IMMERSE 2008: Assignment 4

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1 IMMERSE 2008: Assignment 4 4.) Let A be a ring and set R = A[x,...,x n ]. For each let R a = A x a...xa. Prove that is an -graded ring. a = (a,...,a ) R = a R a Proof: It is necessary to show that (a) each component of the direct sum, R a, is an additive subgroup, (b) the multiplication described in the definition of a graded ring holds, and that (c) R = a R a. (a) First we will show that R a is nonempty, closed under addition, and contains inverses. Since A is a ring, 0 A. Thus Hence R a is nonempty. ow assume that r a and s a R a, and let 0 x a 2 xa = 0 R a. r a = r x a 2 xa and s a = s x a 2 xa, where r, s A. Thus, since A is a ring, r + s A. Hence r a + s a = (r + s)(x a 2 xa ) R a. It remains to show that for all r a = r x a 2 xa R a, there exists s a R a such that r a + s a = 0. Since r R, there exists an additive inverse s R. Let s a = s x a 2 xa R a. Then r a + s a = (r + a)(x a 2 xa ) = 0 x a 2 xa = 0 Therefore R a contains additive inverses. Therefore R a is an additive subgroup for a. (b) ext we must show that R a R b R a+b Let r a = r x a 2 xa R a and s b = s x b r a s b R a R b. Then we have x b 2 r a s b = rs x a +b x a 2+b 2 2 x a +b Hence the multiplication property of graded rings holds, so a R a 2 x b R b, where r, s A, then R a+b

2 is an -graded ring. (c) Finally, we will show that A[x,..., x n ] = a R a by demonstrating set inclusions and that R a R b = 0 for a b. Let n (α i x j i ) i,j= be an element of R. Since α i A for all i and x j i R a j for some a j, we have α i x j i R a j. Therefore A[x,..., x n ] ext consider an element r in a R a. Thus a R a r = n i= A x a i i, which is an element of the ring A[x,..., x n ]. Therefore a R a A[x,..., x n ]. ow let a, b such that a b. Moreover, a i b i for i. Thus R a R b = 0. Hence R = A[x,..., x n ] = a R a is an -graded ring. 4.2) Let R be a graded ring. Prove that if I is a homogeneous ideal of R, then R/I is a homogeneous R-module. That is, show that R/I is generated by homogeneous elements and is hence graded with the inherited grading. Proof: Let R be a graded ring such that R = i=0 R i. Let I be a homogeneous ideal of R. We want to show that R = i=0 R i/r i I = i=0 R i (i.e. that R/I is graded with the inherited grading from R). To do this we must prove four things: )Each R i is a subgroup of R; 2) R = i=0 R i; 3) R i R j = {0}. 4) R m R n = R m+n Proof of () ote that I is an ideal of R a ring, so I must be an abelian group. Also, R i is a abelian subgroup of R. So R i I must also be an abelian group. So R i /Ri I is also an abelian group. Therefore each R i is an abelian subgroup of R. Proof of (2) 2

3 ( ) Let x R = R/I. Then x R so x i=0 r i for some r i R i. Then x = i=0 r i = i=0 r i. So x i=0 R i. Hence, R i=0 R i. ( ) Let x i=0 R i. Then x = i=0 r i for some r i R i. So x = i=0 r i which implies that x = i=0 r i. So x R. So x R. Hence R i=0 R i. Therefore R = i=0 R i. Proof of (3) Seeking a contradiction, assume that there exists some x R i R j. Then x R i and x R j. So x R i and x R j. So x R i Rj. But this contradicts the assumption that R is a graded ring. Hence there is no x R i R j. Therefore R i R j = {0}. Proof of (4) ote that R m R n R m+n since R is a graded ring. So let s look at an element from R m /R m I R n /R n I, s = r n r m R m /R m I R n /R n I. Then r n r m = {(r n + i n )(r m + i m ) i n I n, i m I m } = {(r n r m + i n r m + r n i m + i n i m i n I n, i m I m }. So r n r m R n+m /I n+m. So R m R n R m+n. Therefore, R/I is graded. 4.3) Let K be a field and R = K[x,...,x n, y,...,y n ] where we set deg(x i ) = (, 0) and deg(y j ) = (0, ). Let I be an ideal generated by finitely many monomials. By the previous exercise, A = R/I is a graded R-module. Prove that the monomials of degree (λ, ν) form a basis for A (λ,ν) over K. 4.4) Let S = k[x,...,x n ] be a standard graded ring and f,...,f d be homogeneous elements of S of degrees α,...,α d respectively. Prove that R = S 0 [f,...,f d ] is an -graded ring where R n = r m f m f m d d : r m S 0 and α m +... α d m d = n. m d Proof: For ease of notation, let f α, β = x α xα 2 2 xαn n y β yβn n. To begin, notice that the set of monomials in R of degree (λ, ν) are the set { } n n F = f α, β such that α i = λ, β i = v. Finally, let F = { f α, β R/I such that i= n α i = λ, i= i= } n β i = v and f α, β / I. We will show that F is a basis for A (λ,ν). To do this, we must show that F generates A (λ,ν) and is linearly independent. i= 3

4 Any element in A (λ,ν) is some h where h is in R (λ,ν). otice that F is a finite set, and that it is all the monomials from R (λ,ν) that are not sent to zero in the quotient module R/I. Thus, since F generates R (λ,ν), F spans A (λ,ν). To show that the set is linearly independent, let c i f i = 0 for a set of coefficients C i R and we will show that c i = 0 for every i. We then know that c i f i = 0, so ci f i R (λ,ν) I. Obviously, c i f i R (λ,ν). However, since it is also an element of I, a finitely generated ideal, given G = {g g m } a generating set for I, there exist b i R such that ci f i = b i g i and so ci f i b i g i = 0. However, by the way we defined the set F, the f i s we have above cannot be in I, and so f i g j for any i and j. Thus, all coefficients are 0, and particularly, c i = 0 for every i. Thus F is linearly independent and generates A (λ,ν), so it is a basis. 4.5) Let k be a field and R = k[x]. Set for all n. R n = {c(x ) n : c k} (a) Prove that R is an -graded ring. ote: This looks like a monomial ideal; however, it is not with this grading. Proof: For R to be an -graded ring, we must show that it has additive subgroups R n such that R = n R n, and R n R m R m+n. First, we will show that R n as defined is an additive subgroup of R. To show R n is an additive subgroup of R, we must show that is nonempty, closed under addition, and has inverses. Since k is a field, 0 k, so 0(x ) n = 0 R n. Where a k and b k, a(x ) n + b(x ) n = (a + b)(x ) n, and since (a+b) is in k, (a+b)(x ) n R n, thus R n is closed under addition. Again, because k is a field, each c k has an inverse c, and c(x ) n + ( c(x ) n ) = c( x) n c( x) n = 0. So everything in R n has an inverse. Thus R n is an additive subgroup of R. We next show that R = n R n, by subset inclusion. We know that n R n is a polynomial in x over elements of k, so it is clearly contained in R. We also know that c( x) 0 = c R 0, and ( x) = (x ) R, and since R 0 and (x ) R, (x ) + = x R 0 R. Given any polynomial in R, we can find a linear combination of c and x to mirror it in n R n. Thus R is contained in n R n, and by subset inclusion, R = n R n. 4

5 Finally, we will show that R n R m R m+n. Let c( x) n R n and d( x) m R m. Then c(x ) n d( x) m = c d(x ) m+n R n R m. But since c d k, c d( x) m+n R m+n. Thus R n R m R m+n. Since there exist additive subgroups R n of R such that R = n R n and R n R m R m+n, R is an -graded ring. (b) Prove that I = (x) is not an homogeneous ideal of R. ote: This looks like a monomial ideal; however, it is not with this grading. Proof: An ideal is homogeneous if and only if for every x I the homogeneous components of x are contained in I. Since (x ) = ( x) R and (x ) 0 = R 0, and (x ) + = x (x), it follows that x I, but this is not the case. Therefore (x) is not a homogeneous ideal of R. 4.6) Assuming that all units in a Z-graded domain are homogeneous, prove that if R is a Z-graded field, then R is concentrated in degree 0, meaning R 0 = R and R n = 0 for all n. 4.7) Let R be a Z-graded ring and I be an ideal of R 0. Prove that IR R 0 = I. Solution. Since R is a Z-graded ring, we have R = n Z R n and R n R m R n+m. Let x IR R 0. Then x IR and x R 0. Since x IR, x is a finite sum of elements of the form ir where i I, r R. So we start by considering elements of the form ir. Since R is Z-graded, we can write x = i(...r 2, r, r 0, r, r 2,...) where r i R i. So x =...ir 2 +ir +ir 0 +ir +ir ote that since any element of I has degree 0, ir i only has degree 0 when r i R 0. Since x R 0, x = ir 0 because ir i = 0 for i 0. Thus x I since I is an ideal of R 0. ow we consider elements of the form x = i n r n where the sum is finite. Since each i n r n I, x = i n r n I. ow let x I. Since R is a commutative ring with unity, we can write x = x IR. Since I is an ideal of R 0, I R 0, so x R 0. Therefore, x IR R 0. This proves that IR R 0 = I. 4.8) Let R be a nonnegatively graded ring and I 0 an ideal of R 0. Prove that I = I 0 R R 2 is an ideal of R. Also, show that M is a homogeneous maximal ideal of R if and only if M = m R R 2 for some maximal ideal m of R 0. 5

6 Proof: Let x = i x i, y = i y i be elements of I where x 0, y 0 are in I 0 and x i, y i are in R i for i > 0 and let r be an element of R. Since I 0 is an ideal of R 0 and R i is an abelian group we know that x 0 +y 0 is in I 0 and x i +y i is in R i for i > 0 and hence x + y = i x i + y i is an element of I. ow consider the product rx. Given that we have shown I to be closed under addition we may assume that r is homogeneous and contained in R n. If n = 0 then since I 0 is an ideal of R 0 we know rx 0 is an element of I 0 and rx i is an element of R i so rx = i rx i is in I. If n > then rx i R n+i R 0 so rx = i rx i is in I as well. Thus I is an ideal. ow suppose M is of the form above, then consider the quotient R/M = R i M i where M i is the graded component of M. If we notice that R i M i = R0 /m R /R R 2 /R 2 then R/M is a field since R i /R i = 0 and R0 /m is a field because m is maximal in R 0. We conclude that M is maximal in R. Conversely suppose that M is a homogeneous maximal ideal in R. Since I is homogeneous we can write M = R 0 M R M. But then we know R/M is a field since M is maximal and from a previous problems (4.2, 4.6) we know that R/M has an induced grading from R and hence the quotient ring is concentrated in degree zero, that is R/M = R 0 /(R 0 M). We can then conclude that the zero degree component of M, R 0 M, is maximal in R 0 and that positive degree components are just R i M = R i i > ) Let f : Z Z be the integer function defined by f(n) = n! for n > and f(n) = 0 for n 0. Show that f is not of polynomial type. 4.0) Let k be a field. Suppose the following rings have the standard grading. (a) If R = k[x, y, z], compute HF R (n) for all n 0. (b) If R = k[x, y, z, w], compute HF R (n) for all n 0. (c) If R = k[x,...,x i ], compute HF R (n) for all n 0. For each of the cases above, what is the respective Hilbert polynomial and Hilbert series? 6

7 Solution.: In order to see examples, we will pick our favorite field to work with and the following are the commands that should be entered into MacCauly 2: a)k = ZZ/5 R = ZZ/5[x, y, z] hilbertfunction(0,r) hilbertpolynomial(r) hilbertseries(r) note: The function command returns how many generators are there for that degree and you can find another degree, say 2, by replacing the 0. The hilbertseries is ( t) 3 and the hilbertpolynomial is 2 i i +. b)k = ZZ/5 R = ZZ/5[x, y, z, w] otice the only thing changing in the commands is that there are four variables now, but the commands will be the exact same! The hilbertseries is and the ( t) 4 hilbertpolynomial is 6 i3 + i i +. c)this one can t be done in MacCauly 2 since it doesn t read an infinite number of variables, but you play around and input all different sizes of variables to try to catch the pattern. The hilbertseries will be where is the number of variables ( t) i you input. 4.) Let k be a field. Suppose the following rings have the standard grading. (a) If R = k[x 3 ], compute HF R (n) for all n 0. (b) If R = k[x 3, x 5 ], compute HF R (n) for all n 0. (c) If R = k[x, y 2 ], compute HF R (n) for all n 0. For each of the cases above, what is the respective Hilbert series? Solution.: In this problem, we use Macaulay 2 to help find solutions. We have to designate k to be a particular field; we used ZZ/2. The command for computing the Hilbert function is hilbertfunction(n), where the polynomial ring is assumed, and n is the degree. (a) In this ring, since we have the standard grading, all elements will have degree which is a multiple of 3. Using Macaulay 2 to test our intuition, we find that we are correct, and { n 0 (mod 3) HF R (n) = 0 else. In computing the Hilbert series, we input hilbertseries(r) into Macaulay 2, and what we get is: HS(R) = ( t 3 ). ote that in problem 0, we computed the Hilbert polynomial to find a general formula for the Hilbert function. That only works in Macaulay 2 when the variables have degree one. 7

8 (b) To compute the Hilbert series, first input the ring: R = ZZ/2[x,x,Degrees=>3,5]. Then, compute the Hilbert series: HS = hilbertseries(r) to obtain HS(R) = ( t 3 )( t 5 ). (c) We input this ring into Macaulay 2 using the command R = ZZ/2[x,y,Degrees=>,2] to designate the degrees of the variables. We then compute the Hilbert function for several values of n, and see a pattern begin to emerge. We get the equation n HF R (n) = +, 2 where x denotes the greatest integer less than or equal to x. 4.2) Let R be a graded ring and M = i= M n a finitely generated graded R-module. Prove Ann(M) is a homogeneous ideal. Proof: Let R be a graded ring and M = i= M n a finitely generated graded R-module. Then there exists a homogeneous generating set (m, m 2,..., m l ), where m i R i. From a theorem in class it is enough to show that if x Ann(M) and we write x = t i=0 x i, where all x i R i, Ann(M) is a homogeneous ideal if x i Ann(M) for all 0 i t. ote that x must be composed of only finitely many x i R i, since M is an infinite sum and not product. ext let x Ann(M). Also note that xm i = 0 for all the m i in the homogeneous generating set of M. So let s take the first generator m 0, xm 0 = (x 0 m 0 + x m x t m 0 ) = 0, with x i R i, and m 0 M 0. If we look at the degree of each part of the sum we see deg(x i m 0 ) < deg(x i m 0 ) for all i t. Therefore x i m 0 = 0 for all 0 i t, and hence x i Ann(M) for all 0 i t. 4.3) Let H(t) = n=0 a nt n be an infinite series with nonnegative integer coefficients, and assume that H(t) = L(t), where L() 0 and L(t) = b ( t) d s t s + b s+ t s+ + + b r t r, with each b i Z, b s 0, b r 0. Prove that a n = 0 for all n < s and there exists a polynomial P(t) such that P(n) = a n for all n r. 4.4) Let R be an -graded ring that is generated in degree one. For an ideal I of R, let I denote the ideal of R generated by the homogeneous elements of I. Prove that if P is a prime ideal then P is a prime ideal. 4.5) Let R be a graded ring and 0 M k M k M 0 0 8

9 an exact sequence of graded R-modules with degree 0 maps. Prove that i ( )i HS Mi (t) = 0. IMMERSE 2008: Extras 4 4.6) Prove that all units in a Z-graded domain are homogeneous. 4.7) Suppose I is a homogeneous ideal of a Z-graded ring R. Prove that I is homogeneous. 9