Manipulation of Ideals

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1 Manipulation of Ideals 1 Radical ideals the radical ideal membership problem 2 Independence Varieties a problem in algebraic statistics 3 Zariski closure and Intersection of Ideals Zariski closure computing the generators of the intersection 4 Quotient of Ideals computing a basis for the quotient and saturation MCS 563 Lecture 33 Analytic Symbolic Computation Jan Verschelde, 7 April 2014 Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

2 Manipulation of Ideals 1 Radical ideals the radical ideal membership problem 2 Independence Varieties a problem in algebraic statistics 3 Zariski closure and Intersection of Ideals Zariski closure computing the generators of the intersection 4 Quotient of Ideals computing a basis for the quotient and saturation Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

3 polynomial ideals An ideal I generated by N polynomials f i K[x 1, x 2,...,x n ] = K[x], i = 1, 2,...,N, where K is an algebraically closed field, is defined as I = f 1, f 2,..., f N = { a 1 f 1 + a 2 f 2 + +a N f N a i K[x] }. The variables involved in the polynomial ring matter. An ideal I in K[x] is not an ideal in K[x, y] because it is not closed under multiplication with variables in y. Introducing a new variable is a useful technique, for example to solve the radical ideal membership problem. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

4 radical of an ideal For any ideal I, its radical is I = { p K[x] p k I for some k N }. One motivation for Gröbner bases is to solve the ideal membership problem, to decide whether f I. The radical ideal membership problem: f? I. For I = I, we solve the ideal membership problem via the division (or normal form) algorithm if we have a Gröbner basis for the ideal. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

5 radical ideal membership Theorem 1 (radical ideal membership) For I = f 1, f 2,..., f N an ideal in K[x] and f K[x]: f I 1 J := f 1, f 2,...,f N, 1 yf. Proof. To show the direction, we have f I: there is a power k 1: f k I. As I J: f k J. We use k to show that 1 J: 1 = 1 y k f k + y k f k = (1 yf)(1+yf + +y k 1 f k 1 )+y k }{{}}{{} f k J J J. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

6 proof continued For the direction, we have 1 J: 1 = = N p i (x, y)f i (x)+b(x, y)(1 yf(x)), y = 1/f i=1 N p i (x, 1/f)f i (x). i=1 To clear all denominators of p i (x, 1/f), use a power k of f. Denoting the polynomials with cleared denominators as P i (x): f k = N P i (x)f i (x). i=1 Thus f I. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

7 an algorithm To verify that an ideal is radical, we use the following Proposition 1 Let I be an ideal and < be any term order. If the initial monomial ideal in < (I) is square free, then I is a radical ideal. One definition of a Gröbner basis G is that in < (G) generates the initial ideal in < (I) so if all polynomials g G are square free, then I is radical. So a Gröbner basis provides an effective version of the proposition. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

8 Manipulation of Ideals 1 Radical ideals the radical ideal membership problem 2 Independence Varieties a problem in algebraic statistics 3 Zariski closure and Intersection of Ideals Zariski closure computing the generators of the intersection 4 Quotient of Ideals computing a basis for the quotient and saturation Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

9 conditional independence We consider a conditional independence statement: A B C meaning: A is independent of B given C. An independence statement translates into a set of quadratic polynomials. We illustrate this with A B C. Suppose that A, B, and C are discrete random variables with outcomes in {0, 1}. The statement for all i, j, k {0, 1} : Prob(A = i, B = j C = k) = Prob(A = i C = k) Prob(B = j C = k) says that the probability of A = i and B = j, given C = k, is the product of the probabilities that A and B separately equal i and j respectively, given C = k. This probability statement expresses A B C. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

10 deriving equations Denoting p ijk = Prob(A = i, B = j, C = k), for i, j, k {0, 1} introduces eight indeterminates. Via Prob(A = i C = k) = Prob(A = i, C = k)/prob(c = k) we remove the conditional probabilities. The statement A is independent of B given C gives rise to the prime ideal I A B C = p 000 p 011 p 001 p 010, p 100 p 111 p 101 p 110. Its variety V(I A B C ) is an independence variety. The generators of I A B C are 2-by-2 minors of [ p000 p 010 p 100 p 110 p 001 p 011 p 101 p 111 ]. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

11 Manipulation of Ideals 1 Radical ideals the radical ideal membership problem 2 Independence Varieties a problem in algebraic statistics 3 Zariski closure and Intersection of Ideals Zariski closure computing the generators of the intersection 4 Quotient of Ideals computing a basis for the quotient and saturation Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

12 Zariski closure Given an ideal I = f 1, f 2,...,f N of K[x] and we consider V = V(I) as the solution set of the system f(x) = 0, defined by the polynomials f i, i = 1, 2,...,N. In reverse, we consider as given a set S K n and consider I(S) = { f K[x] f(z) = 0 for all z S }. For a set S, we define S = V(I(S)) as the Zariski closure of S. S is the smallest variety that contains S. Proposition 2 For S K n, V(I(S)) is the smallest variety that contains S. Note that for W S: I(W) I(S) and I(S) is a radical ideal. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

13 elimination ideals Elimination ideals are natural examples: Theorem 2 Let I = f 1, f 2,...,f N be an ideal in K[x] with solution set V = V(I). Let π l : K n K n l be the projection onto the last n l components. Let I l = I K[x l+1,...,x n ] be the lth elimination ideal. Then V(I l ) is the Zariski closure of π l (V). A Gröbner basis with a lexicographical term order provides a basis for the elimination ideal. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

14 Manipulation of Ideals 1 Radical ideals the radical ideal membership problem 2 Independence Varieties a problem in algebraic statistics 3 Zariski closure and Intersection of Ideals Zariski closure computing the generators of the intersection 4 Quotient of Ideals computing a basis for the quotient and saturation Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

15 sum, product, intersection Given two ideals I and J in K[x], we define their sum, product and intersection respectively as I + J = { f + g f I and g J } I J = { f g f I and g J } I J = { f K[x] f I and f J }. We leave it as an exercise to show that I + J, I J, and I J are ideals in K[x]. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

16 examples Good examples are monomial ideals. For example, consider I = x 2 y. It is not too hard to see think about the staircase representation of monomial ideals that I = x 2 y. For J = xy 2, I J = x 2 y, xy 2 and I J = x 2 y 2. For general exponents a, b N n : x a x b = x c, where c is the least common multiple of a and b. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

17 generators for I J Given generators for I and J, we consider the problem of computing generators for I J. Theorem 3 For I and J ideals in K[x]: I J = (ti +(1 t)j) K[x]. Proof. We show the equality = via containments and. f I J implies f I and f J and thus also tf ti and (1 t)f (1 t)j. Thus we can write f as f = tf +(1 t)f tf +(1 t)j. Since I and J are ideals in K[x], we have f (ti +(1 t)j) K[x]. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

18 proof continued f (ti +(1 t)j) K[x] implies f(x) = g(x, t)+h(x, t), for g ti and h (1 t)j. We show that f J by setting t = 0. Because every element in ti is a multiple of t: g(x, 0) = 0 and therefore: f(x) = h(x, 0). Now we have to show that h(x, 0) J. Let J = j 1, j 2,...,j N, then for any h (1 t)j: h(x, t) = (1 t)(a 1 (x)j 1 (x)+a 2 (x)j 2 (x)+ +a N (x)j N (x)), for a i K[x]. As h(x, 0) = a 1 (x)j 1 (x)+a 2 (x)j 2 (x)+ +a N (x)j N (x) J, we have f J. Similarly, we can show that f I by setting t = 1. Thus f I J. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

19 an algorithm The importance of the theorem is that it leads to an algorithm to compute the generators of I J. Let I = i 1, i 2,..., i r and J = j 1, j 2,...,j s be ideals in K[x]. Then consider K = ti 1, ti 2,..., ti r,(1 t)j 1,(1 t)j 2,...,(1 t)j s K[x, t]. A Gröbner basis of K with respect to any lexicographical order for which t is greater than any variable in x will give a Gröbner basis for I J. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

20 Macaulay 2 In Macaulay 2, the command to intersect ideals isintersect. From the documentation, we copy the input statements: i1 : R = QQ[a..d]; i2 : intersect(ideal(a,b),ideal(c*d,a*b), ideal(b*d,a*c)) o3 = ideal (b*c*d, a*c*d, a*b*d, a*b*c) Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

21 solution set of I J The intersection turns into a union. Theorem 4 For I and J ideals in K[x]: V(I J) = V(I) V(J). Proof. We prove = in 2 steps: first and then. z V(I) V(J) z V(I) or z V(J) f(z), f I or f(z), f J So for f I J: f(z) = 0, thus V(I) V(J) V(I J). For, observe V(I J) = V(I) V(J). We show I J I J. For f I J: f = gh, g I and h J. Because I is an ideal, g I and h K[x] implies gh = f I. Likewise, because J is an ideal, f J, so we have f I J. As I J I J V(I J) V(I J), the follows. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

22 Manipulation of Ideals 1 Radical ideals the radical ideal membership problem 2 Independence Varieties a problem in algebraic statistics 3 Zariski closure and Intersection of Ideals Zariski closure computing the generators of the intersection 4 Quotient of Ideals computing a basis for the quotient and saturation Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

23 ideal quotient Let I and J be ideals in K[x], the ideal quotient of I and J (or colon ideal) is I : J = { f K[x] : fg I for all g J }. We leave it as an exercise that I : J is an ideal in K[x] and that I I : J. An example with monomial ideals justifies the name quotient: xz, yz : z = { f K[x, y, z] : f z xz, yz }, = { f K[x, y, z] : f z = a 1 xz + a 2 yz, a 1, a 2 K[x, y, z] }, = { f K[x, y, z] : f = a 1 x + a 2 y, a 1, a 2 K[x, y, z] }, = x, y. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

24 using generators The quotient I : J can be computed via the intersection of quotient of I with the generators of J, formally denoted as I : f 1, f 2,..., f N = N (I : f i ). To justify this formula, we need to show that I : (J + K) = I : J I : K for ideals I, J, and K. i=1 Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

25 computing I : f Theorem 4 Let I be an ideal in K[x] and f K[x]. If {g 1, g 2,...,g s } is a basis of I f, then {g 1 /f, g 2 /f,..., g s /f} is a basis of I : f. g1 Proof. We first show that any element in f, g 2 f,..., g s f I : f. We have: q g1 f, g 2 f,..., g s f belongs to fq g 1, g 2,...,g s = I f. Applying I : f = { a K[x] : ap I, for all p f }. For q to belong in I : f, we must have that qp I for all p f. As p f, p = bf, for some b K[x]. So qp = qbf and q I f I. Thus q I : f. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

26 proof continued We { still have to show that every q I : f is a polynomial combination g1 of f, g 2 f,..., g } s. f For q I : f, we have: fq I. Since fq f, we have fq I f, so fq = a 1 g 1 + a 2 g 2 + +a s g s, for a i K[x], i = 1, 2,...,s, as g 1, g 2,...,g s = I f. Because each g i f, each g i /f is a polynomial, and we have g 1 g 2 g s q = a 1 + a 2 + +a s f f f. The theorem gives an algorithm to compute I : J for J = j 1, j 2,..., j s, s using I : j 1, j 2,..., j s = (I : j i ). i=1 Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

27 Macaulay 2 In Macaulay 2, copying from the online documentation, we compute the quotient of two ideals via the colon operator (the command quotient allows the user to give options): i1 : R = QQ[a..d]; i2 : I = ideal(a^2*b-c^2,a*b^2-d^3,c^5-d); i3 : J = ideal(a^2,b^2,c^2,d^2); i4 : I:J o4 = ideal (a*b - d, a b - c, c - d) Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

28 saturation The saturation of I by p is (I : p ) = { q K[x] qp N I, for some N }. Geometrically, the components of V(I : p ) are those components of V(I) which do not lie on the hypersurface p 1 (0). Via Gröbner bases we may compute (I : p ) as follows. Let J = I +(tp 1) K[t, x], then (I : p ) = J K[x]. Gröbner bases with a lexicographical term order eliminate. The Macaulay 2 command for saturation issaturate. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

29 An Algebra-Geometry Dictionary algebra geometry radical ideal I = I V(I) variety I(V) ideal of a set solution set V sum of ideals I + J V(I) V(J) intersection of varieties I(V)+I(W) radical of sum intersection of sets V W product of ideals IJ V(I) V(J) union of varieties I(V)I(W) radical of product union of sets V W intersection of ideals I J V(I) V(J) union of varieties I(V) I(W) union of sets V W quotient of ideals I : J V(I) V(J) difference of varieties I(V) : I(W) difference of sets V W elimination I C[x k+1,...,x n ] π k (V(I)) projection of varieties Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

30 Summary + Exercises Gröbner bases provide algorithms to manipulate ideals. Exercises: 1 For ideals I and J in K[x], show that I + J, I J and I J are ideals in K[x]. 2 For I = J = x, y, show that I J I J. 3 Let I = x 1 x 4 + x 2 x 3, x 1 x 3, x 2 x 4. Compute (I : x4 2). 4 Show that V(I J) = V(I) V(J) for ideals I and J in K[x]. Analytic Symbolic Computation (MCS 563) Manipulation of Ideals L-33 7 April / 30

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