Discrete valuation rings. Suppose F is a field. A discrete valuation on F is a function v : F {0} Z such that:

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1 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 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

2 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

3 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 r R. To show the Euclidean property, suppose a, b R and 0 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

4 Example: Suppose p is a rational prime. There is a discrete valuation defined by letting v p : Q Z v p ( a b ) = ord p(a) ord p (b) for all non-zero integers a, b, where ord p (a) is the exponent of the highest power of p which divides a. The valuation ring of v p is just the localization Z (p) = { a b Q : a, b Z, p b} It will be a homework problem on the next homework assignment to check that every discrete valuation of Q has the form v p for some prime p. 4

5 Example: Suppose F = C(t) is the rational function field in one variable. If α C, we can define a discrete valuation by if v α : C(t) Z v α (f(x)) = n f(x) = (x α) n s(x) t(x) with s(α) 0 t(α). The discrete valuation ring of v α is the localization C[x] (x α) = { h(x) g(x) : h(x), g(x) C[x], g(α) 0} and all the discrete valuations of C(t) have the form v α for some α C. 5

6 Example: If p is a prime, the p-adic integers Z p form the valuation ring of the discrete valuation v p : Q p Z defined in the following way. Let v p (r) = n if rp n is a unit of Z p. This is equivalent to saying that r can be represented in the form r = p n i=0 a i p i with a i {0,..., p 1} for all i and a

7 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 = dd for some d R, and d b. 2. If d is any other element of R such that d a and d b then d 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

8 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 d a and d b then Ra + Rd Rd. So Rd = Ra + Rb Rd and d = fd for some f R. This means d 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 Rd. It s clear that 1 a and 1 b. So to show 1 gcd(a, b) it will suffice to show that if d a and d b in R, then d 1, which is the same as saying d R. If d 2, then d must be a constant polynomial, and d {±1, ±2}. But d x implies d {±2}. So d = ±1 is a unit in R, and 1 gcd(2, x). 8

9 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 d is some other element of gcd(a, b), then d d and d d. So d = ud and d = vd for some u, v R. Then d = uvd so d(1 uv) = 0. Since d 0 and R is an integral domain, this forces uv = 1, so u, v R. 9

10 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 b. Construct a sequence of pairs (q i, r i ) of elements of R using the Euclidean property of N: a = qb + r 1 with r 1 = 0 or N(r 1 ) < N(b) b = q 1 r 1 + r 2 with r 2 = 0 or N(r 2 ) < N(r 1 ) r 1 = q 2 r 2 +r 3 with r 3 = 0 or N(r 3 ) < N(r 2 ) r k = q k+1 r k The last non-zero remainder r k+1 in this sequence is in gcd(a, b). Note: The sequence terminates because the norms N(r i ) form a decreasing sequence of non-negative integers. 10

11 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 r 1 = a qb. So by induction, gcd(a, b) = gcd(b, r 1 ) = gcl(r 1, r 2 ) = gcd(r k, r k+1 ) However, by assumption, r k+1 divides r k, so we find r k+1 gcd(r k, r k+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 = r k+1. For example, r k+1 = r k 1 q k r k = r k 1 q k (r k 2 q k 1 r k 1 ) = q k r k 2 + (1 + q k q k 1 )r k 1 = q k r k 2 + (1 + q k q k 1 )(r k 3 q k 2 r k 2 ) = ca + db 11

12 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 ( ) 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

13 Proof of the Theorem: Let P be a non-zero prime ideal. Then P = Ra for some 0 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 m for some m R. If we can show m, m P, this will contradict the fact that P is prime and we ll be done. We have m P since P = Ra Rm = M. If m P then m = za for some z R, and a = m m = mza. But then a (1 mz) = 0 so a 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

14 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

15 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

16 Unique Factorization Domains Def: Suppose R is an integral domain. 1. An element r R is irreducible if r 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 0 since r 0, so 1 ub = 0 and b is a unit. 16

17 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 p 1,, p n of R and a unit u R such that r = u n p i i=1 2. The factorization in (1) is unique up to rearranging the p i, replacing them by associate elements of R and replacing u by a different unit u. Example: Z is a U.F.D.. Each irreducible is a prime element; these have the form ±p. 17

18 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 r 1 be an element of {s, t} which is not a finite product of irreducibles. Then Rr is properly contained in Rr 1. On replacing Rr by Rr 1 and continuing this argument, we produce an infinite chain of ideals Rr Rr 1 Rr 2 such that Rr i Rr i+1 for all i. 18

19 The ideal i Rr i is principal, since R is a P.I.D. Writing i Rr i = Rd for some d, we see that d Rr j for some j, But then Rr j+1 Rd Rr j 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 = up 1 p n = vq 1 q m (1) for some integers n, m 0 and some irreducibles p i and q j and some units u and v. We claim Rp 1 is a maximal ideal. If not, then Rp 1 is properly contained in M = Rm for some maximal ideal M and some m R. But then p 1 = mm and neither m nor m is a unit, so p 1 is not irreducible. We ve thus shown Rp 1 is maximal, and hence a prime ideal. 19

20 Now (1) shows q 1 q m Rr Rp 1 so one of the q i must be in Rp 1. Then p 1 divides q i for some i, but this implies p 1 and q i are associates since they are irreducible. Because R is an integral domain, we can then cancel p 1 and q i from the left and right sides of up 1 p n = vq 1 q m 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

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