# MATH 326: RINGS AND MODULES STEFAN GILLE

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1 MATH 326: RINGS AND MODULES STEFAN GILLE 1

2 2 STEFAN GILLE 1. Rings We recall first the definition of a group Definition. Let G be a non empty set. The set G is called a group if there is a map called multiplication G G G, (g, h) g h subject to the following axioms: (Associative law) g (h l) = (g h) l for all g, h, l G. (Existence of a left neutral element) There exists e G, such that e g = g for all g G. (Existence of a left inverse) For all g G there exists h G, such that If moreover (Commutative law) h g = e. g h = h g for all g, h G then the group G is called abelian or commutative. In this case we write the multiplication in G additively and call it addition, i.e. we write g + h instead of g h Examples. (a) The integers, rational-, real-, and complex numbers (denoted Z, Q, R, and C, respectively) with the usual addition are an abelian group. (b) The general linear group GL n (R), i.e. the set of invertible n n-matrices over R, with the usual matrix multiplication is a group, which is for n 2 not commutative. (c) If X is set then Aut(X), the set of bijective, i.e. one-to-one and onto, mappings from X into itself is a group with composition of maps as multiplication. The neutral element is the identity mapping id X : X X, x x, and as left inverse of a map the usual inverse of the map. (d) A vector space over the real- or complex- numbers with the addition as operation is an abelian group Some consequences of the axioms. Let G be a group. (1) If h g = e, where e is a left neutral element, then also g h = e. In fact, it follows from h g that h g h = h. Multiplying this equation on the left by an left inverse k of h gives k (h g h) = k h = e and so e = k (h g h) = (k h) g h = g h as claimed. It follows also form this that g is a left inverse of its left inverse h. (2) If e g = g then also g e = g, i.e. a left neutral element is also right neutral. In fact, let h be a left- and so by (1) also right inverse of g. Then g = e g = (g h) g = g (h g) = g e.

3 MATH 326: RINGS AND MODULES 3 (3) There is only one element e G, such that e g = g for all g G. This element is called the neutral element of G and is also denoted by 1 and called the one of G, or if G is abelian and the multiplication is written additively denoted by 0 and called the zero of G. Proof of claim: Assume that ẽ is another left neutral element. Since by (2) the left neutral element e is also right neutral this gives ẽ = ẽ e = e. (4) Every g G has a unique left inverse, called the inverse of g and denoted by g 1, or if G is abelian and the multiplication is written additively by g. In fact, if h g = e = h g then multiplying this equation on the right by a right inverse of g gives h = h Definition. A homomorphism of groups is a map α : G H between groups, such that α(g 1 g 2 ) = α(g 1 ) α(g 2 ) for all g 1, g 2 G. Denote by e G and e H the neutral elements of G and H, respectively. Then α(e G ) = e H and α(g 1 ) = (α(g)) 1. In fact, we have α(e G ) = α(e G e G ) = α(e G ) α(e G ) multiplying on the left by (α(e G )) 1 gives e H = α(e G ). Therefore we have e H = α(e G ) = α(g 1 g) = α(g 1 ) α(g) and so the uniqueness of the inverse implies α(g 1 ) = (α(g)) Examples. (a) Let V, W be a vector spaces considered as abelian groups as in 1.2 (d). Then a linear map α : V W is a homomorphism of groups. (b) Let r Z be an integer. Then Z, Z, n r n is a homomorphism of the abelian group Z into itself. (c) The positive real numbers R >0 := { r R r > 0 } with the usual multiplication are an abelian group. The exponential function exp : R R >0, t exp(t) is a homomorphism of groups from the additive group of real numbers into this group since exp(s + t) = exp(s) exp(t). We come now to the definition of a ring Definition. A ring is a non empty set R equipped with two operations called addition, and R R R, (r, s) r + s R R R, (r, s) r s called multiplication, which are subject to the following axioms: The set R with the operation + is an abelian group. The neutral element of this operation will be denoted by 0, or 0 R to indicate the ring R, and called the zero of R.

4 4 STEFAN GILLE (Associative law for the multiplication) r (s t) = (r s) t for all r, s, t R. (Existence of a neutral element for the multiplication) There exits an element in R called one and denoted by 1, or 1 R to indicate the ring R, which is 0, such that 1 r = r = r 1 for all r R. (Distributive laws) r (s + t) = r s + r t and (r + s) t = r t + s t for all r, s, t R. If moreover the elements of R satisfy the (commutative law) r s = s r for all r, s R then R is called a commutative ring. Note the following consequences of these axioms: (i) 0 r = 0 = r 0 for all r R, since 0 r = (0 + 0) r = 0 r + 0 r, and subtraction from both sides of this equation the inverse of 0 r gives then 0 = 0 r. Analogous, one shows r 0 = 0. (ii) The one is unique: If 1 is another element satisfying 1 r = r = r 1 for all r R then 1 = 1 1 = Examples. (a) The integers Z with the usual addition and multiplication are a ring. (b) The rational, real, and complex numbers with the usual addition and multiplication are rings. We denote them by Q, R, and C, respectively. (c) The n n-matrices over R or C with the usual matrix addition and multiplication are a ring. We denote this ring by M n (R) respectively by M n (C). (d) Let V be a vector space over the real numbers. Then End R (V ) the set of all R-linear maps from V into itself is ring with addition (f + g)(x) := f(x) + g(x) and multiplication the composition of maps f g. (e) Let X be set and Maps(X, R) the set of maps X R (or more general Maps(X, R) the set of maps X R, where R is a ring). This is a ring with addition (f + g)(x) = f(x) + g(x) and multiplication (f g)(x) = f(x) g(x). The neutral elements for the addition and multiplication are the constant functions x 0, respectively x 1 for all x X Definition. Let R be a ring. A subring of R is a subset S R, such that 0 and 1 are in S, and given s, t S then (a) s t = s + ( t) S, and

5 MATH 326: RINGS AND MODULES 5 (b) s t S. This implies that S is also a ring, in other words, one can define a subring of a ring R also as subset of R, which contains 0 and 1, and is a ring with respect to the addition and multiplication of R. Note that if T is a subring of S and S is a subring of R then T is also a subring of R Examples. (a) Z is a subring of Q, which in turn is a subring of R. (b) The set of real valued diagonal matrices is a subring of M n (R). (c) The set of continuous functions and the set of differentiable functions are subrings of Maps(R, R) Definition. Let R be a ring. An element r R is called a unit, or an invertible element of R if there exists an element s R, such that r s = 1 = s r. The element s with this property is unique. In fact, if s is another such element, i.e. r s = s r = 1 then s = s 1 = s (r s) = (s r) s = 1 s = s. We call this unique element an inverse of r and denote it by r 1. Lemma. Let R be a ring. The set R of all units in R is a group with the ring multiplication as multiplication. It is abelian if R is commutative. Proof. To show that R is a group we observe first that if r, s R then (r s) (s 1 r 1 ) = r (s s 1 ) r 1 = r 1 r 1 = 1 and similar (s 1 r 1 ) (r s) = 1, i.e. r s is also a unit (and so in R ) with inverse s 1 r 1. Hence the multiplication in R defines an operation on the set R : R R R, (r, s) r s. This operation is associative, has 1 as neutral element, and r 1 as inverse of r R. Hence R is a group with this multiplication, which is clearly commutative, i.e. abelian, if R is a commutative ring. The group R is called the multiplicative group of the ring R Examples. (a) The 1 and 1 are all units of Z, i.e. Z = { 1, 1}. (b) In Q, R, and C all non zero elements are units. (c) The general linear group GL n (R) of invertible matrices is the group of units in M n (R) Definition. A ring R, such that every non zero element is a unit, i.e. R = R \ {0}, is called a skew-field. If R is moreover commutative we say that R is a field. The rational-, real-, and complex numbers are fields. An example of a skew-field is the quaternion algebra over the real numbers R.

6 6 STEFAN GILLE The quaternion algebra. Let H be a 4-dimensional real vector space with basis 1, i, j, k, i.e. every z H can be written z = a 1 + b i + c j + d k with unique real numbers a, b, c, d. Define a multiplication on H by setting letting 1 acts as 1, i.e. 1 h = h 1 = h for all h H, and set i i = 1, j j = 1, and i j = j i = k. This implies i k = j, k i = j, j k = i, k j = i, and k 2 = 1. One extends this linearly to a multiplication on H as follows: (a 1 + b i + c j + d k) (e 1 + f i + g j + h k) := (ae bf cg dh) 1 + (af + be + ch dg) i +(ag bh + ce + df) j + (ah + bg cf + de) k. With the vector addition and this multiplication H is a ring with the zero vector as 0 and 1 as one. To see that it is a skew-field consider the linear map x x which is defined by 1 = 1, ī = i, jh = j, and A straightforward computation shows that kh = k. (a 1 + b i + c j + d k) (a 1 + b i + c j + d k) = a 2 + b 2 + c 2 + d 2, which is a positive real number except if a = b = c = d = 0. In particular, for all x 0. Let x 0. Then N(x) := x x 0 (N(x) 1 x) x = 1 = x (N(x) 1 x), and so every non zero x in H is a unit with inverse x 1 = N(x) 1 x. Since H is not commutative (e.g. i j j i), it is a skew-field called the quaternion algebra. Another way to define H is as follows. Let M 2 (C) be the set of complex 2 2- matrices. This is a 4-dimensional complex vector space, and by restricting the scalars to the real numbers also a real vector space (of dimension 8). Consider in M 2 (C) the matrices ( ) ( ) ( ) 1 i = 0, j =, and k = 1 1 0, 1 0 where 1 C denotes a square root of 1. Then H can be identified( with the ) dimensional R-vector space of M 2 (C) generated by the identity matrix 0 1 and the matrices i, j, and k The group ring. Let G be a group and R a ring. The group ring of G over R is the set of formal sums r g g with r g R, and all but finitely many r g = 0. The addition is defined by g G g G r g g + g G s g g := g G (r g + s g ) g,

7 MATH 326: RINGS AND MODULES 7 and the multiplication is defined by r g g g G g G s g g := g G ( h k=g r h s k ) g. This is a ring with 1 e + 0 g as one, where e is the neutral element of G, and g e 0 g as zero. g G

8 8 STEFAN GILLE 2. Commutative rings and polynomial rings Convention. In the following sections all rings are understood unless otherwise said to be commutative Notation. In the following the in a product will be often suppressed, i.e. rs instead of r s. Let R be a ring. For an integer n and r R one defines n r as follows: n = 0: 0 r := 0; n > 0: 1 r := r, 2 r := r + r, 3 r := r + r + r, and so on, i.e. inductively n r := (n 1) r + r. n < 0: n r := [( n) r]. Then (m + n) r = m r + n r and n r = (n 1 R ) r for all n, m Z. The definition of r n is for n 0 is analogous: r 0 := 1, r 1 := r, and for n 2 inductively by r n := r n 1 r. If r is a unit and so also r m for all integers m 0 as one proves by induction, r n is also defined for negative n by r n := (r n ) 1. Then r m+n = r m r n for all integers m, n 0, and if r is a unit also for all integers m, n Definition. Let R and S be rings. A homomorphism of rings, or (shorter) morphism of rings is a map α : R S, such that α(1 R ) = 1 S. α(r 1 + r 2 ) = α(r 1 ) + α(r 2 ); α(r 1 r 2 ) = α(r 1 ) α(r 2 ); and These axioms imply: (1) α(0 R ) = 0 S since α(0 R ) = α(0 R + 0 R ) = α(0 R ) + α(0 R ). (2) α( r) = α(r) since by (1) one has 0 S = α(0 R ) = α(r + ( r)) = α(r) + α( r), i.e. α( r) is the additive inverse of α(r). (3) Analogous, if r is a unit in R then α(r 1 ) = α(r) 1 since 1 S = α(1 R ) = α(r r 1 ) = α(r) α(r 1 ), i.e. α(r 1 ) is the multiplicative inverse of α(r) Example. For every ring R there is a unique homomorphism of rings π R : Z R, which is given by π R := n 1 R. It follows from the formulas in 2.1 that this is a homomorphism of rings. Since every homomorphism Z R has to map 1 onto 1 R it follows that every such homomorphism maps the integer n onto n 1 R and so π R is the unique homomorphism from Z to R Definition. The kernel, denoted Ker α, of a homomorphism of rings α : R S is the set of all r R. such that α(r) = 0 S. This set has the following properties: 0 R Ker α, if r 1, r 2 Ker α then also r 1 r 2 Ker α, and for all r Ker α and t R the product t r is also in Ker α. (Proof: If r 1, r 2 are in the kernel of α then α(r 1 r 2 ) = α(r 1 ) α(r 2 ) = 0; and if r Ker α and t R then α(t r) = α(t) α(r) = α(t) 0 S = 0 S.) In other words, Ker α is an example of an ideal Definition. Let R be a ring. An ideal of the ring R is a subset I R, such that

9 MATH 326: RINGS AND MODULES 9 0 R I; r 1 r 2 I for all r 1, r 2 I; and t r I for all t R and r I. Note that an ideal is in general not a subring of R since usually the one 1 R is not contained in the ideal. Every ring has at least two ideals, the zero ideal {0} which is denoted by (0) and the full ring are ideals. These are also called the trivial ideals. An ideal I in the ring R is called finitely generated if there are x 1,..., x n I, such that n I = R x i := { n } r i x i r1,..., r n R. The elements x 1,..., x n are called generators of I. If I is generated by one element then I is called a principal ideal, i.e. I is of the form I = R x { r x r R } for some x R. Note that for all x, x 1,..., x n in the ring R the sets R x and n R x i are ideals Lemma. Let R be a ring. (i) An element x R is a unit if and only if R x = R. (ii) The ring R is a field if and only if (0) and R are the only ideals of R. (iii) If R is an integral domain and R x = R y then there exists a unit u R, such that x = u y. Proof. (i): If x is a unit then r = (r x 1 ) x and so all r R are in R x, i.e. R = R x. On the other hand, if R x = R then there exists y R, such that y x = 1 R, i.e. x is a unit. (ii): If R is a field and I (0) an ideal in R then there is 0 x in I. This x is a unit and so by part (i) R x = R. Since R x I this implies I = R. The reverse direction follows also from part (i): Since for x 0 in R the ideal R x is not the zero ideal, and so has to be the full ring R, which by (i) implies that x is a unit. (iii): If R x = R y then y R x and x R y. Hence there are u, v R, such that y = v x and x = u y. Inserting the second into the first of these equations gives y = v (u y) = (v u) y, which is equivalent by the distributive law to (1 v u) y = 0. Since R is an integral domain it follows 1 = v u, i.e. u is a unit Example. All ideals of the integers Z are given by { Z n := r n } r Z for an integer n 0. In fact, if I is an ideal of Z, which is not the zero ideal then there exists an integer s 0 in I. Since I is an ideal also s I. Hence there exists a positive integer in I. Let n be the smallest positive integer in I. Then since I is an ideal Z n = {t n t Z} I. To show the reverse inclusion I Z n, let m 0 be an element of I. By the Euclidean algorithm one can write m = t n + r with integers r, t, such that 0 r < n. Since I is an ideal also r = m t n I. This implies r = 0 as n is the smallest positive integer in I, and therefore m = t n Z n as claimed.

10 10 STEFAN GILLE Hence all ideals of Z are principal ideals, and since Z has also no zero divisors the ring Z is an example of a principal ideal domain Definition. Let R be a ring. A zero divisor in R is a non zero element r, such that there exists an element s 0 with r s = 0. The ring R is called an integral domain if there are no zero divisors in R. If all ideals in an integral domain R are principal then R is called a principal ideal domain. Examples of integral domains are fields (note that a unit can not be a zero divisor), or the integers Z, which is in fact as seen above a principal ideal domain. An element r R is called nilpotent if there exists an integer n 1, such that r n = Lemma. Let R be a ring. Then the set of nilpotent elements in R is an ideal. Proof. Let r, s be nilpotent, say r m = 0 and s n = 0, and t R be arbitrary. Since R is commutative (t r) m = t m r m = 0 and so also t r is nilpotent. To show that also r s is nilpotent one uses the binomial formula: m+n ( ) m + n (r s) m+n = ( 1) m+n i r i s m+n i. i If 0 i m 1 the sum m + n i is bigger or equal n and so s m+n i = 0, and if i m then r i = 0. Hence (r s) m+n = 0 and so also r s is nilpotent. Therefore the set of nilpotent elements in R is an ideal. This ideal is called the prime radical of R and is denoted by R The characteristic of a ring. Let R be a ring and π R : Z R be the unique homomorphism from the integers into R, see Example 2.3. The kernel of π R is equal an ideal Z n, n 0. This integer n is called the characteristic of R, denoted char R. In other words, the characteristic of a ring R is either 0, if n 1 R 0 R for all 0 n Z, or the smallest positive integer n, such that n 1 R = 0. Lemma. If R is an integral domain then either char R = 0 or char R is a prime number. Proof. Let n := char R. If n = 0 there is nothing to prove. So assume n 0. To show that n is prime number consider a factorization n = a b with 1 a, b n. Then 0 R = n 1 R = (a b) 1 R = (a 1 R ) (b 1 R ), and so since R is an integral domain a 1 R = 0 or b 1 R = 0. But 1 a, b n and so since n is the smallest positive integer, such that n 1 R = 0, this implies a = n or b = n. Therefore n is a prime number Definition. A homomorphism of rings α : R S is called an isomorphism if it is one-to-one and onto. The inverse map α 1 : S R of an isomorphism of ring α : R S is also a homomorphism of rings: Obviously α 1 (1 S ) = 1 R as 1 R is the only element mapped to 1 S by α. To show that α 1 (s 1 + s 2 ) = α 1 (s 1 ) + α 1 (s 2 ) let r i R be

11 MATH 326: RINGS AND MODULES 11 elements, such that α(r i ) = s i and so α 1 (s i ) = r i for i = 1, 2 (these exists since α is onto). Then: α 1 (s 1 + s 2 ) = α 1 (α(r 1 ) + α(r 2 )) = α 1 (α(r 1 + r 2 )) α is a homomorphism = r 1 + r 2 α 1 α = id R = α 1 (s 1 ) + α 1 (s 2 ). The equation α 1 (s 1 s 2 ) = α 1 (s 1 ) α 1 (s 2 ) follows analogous Operations on ideals. Let R be a ring and I i, i K, ideals in R, where K is an index set. The intersection i K is also an ideal (exercise), but the union in general not. However if for all i, j K there exists k K, such that I i I j I k then J := is an ideal. In fact, 0 R is in all I i s and so also in J. If r 1, r 2 J then there exists indices i, j, such that r 1 I i and r 2 I j. Let k K, such that I i I j I k. Then r 1, r 2 I k and so also r 1 r 2 I k J since I k is an ideal. Finally, t r J for all r J and t R: The element r is in I i for some i K and so since I i is an ideal also t r I i J. Let now s j, j K, where K is an index set, elements in the ring R. Then the smallest ideal containing the set S := {s j j K} is called the ideal generated by this set. It exists, since there is at least one ideal which contains this set (the ring R), and certainly it has to be contained in all ideals which contain S, i.e. it is equal the intersection of all ideals which contain S. Equivalently, it is the set of all finite sums m r ji s ji, I i i K where r j1,..., r jm R and j 1,..., j m are indices in K. Let now I 1,..., I l be finitely many ideals in R. The set l { l } I i = I I l := a i ai I i, for all 1 i l is an ideal in R called the sum of the ideals I 1,..., I l. For instance if r 1,..., r l R then the ideal generated by these elements is the sum of the ideals R r 1,..., R r l. The product of the ideals I 1,..., I l, which is denoted by I i l I l = I 1... I l, is the ideal generated by all products l r i with r i I i for all 1 i l.

12 12 STEFAN GILLE Here are some properties of the addition and multiplication of ideals. These follow all form the analogous properties of the addition and multiplication in the ring R. To state them let I, J, and K be ideals in R, and a, b R elements. Then: I (J K) = (I J) K and I J = J I; (R a) (R b) = R (ab); I J I J (since for x I and y J the product x y = y x is in I as well as in J); and I (J + K) = I J + I K Prime ideals. Let R be a ring. An ideal P of R is called a prime ideal if P R and if a b P a or b are in P for all a, b R. Examples. (a) If R is an integral domain the (0) is a prime ideal, since a b = 0 in R if and only if a = 0 or b = 0. (b) By (a) the zero ideal is a prime ideal in Z. The other prime ideals are given by Z p, where p is a prime number. Further examples of prime ideals are the so called maximal ideals Definition. Let R be a ring. An ideal m of R is called a maximal ideal, if (1) m R, and (2) if I R is an ideal with I m then I = m. In other words, a maximal ideal is an ideal R, such that there is except for R no ideal I m with I m Lemma. Let R be a ring. Then: (i) Maximal ideals in R are prime ideals. (The converse is not true, e.g. (0) is a prime ideal in Z but not a maximal ideal.) (ii) Let P be a prime ideal of R and x R an element, such that x n P for some n 1. Then x P. Proof. (i): Let m be a maximal ideal and a, b R with a b m. If a m then m + R a is an ideal which contains m and is bigger than m. Hence R = m + R a. Then there exists r R and x m, such that 1 R = x + r a. This implies b = b 1 R = b x + r (a b) m. (ii): Assume x P. Let 1 l < n be maximal, such that x l P. Then the element x l+1 = x l x is in P, and so since P is a prime ideal x P or x l P, a contradiction. To show that every ring has at least on maximal- and so also one prime ideal Zorn s lemma is used Zorn s lemma. A partial ordering on a set K is a relation x y defined between some (not always all) elements of K, satisfying x x; x y and y x implies x = y; and x y and y z implies x z for all x, y, z K. For instance, the set of all subsets of a set T are ordered by inclusion, i.e. U V if and only if U V.

13 MATH 326: RINGS AND MODULES 13 The set K with its partial ordering is called totally ordered if x y or y x for all x, y K. For instance, the real numbers with are totally ordered. Let T be a subset of K. An upper bound for T is an element x K, such that t x for all t T. A maximal element of K is an element x K, such that x y implies y = x, i.e. there is not element bigger than x in K. Note that x a maximal element of K does not imply that y x for all y K as there may be elements y, such that neither y x nor x y. In particular a maximal element has not to be an upper bound. The partially ordered set K is called inductive partially ordered if every totally ordered subset of K has an upper bound. Zorn s lemma asserts: Every non empty inductive partially ordered set has a maximal element. This assertion is equivalent to the axiom of choice and will therefore be taken as an axiom. Note that the maximal element is not unique Theorem. Let R be a ring and I 0 R an ideal in R which is R. Then there exists a maximal ideal m I 0. In particular, since (0) is an ideal the ring R has a maximal ideal. Proof. Let K := { J } J I0 and J is an ideal R. This set is partially ordered by inclusion, non empty as I 0 K. To show that it is also inductive partially ordered let T be a totally ordered subset of K. Then J = I is an ideal of R, see 2.12, with I J for all I T. As I R for all I K I T and so also for all I T the element 1 R is not in I for all I T, and so 1 R J. In particular J R and so J is in K and an upper bound for T. It follows that K is inductively ordered, and so has a maximal element m by Zorn s lemma. This ideal m contains I and is R. Moreover it is clearly maximal since if H m is an ideal R then H is in K and has therefore to be equal m as m is maximal in K. The following result is another application of Zorn s lemma Theorem. Let R be a ring. Then R = P. P prime ideal Proof. If x is nilpotent then there exists n 0, such that x n = 0 P for all prime ideals P, and so by Lemma 2.15 x P for all prime ideals P. To show the reverse inclusion R P it is enough to show that P prime ideal given an element x R which is not nilpotent then there exists a prime ideal P, such that x P. Let x R be not nilpotent. Consider the set S = { x n n N }. Let further { L := I } I is an ideal in R with I S =.

14 14 STEFAN GILLE The set L is not empty since the ideal (0) is in L, and partially ordered by inclusion. If T L is a totally ordered subset then J := I is an upper bound for T. It is I T an ideal, see 2.12, and ( ) J S = I S = (I S) =, I T and so J is also in L. Hence L is inductive partially ordered and so by Zorn s lemma there is a maximal element P in L. This is a prime ideal. In fact, if a, b P then the ideals P + R a and P + R b are both bigger than P and so not in L. Hence there are integers m, n 1, such that x m P + R a and x n P + R b. Then x n+m (P + R a) (P + R b) = P 2 + (R a) P + P (R b) + R ab I T P + R (ab). Since P is maximal with the property that S P = this implies that P + R (ab) is strictly bigger than P and so a b P. Hence P is a prime ideal with x P Definition. Let R be a ring. (i) The Jacobson radical of the ring R is the intersection of all maximal ideals of R. As intersection of ideals this is an ideal, which is denoted by Jac(R). (ii) The ring R is called local if it has only one maximal ideal m, or equivalently if Jac(R) is a maximal ideal. There is the following characterization of local rings Lemma. A ring R is local if and only if the set of non units is an ideal (which is then the only maximal ideal of R). Proof. Let R be a local ring with maximal ideal m. If x m then R x m, and so R x R. Therefore x is not a unit in R by Lemma 2.6. It follows that m R\R. On the other hand if x is not a unit then by Lemma 2.6 again R x R and so there exists a maximal ideal J with J R x by Theorem Since m is the only maximal ideal it follows J = m and so x m. Hence if R is local m = R \ R and so in particular the set of non units is an ideal. For the other direction, the set of non units is then an ideal by assumption and R since 1 R is a unit. If I R is an ideal then R x R for all x I and so x R for all x I by Lemma 2.6. It follows I R \ R for all ideals I R. Therefore the set of non units R \ R is a maximal ideal Definition. A ring R is called noetherian if all non zero ideals of R are finitely generated. For instance principal ideal domains are noetherian (every ideal is generated by a single element), and also fields are noetherian Theorem. The following is equivalent for a ring R: (a) R is noetherian. (b) R satisfies the ascending chain condition for ideals, i.e. there is no infinite strictly ascending chain of ideals I 0 I 1 I 2 I 3... in R, or equivalently, if I 0 I 1 I 2... is an ascending chain of ideals then there is an integer k 1, such that I m = I k for all m k.

15 MATH 326: RINGS AND MODULES 15 (c) Every non empty subset of ideals in R has a maximal element. (Note that this is not a consequence of Zorn s lemma. Such a subset is partially ordered by inclusion but this partial ordering is in general not inductive.) Proof. (a) = (b): Let I 0 I 1, I 2... be an infinite chain of ideals. Let I := I i be their union. As seen in 2.12 this is an ideal and so since R is noetherian i 0 it is finitely generated, say by r 1,..., r m. Then r j I ij for some integer i j. Let k := max { i j 1 i m }. Then the generators r 1,..., r m are in I k and so I = I k. Hence the ascending chain of ideals becomes stable at k, i.e. I k = I k+1 = I k+2 =.... (b) = (c): Let K be a subset of ideals in R. If there is no maximal element one can construct inductively an infinite strictly ascending chain of ideals in K, contradicting the assumption that such a chain of ideals does not exists: Start with any ideal I 0 K; it is not maximal and so there exists a strictly bigger I 1 in K; and if I 0 I 1... I n in K are chosen, I n is also not maximal in K and so there is an ideal I n+1 K which strictly contains I n, giving the strictly ascending chain I 0 I 1... I n I n+1. Continuing in this way one gets an infinite strictly ascending chain of ideals. (c) = (a): Let I (0) be an ideal in R and K the subset of finitely generated ideals J with J I. This set is not empty. In fact, since I (0) there is 0 x I. The ideal R x lies then in K. Let J be a maximal element of K. Then J is a finitely generated ideal I. Showing I = J finishes the proof. In fact, if x I then J + R x is a finitely generated ideal contained in I, and so since J is maximal in K this implies J + R x = J. It follows x J, and therefore I = J. Replacing ascending by descending gives the notion of an artinian ring Definition. A ring R is a called artinian if it satisfies the descending chain condition for ideals, i.e. there is no infinite strictly descending chain of ideals I 0 I 1 I 2 I 3... in R, or equivalently, it J 0 J 1 J 2... is a descending chain of ideals then there is an integer k 1, such that I m = I k for all m k. As for ascending one shows that the descending chain condition for ideals in R is equivalent to the assertion that every non empty subset K of ideals in R has a minimal element, i.e. there is J K, such that if I K and I J then I = J Polynomial rings in one variable. Let R be a ring. The polynomial ring R[T ] in the variable T over R is the set of all formal finite sums, the polynomials over R, m F (T ) = r i T i, with r 0,..., r m R. This is a ring with addition m r i T i + n s i T j := j=0 max(m,n) k=0 (r k + s k ) T k, where r k = 0 if k > m and s k = 0 if k > n is understood, and multiplication m r i T i n s i T j := j=0 m+n k=0 ( k r i s k i ) T k.

16 16 STEFAN GILLE The zero of this ring is the zero polynomial 0 = T + 0 T , and the one is the polynomial T + 0 T The degree deg f(t ) of the polynomial f(t ) = m r i T i R[T ] is the biggest integer j m, such that r j 0. By definition the zero polynomial has degree 1. A polynomial of degree 0 is called constant. The set of constant polynomials are a subring of R[T ], which is isomorphic to R via the one-to-one homomorphism ι R : R R[T ], a a T 0. We identify this subring with R and denote the polynomial a T 0 also by a only. Note that deg ( f(t )+g(t ) ) = max { deg f(t ), deg g(t ) }, and if R is an integral domain deg ( f(t ) g(t ) ) = deg f(t ) + deg g(t ). The latter equation implies: Lemma. If R is an integral domain then also R[T ] is an integral domain. If f(t ) = n a i T i is a polynomial of degree n then a n is called the leading coefficient of f(t ). If the leading coefficient of f(t ) is 1 the polynomial is called monic. Notations. Polynomials of degree 1 are called linear, of degree 2 quadratic, and of degree 3 cubic The substitution principle. Let γ : R S be a homomorphism of rings and s S. Consider the following map of rings Γ s : R[T ] S, m a i T i m γ(a i )s i. A straightforward computation shows that this is a homomorphism of rings. It is the only homomorphism of rings R[T ] S, such that the composition with ι R is equal γ, and which maps T onto s S. In words, Γ s substitutes for the coefficients of a polynomial the images of them under the homomorphism γ and for the variable T the element s in S. If f = id R : R R the image Γ s (f(t )) of f(t ) is also denoted by f(s). The following lemma will be used to show that the polynomial ring over a field is a principal ideal domain Lemma. Let R be a ring and f(t ) a non constant polynomial in R[T ] whose leading coefficient is a unit in R. Then given g(t ) R[T ] there exist polynomials h(t ) and r(t ) with deg r(t ) < deg f(t ), such that g(t ) = h(t ) f(t ) + r(t ). Proof. Let f(t ) = m a i T i be a polynomial of degree m and g(t ) = one of degree n. The lemma is proven by induction on n = deg g(t ). n j=0 b j T j

17 MATH 326: RINGS AND MODULES 17 If n < m = deg f(t ) set r(t ) = g(t ) and h(t ) = 0. So let n m 0 and consider k(t ) := g(t ) (b n a 1 m T n m ) f(t ) = b n T n (a 1 m b n T n m ) a m T m + n 1 b j T j m 1 = n 1 j=0 b j T j m 1 (a 1 m b n a i )T n m+i. j=0 (a 1 m b n a i )T n m+i Hence deg k(t ) n 1 and so by induction there exists polynomials l(t ) and r(t ) with deg r(t ) < m = deg f(t ), such that Inserting this in (1) gives k(t ) = l(t ) f(t ) + r(t ). g(t ) = k(t ) + (b n a 1 m )T n m f(t ) = l(t ) f(t ) + r(t ) + (b n a 1 m )T n m f(t ) = (l(t ) + b n a 1 m T n m ) f(t ) + r(t ), and so h(t ) = l(t ) + b n a 1 m T n m together with r(t ) do the job Roots of polynomials. Let R be a ring and f(t ) = n a i T i R[T ]. An element x R is called a root or zero of the polynomial f(t ) if n 0 = f(x) = a i x i. Let x be a root of f(t ). Consider the polynomial T x of degree 1. By the lemma above there are polynomials h(t ) and r(t ) in R[T ] with deg r(t ) < deg(t x) = 1, such that f(t ) = h(t ) (T x) + r(t ). Then 0 = f(x) = h(x) (x x) + r(x) = r(x). Hence x is a root of the constant polynomial r(t ). But this implies that r(t ) is the zero polynomial and so f(t ) = h(t ) (T x) for some polynomial h(t ). Hence: Lemma. Let f(t ) be a non zero polynomial in R[T ]. Then x R is a root of f(t ) if and only if there exists h(t ) R[T ], such that f(t ) = h(t ) (T x) Theorem. Let F be a field. Then the polynomial ring in one variable F [T ] over F is a principal ideal domain. Proof. (Compare the arguments below with the proof that Z is an integral domain in Example 2.7.) As F is an integral domain also F [T ] is an integral domain, and so one has only to show that given an ideal (0) J F [T ] then there exists a polynomial 0 f(t ), such that { J = F [T ] f(t ) = h(t ) f(t ) } h(t ) F [T ]. (1)

18 18 STEFAN GILLE By the assumption J (0) there exist a polynomial of degree 0 in J. Let f(t ) 0 be a polynomial of minimal degree 0 in J. Then F [T ] f(t ) J. To show J F [T ] f(t ) let g(t ) J. By Lemma 2.26 above there exists polynomials h(t ) and r(t ) with deg r(t ) < deg f(t ), such that g(t ) = h(t ) f(t ) + r(t ). Then r(t ) = g(t ) h(t ) f(t ) is also in J. Since f(t ) has minimal positive degree in J it follows that deg r(t ) = 1, i.e. r(t ) = 0. Therefore g(t ) = h(t ) f(t ) F [T ] f(t ) Polynomial rings in several variables. The polynomial ring R[T 1,..., T n ] in the n variables T 1,..., T n over the ring R is defined inductively as R[T 1,..., T n ] = ( R[T 1,..., T n 1 ] ) [T n ]. A element in this ring is called a polynomial the n variables T 1,..., T n, denoted f(t 1,..., T n ). It can be uniquely written as d f(t 1,..., T n ) = g i (T 1,..., T n 1 ) Tn i, where g i (T 1,..., T n 1 ) R[T 1,..., T n 1 ] for all 0 i n 1. Writing the g i (T 1,..., T n 1 ) s as polynomials over R[T 1,..., T n 2 ] and so on one sees that f(t 1,..., T n ) = µ a µ T µ, where the sum is over all multi-indices µ = (µ 1,..., µ n ), the a µ s are elements of the ring R with all but finitely many zero, and n T µ := for µ = (µ 1,..., µ n ). The expression n is called a monomial in the n variables T 1,..., T n. The sum n j=1 T µj j j=1 T µj j µ i is called the degree of the monomial. With this notation the sum of two polynomials in n variables is given by a µ T µ + b µ T µ := (a µ + b µ ) T µ, µ µ µ and the product by a µ T µ b µ T µ := µ µ ϕ µ+ν=ϕ a µ b ν T ϕ, where (µ 1,..., µ n ) + (ν 1,..., ν n ) = (µ 1 + ν 1,..., µ n + ν n ). The degree of the polynomial a µ T µ is the maximum of the degrees of the monomials T µ whose coef- µ ficient a µ is not zero. A polynomial f(t 1,..., T n ) = µ a µ T µ is called homogeneous of degree m if it is sum of monomials of degree m Examples.

19 MATH 326: RINGS AND MODULES 19 (a) f(s, T ) = S 2 + ST + 3T 2 ST 7 S 17 T is a polynomial of degree 19 in the two variables S, T over Z. One can consider this as polynomial in the one variable T over Z[S]: f(x, T ) = ( S) T 7 + ( S ) T 2 + S T + (S 2 + 5). (b) The polynomial f(s, T, U) = U 3 6 ST U + ST T U 2 is a homogeneous polynomial of degree 3 over R Hilbert basis Theorem. If R is a noetherian ring then also the polynomial ring in n variables R[T 1,..., T n ] is noetherian. Proof. As R[T 1,..., T n ] = (R[T 1,..., T n 1 ])[T n ] it is by induction enough to show this for n = 1, i.e. to show that S = R[T ] is noetherian. Let J S be a non zero ideal and I i the set of leading coefficients of polynomials of degree i 1 in J together with 0. These are ideals in R. In fact, if a, b are the leading coefficients of polynomials f(t ) and g(t ), respectively, of degree i in J then either a b is 0 or the leading coefficient of f(t ) g(t ), which is also a polynomial of degree i in J. And if r R then r a is 0 or the leading coefficient of the polynomial r f(t ) of degree i, which is also in J. The ideals I i, i 1, constitute an ascending chain I 1 I 0 I 1 I In fact, if a I i is the leading coefficient of a polynomial f J of degree i then a is also the leading coefficient of the polynomial T f(t ) of degree i + 1, which is also in J as J is an ideal. Hence a I i+1 and so I i I i+1. Since R is noetherian there is k 1, such that I m = I k for all m k and also all ideals I i are finitely generated. Let a 1,i, a 2,i,..., a ni,i be generators of I i for 0 i k. Denote by f s,t (T ) J a polynomial of degree s in J with leading coefficient a rs. Claim: The polynomials f s,t (T ), 0 t k, 1 s n t, generated the ideal J, i.e. J is finitely generated. To show this, let f(t ) J be a non zero polynomial with leading coefficient a R. It is shown by induction on the degree d of f that f is in the ideal generated by the f s,t s. If d = 0 then f(t ) = at 0 is constant, and a I 0. The ideal I 0 is generated by a 1,0,..., a n0,0 and so f(t ) = n0 r i f i,0 (T ). Let now d 1. If d k then a I d = I k, which is generated by a 1,k,..., a nk,k. Hence n k a = r j,k a j,k for some r j,i R, and so j=0 n k g(t ) := f(t ) r j,k (T d k f j,k (T )) j=0

20 20 STEFAN GILLE is a polynomial of degree d 1. It is also in J since f(t ) and all f j,k (T ) are in J. Hence by induction k n 1 g(t ) = h r,i (T ) f r,i (T ) r=1 for some polynomials h r,i (T ) R[T ]. Therefore k n 1 n k f(t ) = h r,i (T ) f r,i (T ) + r j,k (T d k f j,k (T )) r=1 j=0 is in the ideal generated by the polynomials f i,j (T ). If d < k then a I d and so a = n d r j,d a j,d for some r j,d R, and so j=0 n d l(t ) := f(t ) r j,d f j,k (T ) is a polynomial of degree d 1. It is also in J since f(t ) and all f j,k (T ) are in J. Hence by induction l(t ) = k n 1 h r,i (T ) f r,i (T ) for some polynomials j=0 r=1 h r,i (T ) R[T ]. Therefore k n 1 n d f(t ) = h r,i (T ) f r,i (T ) + r j,d f j,k (T ) r=1 j=0 is in this case also in the ideal generated by the polynomials f i,j (T ).

21 MATH 326: RINGS AND MODULES Principal ideal- and unique factorization domains 3.1. Definition. Let R be a ring. One says that an element b of R divides a R if there exists c R, such that a = b c. Notation: b a. Note that b divides a if and only if R a R b. If R is an integral domain and a b as well as b a for some non zero elements a, b R then a = b u for a unit u. In fact, by the assumptions one has a v = b and b u = a for some u, v R, and therefore a (v u) = a, or equivalently a (vu 1) = 0. Hence since a is not 0 and R an integral domain this implies v u = 1, i.e. u, v R. Elements a and b in a ring R are called associates if a = u b for some unit u. Note that a is associate to b is an equivalence relation. Clearly, if a and b are associates then R a = R b. The converse holds, i.e. R a = R b implies a and b are associates, in an integral domain, see Lemma Irreducible elements. Let a be a non zero element in a ring R. A factorization of a is an expression a = b c with b, c R. Such a factorization is called trivial if b or c is a unit, otherwise it is called non trivial, or proper. A non zero element a R is called irreducible if it is not a unit and has only trivial factorizations. If R is an integral domain and a = b c is a proper factorization of a R \ {0} then the ideal R a is a proper subset of the ideals R b and R c. Therefore a non zero element a in the integral domain R is irreducible if and only if there is no principal ideal R b R which properly contains R a Lemma. Let R be a noetherian integral domain. Then every non unit a R \ {0} can be written n a = with b 1,..., b n irreducible elements in R. Proof. Assume the contrary. Let A R be the set of non units in R, which can not be written as product of irreducible elements. By assumption this set is not empty and R a R for all a A. Set b i K := { R a a A }. Since R is noetherian there exists a maximal element R a 0 in K. The element a 0 A can not be irreducible since then a 0 = a 0 would be a factorization into irreducible elements. Hence there are b, c R \R, such that a 0 = b c. But then R b and R c properly contain R a 0, and so these ideals are not in K. Therefore c, b A, and so b and c have a factorization into irreducible elements: b = d 1... d l and c = e 1... e m for some m, l 1 and irreducible elements d 1,..., d l, e 1,..., e m. But then a 0 = d 1... d l e 1... e m, a contradiction Prime elements. A prime element of the ring R is an element p 0 in R, which is not a unit, such that if p divides a product a b in R then p divides a or b. In an integral domain prime elements are irreducible: If p = a b then p divides one

22 22 STEFAN GILLE of a or b, say a = p c. Then (as above) one gets c b = 1 and so b is a unit, i.e. the factorization p = a b is trivial Lemma. Let R be a ring and p a prime element in R. If p divides a product a 1... a n 1 a n then p divides on of the factors a i. Proof. By induction on n. If n = 2 this is the definition of prime element, so let n 3. The prime element p divides (a 1... a n 1 ) a n. Hence p divides a 1... a n 1 or a n. In the latter case p divides the factor a n of the product, and in the former it divides one of the elements a 1,..., a n 1 by induction Greatest common divisor. Let R be an integral domain. An element d R is called a greatest common divisor (short: gcd) of the non zero elements r 1,..., r l R if (and only if) d divides r i for all 1 i l, and if t is another element in R which divides all r i s then t divides d Lemma. Let R be a principal ideal domain. Then d R is a greatest common divisor of the non zero r 1,..., r l if and only if R d = l R r i. In particular, if d and e are greatest common divisors of r 1,..., r l then there exists a unit u R, such that d = u e. Proof. Assume that d is a gcd of r 1,..., r l. Then d divides all r i, say r i = a i d for some a i R, i = 1,..., l. One has then: l s i r i = for all s 1,..., s l R, and so I := l ( l ) s i (a i d) = s i a i d l R r i R d. On the other hand since R is a principal ideal domain there is t R, such that I = R t. Then t divides all r i s since these are in I, and so t divides the greatest common divisor d. But this implies R d R t and so i.e. R d = I as claimed. R d R t = I R d, For the other direction, assume R d = I = l R r i. Then d divides all r i s since r i I = R d for all 1 i l, and if t R also divides all r i s then (by the arguments above) I = R d R t and so t divides d, i.e. d is a gcd of r 1,..., r l. For the last assertion, if d and e are greatest common divisors of r 1,..., r l then Z d = Z e and so by Lemma 2.6 (iii) d = u e for some unit u in R. In other words, a greatest common divisor of the elements r 1,..., r l can be also defined as a generator of the ideal generated by the r i s. This implies the following corollary.

23 MATH 326: RINGS AND MODULES Corollary. Let R be a principal ideal domain and r 1,..., r n R \ {0}. Then an element d R, which divides all r 1,..., r n, is a gcd of r 1,..., r n R if and only if there are s 1,..., s n R, such that d = n s i r i Remark. In Z the only units are ±1. Hence if e and d are greatest common divisors of some integers then e = ±d. Therefore one can force the gcd to be unique by requiring that it is positive. Another consequence of Lemma 3.7 is the fact that in a principal ideal domain irreducible elements and prime elements are the same Theorem. Let R be a principal ideal domain. Then: (i) x R is irreducible if and only if it is a prime element. (ii) The ideal R x (0) is a prime ideal if and only if x is a prime element. (iii) Every prime ideal (0) is a maximal ideal. Proof. (i): As seen above, a prime element is irreducible. For the converse, let x R be irreducible. Assume that x divides the product a b and not a. To show that then x divides b let d be the gcd of x and a. Since x is irreducible and does not divide a the gcd d has to be 1. By the corollary above there exists then r, s R, such that 1 = ra + sx.multiplying this equation by b gives b = rab + sbx and so since x ab that x divides b as claimed. (ii): The ideal R x is prime if and only if a b implies a R x or b R x. But y R x if and only if x divides y, and so x being a prime element is equivalent to R x be a prime ideal. (iii): Let R x be a prime ideal and I R x another ideal. Then x is a prime element by (ii) and I = R y for some y R since R is a principal ideal domain. Hence y divides x and so since x is irreducible either y is a unit and so I = R, or x and y are associates and so I = R y = R x. Hence R x is a maximal ideal Definition. An integral domain R is called a unique factorization domain if every non unit 0 x R can be written as a product x = p 1... p m with p 1,..., p n irreducible, and this factorization is unique, i.e. if x = q 1... q n another factorization with q 1,..., q n irreducible then m = n and there exists a one-to-one and onto map σ of {1,..., n} into itself (a permutation), such that q i and p σ(i) are associates for all 1 i n. The relation p is associate of q is an equivalence relation on the set of irreducible elements. One can therefore choose a representative for every irreducible element of a ring R, i.e. there exists a set P R of elements, such that (1) if q R is irreducible then q is associate of exactly one p P, and (2) if p 1 p 2 in P then p 1 and p 2 are not associates. Such a set P will be called a set of representatives of the irreducible elements of R. If now R is a unique factorization domain, and 0 r R is a non unit then r = l j=1 q j for some irreducible q j, 1 j l. There is then p j P,

24 24 STEFAN GILLE such that q j = u j p j for some unit u j R for all 1 j l. Then r = l l (u j p j ) = u p j, j=1 j=1 where u := l j=1 u j. Hence if one chooses in a unique factorization domain R a set P of representatives for the irreducible elements then one has a factorization r = u p np(r), p P where u R is a unit and n p (r) 0 are integers which are 0 except for finitely many p s in P. This factorization is unique, i.e. if u l j=1 p mj j = v l j=1 p nj j, (2) where {p 1,..., p l } is a set of (different) elements in P, u, v units in R, and n j, m j integers 0, then u = v and m j = n j for all 1 j l. This is proven by induction on l 0. If l = 0 then u = v and nothing is to prove, so let l 1. Assume m 1 n 1. Then by (2) one has ( 0 = p m1 1 u l j=2 p mj j p n1 m1 1 v l j=2 p nj j ), and so since R is an integral domain u l j=2 p mj j = v p n1 m1 1 l j=2 p nj j. If now n 1 m 1 > 0 then by the uniqueness of the factorization into irreducible elements in a unique factorization domain this equation would imply that p 1 is associate of l l one of the p 2,..., p l, a contraction. Hence n 1 = m 1 and u p mj j = v p nj j. The claim follows now by induction Lemma. Let R be a unique factorization domain. Then: (i) A element p R is irreducible if and only if it is prime. (ii) For all non zero elements r 1,..., r n R the gcd exists. Proof. (i): That prime elements are irreducible has been shown above. For the other direction, let p be an irreducible element which divides a product a b, say ab = p c. One can write a = p 1... p l and b = q 1... q m for some irreducible p 1,..., p l, q 1,..., q m. Factoring also c into irreducible factors: c = o 1... o n gives p o 1,... o n = p 1... p l q 1... q m. Since the factorization into irreducible elements is unique this implies that p is associate of one of the elements p 1,..., p l, q 1,..., q m and hence p divides either a or b. It follows that p is a prime element. j=2 j=2

25 MATH 326: RINGS AND MODULES 25 For (ii), since R is a unique factorization domain one can choose, as shown above, different irreducible (and so by (i) prime) elements p 1,..., p l, such that r i = u i for some integers n ij 0 and units u i in R. Set m j := min { n ij 1 i n }. Then l j=1 p nij j is a gcd of r 1,..., r n. d := l j=1 p mj j Theorem. Let R be a noetherian integral domain. Then R is a unique factorization domain if and only if every irreducible element in R is a prime element. Proof. One direction is part (i) of the lemma above. So assume that every irreducible element in the noetherian integral domain R is prime. Since R is noetherian it follows from Lemma 3.3 that every element of R can be written as a product of irreducible elements. Therefore to show that R is a unique factorization domain one has only to prove that this factorization is unique, i.e. if m n 0 x = p i = for some irreducible elements p 1,..., p m, q 1,..., q n then m = n and there exists a permutation σ of {1,..., m}, such that p i and q σ(i) are associates. This is proven by induction on m 1. If m = 1 it is clear since then x = p 1 is irreducible, so let m 2. Let x = n q j be another factorization with q j irreducible for 1 j n. j=1 As x is not irreducible n 2. By assumption the irreducible elements p 1,..., p m and q 1,..., q n are prime, and so by Lemma 3.5 the prime element p 1 divides one of the q j s. After renumbering one can assume p 1 divides q 1. Then q 1 = u p 1 for some unit u as both are prime elements. In particular, p 1 and q 1 are associated. Dividing now the equation m p i = n q i by p 1 gives the equation i=2 j=1 m n p i = u q i. By induction one gets m = n, and p i and q σ(i) are associates, where σ is a permutation of {2,..., n} Corollary. A principal ideal domain is a unique factorization domain. i=2 q j Proof. This follows from the theorem above and Theorem 3.10 (i). This implies in particular that a polynomial ring over a field is a unique factorization domains since it is a principal ideal domain by Theorem A prime element in a polynomial ring over a field is also called an irreducible polynomial.

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