PROBLEMS AND SOLUTIONS IN COMMUTATIVE ALGEBRA

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1 PROBLEMS AND SOLUTIONS IN COMMUTATIVE ALGEBRA Mahir Bilen Can Disclaimer: This file contains some problems and solutions in commutative algebra as well as in field theory. About first hundred problems are those that we encountered at some point probably between years 2003 and We do not claim correctness of those solutions (neither of the other solutions). Read them at your own risk. However, we do appreciate if you send us corrections and suggest new problems and solutions. 1 Notation: Unless otherwise stated all rings are assumed to be commutative with unity. 1. Find two ideals I and J in a ring R such that I J I J. Let I = J be the ideal generated by x in the polynomial ring R = k[x]. Then I J = I = (x). Since the product of two ideals consists of finite sum of products of elements of I and J, the ideal product I J is equal to (x 2 ) which is different from (x). 2. Definition: ideals I and J from R are called co-prime, if their sum I + J is equal to R. Show that if I and J are two co-prime ideals in a ring R, then I J = I J. For all ideals I and J the inclusion I J I J is clear. To prove the other inclusion we observe the simple fact that, for any ideal K of R, the following equality KR = K is true. Hence, assuming I and J are co-prime, on the one hand we have I J = (I J)(I + J) = (I J) I + (I J) J. On the other hand, (I J) I J I and (I J) J I J. Equality is now obvious. 3. Definition: A multiplicative subset S of a ring R is a multiplicative submonoid of R. Let S be a multiplicative subset in a ring R and I be an ideal. 1 We thank Professor Lex Renner for his comments and critical eye on some of the problems with faulty solutions. We thank Şafak Özden, also. 1

2 (a) Show that S 1 I := {a/s : a I, s S} is an ideal in the localized ring S 1 R = {r/s : r R, s S}. (b) Show that the localization commutes with quotients: S 1 R/S 1 I = S 1 (R/I). Here we are abusing the notation on the right: of course R/I is localized at the image of S in R/I. (a) Let a 1 /s 1 and a 2 /s 2 be two elements from S 1 I. And let r/s S 1 R. Then r/s a 1 /s 1 + a 2 /s 2 = (s 2 ra 1 s 1 sa 2 )/s 1 s 2 S 1 I. Therefore S 1 I is an ideal. (b) Elements of S 1 (R/I) are of the form r/ s where bar denotes the images of elements of S in R/I. If we start with an element r/s of S 1 R, then r/ s makes sense. So we can define the homomorphism φ : S 1 R S 1 (R/I) by φ(r/s) = r/ s. By its construction φ is surjective. How about its kernel? Suppose r/ s = 0 in S 1 (R/I). Then there exists s in the image of S in R/I such that r s = 0 that is rs I. But it is always the case that rs /ss = r/s in S 1 R. Therefore r/s S 1 I. This shows that the kernel of φ is S 1 I, hence we get the desired isomorphism. Remark 0.1. Let S be the complement of a prime ideal P in a ring R and I be an ideal contained in P so that I S =. By the above proven fact, R P /I P = (R/I)P. Some authors write IR P for what we are calling I P ; the ideal generated by the image of I in R P. 4. Let φ : A B be a ring homomorphism by which B has a finitely generated A-module structure. It is easy to verify that for any multiplicative submonoid S A the image φ(s) B is a multiplicative submonoid also. Show that the induced homomorphism φ S : A S B φ(s) gives B φ(s) a finitely generated A S -module structure. (Here, without loss of generality we assume that 0 / φ(s). Otherwise, B φ(s) = 0, which is a finitely generated A S -module.) Let {b 1,..., b r } be a generating set for B as an A-module. Let a/b be an element of the localized ring A S. For c/φ(d) B φ(s) with c B, d S the action of A S on B φ(s) is defined by a b c φ(d) = φ(a) c φ(bd). 2

3 Since B is generated by {b 1,..., b r } as an A-module, c is of the form a 1 b 1 + a r b r = φ(a 1 )b φ(a r )b r for some a 1,..., a r A. Therefore, c φ(d) = φ(a 1)b φ(a r )b r r a i = φ(d) d bi 1 proving that {b 1 /1, b 2 /1,..., b r /1} is a generating set for B φ(s) as an A S -module. i=1 5. Let P 1,..., P m be a finite set points from C n such that P 1 P j for all j {2,..., m}. Find an explicit polynomial F (x 1,..., x n ) C[x 1,..., x n ] which takes constant value 1 on P 1 and 0 on P j for all j {2,..., m}. Let (a k1,..., a kn ) C n denote the coordinates of P k for k = 1,..., n. Since P 1 P j, there exists smallest index r j {1,..., n} such that a jrj a 1rj. Define g j (x 1,..., x n ) by g j (x 1,..., x n ) := x r j a jrj a 1rj a jrj It is clear that g j (P j ) = 0 and g j (P 1 ) = 1. The product of g j s for j = 2,..., m is the desired polynomial F. 6. Definitions: Given two ideals I and J, the ideal quotient I : J is defined to be the ideal {h R : hj I}. The radical of an ideal I, denoted by rad(i), is the ideal consisting of elements r R such that some power r n, n N of r lies in I. Logically, we call an ideal I radical if rad(i) = I. Notation: Given an ideal I of a polynomial ring k[x 1,..., x n ], V (I) k n denotes the set of points a k n such that all polynomials from I vanishes on a. Let I and J be two radical ideals. Show that the ideal of the Zariski closure V (I) \ V (J) coincides with the quotient ideal I : J. Remark 0.2. (a) Hilbert s Nullstellensatz in commutative algebra says that for an algebraically closed field k, and for any finitely generated polynomial ideal J the ideal of the vanishing locus of J is equal to radical of J. In other words, I(V (J)) = rad(j). (b) The complement V (I) \ V (J) of V (J) in V (I) need not to be an algebraic set (it is an open subset of V (I)). It doesn t make sense to talk about the ideal of an open subset. We must take its closure so that we can talk about the ideal of the closed set. 3

4 (c) The quotient I : J need not be a radical ideal in general. If I is a radical ideal, then so is I : J; suppose f n I : J for some n. Then f n h n I for any h J. But I being radical, fh I hence f I : J. (d) f R is not a zero divisor in R/I if and only if I = I : f. In this case, the variety of the ideal generated by f and I has dimension one less than V (I). For obvious reasons we assume that V (I) is not equal to V (J) (otherwise there is nothing to prove). Let α V (I) \ V (J) be a point, hence there exists a polynomial f in J such that f(α) 0. We claim that each element h of I : J vanishes on α, that is to say h(α) = 0. Indeed, hf I. But f(α) 0, so every element h of I : J vanishes on α V (I) \ V (J). It follows that I : J lies in the ideal of the closure of the complement. Conversely, if we take a polynomial f vanishing on the closure of V (I) \ V (J), then obviously it vanishes on V (I) \ V (J). Then for any h J, fh vanishes on all of V (I); f vanishes on the complement and h vanishes on V (J). Thus fh I since I is radical. But then f belongs to the quotient ideal I : J. 7. Let M be a finitely generated R-module and a R an ideal. Suppose ψ : M M is an R-module map such that ψ(m) am. Find a monic polynomial p(t) R[t] with coefficients from a such that p(ψ) = 0. Here, am is the module consisting of all finite sums of elements of the form bm, where b a and m M. The solution technique is important here. Let {x 1,..., x m } be a generating set for M as an R-module. By hypothesis, for each i = 1,..., m we have ψ(x i ) = a i,j x j, (1) for some a i,j a. We define A i,j to be the operator δ i,j ψ a i,j e, where e is the identity endomorphism of M and δ i,j is the Kronecker s delta function. It is clear from (1) that j=1 A i,jx j = 0 for all i = 1,..., m. In other words, the matrix of operators A := (A i,j ) m i,j=1 annihilates the column vector v = (x j ) m j=1. Notice that we can consider M as an R[ψ]-module, and that A i,j R[ψ]. Thus A is a matrix over R[ψ]. Therefore, its adjugate makes sense and multiplying Av = 0 on the left by the adjugate gives us (det A)x j = 0 for all j = 1,..., m. Consequently, det A annihilates all of M. Expanding the determinant we obtain a monic polynomial p in ψ with entries from a. Furthermore, p(ψ) = 0 on M. 4

5 8. If M is a finitely generated R-module such that am = M for some ideal a, then there exists x R such that 1 x a and xm = 0. By the previous problem we observe that the identity operator id on M satisfies a monoic polynomial: p(id) = id r + a 1 id r a r id = 0 for some a j a. Therefore, if we define x = 1 + a a r, then x 1 a and furthermore xm = 0 for all m M. 9. If a R is an ideal such that every element of 1 + a is invertible, M is a finitely generated R-module, and am = M, then M = {0}. Let x R be as in the previous problem; 1 x a and xm = {0}. In particular x 1 a, hence x = 1 + x 1 is invertible. It follows from xm = {0} that M = {0}. 10. Let Jac(R) denote the intersection of all maximal ideals in R. (Jac(R) is called the Jacobson radical of R.) Show that x Jac(R) then 1 xy is invertible for any y R. Conversely, if 1 xy is invertible for all y R, then x belongs to all maximal ideals. Suppose x is from Jac(R). If 1 xy is not invertible, then it is contained in a maximal ideal m of R. In particular, since x is from m, we see that 1 m which is a contradiction. Conversely, suppose that 1 xy is invertible for all y R. If x does not lie in a maximal ideal m, then the ideal generated by x and m is equal to R. Hence, xy + m = 1 for some y R and m m. In this case, m = 1 xy m, so it is not a unit, contradicting with our initial assumption. Therefore, x m. Definition: Have a taste of Zorn s lemma: A Noetherian ring is a ring in which every non-empty set of ideals has a maximal element. For other definitions and properties of Noetherian rings see Fact 0.3. Artin-Rees Lemma: Let a be an ideal in a Noetherian ring R and let M be a finitely generated R-module. If N M is a submodule, then there exists a positive integer k such that for all n k, a n M N = a n k ((a k M) N). 11. Let R be a Noetherian ring and a be an ideal such that every element of 1 + a is invertible in R. Show that n>0 a n = (0). 5

6 Let M denote n>0 a n. Obviously, M is a R-submodule of a. By Artin-Rees lemma, there exists n such that a n+i M = a i (M a n ) for all i 0. Since set theoretically we have M a n+i = M for any n + i, we get M = a i M for any i 0. By Problem 9 above, we get M = If a is an ideal such that every element of 1 + a is invertible, M a fininitely generated R-module and M M a submodule, then M + am = M implies that M = M. We consider the R-module M/M. Our assumption M + am = M implies that am/m = M/M. Hence, by Problem 9 M/M = {0}, or equivalently, M = M. 13. If a is an ideal such that every element of 1 + a is invertible, M a fininitely generated R-module. Show that the elements u 1,..., u n generate M if and only if the images u 1,..., u n generate M/aM as an R-module The implication ( ) is obvious. We prove the converse. Suppose u 1,..., u n generate M/aM as an R-module. Let {u 1,..., u n } denote a set of preimages of u i s, and let M denote the submodule generated by u i s. It is clear that M + am = M, hence by Problem 12 it follows that M = M. Definition: A local ring is a ring with unique maximal ideal. 14. Let (R, m) be a Noetherian local ring and suppose that the images of the elements a 1,..., a n m generate m/m 2 as a vector space. Show that a 1,..., a n generate m as an ideal. We denote by M the maximal ideal m viewed as an R-module, and denote by a the maximal ideal m viewed as an ideal. The solution is now an application of Problem In the notation of the previous problem, a 1,..., a n generates m/m 2 as a vector space, then a 1,..., a n generates m minimally, that is to say none of a i s is redundant. 6

7 Towards a contradiction, without loss of generality, assume that a 1 redundant; a 1 = r 2 a r n a n for some r i R. Then, modulo m 2, a 1,..., a n are linearly dependent which is a contradiction. Definition. An ideal I (1) is primary if fg I implies either f I, or g m I for some m N. 16. Prove that if Q is primary, then rad(q) is a prime ideal. Furthermore, in this case, rad(q) is the smallest prime ideal containing Q. Notation: If P denotes the prime ideal rad(q), then Q is called P -primary. Let fg rad(q), hence (fg) n Q for some n N. Since Q is primary, either f n Q, or g nm Q for some m N. In other words, either f rad(q), or g rad(q) implying that rad(q) is a prime ideal. If M is a prime ideal such that Q M, then rad(q) M because of the following two things: First, for any two ideals I and J, I J implies rad(i) rad(j). Indeed, f rad(i), then f n I for some n N, hence f n J. In particular, f rad(j). Secondly, if an ideal J is prime, then J is equal to its own radical. To see this it is enough to show that rad(j) J whenever J is prime. Let f rad(j), hence f n J for some n N. It follows primeness that f J. We apply this observation to our original problem. If M is a prime ideal containing Q, then rad(q) M. Therefore, we conclude that rad(q) is the smallest prime ideal containing Q whenever Q is primary. 17. Let M be a finitely generated R-module and S R be a multiplicative subset of R. Show that M S = 0 if and only if an element s of S annihilates M, that is to say, sm = 0. If s annihilates M, then for any m/r M S, we have m/r = sm/sr = 0/sr = 0, thus M S = 0. Conversely, assume that M S = 0, that is m/r = 0 for every m M and r S. By definition this holds if there is an s R such that s(1 m 0 r) = 0. Thus s m = 0. Now, this s is specific to m. Since M is finitely generated, there exists a finite generating set {m 1,..., m n } of M, and there exists a corresponding set {s 1,..., s n } of annihilators. Since R is commutative, the product s 1 s n annihilates all m i s, hence it annihilates whole of M. 7

8 Definition: Support of an R-module M is the set of all prime ideal p such that M p is non-trivial. Here, M p is the localization of the module at the multiplicative subset R p. 18. Let M be a finitely generated R-module. Show that a prime ideal p is in the support of M if and only if the annihilator ideal ann(m) of M is contained in p. By Problem 17 we see that M p = 0 if and only if there exists s R p such that sm = 0. But then ann(m) cannot be contained in p. This proves one implication. For the converse, suppose that p is in the support of M. Then by the previous problem again, there cannot be any s R \ p such that sm = 0. Hence ann(m) must be contained in p. 19. Show that a short exact sequence of R-modules: gives rise to a left exact sequence α 0 A 1 α 1 2 A2 A3 0 (2) 0 Hom R (A 3, N) α 2 Hom R (A 2, N) α 1 Hom R (A 1, N) (3) Given a homomorphism f : A 3 N, we pull it back to a homomorphism from A 2 to N by fα 2 : A 2 N. α Hence we get a homomorphism Hom R (A 3, N) 2 Hom R (A 2, N). Next, let us see that α2 is injective: suppose α2f = α2g for some f and g from Hom R (A 3, N). Then, α 2 f = α 2 g. But α 2 is surjective by exactness (1). Thus, f and g agree on every point of A 3 showing that they are the same functions. Therefore α2 is injective. Next, let us see that im(α 2) ker(α 1), namely α 1α 2 = 0. Let f Hom R (A 3, N). Then α 1α 2f = fα 2 α 1 = f 0 = 0 again by exactness of (1). Therefore im(α 2) ker(α 1). Finally, let us see that im(α 2) ker(α 1): let f : A 2 N be in the kernel of α 1, namely fα 1 = 0. Define g : A 3 N as follows, let g(a 3 ) be the value f(a 2 ) for any a 2 A 2 such that a 2 and g is well-defined. Notice that gα 2 = f. Therefore im(α 2) = ker(α 1) and (2) is left exact. 8

9 20. Give an example of a module N and a short exact sequence such that Hom R (, N) does not give a short exact sequence. Let p Z be a prime and consider the following exact sequence of Z-modules: 0 Z p Z Z/p 0 Apply Hom Z (, Z) and check that the result is not an exact sequence. 21. Let G be an abelian group and write G Z n G torsion. Show that Hom Z (G, Z) = Z n. Let f Hom Z (G, Z), then f is determined by the images to the generators of the copies of Z in G. The reason for not affected by the torsion part of G is the following: if a G torsion, then n a = 0 for some n Z. Then 0 = f(0) = f(n a) = nf(a) since f is a Z-module homomorphism. Therefore effect of the torsion part of G is 0 showing that an f is determined by the images of the generators of the copies of Z in G. So, the result follows. 22. In the category of R = k[x 1,..., x n ]-modules show that Hom R (R( a), R) R(a). Note that 1 in R( a) is an element of degree a. Therefore, a homomorphism f Hom R (R( a), R) R(a) of graded R-modules maps 1 to a degree a element in R that is f(1) is of degree a. Now, since an R-module homomorphism R is determined by its value on 1; we have an isomorphism of R-modules f Hom R (R( a), R) R via f f(1). Thus if we declare 1 in R to be of degree a, namely, if we regard the image as the graded module R( a), then we get an isomorphism of graded R-modules. 23. What is Z/2 Z Z/3? Let a b Z/2 Z Z/3. If b is 0 or 2, then a b = 0 0. If b = 1, then we can replace it by 4 (since 4 1 mod 3) and get a b = 0 0 again. Therefore Z/2 Z Z/3 = 0. 9

10 24. More generally, show that Z/a Z Z/b Z/ gcd(a, b). First of all remember that gcd(a, b) of two integers is the largest (hence unique) integer that divides both a and b. Furthermore, there are integers x and y such that gcd(a, b) = ax + by. Now, that said, define φ : Z/a Z Z/b Z/ gcd(a, b) by r s rs mod gcd(a, b). This map is a well-defined homomorphism of Z-modules. It is injective because: if rs = 0 mod gcd(a, b), then rs = n(ax + by) for some n Z. Thus r s = 1 rs = 1 n(ax + by) = 1 nax = nax 1 = 0 1 = 0. Surjectivity is clear by 1 r r mod gcd(a, b). Therefore φ is an isomorphism of Z-modules. 25. Show that R M is a right exact functor on the category of R-modules, but it need not be an exact functor. Let α 0 A 1 α 1 2 A2 A3 0 be a short exact sequence of R-modules and let M be some other R-module. We want to show that the following is a right exact sequence: A 1 M A 2 M A 3 M 0 Since the maps are defined by a i m α i (a i ) m, right exactness is straightforward to check. A counter example to exactness of tensoring is the following: 0 Z p Z Z/p 0 This is an exact sequence of Z-modules. Assume that Z Z/p gives us an exact sequence: 0 Z Z Z/p p Z Z Z/p Z/p Z Z/p 0 But Z Z Z/p = Z/p p Z Z Z/p = Z/p is the 0 map which, on the contrary, was supposed to be an injection. Contradiction. 10

11 26. For R-modules M, N, and P, prove that Hom R (M R N, P ) Hom R (M, Hom R (N, P )). φ Consider the map f Hom R (M R N, P ) Hom R (M, Hom R (N, P )) φ(f), defined by φ(f)(m) = f(m ) Hom R (N, P ). Here, it should be clear that f(m ) is a homomorphism from N to P for every m M. Let us prove that φ is an isomorphism of R-modules: If φ(f) = 0, then f(m n) = 0 for every m n M R N, hence f is the zero homomorphism in Hom R (M R N, P ). So, φ is injective. If g Hom R (M, Hom R (N, P )), then for every m M, g(m) is an R-homomorphism from N to P. So, define f : M N P by f(m n) = g(m)(n). It is clear that we get a homomorphism. Also, by definition, φ(f) = g. So, φ is surjective, hence an isomorphism of R-modules 27. It is easy to see that every direct sum of modules gives an exact sequence. Prove that the converse is not true; there is an exact sequence such that B C A Consider 0 A B C 0 0 Z/2 Z/4 Z/2 0 with obvious maps. This is an exact sequence. However, Z/4 Z/2 Z/ For a short exact sequence of R-modules α 0 A 1 α 1 2 A2 A3 0 show that A 2 = A1 A 3 if and only if there is a homomorphism β 2 (or a homomorphism β 1 ) with α 2 β 2 = id A3 (with α 1 β 1 = id A1 ). φ Let A 2 A 1 A 3 be an isomorphism. We have the following commutative diagram: 11

12 α 1 0 A 1 A 2 A 3 0 id φ id 0 A 1 i A 1 A 3 π A 3 0 α 2 An element of A 1 A 3 is of the form (a 1, a 3 ) for unique a 1 A 1 and a 3 A 3. Define β 2 : A 3 A 2 as follows: given a 3 A 3 there exists unique a 2 A 2 such that φ(a 2 ) = (0, a 3 ) since φ is an isomorphism. Set β 2 (a 3 ) = a 2. Then by the commutativity of the diagram it is clear that α 2 β 2 (a 3 ) = a 3. Conversely, suppose that we have a homomorphism β 2 : A 3 A 2 such that α 2 β 2 = id A3. Then define ψ : A 1 A 3 A 2 by ψ((a 1, a 3 )) = α 1 (a 1 ) + β 2 (a 3 ). This is a well defined R-module homomorphism. Let us see why it is an isomorphism. Suppose ψ((a 1, a 3 )) = α 1 (a 1 ) + β 2 (a 3 ) = 0. Then α 1 ( a 1 ) = β 2 (a 3 ). Compose this equality with α 2 to get 0 = a 3. Then α 1 ( a 1 ) = 0. But α 1 is injective, so a 1 = 0. Therefore ψ is injective. Now, let a 2 A 2. Then α 2 (a 2 β 2 α 2 (a 2 )) = α 2 (a 2 ) α 2 (a 2 ) = 0, therefore a 2 β 2 α 2 (a 2 ) im(α 1 ), say, α 1 (a 1 ) = a 2 β 2 α 2 (a 2 ). Then ψ((a 1, α 2 (a 2 )) = α 1 (a 1 ) + ψ β 2 (α 2 (a 2 )) = a 2, therefore A 1 A 3 A2 is surjective and hence an isomorphism. The case of β 1 is proven similarly. Definition 0.4. We recall some very basic definitions from homological algebra. A chain complex is a sequence of abelian groups with homomorphisms between them: satisfying 4 3 C3 2 C2 1 C1 0 C0 0 n+1 n = 0 for all n 0. Last condition ensures that the image of n+1 lies in the kernel of n, hence next definition makes sense: ith Homology group of the chain complex (C, ) is defined by H i (C) = ker i /im i+1. These groups measure how far the chain complex from being exact. Note that the chain complex (C, ) does not need to be of infinite length, however, a finite length chain complex can be extended by adding trivial groups and trivial connecting homomorphisms in between them. 12

13 φ 29. Compute the homology of the complex 0 V 1 V0 0 where V 1 = V 0 = k 3 and φ is: a a c = 0 1 Kernel of φ is: b V 1 : a + b = 0 which is spanned by 1. So, H 1 is iso- to k. For H 0, note that the image of φ is spanned by the image of the basis c b + c = 0 1 morphic , 1, 0 under φ. It is straightforward to see that the image of this basis is nothing but the same basis. Thus, H 0 is trivial. 30. Prove that for a complex V : V n... V 0 0 of finite dimensional vector spaces, the following equality is always true: ( 1) i dim H i (V ) i=0 ( 1) i dim V i = i=0 This alternating sum is called the Euler characteristic of the complex. It is clear that the complex is exact then its Euler characteristic is 0. This follows from the basic linear algebra fact that if V i+1 and V i are two finite dimensional vector spaces and φ i+1 is a linear map between them, then dim(ker φ i+1 ) + dim(imφ i+1 ) = dim(v i+1 ). Alternating sum of these equations (for i 0) gives us the desired equality. 31. Let k be a field. Suppose k[x 0, x 1,..., x n ] is graded by degree. Show that ith graded component has vector space dimension ( ) n+i i. The vector space basis for the ith graded piece is {x p 0 0 x p x pn n : t p j = i}. So, the problem is equivalent to finding the number of ways of distributing i candies to n + 1 children named x 0, x 1,..., x n. Of course, this is done in ( ) n+1 i different ways. 13

14 Definition 0.5. Let M = n Z M n be a Z-graded R-module. The numerical function H M : Z Z defined as H M (n) = dim k (M n ) is called the Hilbert function of M. The power series is called the Hilbert series of M. P M (z) = H M (t)z t 32. What is the Hilbert series of k[x 1,..., x n ]? Using Problem 31, we see that it is nothing but 1/(1 t) n. 33. Find the Hilbert series of k[x 1, x 2 2,..., x n n]. The set of monomials contained in the ith graded piece is {x i 1 1 x 2i x nin n : i 1 + 2i ni n = i} Therefore, the Hilbert series is given by r t r 34. Let P be a function P : N Z such that the associated difference function P (i) := P (i) P (i 1) is a polynomial with rational coefficients (for sufficiently large i). Show that P itself is a polynomial with rational coefficients and has degree one greater than P. Hint: Use induction on the degree s of the difference polynomial. The base case is trivial. If the leading coefficient of P is a s, then define h = a s s! ( i s+1), and compute h. By construction, P h will have degree s 1. As given in the hint, obviously, h is of degree s 1, and h(i) = a s s! ( ) i s+1 as s! ( i 1 s+1). Computing this we find h(i) = a s s! ( ) i 1 s. Clearly h is of degree s. At the same time, the leading coefficient of h is a s. Thus, the degree of h P = (P h) is s 1. By the induction hypothesis, this says that P h is a polynomial with rational coefficients and of degree s. But h is a polynomial with rational coefficients of degree s + 1, hence P is has the same type. This finishes the induction. 14

15 35. Let M be a finitely generated, Z-graded module. Show that there exists a polynomial f(x) Q[x] such that for i 0, H M (i) = f(i). Definition 0.6. The polynomial f(i) is called the Hilbert polynomial of M, written HP (M, i). Applying induction on the number of variables in the ring over which M is defined, the base case is trivial. So, suppose it is true for n 1 variables {x 1,..., x n 1 }. Using the homomorphism, multiplication by x n, M( 1) xn M, we can build an exact sequence: 0 ker(x n ) M( 1) xn M coker(x n ) 0 Since multiplication by x n kills both ker and coker, they can be regarded as modules over the polynomial ring of n 1 variables. (Of course, they are finitely generated). By exactness, H M (i) H M (i 1) Q[i]. We are done by the previous exercise. Definition 0.7. Let I be a homogenous ideal in R = k[x 0,..., x n ], hence it corresponds to a projective variety Y = V (I) in P n with ring of global sections R/I. Let us write the Hilbert polynomial of Y (or that of I) as the Hilbert polynomial of the coordinate ring R/I: HP (R/I, i) = a m m! im + a m 1 (m 1)! im Then we define the dimension of the projective variety Y P n as m, and its degree as a m. 36. Compute the Hilbert polynomial of P n and find its degree. The coordinate ring of P n is k[x 0,..., x n ]. Thus, for large i s we have HP (P n, i) = H R (i) = dim k (k[x 0,..., x n ] i ) = ( ) ( n+i i = n+i ) n. Obviously this is a polynomial in i with leading term 1 n! in. Recall that for a Hilbert Polynomial as s! is +... the degree is the dimension of the variety and a s is the degree of the variety. Thus, the degree of P n is 1. Definition 0.8. Given n, d > 0, let M 0,..., M N be all the monomials of degree d in the n+1 variables x 0,..., x n where N = ( ) n+d n 1. We define a mapping ρd : P n P N by sending the point P = (a 0,..., a d ) to the point ρ d (P ) = (M 0 (a),..., M N (a)). This is called d-uple embedding of P n in P N. 15

16 37. Let θ : k[y 0,..., y N ] k[x 0,..., x n ] be the homomorphism defined by sending y i to M i (in the notation of the above definition). Let a be the kernel of θ. Show that a is a homogenous prime ideal, and V (a) is an irreducible projective variety in P N. Recall that an ideal is called homogenous if it is generated by homogenous polynomials. A relation among M i s is zero if and only if that relation is a homogenous polynomial. For example, if M 0 = x 2 0, M 1 = x 0 x 1, M 2 = x 2 2, then M 0 M 2 M 2 1 = 0 which is of course y 0 y 2 y 2 1. This is because each M i has the fixed degree d. Therefore, the kernel of θ is generated by homogenous elements. The image of θ is a subring of k[x 0,..., x n ] generated by M i s. Obviously being an integral domain is preserved for the subrings of integral domains. Therefore the quotient k[y 0,..., y N ]/ ker θ is an integral domain showing that ker θ is a prime ideal. 38. Show that the variety, ρ d (P n ) defined in Problem 37 has the ring of global sections equal to k[y 1,..., y N ]/ ker θ, namely it is given as the locus of ker θ. The kernel of θ is generated by the polynomials that are zero when evaluated in the monomials M 0,..., M N in place of y 0,..., y N. Therefore these are exactly the generators of the vanishing ideal on the set of points (M 0 (a);... ; M N (a)) in P N. Therefore the variety ρ d (P n ) is cut out by the kernel of θ. 39. Show that ρ d is a homeomorphism of P n onto V (ker θ). Solution First, let us see that ρ d is an injection. Suppose (M 0 (a);... ; M N (a)) = (M 0 (b);... ; M N (b)), namely ρ d (a) = ρ d (b) for some a, b P n. Observe that a i s cannot be all zero (otherwise M i (a) = 0 for all i). Say a i 0. Among monomials of degree d we have a d i and a d 1 i a j for j = 0,..., n. Since (M 0 (a);... ; M N (a)) = (M 0 (b);... ; M N (b)) in P N, we have (M 0 (a)/a d i,... ; M N (a)/a d i ) = (M 0 (b)/b d i ;... ; M N (b)/b d i ) in A N. Therefore we have a d 1 i a j /a d i = b d 1 i b j /b d i for j = 0,..., n. But then a j /a i = b j /b i hence (a 0 ;... ; a n ) = (b 0 ;... ; b n ) in P n. Thus ρ d is injective. Since a map is surjective onto its image. Therefore ρ d is a bijection. It is clear that it is continuous (on each open piece y i 0 in P N it is given by continuous maps). It is enough to show that it has a maps (M 0 (a);... ; M N (a)) to a. Since at each d ((M 0(a)/a d i,..., M N (a)/a d i )) = (a 0 /a i,..., a n /a i ) and among M j (a)/a d i we have a j /a i for j = 0,..., n, we see that the inverse is basically a projection. Therefore, it is continuous hence ρ d is a homeomorphism. continuous inverse. The inverse map ρ 1 d affine open x i 0, this reduces to ρ 1 16

17 40. Find the degree of d-uple embedding of P n in P N. Solution By definition, we want to compute the Hilbert polynomial of the k-vector space k[y 0,..., y N ]/ ker θ, where θ(y i ) = M i (x). This ring is isomorphic to the subring k[m 0,..., M N ] of k[x 0,..., x n ] generated by all the monomials of degree d. Therefore it suffices to compute the Hilbert polynomial of the graded (by degree) vector space k[m 0,..., M N ] = S i. Obviously S i = 0 if i is not a multiple of d. Let i = rd for some nonzero r, then S i is equal to the ith piece of k[x 0,..., x n ]. Thus the Hilbert polynomial is equal to ( ) ( n+rd rd = n+rd ) n as a function of r = 0, 1,.... This is a polynomial in r; ( ) n + rd = n (n + rd)(n + rd 1) (rd + 1). n! The leading coefficient of this polynomial is d n /n!. Therefore, the degree (leading term times n!) of ρ d (P n ) is d n. Fact 0.9. If partially ordered set S has the property that every chain has an upper bound in S, then the set S contains at least one maximal element. Definition The nilradical of a ring is the set of all nilpotent elements of the ring. 41. Show that a nilradical is an ideal. Furthermore, show that the nilradical is equal to the intersection of all prime ideals of the ring. Let n denote the nilradical of the ring R. Let x and y be two elements from n, and suppose n, m N are such that x n = y m = 0. Binomial theorem implies that (x + y) n+m = 0, hence n is closed under addition. Let a R. Since R is commutative, (ax) n = 0, hence ax n. Therefore, n is an ideal. To prove that n is the intersection of all prime ideals, we first make the following simple observation: If P is prime ideal P R, then the quotient ring R/P (which is an integral domain) does have any nilpotent elements. Therefore, n P. This proves that n P : prime P. For the opposite inclusion, it suffices to show that for any non-nilpotent element x, there is some prime ideal that does not contain x. Towards this contradiction we assume that P : prime P n is non-empty. Let S be the set of all ideals that does not contain any power of x. Since n S, we know that S is non-empty. Furthermore, if I 1 I 2 is a nested sequence of elements from S, then I i S. Hence, by Zorn s lemma (Fact 0.9) S has a maximal element M S. Notice that if we prove M 17

18 is a prime ideal, then we succeed in our goal that there is a prime ideal that does not contain x. Assuming M is not prime, we take two elements a, b R M such that ab M. The ideals M a generated by a and M, and M b, generated by b and M both properly contain M, hence both of them contains a power of x: ay + m 1 = x p and bz + m 2 = x q. Since (ay + m 1 )(bz + m 2 ) = abyz + aym 2 + bzm 1 + m 1 m 2 lies in M we conclude that x p x q = x p+q is contained in M also. This contradiction shows that M is a prime ideal, hence our proof is complete. Fact Recall that the Jacobson radical Jac(R) of a ring R is the intersection of all maximal ideals of R. Given an ideal I Jac(R) and a finitely generated R-module M, Nakayama s Lemma says that IM = M implies M = 0. An R-module P is projective if there exists a module K such that P K F for some free module F. Equivalently, every short exact sequence splits; there exists P 0 N M f P 0 h M such that f h = id P. Remark Being projective is transitive in the following sense: if M is a projective B-module (hence M is a direct summon of a free B-module F ) and if B is an A-algebra with free A-module structure, then M is a projective A-module as well. The reason for this transitivity is that F is a free A-module, hence M is a direct summand of a free A-module. 42. Show that the following are equivalent for a finitely generated Z-module M: (a) M is projective; (b) M is torsion free; (c) M is free. (c) = (b) and (c) = (b) are obvious. Suppose that M is torsion free and let {m 1,..., m r } be a generating set for M. Define ψ : M Z r by ψ(m i ) = e i and extend by linearity, where e i is the ith standard vector in Z r. Since none of the m i s are torsion, this is a well defined map of Z-modules. Furthermore, it is an isomorphism. So (b) = (c). Finally, since M is a direct summand of a free module, it is torsion free; (a) = (b). So, we are done. 18

19 Definition For an integral domain R, a fractional ideal is an R-submodule A of the fraction field K of R such that for some nonzero element d R the following inclusion holds: { } da d A = b : a b A R. Equivalently, A = d 1 I for some ideal I R and a nonzero d R. In particular, any ideal of R is a fractional ideal. The product of two fractional ideals A = d 1 I, B = r 1 J is another fractional ideal AB = (dr) 1 IJ. This operation defines a monoid structure on the set of all fractional ideals of R. Indeed, the identity fractional ideal is the ring R itself. The unit group of this monoid contains all principal ideals. Obviously, the set of all fractional principal ideals forms a subgroup of the unit group of the monoid. Invertible fractional ideals modulo its subgroup of principal fractional ideals is called the class group of R. Thus the class group of an integral domain R measures how far is R from being a PID. 43. Show that in an integral domain R with fraction field K, if a fractional ideal A is invertible, then it is a projective R-module. Assume that A is an invertible fractional ideal. Let A 1 be its inverse. Then a 1 a a n a n = 1 for some a i A and a i A 1 since AA 1 = R. Let S be a free R module of rank n say generated by y 1,..., y n. Define φ : S A by φ(y i ) = a i and extend it by linearity. Define also ψ : A S by ψ(c) = c(a 1y a ny n ). This makes sense because ca i R. Obviously, φψ = id A, so A is a direct summand of F. In other words, A is a projective module. Remark The converse of this problem is true also: if a fractional ideal A is projective, then it is invertible 44. Show that the ideal (2, 1+ 5) of Z[ 5] is a projective Z[ 5]-module but not free. Note that Z[ 5] = 1 Z 5 Z, therefore Z[ 5] is a free Z-module of rank 2. If I := (2, 1 + 5) were free Z[ 5]-submodule of Z[ 5], then it would be of rank one. Thus there would be a single generator; I αz[ 5] for some α I. But it is easy to see that 2 and are Z[ 5]-linearly independent; αz[ 5] cannot generate I. Therefore I cannot be a free Z[ 5]-module. However, I 2 is the ideal (2). 19

20 Since (2) is an invertible ideal with inverse (1/2)Z[ 5], I is an invertible ideal with inverse I (1/2)Z[ 5], hence I is a projective module. 45. Use Nakayama s Lemma to prove that a finitely generated projective module over a local ring is free. Let {w 1,..., w n } be a minimal set of generators for a finite projective module M and let F be a free module of rank n. Consider the following surjection φ : F M: φ( n i=1 a ie i ) = n i=1 a iw i, where e i s are the generators for F. Since we have declared w i s to be the minimal generating set for M, if n i=1 a iw i = 0, then we should have a i s in the maximal ideal m of the local ring. Otherwise, they would be units and that would contradict the minimality of w i s. Thus ker φ mf. Now, for M being a projective module and having a surjection φ on it, we get a splitting of F via an injection ϕ : M F such that φϕ = id M. Therefore we can write F = ϕ(m) ker φ. Since we have the containments mϕ(m) ϕ(m) and ker φ mf and since ker φ is disjoint from ϕ(m), we see that ker φ m ker φ. So, ker φ = m ker φ. Now, by Nakayama s Lemma, ker φ = 0, hence φ is an isomorphism. Definition Left Derived Functors: Let C denote the category of R-modules and suppose F is a right exact, covariant, additive (preserving addition of homomorphisms) functor from C to C. The left derived functors L i F (N) of an R module N are defined as follows: take a projective resolution of N (for instance, a free resolution) and apply the functor F to the exact sequence omitting N. The new sequence is a complex and L i F (N) is defined as the ith homology of this complex. 46. Let C denote the category of graded C[x]-modules and let F stand for the functor C[x] (C/x) on C. Compute the left derived functors L i F of N = C[x]/x 2 for all i. Here is a projective resolution for N : 0 (x 2 ) x2 C[x] N 0 Applying F to it, we get the following complex, say P : 0 C[x] C[x]/x 0 1 (x 2 ) C[x] C[x]/x x2 1 C[x] C[x] C[x]/x 0 20

21 Then L 0 F (N) = H 0 (P ) = (C[x] C[x] C[x]/x)/x 2 C[x] C[x]/x = (C[x]/x)/(x 2 C[x]/x) = (C[x]/x) because (x 2 C[x]/x) = 0. Therefore L 0 F (N) = (C[x]/x). Next, L 1 F (N) = H 1 (P ) = ker( x 2 1)/im(0 1) Of course, L i F (N) = H i (P ) = 0 for all i > 1. = ((x 2 ) C[x] C[x]/x)/0 = (x 2 C[x]/x)/0 = 0. Definition Tor. Left derived functors of the tensor products are called Tor. What we have computed in the previous problems are T orc i [ x] (N, C[x]/x). 47. Let M be an R-Module and let r R. Compute the R-modules T orr i (R/r, M). Consider the free resolution 0 R r R R/r 0 Apply R M to this by omitting R/r to get 0 R R M 1 R R M 0 This is equal to 0 M r M 0. T orr 1 (R/r, M) = {m M : r m = 0}. Therefore T orr i (R/r, M) = M/r M and 48. Let M be an R-module and let r R. Compute the R-modules T orr i (R/r, M). Consider the free resolution 0 R r R R/r 0 Apply R M to this by omitting R/r to get 0 R R M r 1 R R M 0 21

22 This is equal to 0 M r M 0. Therefore T orr 0 (R/r, M) = M/r M and T orr 1 (R/r, M) = {M M : r m = 0}. 49. Prove that for a homogenous polynomial f of degree d and a homogenous ideal I R, there is a graded exact sequence: 0 R( d)/i : f R/I R/ I, f 0. Hint: Clearly, 0 I, f /I R/I R/ I, f is exact. How can you get a graded map from R to I, f /I? Obviously the exact sequence given in the hint is graded. Consider the graded map R( d) f I, f /I. The kernel is {g R : fg I} : f after a glance at the definition of I : J. Thus, R( d)/i : f = I, f /I. Thus, replacing I, f by R( d)/i : f in the exact sequence given in the hint, we get the graded exact sequence: 0 R( d)/i : f R/I R/ I, f Show that a given f R is a nonzero divisor on R/I if and only if I : f = I. Recall that I : J = {h R : hj I}. Suppose now that I : f = I, therefore if for some h R it happens that h f I, then h I. So, f (h mod I) = fh mod I 0 mod I implies that h I. Thus, f is a nonzero divisor. Conversely, if f is a nonzero divisor on R/I then for any h R with the property that h f I we must have h I. Remark We defined the Hilbert polynomial of a ring in Problem 35. Suppose the initial term of the Hilbert polynomial of a quotient ring R/I is given by HP (R/I, i) = a m m! im + If f is a homogenous linear form which is a nonzero divisor on R/I, then we have HP (R/ I, f, i) = HP (R/I, i) HP (R/I, i 1) a m = (m 1)! im We deduce that, by slicing with the hyperplane defined by f the dimension drops by one while preserving the degree. 22

23 Definition Let M be an R module. An element x of R is called M-regular if it is not a zero divisor on M. An ordered sequence of elements x = x 1,..., x n from R is called M-regular if the following two conditions hold: (a) x i is M/(x i,..., x i 1 )-regular for i = 1,, n. (b) M/xM 0 A weak M-regular sequence is defined by requiring only the first condition. 51. Show that given an R-module M over a local ring (R, m), a weak M-regular sequence x m is always M-regular. Let x m be a weak M-regular sequence and assume M/xM = 0 that is M = xm. But then by Nakayama s Lemma M must be trivial. 52. Show that x, y(1 x), z(1 x) is an R-sequence, but y(1 x), z(1 x), x is not where R = k[x, y, z] a polynomial ring. Here, by an R-sequence we mean an R-regular sequence for R regarded as a module over itself. Multiplication by y(1 x) on R/xR is the same as multiplication by y on k[y, z]( = R/xR). So y(1 x) is R/xR regular. And multiplication by z(1 x) on R/(x, y(1 x))r is the same as multiplication by z on k[z]( = R/(x, y(1 x))r. So, z(1 x) is R/(x, y(1 x))r-regular. Furthermore, the ideal I = (x, y(1 x), z(1 x)) is equal to (x, y, z). Therefore R/IR = k 0. Thus x, y(1 x), z(1 x) is an R-sequence. Since (y(1 x), z(1 x), x) is the same ideal with I above, we must show that this sequence fails to be a weak R sequence. Multiplication by z(1 x) on R/(y(1 x))r is not regular: z(1 x) y = 0 in R/(y(1 x))r. Therefore (y(1 x), z(1 x), x) cannot be an R-sequence. 53. Let R be a Noetherian local ring, M a finite R-module, and let x be an M-sequence. Show that every permutation of x is an M-sequence. Since every permutation is the product of adjacent transpositions, it suffices to show that x 1,..., x i 1, x i,, x n is M-regular. Note that for the module M/(x 1,, x i 1 )M, (x i,... x n ) is a regular sequence. Therefore, by induction, it will suffice to handle the case of n = 2. We want to show that x 2, x 1 is M-regular assuming X 1, x 2 is M-regular. Of course, we may assume that x i is not a unit, otherwise x 1 M = M, hence we would 23

24 be done. Now, look at the kernel of multiplication by x 2 on M. Since x 2 must be regular on M/x i M, ker( x 2 ) x i M. If z ker( x 2 ), then z x 1 m for some m M. Since x i (x 2 m) = x 2 (x 1 m) = 0, and x 1 is regular on M, (x 2 m) must be 0. But then m ker( x 2 ). Therefore ker( x 2 ) = x i ker( x 2 ). Since x 1 is a non unit, it is in the maxima ideal, hence any element 1 + (x 1 ) is regular on M. Finally we have to show that x 1 is regular on M/x 2 M. Assume otherwise; there exists m M/x 2 M such that x i m x 2 M hence x i m = x 2 m for some m M. Here m cannot be in x i M otherwise the equation would imply m x 2 M. So, x 2 is not regular on M/x 1 M, contradiction. Thus, x 2, x 1 is a regular sequence too. Definition Let A be a subring of a ring B and b B. Then b is call integral over A if b is a root of a monic polynomial with coefficients from A, that is if there is a relation of the form b n + a 1 b n a n = 0 with a i A. 54. Let A B be rings. Show that an element b B is integral over A if and only if there exists a ring C between A and B such that b C and C is finitely generated as an A-module ( ) Let b B be integral over A. The let C be the ring generated by A and b, that is C := A[b]. Obviously A CsubsetB. Let us see that C is indeed a finitely generated module over A. Obviously any element of C is a polynomial in b with coefficients of A. Since b n = (a 1 b n a n ), any element of C can be written as an A-linear combination of 1, b, b 2,..., b n 1 hence C is finitely generated A-module. ( ) Suppose there exists an intermediate ring C which is a finitely generated A module. Then C = Aw i +... Aw n for some w i C. Let b C. Then w i b = j a ijw j for i = 1,..., n. But then we get a relation b n + a 1 b n a n = 0 (by the Cayley s theorem; expanding det(bδ ij a ij )). 55. Let B be an integral domain and A B a subring such that B is integral over A. Then A is a field if and only if B is a field. ( ) Suppose B is an integral domain which is integral over a field A. Then any element b in B satisfies a monic polynomial over A i.e, b n + a 1 b n a n = 0 for some a i A. Since a n is invertible, from this equation we see that b is invertible. Therefore, every element of B is invertible, hence it is a field. ( ) Suppose B is a field and A B a subring such that B is integral over A. It is enough to show that for a A, the inverse a 1 of a also lies in A. Assume otherwise; b := a 1 A. But then it satisfies a monic polynomial over A, that is b n + a 1 b n

25 + a n = 0 for some a i A. Multiplying this relation by a n 1, we see that b A which is a contradiction. Therefore A is a field, also. 56. Let B be an extension ring of A which is integral over A. Let P B be a prime ideal. Then P is maximal if and only if P A is maximal in A. P A is obviously a prime ideal. The composition A B B/P gives the injection A/A P B/P. Note that B/P is integral over A/A P. Therefore, by the previous problem that A/A B is a field if and only if B/P is a field. Hence A P is maximal in A if and only if P is maximal in B. 57. Let A B be rings and suppose B is integral over A. Let m be a maximal ideal in A. Show that there exists a prime P in B lying over m, that is P A = m. Furthermore, any such P is maximal ideal of B. Let us first see that mb B. Assume contrary that mb B, then there exists b i B and π i m such that n i=1 b iπ i = 1. Set C := A[b 1,...b n ], then C is finite over A and mc = C. Let C = Au 1 + +Au r for some u j C. Then, we get u i = π ij u j for some π ij m. Therefore := det(δ ij π ij ) satisfies u j = 0 for every j. Hence C = 0. But since 1 C, = 0. On the other hand mod m simply by its expansion. Hence 1 m, a contradiction. Now, since mb B, it is contained in a prime ideal, say P. By the previous problem we know that P must be a maximal ideal. 58. Let A B be rings ans suppose B is integral over A. Let p be a prime ideal of A. Show that there exists a prime p B such that P A = p. Furthermore, there is not inclusion between primes in B that lie of p. Localizing the exact sequence 0 A B at p, we get an exact sequence 0 A p B p = B A A p in which B p is an extension ring of A p and integral over A p. Using the following commutative diagram: A p B p A B 25

26 we see that the prime ideals of B lying over p corresponds bijectively with the maximal ideals of B p lying over the maximal ideal pa p A p. Hence by the previous problem we are done. 59. Let A B be rings and suppose B is a finitely generated A-module( hence integral over A.) Show that for a prime ideal of A, there are only finite number of prime ideals in B that lie over p. First we assume A to be a local ring with maximal ideal m. Since B is finitely generated module over A, B = Aw i + + Aw r for some w i B. Then, we the quotient ring B/mB becomes a vector space over A/m with a generating set { w i }. Every prime P B containing mb gives a vector space P/mB B/mB. Also note that by the previous problems we know that if any prime lying over a maximal must be maximal too. Therefore these primes are coprime to each others. But then by the Chines Remainder Theorem we see that B/mB = i P i/mb. Since each P i /mb contributes to the dimension of the vector space, there must be finitely many P i s. So, in the local case we are done. For the general case we make use of the diagram: A p B p A B Since primes containing p A correspond to the maximal ideals of B p over pa p in the local ring A p, we reduce to the local case. Hence we are done. 26

27 Note: Next two problems have been added to the file on January Definition Let p be a prime number and consider the ring of p-adic integers Z p, which is constructed as the inverse limit lim Z/p i Z. Let A i denote Z/p i+1 Z for simplicity. By definition, this is the ring of sequences (a i ) i 0 i=0 A i which satisfy the compatibility condition with respect projections µ ji : A j A i defined by µ ji (a j mod p j+1 ) = a i mod p i+1 for all j > i. 60. Prove that there exists a bijection between Z p and the set of all formal power series of the form α 0 + α 1 p + α 2 p 2 +, where α i {0, 1,..., p 1}. Let (a i ) i=0 be an element from Z p. As a i A i = Z/p i+1 Z, i = 0, 1,..., we assume that a i is a positive integer less than p i+1. Set α 0 = a 0 {0,..., p 1}. Since a 1 = a 0 mod p, a 1 a 0 is divisible by p. Since a 1 < p 2 and a 0 < p, there exists unique α 1 {0,..., p 1} such that a 1 = a 0 + pα 1. Similarly, by using projections µ ji, we have that a 2 = a 1 mod p 2, or equivalently that a 2 a 1 = α 2 p 2 for a unique integer α 2. Since 0 a 2 < p 3, we see that α 2 has to be less than p. Therefore, we see that a 2 = a 1 + pα 2 = α 0 + pα 1 + p 2 α 2. Continuing in this manner, we conclude that the nth term of the sequence (a i ) i=0 is equal to n i=0 α ip i for some non-negative integers 0 α j < p, uniquely determined by a 0, a 1,..., a n 1. Therefore, the data of the element (a i ) Z p is represented by the infinite series α 0 + α 1 p + α 2 p 2 + in a unique way. Let us revert this process to compute α i s in terms of a i s. a 0 = α 0. α 1 = (a 1 a 0 )/p. More generally α n = a n (α 0 + α n 1 p n 1 ) p n. We already know that 61. What are the ring operations on the series representation of p-adic integers? Suppose (a i ), (b i ) are two p-adic integers represented by the series α i p i and β i p i, respectively. We first analyze what happens to addition in the series notation. Let (c i ) = (a i ) + (b i ) = (a i + b i ) be represented by the summation γ i p i. 27

28 Since α 0 = a 0, β 0 = b 0, we see that the constant term γ 0 of the series representation of (c i ) has to be equal to α 0 + β 0 mod p. Note that α 0 + β 0 = γ 0 + pδ 0 for a unique δ 0 {0, 1}. Next we determine γ 1. Since c 1 = a 1 + b 1 = (α 0 + β 0 ) + p(α 1 + β 1 ) Z/p 2 Z is equal to γ 0 + γ 1 p, by the uniqueness of the power series coefficients we see that γ 1 has to be equal to δ 0 +α 1 +β 1 mod p. Since δ 0 +α 1 +β 1 < 2p, we see that δ 0 +α 1 +β 1 = γ 1 +pδ 1 for a unique δ 1 {0, 1}. Thus we write α 1 + β 1 = γ 1 δ 0 + pδ 1. Similarly, a 2 + b 2 = α 0 + β 0 + (α 1 + β 1 )p + (α 2 + β 2 )p 2 mod p 3 = γ 0 + (δ 0 + α 1 + β 1 )p + (α 2 + β 2 )p 2 mod p 3 = γ 0 + (δ 0 + γ 1 δ 0 + δ 1 p)p + (α 2 + β 2 )p 2 mod p 3 = γ 0 + γ 1 p + (δ 1 + α 2 + β 2 )p 2 mod p 3 Similar to the previous case, we write γ 2 for δ 1 + α 2 + β 2 γ 2 + δ 2 p = δ 1 + α 2 + β 2 implies that δ 2 {0, 1}. Thus mod p, hence the equality a 2 + b 2 = γ 0 + γ 1 p + γ 2 p 2 mod p 3, where γ 2 is equal to δ 1 + α 2 + β 2 mod p, and δ 1 is found from δ 0 + α 1 + β 1 = γ 1 + pδ 1. More generally, if a n = n i=0 α ip i and b n = n i=0 β ip i, then a n + b n = n i=0 (α i + β i )p i, and γ n is equal to δ n 1 + α n 1 + β n 1 mod p, where δ n 1 is found (inductively) from δ n 2 + α n 2 + β n 2 = γ n 1 + pδ n 1. Next we look at what happens to the nth term of the product (a i )(b i ) = (a i b i ). Suppose γi p i corresponds to this product. Clearly γ 0 = a 0 b 0 mod p = α 0 β 0 mod p. For γ 1 we look at a 1 b 1 modulo pz, which has to be equal to a 0 b 0 modulo pz. Since a 1 b 1 = α 0 β 0 + (α 1 β 0 + α 0 β 1 )p in Z/p 2, the equality a 1 b 1 = a 0 b 0 mod p is straightforward. On the other hand, to compute γ 1 we need to look at the carry over from α 0 β 0. Indeed, writing α 0 β 0 as γ 0 + pu 1, we see that γ 1 must be α 0 β 1 + α 1 β 0 + u 1 mod p. Similarly, γ 2 must be u i=0 α iβ 2 i mod p, where u 2 is the carry-over from the previous parts. Indeed, a 2 b 2 = α 0 β 0 + (α 0 β 1 + α 1 β 0 )p + (α 0 β 2 + α 1 β 1 + α 2 β 0 )p 2 mod p 3 = γ 0 + (α 0 β 1 + α 1 β 0 + u 1 )p + (α 0 β 2 + α 1 β 1 + α 2 β 0 )p 2 mod p 3 = γ 0 + γ 1 p + (u 2 + α 0 β 2 + α 1 β 1 + α 2 β 0 )p 2 mod p 3, where u 2 is found from the equation α 0 β 1 + α 1 β 0 + u 1 = u 2 p + γ 1. Let γ 2 denote u 2 + α 0 β 2 + α 1 β 1 + α 2 β 0 mod p 2, hence we write γ 2 = u 2 + α 0 β 2 + α 1 β 1 + α 2 β 0 + u 3 p 2. Iterating this process we see how to multiply power series representations of p-adic integers by carry-overs. 28