NOTES FOR NUMBER THEORY COURSE 1. Unique factorization 1.1. All the rings we consider are assumed to have multiplicative unit 1 and almost always they will be commutative. N, Z, Q, R, C will denote the natural numbers, integers, rational numbers, real numbers and complex numbers respectively. A number α 2 C is called an algebraic number, if there exists a polynomial p(x) 2 Q[x] with p(α) = 0. We shall let Q¯ be the set of all algebraic numbers. Fact: \C and Q¯ are algebraically closed". IF R is a ring R[x1; ··· ; xn] will denote the ring of polynomials in n variables with coeffi- cients in R. The letter k will usually denote a field. If R ⊆ S are rings, and α1; ··· ; αk are elements of S, we shall let R[α1; ··· ; αn] be the subring of S generated by R and α1; ··· ; αn. Here are somep examples of rings R of the type we willp be interested in: R = Z, R = k[x], R = Z[i]( i = −1), R = Z[!](! = e2πi=3), R = Z[ 3], or more generally let R = Z[α], where α is an algebraic number. 1.2. First definitions: principal ideals, prime ideals... An element u 2 R is called an unit if there exists v 2 R such that uv = 1. Such a v is necessarily unique (Why?) and is called the inverse of u. The set of units in R will be denoted by U(R). The units in Z are 1 and −1. There are six units in Z[!] (the sixth roots of unity). The units in k[x] are the scalars, i.e. the elements of k. An ideal I in R is called principal if there exists a 2 R such that I = far : r 2 Rg. We say that a is a generator for the principal ideal I and write I = (a) = aR. An element a generates the unit ideal R = (1) if and only if a is an unit. Let p; q 2 R. Say that p divides q if there exists r 2 R such that q = pr. We shall write p j q. Note that p j q () q 2 (p) () (q) ⊆ (p) In general, given two ideal P; Q in R we say P j Q if Q ⊆ P . So for principal ideals (p) j (q) iff p j q. An ideal P ⊆ R is called a prime ideal if ab 2 P implies a 2 P or b 2 P . In other words, if P j (ab) then P j (a) or P j (b). An element p 2 R is a prime if p j ab implies p j a or p j b i.e. (p) is a prime ideal. A ring R is called an integral domain (or simply a domain) if a; b 2 R and ab = 0 im- plies a = 0 or b = 0. Equivalently R is a domain if and only if (0) is a prime ideal. A non-unit x 2 R is called irreducible if x cannot be written as a product of two non- unit elements of R i.e. x = ab implies either a is an unit or b is an unit. Note that in a domain R, if p 2 R is a prime then p is irreducible. 1 Proof: Suppose p = ab. Then p j ab, so p j a or p j b. Without loss suppose p j a. Then a = cp, so p = cpb, implying bc = 1 since we are in a domain, i.e. b is an unit. 1.3. Definition. Euclidean domains are rings where Euclidean algorithm for division works. A domain R is an Euclidean domain if there exists a function λ from the nonzero elements of R to Z≥0 such that if a; b 2 R and b =6 0 there exists c; d 2 R with the property a = cb + d where either d = 0 or λ(d) < λ(b). 1.4. Example. The rings Z, k[x], Z[i], Z[!] are Euclidean domains. Proof. (1) Integer division shows that Z is an Euclidean domain with λ(n) = jnj. More precisely let a; b 2 Z. For simplicity assume they are positive. Let c0 ≥ 1 be the smallest positive integer such that bc0 > a. Let c = c0 − 1 and d = a − bc. Then d < b since otherwise b(c + 1) would be less than a. (2) Long division of polynomials show that k[x] is a integral domain with λ(f) = deg(f) being the degree of the polynomial. For Z[i] and Z[!] see Ireland and Rosen (p: 12-13). 1.5. Definition. A domain R is called a principal ideal domain or a PID if every ideal in R can be generated by one element, i.e. is principal. 1.6. Lemma. Any Euclidean ring is a PID. The rings Z, k[x], Z[i], Z[!] are Euclidean, hence PID. Proof. This is basically the proof that two integers have a greatest common divisor. Let I be an ideal in the Euclidean ring R. Choose an b 2 I such that λ(b) has smallest among all elements of I. For any a 2 I write a = bc + d, where either d = 0 or λ(d) < λ(b). Since a; b 2 I, so is d. Since λ(b) is the smallest among all elements of I, so d must be zero. So I = (b). 1.7. Remark. Call d the g.c.d. of a and b if d divides both a and b and any common divisor of a; b divides d. The theorem shows that any two elements a and b in a PID R has a g.c.d. d, namely, a generator of the ideal (a; b), which is unique upto a unit of R. The elements a and b are relatively prime (i.e. does not have any non-unit common factor), if and only if their g.c.d is 1. 1.8. Lemma. In a PID R, every irreducible element is a prime. (So we shall not distinguish between the concepts of irreducible and prime in a PID.) Proof. let p 2 R be irreducible. Suppose p j ab and p - a. Since p is irreducible and p - a, the only common divisors of a and p are units, so (p; a) = (1). So (pb; ab) = (b). But ab and pb belong to (p), hence (b) ⊆ (p), i.e. p j b. 1.9. Lemma. Let R be a PID. Any increasing sequence of ideals in R stabilizes i.e. has a maximal element. Proof. Let (a1) ⊆ (a2) ⊆ (a3) ⊆ · · · be a increasing sequence of ideals in R. Then I = [(ai) is an ideal, so there exists a 2 R such that I = (a). There is a j ≥ 1 such that a 2 (aj). It follows that (a) = (aj) = (aj+1) = ··· . 1.10. Definition. A domain R is called an unique factorization domain or an UFD if every nonzero element can be written, uniquely upto units as a product of irreducible elements. 2 1.11. Theorem. Every PID is an UFD. Proof. Fix a a 2 R. We want to write a as a product of primes (equivalently irreducibles) and show that such a decomposition is unique upto permutation of the prime factors and upto units. Step 1: Any non-unit a is divisible by an irreducible element. Suppose not. Since a is not irreducible write a = a1b1 where a1; b1 are non-units. Since a1 j a, a1 is not irreducible, so write a1 = a2b2 where a1; b1 are non-units. Continuing this way we get a strictly increasing infinite sequence of ideals (a1) ( (a2) ( (a3) ( ··· which, is not possible by lemma ??. This proves step 1. Step 2: Any a is a product of irreducibles and an unit. Suppose not. By step 1, write a = p1c1 where p1 is an irreducible. Then c1 is not a unit. So write c1 = p2c2 where p2 is irreducible. Continuing this way we get a sequence (c1) ( (c2) ( (c3) ( ··· which, is not possible by lemma ??. So This proves step 2. Step 3: By step 2 we can write a = p1p2 ··· pr where pi are irreducible elements, not necessarily all distinct. Let a = p1p2 ··· pr = q1 ··· qs be two such decompositions. Each qj is a prime and qj j p1 ··· pr, hence qj j pi for some i, hence qj = ujpi for some unit uj. Similarly each pi is equal to some qj upto a unit. If there are more p's than q's then canceling all the q's will yield a product of p 's equal to an unit which is impossible. So r = s and pi and qi are same upto units and upto permutation. 1.12. Remark. The rings Z, k[x], Z[i], Z[!] are all UFD's. This in particular proves that every integer can be written uniquely a a product of positive primes and ±1 and that every polynomial in one variable can be written as a product of irreducible polynomials that are unique upto a scalar. Q 2 e(p) Let R be an UFD and a R. We can write a = p p where the product is over distinct primes of R and almost all e(p) is zero. The numbers e(p) is uniquely determined by a and p. In fact e(p) is the largest integer n such that pn j a. This is because a0 = a=pe(p) is a 0 e(p)+1 product over primes different from p, so p - a , i.e.