MATH 371 Class notes/outline September 24, 2013

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1 MATH 371 Class notes/outline Setember 24, 2013 Rings Armed with what we have looked at for the integers and olynomials over a field, we re in a good osition to take u the general theory of rings. Definitions: A ring is an abelian grou R, q with an additional binary oeration called multilication. For every x, y, z P R: 1. x yq z x y zq 2. There exists an element 1 P R such that 1 x x 1 x 3. x y zq x y x z and y zq x y x z x. (the identity for addition in R is denoted 0, and we ll usually leave out the dot in x y). A set S R is called a subring of R if S is a subgrou of R, q, 1 P S and xy P S if x, y P S. An element x P R is called a zero divisor if there exists y P R with y 0 but xy 0 or yx 0. An element x P R is called a unit if there exists y P R such that xy yx 1. Then we write y x 1. The set of units in R is denoted R. R is called a commutative ring if xy yx for all x, y P R. Examles: Commutative rings: Z, Q, R, C, krx 1,..., x n s, Rrx 1,..., x n s. Non-commutative rings: n n matrices (with coefficients in a commutative ring) for n 2, quaternions H. We will stick to commutative rings from here out, unless otherwise stiulated!! More definitions: A field is a ring with R tr P R r 0u If K L are fields and K is a subring of L then K is called a subfield of L and L is called an extension field of K. An integral domain (or sometimes just domain ) is a ring with no zero divisors. Proosition: In an integral domain R, suose a, x, y P R with a 0. If ax ay then x y (this is called cancellation). Let F be a field. Then F is an integral domain. Examle: The set Qiq ta bi a, b P Qu C is a subring of C. In fact, Qiq is a field so Qiq is a subfield of C. In algebra, if z a bi P C, it is common to call z 2 a 2 b 2 the norm of z and to denote it Nzq. Note that Nz 1 z 2 q Nz 1 qnz 2 q. Within Qiq there is the subring Zris of Gaussian integers. The norm Nzq of a Gaussian integer z P Zris must be a non-negative integer, and then the multilicative roerty of the norm imlies that z P Zris is a unit of Zris if and only if Nzq 1 (Proof?). This yields Zris t 1, iu. Note

2 2 that rime numbers in Z are not necessarily rime in Zris, e.g., 5 2 henomenon later. iq2 iq. More on this Ideals: Just as we defined ideals in krx 1,..., x n s, we can define ideals in any ring. An ideal in R is a subgrou I of R, q such that rx P I for every a P R and x P I. Note: R is an ideal of itself, and I R if and only if 1 P I. For r 1,..., r s P R we can define xr 1,..., r s y ta 1 r 1 a s r s a 1,..., a s P Ru to be the ideal generated by r 1,..., r s. If an ideal is generated by a finite set, it is called finitely generated. You can also talk about ideals generated by infinite sets (the sums are still finite, but the set from which the r s are chosen can be infinite). If I and J are ideals in R, then I X J and I J ti j i P I, j P Ju are also ideals in R. The roduct IJ is defined to be the ideal generated by tij i P I, j P Ju. Note that IJ I X J. The only ideals of a field F are t0u and F itself. An ideal generated by one element xry is called a rincial ideal. If R is an integral domain, and every ideal of R is rincial then R is called a rincial ideal domain. We know that Z and krxs (olynomials in one variable over a field) are rincial ideal domains. Perhas surrising: Theorem: The ring of Gaussian integers Zris is a rincial ideal domain. (Idea of roof: If I is a non-zero ideal in Zris, then let d P I have minimal norm. For any z P I consider z{d (in Qiq), and let q P Zris with Nq z{dq 1. Multily this by Ndq to see that Nqd zq Ndq but qz d P I, a contradiction unless d qz.) For the record, krx, ys, Zrxs and Zr? 5s are not rincial ideal domains. Quotient rings: If I is an ideal of R, then the set R{I trxs x P Ru of (left) cosets rxs x I of I (thinking of I as a subgrou of the abelian grou R, q) can be made into a ring in the obvious way (recall rxs rys means x y P I): rxs rys rx ys and rxsrys rxys for all rxs, rys P R{I. This new ring is called the quotient ring of R by I. Note r0s is the additive identity and r1s the multilicative identity in R{I, and rxs r0s if and only if x P I. We ve already met the quotient rings Z{xdy this is a field if and only if d is rime and otherwise has zero divisors (unless d 0 in which case you still have Z). If is a rime number, we ll write F for the field Z{xy. Prime and maximal ideals: By analogy with the basic roerty of rime numbers, namely that if ab then either a or b, we define a rime ideal I R to be an ideal such that if xy P I then either x P I or y P I (or both). If I is a rime ideal and I R then R{I is an integral domain (and conversely). A maximal ideal is one not roerly contained in any other roer ideal, i.e., I is maximal if for any other ideal J, the containment I J imlies that either J I or J R. An ideal I R is maximal if and only if R{I is a field. Homomorhisms of rings: A maing f : R Ñ S from one ring to another is called a homomorhism (or, more recisely, a ring homomorhism) if it is a grou homomorhism from R, q

3 3 to S, q (i.e., if fx yq fxq fyq for all x, y P R) and if fxyq fxqfyq for all x, y P R. It s called an isomorhism if it s a bijection (and so is invertible), and then R and S are isomorhic (denoted R S). The ma R Ñ R{I given by x ÞÑ rxs is the canonical examle of a ring homomorhism (it s surjective but not injective unless I t0u). The kernel kerfq tr P R frq 0u of a ring homomorhism is an ideal of R and the image is a subring of S. The standard isomorhism theorem is almost immediate, that if f : R Ñ S is a homomorhism with kerfq K, then f : R{K Ñ frq (where fr Kq frq is well-defined and an isomorhism. For every ring R there is a unique homomorhism f : Z Ñ R. Using this homomorhism one can define nr for n P Z and r P R as nr fnqr r r r n timesq. The generator of the kernel of this homomorhism is the characteristic of the ring R. If R is an integral domain, then this generator is a rime number. If moreover R is finite then it is a field. The binomial theorem is true in any ring: a bq n a n n 1 a n 1 b n ab n 1 n 1 b n for n P N (roof by induction). characteristic, we have From this get the freshman s dream : In a ring R of rime x yq e x e y e for all x, y P R and e P N. (Induct on e and use that i for 1 i 1 from homework 1). Use this to show that the Frobenius ma F : R Ñ R given by F xq x is a ring homomorhism. Fraction field of an integral domain: Just as you make Q from Z, given any integral domain R you can make a natural field Q containing R, such that Q is the smallest field containing R (in the sense that if F is any other field and there is an injective homomorhism R Ñ F, then there is an injective homomorhism Q Ñ F which extends it.) Q comrises equivalence classes of symbols r{s for r, s P R with s 0 where r{s r 1 {s 1 means that rs 1 r 1 s. And so it goes. For instance, the field of fractions of Zris is Qiq. Factorization In the integers and olynomials over a field, the ideas of divisibility and factorization are imortant. These led to the notion of rime numbers and irreducible olynomials. We generalize these ideas here to arbitrary integral domains. So from here out in these notes, we ll assume all our rings are integral domains (so that in articular, the cancellation law holds). Definitions: Let R be an integral domain. For x, y P R, say y divides x, write y x if x ry for some r P R. We have y x if and only if xxy xyy. We ll have xxy xyy if and only if r P R, in which case we say x and y are associates in R. The element d P R is called a greatest common divisor of a, b P R if d a and d b and for any other x which divides both a and b, we have x d as well. If R is a rincial ideal domain, then xa, by xdy for some d and d is a greatest common divisor of a and b (Proof?).

4 4 A non-unit r in R is called irreducible if r ab for a, b P R imlies that either a or b is a unit. A non-zero non-unit x P R is said to have a factorization into irreducible elements if there are irreducibles 1,..., r P R such that x 1 r And x is said to have unique factorization into irreducible elements if for any other irreducible factorization x q 1 q s we have that every i for i 1,..., r divides q j for some j 1,..., s (which imlies that q j u i where u is a unit). In articular, r s. A domain R such that every non-zero non-unit has unique factorization into irreducible elements is called a unique factorization domain. A non-zero non-unit in R is called a rime element if xy for x, y P R imlies that either x or y. Proosition. A rime element is irreducible. (Idea of roof): If is rime and ab, then either a or b. If a then a r, but then rb. Cancelling the s shows that b is a unit, so is irreducible. Proosition: Let R be a domain for which every non-zero non-unit has a factorization into irreducibles. Every irreducible element is rime if and only if R is a unique factorization domain. (Idea of roof): (ñ) If x has two irreducible factorizations x 1 r q 1 q s then each i must divide some q j and vice versa because they re also rime. (ð) Suose x is irreducible and x ab. Then ab xr, factor both sides into irreducibles (on the right, do this by factoring r into irreducibles, on the left do this by factoring a and b into irreducibles) by uniqueness x must be an associate of one of the factors on the left, which is a factor of either a or b, so x is rime. Examle: In R Zr? 5s ta b? 5 a, b P Zu, the number 2 is irreducible but not rime. See?? this because there are two irreducible factorizations of q1 5q. (Use the norm function Na b? 5q a 2 5b 2 to conclude that Zr? 5s t 1u and that the four factors are irreducible). Our goal now is to show that if R is a rincial ideal domain, then R is also a unique factorization domain. Lemma: Let R be a rincial ideal domain and x P R a non-zero element. Then x has an irreducible factorization. (Idea of roof): If x is irreducible then we re done. If not, factor x 1 q 1 where neither 1 nor q 1 is a unit. But then xxy x 1 y and xxy xq 1 y. If one of 1, q 1 is not irreducible we can further factor. This rocess cannot go on forever since you can t have an infinite increasing sequence of ideals in a rincial ideal domain (Proof?), and it will terminate in an irreducible factorization of x. Proosition: Let R be a rincial ideal domain that is not a field. An ideal xxy R is a maximal ideal if and only if x is irreducible. Proof. If x is irreducible and xxy xyy, then x sy, but then s must be a unit. Thus xxy xyy or xyy R, so xxy is a maximal ideal. On the other hand, if xxy is maximal and x sy then one of s, y must be a unit, otherwise xxy xyy R, contradicting that xxy is maximal. Theorem. A rincial ideal domain R is a unique factorization domain.

5 5 Proof. From the lemma above, we know that every non-zero element in R has a factorization, so we only have to rove uniqueness. We ll aly one of the roositions above and show that every irreducible element is rime. So let P R be irreducible and suose ab but a. Then a R xy therefore xa, y xy. Since xy is maximal by the roosition immediately above, we have xa, y R x1y. Therefore there are λ, µ P R such that λa µ 1, therefore λab µb b. But we know that ab therefore b, and so is rime, and we re done. (Doesn t this roof look familiar? See the roof of corollary 1 on age 3 of the notes on integers.) Examles: The ring Zr? 5s is not a rincial ideal domain since 2 is an irreducible element that?? is not rime. In fact I x2, 1 5y is not a rincial ideal if x P I then x 2a bq b 5 for a, b P Z and use the norm function from before. The ring of Gaussian integers Zris, being a rincial ideal domain, is a unique factorization domain. One more task before we leave the realm of factorization for a bit: In which rings (integral domains) can we use the Euclidean algorithm (rather than factoring) to find greatest common divisors? The answer is that we have to have a version of the division algorithm so that for every x, d P R with d 0 there exist q, r P R such that x qd r and r is smaller than d in some sense. Definition: A Euclidean domain R is an integral domain on which there is a function N that mas the nonzero elements of R to the non-negative integers and which satisfies: for every x, d P R with d 0 there exist q, r P R such that x qd r with either r 0 or Nrq Ndq. So the integers (with N the absolute value function), olynomials in one variable over a field (with N the degree function) and the Gaussian integers (with N the norm function) are all Euclidean domains. Mimic the roof of the fact that the ring of Gaussian integers is a rincial ideal domain (on age 2) to get Theorem: A Euclidean domain R is a rincial ideal domain (hence a unique factorization domain). And once you have the division algorithm, the Euclidean algorithm follows. There are rincial ideal domains that are not Euclidean domains. Zr1 but that is not so easy to rove.? 19q{2s is one such, The Gaussian integers and number theory The theorem on age 2 shows that the ring of Gaussian integers Zris is a Euclidean domain where the function N is the norm function as we defined it there: Na biq a 2 b 2. So the Gaussian integers are a unique factorization domain. It is interesting to ask what are the rimes in Zris, and to start with asking what haens to the ordinary integer rimes. Proosition: If z P Zris is a Gaussian integer such that Nzq and is a rime number, then z is a rime element of Zris. Idea of roof: Since rimes and irreducibles are the same in a UFD, we ll show that z is irreducible. If z ab then Nzq NaqNbq, so one of Naq and Nbq has to be 1 and the other. But the one with norm 1 is a unit, so z is irreducible.

6 6 Now we know that some rime numbers like 5 2 iq2 iq and iq3 2iq become reducible in Zris. We examine this henomenon a little more closely. Lemma (Lagrange): Let be a rime number (i.e., a rime integer). If 1 mod 4q then the congruence has a solution. x 2 1 mod q Proof: In fact, writing 4n 1 we can take x 2nq!. Use Wilson s theorem (roblem 9(b) on Homework 1), which tells us that 1q! 4nq! 1 mod q. But 4nq! 4nq4n 1q4n 2q 4n 2n 1q2nq2n 1q and we note that 4n 1 mod q, 4n 1 2 mod q,...,4n 2n follows that 2nq!q 2 4nq! 1 mod q. 1q 2n mod q so it Corollary. A rime number 1 mod 4q is not rime in Zris. Because there s an x such that x mod q, i.e., x 2 1 x iqx iq. But x i and x i, since x{ 1{qi R Zris. This leads to a famous theorem of Fermat: Fermat s two-square theorem: A rime number 1 mod 4q is the sum of two uniquely determined squares. Proof. (Uniqueness) Suose a 2 b 2. Then a biqa biq in Zris and since Na biq a 2 b 2 is rime, but the roosition above we have that a bi is irreducible. Now suose also c 2 d 2. Therefore c diqc diq is another irreducible factorization of. Since Zris is a UFD, we must have that c di is a unit times either a bi or a bi, and similarly for c di. But the units of Zris are 1, i, so we must have ta 2, b 2 u tc 2, d 2 u. (Existence) From the corollary above, we know that is not rime in Zris, so write yz where y a bi is rime in Zris. We can t have z a unit, or else would be rime, so Nzq 1. But Nq 2 so the only choice is Nzq and Nyq. But Nyq a 2 b 2 so we have exressed a 2 b 2. Note that this roof doesn t tell us how to find a and b so that a 2 b 2. There is an algorithm for this, but before we can resent it, we need to digress a bit on quadratic residues this has to do with whether one can solve the equation x 2 a mod q for various numbers a. Definition: Let be a rime number. If a then a is called a quadratic residue modulo if there exists x P Z such that a x 2 mod q. Otherwise, a is called a quadratic non-residue modulo. If a then a is considered neither a quadratic residue nor a quadratic non-residue. This definition is encasulated in the Legendre symbol: a $ & % 0 if a 1 if a is a quadratic residue modulo 1 if a is a quadratic non-residue modulo.

7 7 By thinking in the ring Z{xy, we see that if a x 2 mod q for some integer x P Z, then there is a y such that 0 y with a y 2 mod q. Therefore the quadratic residues in Z{xyq are the numbers r1 2 s, r2 2 s,..., r 1s 2, where the brackets mean mod. This tells us that the Legendre symbol satisfies a a k for any k P Z. A little exerimenting will convince you that the following roosition ought to be true: Proosition: If is an odd rime then half the numbers 1, 2,..., 1 are quadratic residues, while the other half are quadratic non-residues mod. Proof. Since x 2 xq 2 mod q, the quadratic residues mod are actually given by the first 1q{2 squares r1 2 s, r2 2 s,..., r 1q{2q 2 s. And these numbers are different, since if ri 2 s rj 2 s we have i 2 j 2 q i jqi jq. But cannot divide i j if 0 i, j 1q{2 so we must have i j. So there are exactly 1q{2 quadratic residues mod, leaving 1q{2 quadratic non-residues. Theorem (Euler): Let be an odd rime and let a be an integer not divisible by. Then a a 1q{2 mod q. Proof: If a is a quadratic residue mod then a x 2 mod q where x for some x P Z. Thus a 1q{2 x 2 q 1q{2 x 1 1 mod q as it should, by Fermat s little theorem. Therefore we have at least 1q{2 different solutions in Z{xy to the congruence X 1q{2 1 0 mod q. But we know that this olynomial can have at most 1q{2 solutions, so all the quadratic non-residues do not satisfy this equation. This means that if a is a quadratic non-residue then a 1q{2 1 mod q. But a 1q{2 q 2 a 1 1 mod q by Fermat s little theorem again, so we must have a 1q{2 1 mod q. Corollary: If is an odd rime, then the Legendre symbols satisfy: ab a b. Corollary: Let be an odd rime. Then 1 is a quadratic residue mod if 1 mod 4q and 1 is a quadratic non-residue mod if 3 mod 4q. An amazing theorem about Legendre symbols is Gauss s famous law of quadratic recirocity, which states that if and q are odd rimes then We ll have more to say about this later. q 1q 1qq 1q{4. q

8 8 Now back to that algorithm for finding Fermat s two squares that add u to the rime if 1 mod 4q. It begins with an alternative roof of Lagrange s lemma above. Another way to get a solution to the congruence x 2 1 mod q is to use Euler s theorem above, as follows: Suose a is quadratic-non-residue mod, then Euler s theorem tells us that a 1q{2 1 mod q. But since 1 mod 4q, we have 1q{4 is an integer so we can take x ra 1q{4 s. Calculating ra 1q{4 s (by reeated squaring) is much easier than calculating 1q{2q! and finding a quadratic non-residue a is easy by trial and error since half the numbers between 1 and 1 work. Now for the algorithm. Given a rime such that 1 mod 4q, choose a solution to the congruence x 2 1 mod q such that 0 x {2 (Why can this always be done?). Then use the Euclidean algorithm on and x. The first two remainders a and b in the rocess such that both a and b are less than? will satisfy a 2 b 2. That s it. Here are a coule of examles: Suose 41. Then x 9 satisfies x 2 1 mod 41q. And the Euclidean algorithm alied to 41 and 9 gives: i r i q i λ i µ i The first two remainders less than? 41 are 5 and 4 and Let 113. Then x 15 satisfies x 2 1 mod 113q. So We conclude that i r i q i λ i µ i Finally, we do an extension of Euclid s roof on infinitely many rimes to rimes congruent to 1 mod 4, using Gaussian integers. Lemma: A rime number 3 mod 4q is a rime element in Zris. Proof. Suose z a bi is a rime element in Zris that divides, so zy for some y P Zris. Since Nq 2 we have Nzq or Nzq 2. But Nzq c 2 d 2 cannot equal since the sum of two squares cannot be congruent to 3 mod 4, but then Nyq 1 so y is a unit and was rime in Zris. Lemma: If is an odd rime number dividing x 2 1 for some x P Z, then 1 mod 4q. This is because x 2 1 x iqx iq in Zris and doesn t divide either factor so can t be rime in Zris. Theorem: There are infinitely many rimes congruent to 1 mod 4.

9 9 If only finitely many, say they are 1,..., n, then form the number M 1 2 n q 2 1. Then M is not a ower of 2 (why?) so it s divisible by an odd rime, which must be congruent to 1 mod 4, but none of the i s divide M. So there must be more of them. This is a really secial case of a celebrated result of Dirichlet on rimes in arithmetic rogressions. We ll rove a less secial case later.