ∈ R, and a = m · m0 = mza. But then a · (1 − mz) = 0 so a 6= 0 implies 1 − mz = 0 since R is an integral domain. But then m is a unit, so M = Rm = R, which is impossible since M is a maximal ideal. 13 Def: Supppose R is a commutative ring. A function N : R → Z≥0 is a Dedekind Hasse norm on R if for all a, b ∈ R, either a ∈ Rb or there is an element r ∈ Ra + Rb such that N (r) < N (b). Comment: Any Euclidean norm is a Dedekind Hasse norm. Theorem: R is a principal ideal domain if and only if it is an integral domain and has a Dedekind Hasse norm. 14 Proof: In one direction, suppose that R has a Dedekind Hasse norm N . Let A be any nonzero ideal of R. Let b be an element of A of minimal norm. We claim A = Rb. If not, there is an a ∈ A not in Rb. The property of D-H norms then says there is an element r ∈ Ra + Rb ⊂ A with N (r) < N (b), contradicting the definition of b. We’ll postpone proving the other part of the Theorem until after we’ve discussed unique factorization domains. 15 Unique Factorization Domains Def: Suppose R is an integral domain. 1. An element r ∈ R is irreducible if r 6= 0, r is not a unit, and whenever r = ab for some a, b ∈ R, one of a or b is a unit. 2. An element r ∈ R is called a prime element if Rr is a prime ideal. 3. If a, b ∈ R − {0} and a = bu for some unit u ∈ R∗, say that a and b are associate. This defines an equivalence relation on R. Note: A non-zero prime element r is irreducible. This is because r = ab and Rr prime implies a ∈ Rr or b ∈ Rr. If a = ur for some u ∈ R, then a = ur = uab, so a(1 − ub) = 0. But a 6= 0 since r 6= 0, so 1 − ub = 0 and b is a unit. 16 Def: An integral domain R is a U.F.D. (unique factorization domain) if the following is true for each non-zero r ∈ R: 1. There are (possibly not distinct) irreducible elements p1, · · · , pn of R and a unit u ∈ R∗ such that r=u n Y pi i=1 2. The factorization in (1) is unique up to rearranging the pi, replacing them by associate elements of R and replacing u by a different unit u0. Example: Z is a U.F.D.. Each irreducible is a prime element; these have the form ±p. 17 Theorem: If R is a P.I.D. then it is a U.F.D.. Proof: We first have to show that each nonzero r ∈ R has a factorization into a finite product of irreducible elements. If r can’t be written as a finite product of irreducibles, then there has to be a factorization r = st in which neither s or t are units. Furthermore one of s or t can’t be written as a finite product of irreducibles. The ideal Rr is properly contained in each of Rs and Rt. For example, if Rr = R(st) = Rs, we would get s = stu for some u, and s(1 − tu) = 0 would imply t is a unit, a contradiction. Let r1 be an element of {s, t} which is not a finite product of irreducibles. Then Rr is properly contained in Rr1. On replacing Rr by Rr1 and continuing this argument, we produce an infinite chain of ideals Rr ⊂ Rr1 ⊂ Rr2 ⊂ · · · such that Rri 6= Rri+1 for all i. 18 The ideal ∪iRri is principal, since R is a P.I.D. Writing ∪iRri = Rd for some d, we see that d ∈ Rrj for some j, But then Rrj+1 ⊂ Rd ⊂ Rrj a contradiction. So every r ∈ R can be written as a finite product of irreducibles. To prove the essential uniqueness of the factorization of r into irreducibles, suppose that r = up1 · · · pn = vq1 · · · qm (1) for some integers n, m ≥ 0 and some irreducibles pi and qj and some units u and v . We claim Rp1 is a maximal ideal. If not, then Rp1 is properly contained in M = Rm for some maximal ideal M and some m ∈ R. But then p1 = mm0 and neither m nor m0 is a unit, so p1 is not irreducible. We’ve thus shown Rp1 is maximal, and hence a prime ideal. 19 Now (1) shows q1 · · · qm ∈ Rr ⊂ Rp1 so one of the qi must be in Rp1. Then p1 divides qi for some i, but this implies p1 and qi are associates since they are irreducible. Because R is an integral domain, we can then cancel p1 and qi from the left and right sides of up1 · · · pn = vq1 · · · qm after adjusting the units u and v. We now continue by downward induction on n + m; this proves the essential uniqueness of factorizations into irreducibles. 20