# ALGEBRA HANDOUT 2.3: FACTORIZATION IN INTEGRAL DOMAINS. In this handout we wish to describe some aspects of the theory of factorization.

Size: px
Start display at page:

Download "ALGEBRA HANDOUT 2.3: FACTORIZATION IN INTEGRAL DOMAINS. In this handout we wish to describe some aspects of the theory of factorization."

Transcription

1 ALGEBRA HANDOUT 2.3: FACTORIZATION IN INTEGRAL DOMAINS PETE L. CLARK In this handout we wish to describe some aspects of the theory of factorization. The first goal is to state what it means for an arbitrary integral domain R to have the property of unique factorization. To see what subtleties are necessary, let us consider first how to express unique factorization in the ring of all integers Z, rather than just restricting to the positive integers. 1. Irreducible elements Any nonzero integer has both positive and negative divisors. Our definition of factorization must be sufficiently enlightened so as to not count factorizations like 3 = ( 1) ( 3) against 3 being prime. Notice indeed that any integer is divisible not just by 1 but also by 1. Moreover, in any integral domain, every element x is divisible by every unit u of the ring. Indeed, the definition of a unit is an element which has a multiplicative inverse say, u, such that uu = 1 and then x = u (u x) for all x. Thus the first key observation is to regard a factorization x = yz as trivial if one of y or z is a unit. We say a nonzero, nonunit element x of R is irreducible if has only trivial factorizations: that is, if x = yz, then one of y or z is a unit. (Note that it cannot be the case that both y and z are units, for then x would itself be a unit.) Example 1.1: the irreducible elements of Z are those of the form ±p, where p is a prime number. 1 Example 1.2: In Q, or in any field, there are no irreducible elements, because every nonzero element is a unit. Now let a be any nonzero nonunit in an integral domain R. of a is an expression a = x 1 x n, A factorization where each x i is irreducible. In other words, a factorization is an expression of a nonzero nonunit as a product of irreducible elements. 1 You may well be wondering why we don t simply call irreducible elements primes. Because the answer is rather subtle, we delay it for a while. 1

2 2 PETE L. CLARK 2. Factorization into irreducibles: existence Let us say that a domain R is a factorization domain (for short, FD) if every nonzero nonunit element has a factorization into irreducibles. 2 Example 2.1: A field is trivially a FD: it has no nonzero nonunits! Example 2.2: Part a) of the fundamental theorem of arithmetic asserts that Z is a FD. The proof was an easy minimal counterexample argument. We will give two sufficient conditions for a domain R to be a factorization domain. The first one is more number-theoretic in spirit and can be viewed as a generalization of Example 4. The second uses a concept which is seldom discussed in undergraduate algebra, but is of inestimable importance in more advanced ring theory, namely the notion of a Noetherian ring. As for the first condition, the idea is extremely simple: factorization ought to be a process of decomposing more complex objects into simpler ones. If to every nonzero element a of R we can assign a positive integer complexity C(a) such that in any nontrivial factorization a = bc i.e., with b and c nonunits we have 1 C(b), C(c) < C(a) then factorizations lower the complexity so that eventually the process must terminate. To implement this strategy we need a complexity function C : R \ {0} Z + which satisfies the above property. It turns out that there often natural choices of such functions which satisfy further properties, which we axiomatize as follows. A multiplicative norm on a domain R is a function N : R N such that: (MN0) N(x) = 0 x = 0. (MN1) For all x, y R \ {0}, N(xy) = N(x)N(y). (MN2) For all x R \ {0}, N(x) = 1 x R (i.e., x is a unit). Example 2.3: The function z z is a multiplicative norm on the integers Example 2.4: Fix a squarefree integer d not equal to 0 or 1. Consider the quadratic ring Z[ d], the subring of C consisting of complex numbers of the form a + b d, with a, b Z. In the main text we study some aspects of the arithmetic of quadratic rings in depth, 3 and our first key observation is that we can define a multiplicative norm on Z[ d] by N(a + b d) = a 2 db 2. (In more advanced number theory, one learns that a similar multiplicative norm can be defined for the ring O K of algebraic integers in any number field K.) Proposition 1. Let R be an integral domain. Suppose that there exists a multiplicative norm N on R. Then R is a factorization domain. 2 The term atomic domain is more commonly used, but is still not common enough to be familiar to a general mathematical audience, so we will go with our more transparent terminology. 3 Moreover one can teach an entire course on this subject: see pete/primesoftheform.html.

3 ALGEBRA HANDOUT 2.3: FACTORIZATION IN INTEGRAL DOMAINS 3 Proof: Suppose for a contradiction that the set of nonzero nonunits in R which do not admit factorizations into irreducibles is nonempty; then, among all such elements there must exist (at least) one x, with N(x) minimal. Such an x certainly is not irreducible, so it can be factored at x = yz, with both y, z nonunits. Then N(x) = N(y)N(z) Z + and N(y), N(z) > 1, so that we must have N(y), N(z) < N(x). Therefore y and z, having strictly smaller norm than x, each have irreducible factorizations, say y = y 1 y r and z = z 1 z s and therefore x = y 1 y r z 1 z s is an irreducible factorization of x. Now for the second condition which ensures factorization into irreducibles: Proposition 2. For a commutative ring R, the following conditions are equivalent: (i) Every nonempty set S of ideals of R has a maximal element, i.e., an element I S such that I is not properly contained in any other ideal J of S. (ii) (ACC) In any infinite sequence of ideals I 1 I 2... I n... we have equality from some point onward: there exists N Z + such that for all k 0, I N+k = I n. (iii) Every ideal I of R is finitely generated: there exist finitely many elements x 1,..., x n in R such that I = x 1,..., x n = {r 1 x r n x n r i R}. A ring satisfying these equivalent properties is caled Noetherian. Proof: (i) (ii): For any nonempty family F of subsets of a given set R, the condition that that any infinite sequence I 1 I 2... of elements of F is equivalent to the condition that every nonempty subset of F has a maximal element: if (i) does not hold, then there exists a sequence I 1 I 2 I 3 neq, and then {I n } n=1 has no maximal element. Conversely, if (ii) does not hold, then there exists I 1 F; since I 1 is not maximal, so there exists I 2 F such that I 2 I 1, since I 2 is not maximal, there exists I 3 F such that I 3 I 2 : continuing in this way, we build an infinite strictly ascending chain. (ii) = (iii): If there exists an ideal I which is not finitely generated, then for any x 1 I, I 1 := x 1 I. Since I 1 is finitely generated and I is not, there exists x 2 I \ I 1. Put I 2 = x 1, x 2, so I 2 I. Again, because I is not finitely generated, there exists x 3 I \I 2. In this way we construct an infinite strictly ascending chain I 1 I 2 I 3, contradicting (ii). (iii) = (ii): Let I 1 I 2... be an infinite sequence of ideals. Then the union I := i=1 I i is again an ideal. By assumption, I is finitely generated, so there exist x 1,..., x n R with I = x 1,..., x n. But since I is the union of the I i s, for each 1 i n, there exists k i Z + such that x i I ki. Put k = max(k 1,..., k n ); then x 1,..., x n are all in I k, so I = I k, which forces I k = I k+1 =... = I. Proposition 3. A principal ideal domain is a Noetherian domain. Proof: This is an immediate consequence of the definitions: a PID is a domain in which each ideal can be generated by a single element, whereas a Noetherian ring is one in which each ideal can be generated by finitely many elements. It turns out that we do not need the full strength of the Noetherian condition.

4 4 PETE L. CLARK Rather, we will need only that there are no infinite strictly acending chains I 1 I 2... I n... of principal ideals I i = (x i ). Let us call this condition on a ring ACCP. 4 Again, we stress that ACCP is a weaker condition than Noetherianity. Remark 2.1: ACCP holds in a ring iff every nonempty set of principal ideals has a maximal element. The proof is the same as that of (i) (ii) in Proposition 2. Proposition 4. An integral domain satisfying ACCP is a factorization domain. Proof: Let S be the set of all nonzero nonunit elements of R which cannot be factored into irreducibles. Assume, for a contradiction, that S is nonempty. Then the corresponding set S = {(x) x S } of principal ideals generated by elements of S is also nonempty. By ACCP and Remark 1, there exists a maximal element (x) of S. Now just follow your nose: by definition of x, it is not irreducible, so can be written as x = yz with y and z nonunits. This means that the principal ideals (y) and (z) each strictly contain the principal ideal (x), so by the assumed maximality of (x), both y and z can be factored into irreducibles: y = y 1 y r, z = z 1 z s, so (as usual!) we get x = y 1 y r z 1 z s so x has an irreducible factorization after all, contradiction. Remark 2.2: In a widely read 1968 paper, P.M. Cohn claimed, without proof, that any factorization domain satisfies ACCP. It was therefore very surprising to the algebra community when in 1974 Anne Grams constructed a factorization domain which does not satisfy ACCP. Example 7: The ring Z of all algebraic integers is not a factorization domain. In fact, Z is as far from a factorization as possible, in the following sense: it has many nonzero nonunit elements, but no irreducible elements! We briefly sketch an argument for this: first, there exist nonzero nonunit elements of the ring, for instance the element 2. Its multiplicative inverse in the fraction field Q (of all algebraic numbers) is 1 2, and 1 2 is not an algebraic integer. Second, we claim that there are no irreducible elements in Z. Namely, if x is any nonzero nonunit algebraic integer, then one can check that x is also a nonzero nonunit algebraic integer and x = x x. 3. Factorization into irreducibles: uniqueness In order to give a definition of a unique factorization domain, we must specify when two different factorizations of the same nonzero nonunit x are to be regarded as equivalent. In the case of factorizations of positive integers into prime numbers, we only had to worry about the ordering of the irreducible factors. Of course we still wish to regard two factorizations into irreducibles differing only in the order of the factors as equivalent, but there is more to say. For instance, in Z we have 18 = = (2) ( 3) ( 3), and several other choices for the sign besides. The correct generalization of this to an arbitrary domain comes from the following observation: if x is an irreducible element of R and u is a unit of R, then ux is also an irreducible element of R. 4 This stands for Ascending Chain Condition on Principalideals.

5 ALGEBRA HANDOUT 2.3: FACTORIZATION IN INTEGRAL DOMAINS 5 Similarly, by multiplying by units we can get many different equivalent-looking factorizations, e.g. a = x 1 x r = (ux 1 ) (ux r 1 ) (u 1 r x r ). Thus we need a relation between elements which regards two elements as equivalent iff they differ multiplicatively by a unit. In fact this is itself a well-defined relation: its properties are recorded below. Proposition 5. Let R be a domain, and let x, y R. The following are equivalent: (i) x y and y x. (ii) There exists a unit u R such that y = ux. (iii) We have an equality of principal ideals (x) = (y). If x and y satisfy any (hence all) of the conditions above, we say that x and y are associates and write x y. The proof is entirely a matter of unwinding the definitions, so we leave it to the interested reader. Finally, we can give the key definition: An integral domain R is a unique factorization domain if: (UFD1) Every nonzero nonunit a admits a factorization into irreducibles, and (UFD2) If a = x 1 x r = y 1 y s are two irreducible factorizations of a, then r = s, and there exists a permutation σ of {1,..., r} such that for 1 i r we have x i y σ(i). In other words, after reordering the elements we can pair off each irreducible element in the first factorization with an associate irreducible element in the second factorization. Note that the second condition (UFD2) is a logically complicated one, and hence can be difficult to check directly (although we managed to do so in the HLZ proof of FTA). As with the integers, Euclid s Lemma is a useful intermediate point. Let us say an integral domain R is an EL-domain if for all irreducible elements x of R, if x ab, then x a or x b. Then we have: Theorem 6. Let R be a factorization domain. Then R is a unique factorization domain iff it is an EL-domain. Proof: The proof is exactly as in the special case of R = Z, which we discussed in [Handout: Foundations and the Fundamental Theorem, 2.2], as we invite the reader to check. It turns out that a UFD need not be Noetherian: as we shall see later, one example is a polynomial ring in infinitely many variables over a field. Neither does a factorization domain need to satisfy ACCP, as mentioned above. However: Theorem 7. For an integral domain R, the following are equivalent: (i) R is a unique factorization domain. (ii) R satisfies ACCP and is an EL-domain.

6 6 PETE L. CLARK Proof: (i) = (ii): suppose R is a UFD. By Theorem 6, R is an EL-domain. Moreover, suppose R does not satisfy ACCP: (x 1 ) (x 2 ).... Then x 2 is a nonzero nonunit. Since (x 3 ) strictly contains x 2, there exists a nonzero nonunit y 1 such that x 2 = x 3 y 1. Since x 3 and y 1 are both nonzero nonunits, they have unique factorizations into irreducibles, which means that the unique factorization of x 2 into irreducibles has at least two irreducible factors. Similarly, there exists a nonzero nonunit y 2 such that x 3 = x 4 y 2, so x 2 = x 4 y 2 y 1, so that we now know that the unique factorization of x 2 into irreducibles has at least 3 irreducible factors. Proceeding in this way we can show that the unique factorization of x n into irreducibles has at least n irreducible factors for any n Z +, which is absurd. (ii) = (i): an ACCP domain is a factorization domain, so apply Theorem Greatest common divisors We recall the definition of a greatest common divisor of two elements a and b in an arbitrary integral domain R. It is an element d of R such that for all e in R with e a and e b, we have e d. Of course it is not clear that such an element must exist. Indeed, an integral domain R is called a GCD-domain if any two elements a, b admit at least one greatest common divisor. Remark 4.1: Let R be any integral domain. a) If a = 0 and b = 0, then 0 is a greatest common divisor of a and b. b) If a is arbitrary and b = 0, then a is a greatest common divisor of a and b. c) If a is a unit and b is arbitrary, then 1 is a greatest common divisor of a and b. The uniqueness of greatest common divisors is easier to sort out: Lemma 8. Let R be an integral domain, a, b R, and suppose d is a greatest common divisor of a and b. Then an element x of R is a greatest common divisor of a and b iff x d, i.e., iff x = ud for some unit u R. Proof: Let d and d be greatest common divisors of a and b. Then d a and d b, so d d, and similarly d d. It follows that d d. Conversely, since associate elements have exactly the same divisibility relations, it is clear that any associate of a greatest common divisor is again a greatest common divisor. Example: For two nonzero integers a and b, there are two greatest common divisors: d and d. Of course it is conventional to mean by gcd(a, b) the unique positive greatest common divisor. Proposition 9. A unique factorization domain is a GCD-domain. Proof: Again, this is an immediate generalization of the usual arguments for R = Z. By Remark 2, we know that GCD necessarily exists except possibly when both a and b are nonzero nonunits. Then, let x 1,..., x r be the set of pairwise nonassociate irreducibles such that any irreducible divisor of either a or b is associate to some x i ; we may then write a = x a1 1 xar r, b = x b1 1 xbr r, with a i, b i N. Then d = x min(a1,b1) 1 x min(ar,br) r is a greatest common divisor of a and b.

7 ALGEBRA HANDOUT 2.3: FACTORIZATION IN INTEGRAL DOMAINS 7 Proposition 10. (GCD identities) Let R be a GCD-domain, and let a, b, c R, with d = gcd(a, b). Then: a) gcd(ab, ac) = a gcd(b, c). b) gcd(ad 1, bd 1 ) = 1. c) If gcd(a, b) = gcd(a, c) = 1, then gcd(a, bc) = 1. Proof: a) Let x = gcd(ab, ac). Then a ab and a ac so a x: say ay = x. Since x ab and x ac, y b and y c, so y gcd(b, c). If z b and z c, then az ab and az ac, so az x = ay and z y. Therefore gcd(b, c) = y = 1 a gcd(ab, ac). Part b) follows immediately. As for part c): suppose gcd(a, b) = gcd(a, c) = 1, and let t divide a and bc. Then t divides ab and bc so t gcd(ab, bc) = b gcd(a, c) = b. So t divides gcd(a, b) = 1. Proposition 11. A GCD-domain is an EL-domain. Proof: Suppose x is irreducible and x yz. Assume, for a contradiction, that x y and x z. Then gcd(x, y) = gcd(x, z) = 1, and by Proposition 10c), gcd(x, yz) = 1, which contradicts x yz. Corollary 12. A factorization domain is a UFD iff it is a GCD-domain. Proof: Let R be a factorization domain. Assume first that R is a UFD. Then R is a GCD-domain by Proposition 9. Conversely, assume that R is a GCD-domain. Then it is an EL-domain by Proposition 11, and by Theorem 7 a factorization domain which is an EL-domain is a UFD. 5. More on Principal Ideal Domains Proposition 13. Let a and b be elements in a PID R. Then d = gcd(a, b) exists and moreover can be expressed as a linear combination of a and b: there exist x, y R such that ax + by = d. Proof: The ideal x, y = {xa + yb x, y R} is by assumption principal, i.e., equal to (d) for some d R. As in the case R = Z, we see easily that d is a greatest common divisor of a and b: it is a common divisor since x, y x, y = (d), and if e a, e b, then e ax + by. But ax + by = d, so e d. Corollary 14. Any principal ideal domain is a unique factorization domain. Proof: We need only put together previously proved results. We know: A PID is a Noetherian domain (Proposition 3) A Noetherian domain is a Factorization Domain (Proposition 4). A PID is a GCD-domain (Proposition 13). A GCD-domain which is also a factorization domain is a UFD (Corollary 12). Done! We wish to give a criterion for an integral domain to be a PID which is due to H. Hasse. In fact, Hasse s criterion is in terms of a multiplicative norm N on R which satisfies one additional property. First, consider any multiplicative norm N : R N on an integral domain R. We assert that because of the multiplicativity, there is a unique extension of N to a function from the fraction field, K, of R to the non-negative rational numbers

8 8 PETE L. CLARK such that N(xy) = N(x)N(y) for all x, y K. Indeed, since axiom (MN2) implies N(1) = 1, we must have N( 1 y ) = 1 N(y) and thus ( ) x N = N(x) y N(y). Since a given element of K has many different representations as a quotient of elements of R, we must check that the definition of N is independent of this representation, but this is easy: if x1 y 1 = x2 y 2, then x 1 y 2 = x 2 y 1, so N(x 1 )N(y 2 ) = N(x 1 y 2 ) = N(x 2 y 1 ) = N(x 2 )N(y 1 ), and, since y 1, y 2 0 implies N(y 1 ), N(y 2 ) 0, we may divide in Q to get N(x 1 ) N(y 1 ) = N(x 2) N(y 2 ). For example, the usual absolute value z z on Z extends multiplicatively to the usual absolute value on Q. From now on, we will assume without comment that a multiplicative norm has its domain extended to the fraction field F of R as above. Definition: A multiplicative norm N : F Q on the fraction field of an integral domain R is a Hasse norm if it satisifes the following property: (HN) For all x F \ R, there exist a, b R such that 0 < N(ax b) < 1. Example: The usual absolute value on Q is a Hasse norm. Indeed, for any rational number x which is not an integer, we can take a = 1 and take b to be x, the greatest integer less than or equal to x. Then 0 < x b < 1. Theorem 15. (Hasse) For an integral domain R, TFAE: (i) R admits a Hasse norm N. (ii) R is a PID. Proof: (i) = (ii): Let I be a nonzero ideal of R, so I contains elements of positive norm. Let d I be an element whose norm is positive and is minimal among all elements of I. We wish to show that I = (d). So let i be any element of I and put x := i d. If d i then x R, so assume for a contradiction that x F \ R. Then by assumption there exist a, b R such that Multiplying through by d we get 0 < N ( ) ai d b < 1. 0 < N(ai bd) < N(d). So ai bd I has norm positive and smaller than N(d), contradiction! (ii) = (i): Suppose R is a PID, and define the norm of x R as follows: N(0) = 0, for any unit u, N(u) = 1; otherwise, if x = x 1 x r is a product of r irreducible elements (counting multiplicities), then put N(x) = 2 r. It is immediate from the uniqueness of factorization that N is a multiplicative norm, so it remains to be seen that it is a Hasse norm. But any element x F \ R can be written as x = p q, where p, q R \ {0}, gcd(p, q) = 1 and q is a nonunit, so N(q) > 1. Now,

9 ALGEBRA HANDOUT 2.3: FACTORIZATION IN INTEGRAL DOMAINS 9 applying Proposition 12, we can find elements a, b in R such that ap + b q = 1. Taking b = b, we have ap bq = 1. Dividing through by q we get ax b = 1 q, so 0 < N(ax b) = N( 1 q ) = 1 N(q) < 1. Here is a sample application of this circle of ideas. Proposition 16. Let F be any field. Then the polynomial ring F [t] is a PID. Proof: Every nonzero polynomial P (t) = a n t n a 0 has a degree deg(p ), which is the largest n N such that a n 0. By convention, we decree that the 0 polynomial has degree. It is easy to check that deg(p Q) = deg(p ) + deg(q). Thus the degree is very much like a norm, only instead of being multiplicative, it is multiplicative-to-additive. That can be remedied, however: put N(P ) = 2 deg P, with the convention that N(0) = 2 = 0. We claim that N is a Hasse norm on F [t]. Let s check: N(P ) = 0 2 deg P = 0 P = 0. N(P ) = 1 iff 2 deg P = 1 deg(p ) = 0 iff P is a nonzero constant polynomial. These are indeed precisely the units of F [t]. N(P Q) = 2 deg(p Q) = 2 deg P +deg Q = 2 deg P 2 deg Q = N(P )N(Q). be a rational function with B(t) not dividing A(t). Then by the division algorithm for polynomials, we may write A(t) = Q(t)B(t) + R(t), where deg R(t) < deg B(t), so x(t) = Q(t) + R(t) B(t), and thus ( ) R(t) Let x(t) = A(t) B(t) 0 < N(x(t) Q(t)) = N < 1. B(t) A multiplicative norm N on a domain R is Euclidean if for any a, b R with b 0, there exist q, r R such that a = qb + r and N(r) < N(b). Proposition 17. Any Euclidean norm is a Hasse norm. Proof: Let x = a b F \ R. Since the norm is Euclidean, there exist q, r R with N(r) < N(b), and then x q = a b ( a b r b ) = r b (, so r ) 0 < N(x q) = N < 1. b Let us repackage this result in a form which will be maximally convenient in numbertheoretic applications: Proposition 18. Let R be an integral domain with fraction field K. Suppose that R has a multiplicative norm N with the property that for all x K \ R there exists yinr with N(x y) < 1. Then R is a PID. Remark 5.1: In the literature, one generally calls a Euclidean domain an integral domain for which there exists a Euclidean norm (although the precise definition of a Euclidean norm can be varied somewhat without changing the class of rings one gets: for instance, the multiplicativity is not essential). The advantage of a Euclidean norm is that, as in Proposition 18, it is especially straightforward to check whether a given norm is Euclidean. The advantage of a general Hasse norm

10 10 PETE L. CLARK is that, as we saw, every PID is guaranteed to have one, whereas in general a PID need not admit any Euclidean norm, and whether it does turns out to be a much more subtle question than whether it is a PID. Thus, in my opinion, the idea of a Euclidean domain in abstract algebra is something of a red herring. 6. Prime elements Recall the notion of a prime ideal p in a ring R: this is a proper ideal such that x, y R, xy p implies x p or y p. Let us define a nonzero element x in a domain R to be a prime element if the principal ideal (x) is a prime ideal. Unpacking this, we see that an element x is prime iff x ab implies x a or x b. Lemma 19. a) In any domain R, a prime element is irreducible. b) A domain R is an EL-domain exactly when all irreducible elements are prime. Proof: a) If x = ab with a and b nonunits, then certainly x a and x b. b) This is, of course, the definition of an EL-domain. In particular, since UFD = GCD-domain = EL-domain, in any UFD there is no distinction to be made between irreducible elements and prime elements. Conversely, a factorization domain will fail to be a UFD iff there exist irreducible elements which are not prime. 7. UFDs and polynomial rings Theorem 20. Let R be a domain and R[t] a univariate polynomial ring over R. a) If R satisfies ACCP, so does R[t]. b) If R is a UFD, so is R[t]. Remark 7.1: The most common proof of Theorem 20b) uses Gauss Lemma on primitive polynomials. A second proof due to Nagata uses the technique of localization. The proof we present below is unusually direct: it is closely modelled on the Hasse-Lindemann-Zermelo proof of unique factorization in Z. Proof: a) In an infinite ascending chain (P i ) of principal ideals of R[t], deg(p i ) is a descending chain of non-negative integers, so eventually stabilizes. Therefore for sufficiently large n, we have P n = a n P n+1, where a n R and (a n+1 ) (a n ). Since R satisfies (ACCP) we have (a n ) = (a n+1 ) for sufficiently large n, whence (P n ) = (P n+1 ) for sufficiently large n: R[t] satisfies (ACCP). b) Suppose R is an EL-domain but R[t] is not. Among the set of all elements in R[t] admitting inequivalent irreducible factorizations, let p be one of minimal degree. We may assume p = f 1 f r = g 1 g s, where for all i, j, (f i ) (g j ) and m = deg f 1 deg f 2... deg f r, n = deg g 1 deg g 2... deg g s, with n m > 0. Suppose the leading coefficient of f 1 (resp. g 1 ) is a (resp. b). Put q = ap bf 1 x n m g 2 g s = f 1 (af 2 f r bx n m g 2 g s ) = (ag 1 bf 1 x n m )g 2 g s.

11 ALGEBRA HANDOUT 2.3: FACTORIZATION IN INTEGRAL DOMAINS 11 Thus q = 0 implies ag 1 = bf 1 x n m. If, however, q 0, then deg(ag 1 bf 1 x n m ) < deg g 1, hence deg q < deg p and q has a unique factorization into irreducibles, certainly including g 2,, g s and f 1. But then f 1 must be a factor of ag 1 bf 1 x n m and thus also of ag 1. Either way ag 1 = f 1 h for some h R[t]. Since a is constant and f 1 is irreducible, this implies h = ah 2, so ag 1 = f 1 ah 2, or g 1 = f 1 h 2, contradiction. By induction, we deduce: Corollary 21. Let R be a UFD and n Z +. Then the polynomial ring R[t 1,..., t n ] is a UFD. Theorem 22. Let R be a UFD and let S = R[t 1, t 2,...] be a polynomial ring over R in infinitely many indeterminates. Then R is a non-noetherian UFD. Proof: We show S is non-noetherian by exhibiting an infinite chain of ideals: t 1 t 1, t 2... t 1,..., t n... Suppose that for any n, t n+1 were an element of t 1,..., t n. In other words, there exist polynomials P 1,..., P n such that t n+1 = P 1 t P n t n. Setting t 1 =... = t 0 = 0, t n+1 = 1 gives 1 = 0 in R, a contradiction. By Theorem 6, to show that R is a UFD it suffices to show that it satisfies the ascending chain condition on principal ideals and Euclid s Lemma. The first is almost immediate: any nonzero element is a polynomial in a finite number of variables, say P (t 1,..., t n ). Any divisor Q of P is again a polynomial in only the variables t 1,..., t n, so that an ascending chain (P ) (P 2 )... (P n )... can be viewed as an ascending chain in the UFD R[t 1,..., t n ], so it stabilizes since UFDs satisfy ACCP. Finally, let P be an irreducible element in S. The EL-condition is equivalent to the principal ideal (P ) being a prime ideal, which is equivalent to the quoteint S/(P ) being an integral domain. But as above P is a polynomial in only finitely many variables, say P (t 1,..., t n ) and if P (t 1,..., t n ) = X(t 1,..., t n )Y (t 1,..., t n ) with neither X nor Y a unit in R[t 1,..., t n ] then the factorization remains valid in the larger domain S, and since S = R[t 1,..., t n ] = R, it remains a nontrivial factorization (i.e., neither X nor Y is a unit in S). So P (t 1,..., t n ) is irreducible in R n := R[t 1,..., t n ]; since R n is a UFD, the principal ideal P R n is prime. But S/P S = R n [t n+1, t n+2,...]/p R n [t n+1, t n+2,...] = (R n /P R n )[t n+1, t n+2,...]. Since the last ring in this equation is visibly an integral domain, so is S/P S, so P S is a prime ideal.

### 32 Divisibility Theory in Integral Domains

3 Divisibility Theory in Integral Domains As we have already mentioned, the ring of integers is the prototype of integral domains. There is a divisibility relation on * : an integer b is said to be divisible

### QUADRATIC RINGS PETE L. CLARK

QUADRATIC RINGS PETE L. CLARK 1. Quadratic fields and quadratic rings Let D be a squarefree integer not equal to 0 or 1. Then D is irrational, and Q[ D], the subring of C obtained by adjoining D to Q,

### Chapter 14: Divisibility and factorization

Chapter 14: Divisibility and factorization Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Summer I 2014 M. Macauley (Clemson) Chapter

### Example: This theorem is the easiest way to test an ideal (or an element) is prime. Z[x] (x)

Math 4010/5530 Factorization Theory January 2016 Let R be an integral domain. Recall that s, t R are called associates if they differ by a unit (i.e. there is some c R such that s = ct). Let R be a commutative

### FACTORIZATION IN INTEGRAL DOMAINS

FACTORIZATION IN INTEGRAL DOMAINS PETE L. CLARK Contents Classical Roots The Fundamental theorem of Arithmetic 2 Basic Terminology 5 1. Norm functions 5 1.1. Weak multiplicative norms and multiplicative

### M3P14 LECTURE NOTES 8: QUADRATIC RINGS AND EUCLIDEAN DOMAINS

M3P14 LECTURE NOTES 8: QUADRATIC RINGS AND EUCLIDEAN DOMAINS 1. The Gaussian Integers Definition 1.1. The ring of Gaussian integers, denoted Z[i], is the subring of C consisting of all complex numbers

### 1. Factorization Divisibility in Z.

8 J. E. CREMONA 1.1. Divisibility in Z. 1. Factorization Definition 1.1.1. Let a, b Z. Then we say that a divides b and write a b if b = ac for some c Z: a b c Z : b = ac. Alternatively, we may say that

### Lecture 7.5: Euclidean domains and algebraic integers

Lecture 7.5: Euclidean domains and algebraic integers Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley

### Part IX. Factorization

IX.45. Unique Factorization Domains 1 Part IX. Factorization Section IX.45. Unique Factorization Domains Note. In this section we return to integral domains and concern ourselves with factoring (with respect

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

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:

### 2. THE EUCLIDEAN ALGORITHM More ring essentials

2. THE EUCLIDEAN ALGORITHM More ring essentials In this chapter: rings R commutative with 1. An element b R divides a R, or b is a divisor of a, or a is divisible by b, or a is a multiple of b, if there

### Contents. 4 Arithmetic and Unique Factorization in Integral Domains. 4.1 Euclidean Domains and Principal Ideal Domains

Ring Theory (part 4): Arithmetic and Unique Factorization in Integral Domains (by Evan Dummit, 018, v. 1.00) Contents 4 Arithmetic and Unique Factorization in Integral Domains 1 4.1 Euclidean Domains and

### Factorization in Domains

Last Time Chain Conditions and s Uniqueness of s The Division Algorithm Revisited in Domains Ryan C. Trinity University Modern Algebra II Last Time Chain Conditions and s Uniqueness of s The Division Algorithm

### However another possibility is

19. Special Domains Let R be an integral domain. Recall that an element a 0, of R is said to be prime, if the corresponding principal ideal p is prime and a is not a unit. Definition 19.1. Let a and b

### Part IX ( 45-47) Factorization

Part IX ( 45-47) Factorization Satya Mandal University of Kansas, Lawrence KS 66045 USA January 22 45 Unique Factorization Domain (UFD) Abstract We prove evey PID is an UFD. We also prove if D is a UFD,

### 8. Prime Factorization and Primary Decompositions

70 Andreas Gathmann 8. Prime Factorization and Primary Decompositions 13 When it comes to actual computations, Euclidean domains (or more generally principal ideal domains) are probably the nicest rings

### 4.4 Noetherian Rings

4.4 Noetherian Rings Recall that a ring A is Noetherian if it satisfies the following three equivalent conditions: (1) Every nonempty set of ideals of A has a maximal element (the maximal condition); (2)

### MATH 326: RINGS AND MODULES STEFAN GILLE

MATH 326: RINGS AND MODULES STEFAN GILLE 1 2 STEFAN GILLE 1. Rings We recall first the definition of a group. 1.1. Definition. Let G be a non empty set. The set G is called a group if there is a map called

### 1. multiplication is commutative and associative;

Chapter 4 The Arithmetic of Z In this chapter, we start by introducing the concept of congruences; these are used in our proof (going back to Gauss 1 ) that every integer has a unique prime factorization.

### Math 547, Exam 2 Information.

Math 547, Exam 2 Information. 3/19/10, LC 303B, 10:10-11:00. Exam 2 will be based on: Homework and textbook sections covered by lectures 2/3-3/5. (see http://www.math.sc.edu/ boylan/sccourses/547sp10/547.html)

### Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35

Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35 1. Let R 0 be a commutative ring with 1 and let S R be the subset of nonzero elements which are not zero divisors. (a)

### MODULES OVER A PID. induces an isomorphism

MODULES OVER A PID A module over a PID is an abelian group that also carries multiplication by a particularly convenient ring of scalars. Indeed, when the scalar ring is the integers, the module is precisely

### 6]. (10) (i) Determine the units in the rings Z[i] and Z[ 10]. If n is a squarefree

Quadratic extensions Definition: Let R, S be commutative rings, R S. An extension of rings R S is said to be quadratic there is α S \R and monic polynomial f(x) R[x] of degree such that f(α) = 0 and S

### 12 16 = (12)(16) = 0.

Homework Assignment 5 Homework 5. Due day: 11/6/06 (5A) Do each of the following. (i) Compute the multiplication: (12)(16) in Z 24. (ii) Determine the set of units in Z 5. Can we extend our conclusion

### Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman

Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 17, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Factorization 0.1.1 Factorization of Integers and Polynomials Now we are going

### Rings. Chapter Homomorphisms and ideals

Chapter 2 Rings This chapter should be at least in part a review of stuff you ve seen before. Roughly it is covered in Rotman chapter 3 and sections 6.1 and 6.2. You should *know* well all the material

### Lecture 7.4: Divisibility and factorization

Lecture 7.4: Divisibility and factorization Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley (Clemson)

### CHAPTER I. Rings. Definition A ring R is a set with two binary operations, addition + and

CHAPTER I Rings 1.1 Definitions and Examples Definition 1.1.1. A ring R is a set with two binary operations, addition + and multiplication satisfying the following conditions for all a, b, c in R : (i)

### (1) A frac = b : a, b A, b 0. We can define addition and multiplication of fractions as we normally would. a b + c d

The Algebraic Method 0.1. Integral Domains. Emmy Noether and others quickly realized that the classical algebraic number theory of Dedekind could be abstracted completely. In particular, rings of integers

### Factorization in Polynomial Rings

Factorization in Polynomial Rings Throughout these notes, F denotes a field. 1 Long division with remainder We begin with some basic definitions. Definition 1.1. Let f, g F [x]. We say that f divides g,

### (Rgs) Rings Math 683L (Summer 2003)

(Rgs) Rings Math 683L (Summer 2003) We will first summarise the general results that we will need from the theory of rings. A unital ring, R, is a set equipped with two binary operations + and such that

### NUMBER SYSTEMS. Number theory is the study of the integers. We denote the set of integers by Z:

NUMBER SYSTEMS Number theory is the study of the integers. We denote the set of integers by Z: Z = {..., 3, 2, 1, 0, 1, 2, 3,... }. The integers have two operations defined on them, addition and multiplication,

### Modern Algebra Lecture Notes: Rings and fields set 6, revision 2

Modern Algebra Lecture Notes: Rings and fields set 6, revision 2 Kevin Broughan University of Waikato, Hamilton, New Zealand May 20, 2010 Solving quadratic equations: traditional The procedure Work in

### THUE S LEMMA AND IDONEAL QUADRATIC FORMS

THUE S LEMMA AND IDONEAL QUADRATIC FORMS PETE L. CLARK Introduction Recently I learned about Thue s Lemma, an elementary result on congruences due to the Norwegian mathematician Axel Thue (1863-1922).

### IRREDUCIBILITY TESTS IN Q[T ]

IRREDUCIBILITY TESTS IN Q[T ] KEITH CONRAD 1. Introduction For a general field F there is no simple way to determine if an arbitrary polynomial in F [T ] is irreducible. Here we will focus on the case

### NOTES ON INTEGERS. 1. Integers

NOTES ON INTEGERS STEVEN DALE CUTKOSKY The integers 1. Integers Z = {, 3, 2, 1, 0, 1, 2, 3, } have addition and multiplication which satisfy familar rules. They are ordered (m < n if m is less than n).

### Finite Fields: An introduction through exercises Jonathan Buss Spring 2014

Finite Fields: An introduction through exercises Jonathan Buss Spring 2014 A typical course in abstract algebra starts with groups, and then moves on to rings, vector spaces, fields, etc. This sequence

### Chapter 5. Number Theory. 5.1 Base b representations

Chapter 5 Number Theory The material in this chapter offers a small glimpse of why a lot of facts that you ve probably nown and used for a long time are true. It also offers some exposure to generalization,

### D-MATH Algebra I HS18 Prof. Rahul Pandharipande. Solution 6. Unique Factorization Domains

D-MATH Algebra I HS18 Prof. Rahul Pandharipande Solution 6 Unique Factorization Domains 1. Let R be a UFD. Let that a, b R be coprime elements (that is, gcd(a, b) R ) and c R. Suppose that a c and b c.

### Polynomial Rings. i=0. i=0. n+m. i=0. k=0

Polynomial Rings 1. Definitions and Basic Properties For convenience, the ring will always be a commutative ring with identity. Basic Properties The polynomial ring R[x] in the indeterminate x with coefficients

### 4 Powers of an Element; Cyclic Groups

4 Powers of an Element; Cyclic Groups Notation When considering an abstract group (G, ), we will often simplify notation as follows x y will be expressed as xy (x y) z will be expressed as xyz x (y z)

### g(x) = 1 1 x = 1 + x + x2 + x 3 + is not a polynomial, since it doesn t have finite degree. g(x) is an example of a power series.

6 Polynomial Rings We introduce a class of rings called the polynomial rings, describing computation, factorization and divisibility in such rings For the case where the coefficients come from an integral

### 2 Arithmetic. 2.1 Greatest common divisors. This chapter is about properties of the integers Z = {..., 2, 1, 0, 1, 2,...}.

2 Arithmetic This chapter is about properties of the integers Z = {..., 2, 1, 0, 1, 2,...}. (See [Houston, Chapters 27 & 28]) 2.1 Greatest common divisors Definition 2.16. If a, b are integers, we say

### ALGEBRA. 1. Some elementary number theory 1.1. Primes and divisibility. We denote the collection of integers

ALGEBRA CHRISTIAN REMLING 1. Some elementary number theory 1.1. Primes and divisibility. We denote the collection of integers by Z = {..., 2, 1, 0, 1,...}. Given a, b Z, we write a b if b = ac for some

### Factorization in Integral Domains II

Factorization in Integral Domains II 1 Statement of the main theorem Throughout these notes, unless otherwise specified, R is a UFD with field of quotients F. The main examples will be R = Z, F = Q, and

### A few exercises. 1. Show that f(x) = x 4 x 2 +1 is irreducible in Q[x]. Find its irreducible factorization in

A few exercises 1. Show that f(x) = x 4 x 2 +1 is irreducible in Q[x]. Find its irreducible factorization in F 2 [x]. solution. Since f(x) is a primitive polynomial in Z[x], by Gauss lemma it is enough

### 4.2 Chain Conditions

4.2 Chain Conditions Imposing chain conditions on the or on the poset of submodules of a module, poset of ideals of a ring, makes a module or ring more tractable and facilitates the proofs of deep theorems.

### Quizzes for Math 401

Quizzes for Math 401 QUIZ 1. a) Let a,b be integers such that λa+µb = 1 for some inetegrs λ,µ. Prove that gcd(a,b) = 1. b) Use Euclid s algorithm to compute gcd(803, 154) and find integers λ,µ such that

### Math 2070BC Term 2 Weeks 1 13 Lecture Notes

Math 2070BC 2017 18 Term 2 Weeks 1 13 Lecture Notes Keywords: group operation multiplication associative identity element inverse commutative abelian group Special Linear Group order infinite order cyclic

### THUE S LEMMA AND IDONEAL QUADRATIC FORMS

THUE S LEMMA AND IDONEAL QUADRATIC FORMS PETE L. CLARK Abstract. The representation of integers by binary quadratic forms is one of the oldest problems in number theory, going back at least to Fermat s

### RINGS: SUMMARY OF MATERIAL

RINGS: SUMMARY OF MATERIAL BRIAN OSSERMAN This is a summary of terms used and main results proved in the subject of rings, from Chapters 11-13 of Artin. Definitions not included here may be considered

### MINKOWSKI THEORY AND THE CLASS NUMBER

MINKOWSKI THEORY AND THE CLASS NUMBER BROOKE ULLERY Abstract. This paper gives a basic introduction to Minkowski Theory and the class group, leading up to a proof that the class number (the order of the

### 0 Sets and Induction. Sets

0 Sets and Induction Sets A set is an unordered collection of objects, called elements or members of the set. A set is said to contain its elements. We write a A to denote that a is an element of the set

### 2. Prime and Maximal Ideals

18 Andreas Gathmann 2. Prime and Maximal Ideals There are two special kinds of ideals that are of particular importance, both algebraically and geometrically: the so-called prime and maximal ideals. Let

### Computations/Applications

Computations/Applications 1. Find the inverse of x + 1 in the ring F 5 [x]/(x 3 1). Solution: We use the Euclidean Algorithm: x 3 1 (x + 1)(x + 4x + 1) + 3 (x + 1) 3(x + ) + 0. Thus 3 (x 3 1) + (x + 1)(4x

### x 3 2x = (x 2) (x 2 2x + 1) + (x 2) x 2 2x + 1 = (x 4) (x + 2) + 9 (x + 2) = ( 1 9 x ) (9) + 0

1. (a) i. State and prove Wilson's Theorem. ii. Show that, if p is a prime number congruent to 1 modulo 4, then there exists a solution to the congruence x 2 1 mod p. (b) i. Let p(x), q(x) be polynomials

### be any ring homomorphism and let s S be any element of S. Then there is a unique ring homomorphism

21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UFD. Therefore

### Factorization in Polynomial Rings

Factorization in Polynomial Rings These notes are a summary of some of the important points on divisibility in polynomial rings from 17 and 18. PIDs Definition 1 A principal ideal domain (PID) is an integral

### Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra

Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................

### Arithmetic Funtions Over Rings with Zero Divisors

BULLETIN of the Bull Malaysian Math Sc Soc (Second Series) 24 (200 81-91 MALAYSIAN MATHEMATICAL SCIENCES SOCIETY Arithmetic Funtions Over Rings with Zero Divisors 1 PATTIRA RUANGSINSAP, 1 VICHIAN LAOHAKOSOL

### NOTES ON SIMPLE NUMBER THEORY

NOTES ON SIMPLE NUMBER THEORY DAMIEN PITMAN 1. Definitions & Theorems Definition: We say d divides m iff d is positive integer and m is an integer and there is an integer q such that m = dq. In this case,

### A Few Primality Testing Algorithms

A Few Primality Testing Algorithms Donald Brower April 2, 2006 0.1 Introduction These notes will cover a few primality testing algorithms. There are many such, some prove that a number is prime, others

### Restoring Factorization in Integral Domains

Virginia Commonwealth University VCU Scholars Compass Theses and Dissertations Graduate School 015 Restoring Factorization in Integral Domains Susan Kirk Virginia Commonwealth University Follow this and

### MATH 501 Discrete Mathematics. Lecture 6: Number theory. German University Cairo, Department of Media Engineering and Technology.

MATH 501 Discrete Mathematics Lecture 6: Number theory Prof. Dr. Slim Abdennadher, slim.abdennadher@guc.edu.eg German University Cairo, Department of Media Engineering and Technology 1 Number theory Number

### School of Mathematics and Statistics. MT5836 Galois Theory. Handout 0: Course Information

MRQ 2017 School of Mathematics and Statistics MT5836 Galois Theory Handout 0: Course Information Lecturer: Martyn Quick, Room 326. Prerequisite: MT3505 (or MT4517) Rings & Fields Lectures: Tutorials: Mon

### NOTES ON FINITE FIELDS

NOTES ON FINITE FIELDS AARON LANDESMAN CONTENTS 1. Introduction to finite fields 2 2. Definition and constructions of fields 3 2.1. The definition of a field 3 2.2. Constructing field extensions by adjoining

### Places of Number Fields and Function Fields MATH 681, Spring 2018

Places of Number Fields and Function Fields MATH 681, Spring 2018 From now on we will denote the field Z/pZ for a prime p more compactly by F p. More generally, for q a power of a prime p, F q will denote

### Solutions to Homework 1. All rings are commutative with identity!

Solutions to Homework 1. All rings are commutative with identity! (1) [4pts] Let R be a finite ring. Show that R = NZD(R). Proof. Let a NZD(R) and t a : R R the map defined by t a (r) = ar for all r R.

### 18 Divisibility. and 0 r < d. Lemma Let n,d Z with d 0. If n = qd+r = q d+r with 0 r,r < d, then q = q and r = r.

118 18. DIVISIBILITY 18 Divisibility Chapter V Theory of the Integers One of the oldest surviving mathematical texts is Euclid s Elements, a collection of 13 books. This book, dating back to several hundred

### Basic Algebra. Final Version, August, 2006 For Publication by Birkhäuser Boston Along with a Companion Volume Advanced Algebra In the Series

Basic Algebra Final Version, August, 2006 For Publication by Birkhäuser Boston Along with a Companion Volume Advanced Algebra In the Series Cornerstones Selected Pages from Chapter I: pp. 1 15 Anthony

### Section III.6. Factorization in Polynomial Rings

III.6. Factorization in Polynomial Rings 1 Section III.6. Factorization in Polynomial Rings Note. We push several of the results in Section III.3 (such as divisibility, irreducibility, and unique factorization)

### Math 120 HW 9 Solutions

Math 120 HW 9 Solutions June 8, 2018 Question 1 Write down a ring homomorphism (no proof required) f from R = Z[ 11] = {a + b 11 a, b Z} to S = Z/35Z. The main difficulty is to find an element x Z/35Z

### are primes, and are ordered so that

Chapter Factorization In this chapter, we will prove the fundamental theorem of arithmetic, i.e., the uniqueness of prime factorization for natural numbers. However, we will set up the framework for this

### Introduction Non-uniqueness of factorization in A[x]... 66

Abstract In this work, we study the factorization in A[x], where A is an Artinian local principal ideal ring (briefly SPIR), whose maximal ideal, (t), has nilpotency h: this is not a Unique Factorization

### WORKSHEET ON NUMBERS, MATH 215 FALL. We start our study of numbers with the integers: N = {1, 2, 3,...}

WORKSHEET ON NUMBERS, MATH 215 FALL 18(WHYTE) We start our study of numbers with the integers: Z = {..., 2, 1, 0, 1, 2, 3,... } and their subset of natural numbers: N = {1, 2, 3,...} For now we will not

### 18. Cyclotomic polynomials II

18. Cyclotomic polynomials II 18.1 Cyclotomic polynomials over Z 18.2 Worked examples Now that we have Gauss lemma in hand we can look at cyclotomic polynomials again, not as polynomials with coefficients

### Finitely Generated Modules over a PID, I

Finitely Generated Modules over a PID, I A will throughout be a fixed PID. We will develop the structure theory for finitely generated A-modules. Lemma 1 Any submodule M F of a free A-module is itself

### Subsets of Euclidean domains possessing a unique division algorithm

Subsets of Euclidean domains possessing a unique division algorithm Andrew D. Lewis 2009/03/16 Abstract Subsets of a Euclidean domain are characterised with the following objectives: (1) ensuring uniqueness

### PRACTICE PROBLEMS: SET 1

PRACTICE PROBLEMS: SET MATH 437/537: PROF. DRAGOS GHIOCA. Problems Problem. Let a, b N. Show that if gcd(a, b) = lcm[a, b], then a = b. Problem. Let n, k N with n. Prove that (n ) (n k ) if and only if

### Chapter 5: The Integers

c Dr Oksana Shatalov, Fall 2014 1 Chapter 5: The Integers 5.1: Axioms and Basic Properties Operations on the set of integers, Z: addition and multiplication with the following properties: A1. Addition

### Commutative Algebra and Algebraic Geometry. Robert Friedman

Commutative Algebra and Algebraic Geometry Robert Friedman August 1, 2006 2 Disclaimer: These are rough notes for a course on commutative algebra and algebraic geometry. I would appreciate all suggestions

### 2 Elementary number theory

2 Elementary number theory 2.1 Introduction Elementary number theory is concerned with properties of the integers. Hence we shall be interested in the following sets: The set if integers {... 2, 1,0,1,2,3,...},

### Math 412, Introduction to abstract algebra. Overview of algebra.

Math 412, Introduction to abstract algebra. Overview of algebra. A study of algebraic objects and functions between them; an algebraic object is typically a set with one or more operations which satisfies

### Euclidean Domains. Kevin James

Suppose that R is an integral domain. Any function N : R N {0} with N(0) = 0 is a norm. If N(a) > 0, a R \ {0 R }, then N is called a positive norm. Suppose that R is an integral domain. Any function N

### Polynomials, Ideals, and Gröbner Bases

Polynomials, Ideals, and Gröbner Bases Notes by Bernd Sturmfels for the lecture on April 10, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra We fix a field K. Some examples of fields

### The following is an informal description of Euclid s algorithm for finding the greatest common divisor of a pair of numbers:

Divisibility Euclid s algorithm The following is an informal description of Euclid s algorithm for finding the greatest common divisor of a pair of numbers: Divide the smaller number into the larger, and

### IRREDUCIBLES AND PRIMES IN COMPUTABLE INTEGRAL DOMAINS

IRREDUCIBLES AND PRIMES IN COMPUTABLE INTEGRAL DOMAINS LEIGH EVRON, JOSEPH R. MILETI, AND ETHAN RATLIFF-CRAIN Abstract. A computable ring is a ring equipped with mechanical procedure to add and multiply

### * 8 Groups, with Appendix containing Rings and Fields.

* 8 Groups, with Appendix containing Rings and Fields Binary Operations Definition We say that is a binary operation on a set S if, and only if, a, b, a b S Implicit in this definition is the idea that

### LEHMER S TOTIENT PROBLEM AND CARMICHAEL NUMBERS IN A PID

LEHMER S TOTIENT PROBLEM AND CARMICHAEL NUMBERS IN A PID JORDAN SCHETTLER Abstract. Lehmer s totient problem consists of determining the set of positive integers n such that ϕ(n) n 1 where ϕ is Euler s

### The primitive root theorem

The primitive root theorem Mar Steinberger First recall that if R is a ring, then a R is a unit if there exists b R with ab = ba = 1. The collection of all units in R is denoted R and forms a group under

### Definitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations

Page 1 Definitions Tuesday, May 8, 2018 12:23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of

### Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35

Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime

### 36 Rings of fractions

36 Rings of fractions Recall. If R is a PID then R is a UFD. In particular Z is a UFD if F is a field then F[x] is a UFD. Goal. If R is a UFD then so is R[x]. Idea of proof. 1) Find an embedding R F where

### IF A PRIME DIVIDES A PRODUCT... ζ(s) = n s. ; p s

IF A PRIME DIVIDES A PRODUCT... STEVEN J. MILLER AND CESAR E. SILVA ABSTRACT. One of the greatest difficulties encountered by all in their first proof intensive class is subtly assuming an unproven fact

### Math 511, Algebraic Systems, Fall 2017 July 20, 2017 Edition. Todd Cochrane

Math 511, Algebraic Systems, Fall 2017 July 20, 2017 Edition Todd Cochrane Department of Mathematics Kansas State University Contents Notation v Chapter 0. Axioms for the set of Integers Z. 1 Chapter 1.

### Homework 6 Solution. Math 113 Summer 2016.

Homework 6 Solution. Math 113 Summer 2016. 1. For each of the following ideals, say whether they are prime, maximal (hence also prime), or neither (a) (x 4 + 2x 2 + 1) C[x] (b) (x 5 + 24x 3 54x 2 + 6x

### Selected Math 553 Homework Solutions

Selected Math 553 Homework Solutions HW6, 1. Let α and β be rational numbers, with α 1/2, and let m > 0 be an integer such that α 2 mβ 2 = 1 δ where 0 δ < 1. Set ǫ:= 1 if α 0 and 1 if α < 0. Show that

### 5: The Integers (An introduction to Number Theory)

c Oksana Shatalov, Spring 2017 1 5: The Integers (An introduction to Number Theory) The Well Ordering Principle: Every nonempty subset on Z + has a smallest element; that is, if S is a nonempty subset