CS 829 Polynomial systems: geometry and algorithms Lecture 3: Euclid, resultant and 2 2 systems Éric Schost
|
|
- Kelly Garrett
- 5 years ago
- Views:
Transcription
1 CS 829 Polynomial systems: geometry and algorithms Lecture 3: Euclid, resultant and 2 2 systems Éric Schost eschost@uwo.ca
2 Summary In this lecture, we start actual computations (as opposed to Lectures 1 and 2, where we discussed properties of univariate representations, but no any actual way to compute them). We discuss systems of 2 equations in 2 unknowns. These systems can be dealt with using resultants. Resultants can be computed using extensions of the Euclidean algorithm. The cost of resolution is O(d 2 M(d) log(d)), where M represents the cost of multiplying univariate polynomials, and d is the total degree of the input.
3 An overview of Euclid s algorithm
4 Euclid s algorithm Euclid s algorithm is used to compute GCD s in either Z of K[Y ]. By running it for random examples in K[X][Y ], we observe: that the remainders are in K(X)[Y ]; there are a lot of repetitions as denominators / common factors in numerators; once all the junk has been cleaned, we get most of the information that we need to solve the system. The resultant (and subresultants) are what is left once all the junk has been cleaned.
5 Euclid s algorithm Euclid s algorithm is used to compute GCD s in either Z of K[Y ]. By running it for random examples in K[X][Y ], we observe: that the remainders are in K(X)[Y ]; there are a lot of repetitions as denominators / common factors in numerators; once all the junk has been cleaned, we get most of the information that we need to solve the system. The resultant (and subresultants) are what is left once all the junk has been cleaned. In K[Y ], the complexity for degree-d inputs is O(d 2 ). This can be made faster, but the extension to K[X][Y ] will add a new layer of difficulties.
6 Intersection of plane curves
7 Common factors of univariate polynomials Let K be a field and let F, G be in K[Y ], with deg(f) = m, deg(g) = n. F and G have a common factor F G = B, deg(b) < m, deg(a) < n A
8 Common factors of univariate polynomials Let K be a field and let F, G be in K[X], with deg(f) = m, deg(g) = n. F and G have a common factor F G = B, deg(b) < m, deg(a) < n A AF BG = 0
9 Common factors of univariate polynomials Let K be a field and let F, G be in K[X], with deg(f) = m, deg(g) = n. F and G have a common factor F G = B, deg(b) < m, deg(a) < n A AF BG = 0 In other words, gcd(f, G) 1 if and only if the map φ : K[Y ] <n K[Y ] <m K[Y ] <m+n (A, B) AF + BG has a non-trivial kernel, where K[Y ] <l = {a a l 1 Y l 1 } K l.
10 The Sylvester matrix The Sylvester matrix represents the previous application. With F = f m Y m + + f 0, G = g n Y n + + g 0, their Sylvester matrix is Syl(F, G) = n {}}{ f ṃ. f m. f m f f 0... m {}}{ g ṇ. g n. g n g g 0... f 0 g 0
11 Resultant Definition The resultant res(f, G) of F, G K[Y ] is the determinant of their Sylvester matrix.
12 Resultant Definition The resultant res(f, G) of F, G K[Y ] is the determinant of their Sylvester matrix. Proposition res(f, G) = 0 gcd(f, G) 1.
13 Resultant Definition The resultant res(f, G) of F, G K[Y ] is the determinant of their Sylvester matrix. Proposition res(f, G) = 0 gcd(f, G) 1. We extend the definition to polynomials F, G with coefficients in a ring R. (warning: GCD s do not really make sense over a general ring).
14 Resultant Definition The resultant res(f, G) of F, G K[Y ] is the determinant of their Sylvester matrix. Proposition res(f, G) = 0 gcd(f, G) 1. We extend the definition to polynomials F, G with coefficients in a ring R. (warning: GCD s do not really make sense over a general ring). Useful proposition Let ϕ be a morphism from a ring R to a ring S: ϕ(1) = 1, ϕ(a + b) = ϕ(a) + ϕ(b), ϕ(ab) = ϕ(a)ϕ(b).
15 Resultant Definition The resultant res(f, G) of F, G K[Y ] is the determinant of their Sylvester matrix. Proposition res(f, G) = 0 gcd(f, G) 1. We extend the definition to polynomials F, G with coefficients in a ring R. (warning: GCD s do not really make sense over a general ring). Useful proposition Let ϕ be a morphism from a ring R to a ring S: ϕ(1) = 1, ϕ(a + b) = ϕ(a) + ϕ(b), ϕ(ab) = ϕ(a)ϕ(b). If ϕ(f m )ϕ(g n ) 0, ϕ(res(f, G)) = res(ϕ(f), ϕ(g)).
16 Resultant Definition The resultant res(f, G) of F, G K[Y ] is the determinant of their Sylvester matrix. Proposition res(f, G) = 0 gcd(f, G) 1. We extend the definition to polynomials F, G with coefficients in a ring R. (warning: GCD s do not really make sense over a general ring). Useful proposition Let ϕ be a morphism from a ring R to a ring S: ϕ(1) = 1, ϕ(a + b) = ϕ(a) + ϕ(b), ϕ(ab) = ϕ(a)ϕ(b). If ϕ(f m )ϕ(g n ) 0 then ϕ(res(f, G)) = res(ϕ(f), ϕ(g)). If ϕ(f m ) 0 then ϕ(res(f, G)) = ϕ(f m ) deg(g) deg(ϕ(g)) res(ϕ(f), ϕ(g)).
17 Resultant Definition The resultant res(f, G) of F, G K[Y ] is the determinant of their Sylvester matrix. Proposition res(f, G) = 0 gcd(f, G) 1. We extend the definition to polynomials F, G with coefficients in a ring R. (warning: GCD s do not really make sense over a general ring). Useful proposition Let ϕ be a morphism from a ring R to a ring S: ϕ(1) = 1, ϕ(a + b) = ϕ(a) + ϕ(b), ϕ(ab) = ϕ(a)ϕ(b). If ϕ(f m )ϕ(g n ) 0 then ϕ(res(f, G)) = res(ϕ(f), ϕ(g)). If ϕ(f m ) 0 then ϕ(res(f, G)) = ϕ(f m ) deg(g) deg(ϕ(g)) res(ϕ(f), ϕ(g)). Examples R = Z and ϕ(k) = K mod p,
18 Resultant Definition The resultant res(f, G) of F, G K[Y ] is the determinant of their Sylvester matrix. Proposition res(f, G) = 0 gcd(f, G) 1. We extend the definition to polynomials F, G with coefficients in a ring R. (warning: GCD s do not really make sense over a general ring). Useful proposition Let ϕ be a morphism from a ring R to a ring S: ϕ(1) = 1, ϕ(a + b) = ϕ(a) + ϕ(b), ϕ(ab) = ϕ(a)ϕ(b). If ϕ(f m )ϕ(g n ) 0 then ϕ(res(f, G)) = res(ϕ(f), ϕ(g)). If ϕ(f m ) 0 then ϕ(res(f, G)) = ϕ(f m ) deg(g) deg(ϕ(g)) res(ϕ(f), ϕ(g)). Examples R = Z and ϕ(k) = K mod p, R = K[X] and ϕ(p(x)) = P(x) = P mod (X x).
19 Let F, G be in K[X, Y ]: Application to the intersection of curves F = m f i (X)Y i G = i=0 n g i (X)Y i. i=0
20 Let F, G be in K[X, Y ]: Application to the intersection of curves F = m f i (X)Y i G = i=0 n g i (X)Y i. i=0 If they have no common factor in K[X, Y ], then in K [Y ], with K = K(X), we have gcd(f, G) = 1. Hence, R = res Y (F, G) K[X] is not zero.
21 Let F, G be in K[X, Y ]: Application to the intersection of curves F = m f i (X)Y i G = i=0 n g i (X)Y i. i=0 If they have no common factor in K[X, Y ], then in K [Y ], with K = K(X), we have gcd(f, G) = 1. Hence, R = res Y (F, G) K[X] is not zero. Proposition. Let x be in K. Then R(x) = 0 if and only if f m (x) = g n (x) = 0 or there exists y such that F(x, y) = G(x, y) = 0.
22 Let F, G be in K[X, Y ]: Application to the intersection of curves F = m f i (X)Y i G = i=0 n g i (X)Y i. i=0 If they have no common factor in K[X, Y ], then in K [Y ], with K = K(X), we have gcd(f, G) = 1. Hence, R = res Y (F, G) K[X] is not zero. Proposition. Let x be in K. Then R(x) = 0 if and only if f m (x) = g n (x) = 0 or there exists y such that F(x, y) = G(x, y) = 0. Proof. If f m (x) = g n (x) = 0, R(x) = 0. Suppose now that e.g. f m (x) 0. Then R(x) = f m (x) k res(f(x, Y ), G(x, Y )), so R(x) = 0 if and only if F(x, Y ) and G(x, Y ) have a common factor.
23 A degenerate example Consider F = X 2 Y + X + 1, G = XY 1 whose resultant with respect to Y is R = X(2X + 1). The root x = 0 of R does not correspond to any root of the system F(X, Y ) = G(X, Y ) = 0 (or so it seems...)
24 Let F 0 be in K[X, Y ]: Curves in generic position F = F 0 (X) + + F m (X)Y m. The curve C = V (F) is in Noether position (for the order Y > X) if F m is a constant. NO YES When not in Noether position, values of x where F m (x) = 0 correspond to points going to infinity above x (more on this later).
25 Let F 0 be in K[X, Y ]: Curves in generic position F = F 0 (X) + + F m (X)Y m. The curve C = V (F) is in Noether position (for the order Y > X) if F m is a constant. Proposition. For a generic change of variable X = X + αy, the curve C is in Noether position (but the degree in Y may change!)
26 Let F 0 be in K[X, Y ]: Curves in generic position F = F 0 (X) + + F m (X)Y m. The curve C = V (F) is in Noether position (for the order Y > X) if F m is a constant. Proposition. For a generic change of variable X = X + αy, the curve C is in Noether position (but the degree in Y may change!) Proof. Let d be the total degree of F and let H be the homogeneous part of degree d of F. Write H = h i X i Y d i.
27 Let F 0 be in K[X, Y ]: Curves in generic position F = F 0 (X) + + F m (X)Y m. The curve C = V (F) is in Noether position (for the order Y > X) if F m is a constant. Proposition. For a generic change of variable X = X + αy, the curve C is in Noether position (but the degree in Y may change!) Proof. Let d be the total degree of F and let H be the homogeneous part of degree d of F. Write H = h i X i Y d i. Then H(X + αy, Y ) = h i (X + αy ) i Y d i = Y d h i α i +
28 Let F 0 be in K[X, Y ]: Curves in generic position F = F 0 (X) + + F m (X)Y m. The curve C = V (F) is in Noether position (for the order Y > X) if F m is a constant. Proposition. For a generic change of variable X = X + αy, the curve C is in Noether position (but the degree in Y may change!) Proof. Let d be the total degree of F and let H be the homogeneous part of degree d of F. Write H = h i X i Y d i. Then H(X + αy, Y ) = h i (X + αy ) i Y d i = Y d h i α i + So it suffices that α is not a root of h i A i.
29 Let F 0 be in K[X, Y ]: Curves in generic position F = F 0 (X) + + F m (X)Y m. The curve C = V (F) is in Noether position (for the order Y > X) if F m is a constant. Proposition. For a generic change of variable X = X + αy, the curve C is in Noether position (but the degree in Y may change!) Proposition. When this is the case, the roots of the resultant of F, G are exactly the X -coordinates of the solutions.
30 Finiteness of the solution set Proposition. Let F, G be in K[X, Y ] with a common factor. Then V (F, G) is finite.
31 Finiteness of the solution set Proposition. Let F, G be in K[X, Y ] with a common factor. Then V (F, G) is finite. Proof. Suppose that the curves C = V (F) and C = V (G) are both in Noether position.
32 Finiteness of the solution set Proposition. Let F, G be in K[X, Y ] with a common factor. Then V (F, G) is finite. Proof. Suppose that the curves C = V (F) and C = V (G) are both in Noether position. Let R K[X] be the resultant of F and G with respect to Y, so that R 0. Each solution of F(x, y) = G(x, y) = 0 satisfies R(x) = 0.
33 Finiteness of the solution set Proposition. Let F, G be in K[X, Y ] with a common factor. Then V (F, G) is finite. Proof. Suppose that the curves C = V (F) and C = V (G) are both in Noether position. Let R K[X] be the resultant of F and G with respect to Y, so that R 0. Each solution of F(x, y) = G(x, y) = 0 satisfies R(x) = 0. R has a finite number of roots.
34 Finiteness of the solution set Proposition. Let F, G be in K[X, Y ] with a common factor. Then V (F, G) is finite. Proof. Suppose that the curves C = V (F) and C = V (G) are both in Noether position. Let R K[X] be the resultant of F and G with respect to Y, so that R 0. Each solution of F(x, y) = G(x, y) = 0 satisfies R(x) = 0. R has a finite number of roots. For any root x of R, there is a finite number of y such that F(x, y) = 0.
35 Curves in generic position, continued Proposition. Let F, G be in K[X, Y ] with a common factor, of total degrees at most d. For a generic choice of X = X αy, X is a separating element for V = V (F, G).
36 Curves in generic position, continued Proposition. Let F, G be in K[X, Y ] with a common factor, of total degrees at most d. For a generic choice of X = X αy, X is a separating element for V = V (F, G). Proof. Let {(x i, y i )} i N be the finite set of common solutions.
37 Curves in generic position, continued Proposition. Let F, G be in K[X, Y ] with a common factor, of total degrees at most d. For a generic choice of X = X αy, X is a separating element for V = V (F, G). Proof. Let {(x i, y i )} i N be the finite set of common solutions. Through the change of variables X = X αy, the solution set becomes {x i αy i, y i } i N.
38 Curves in generic position, continued Proposition. Let F, G be in K[X, Y ] with a common factor, of total degrees at most d. For a generic choice of X = X αy, X is a separating element for V = V (F, G). Proof. Let {(x i, y i )} i N be the finite set of common solutions. Through the change of variables X = X αy, the solution set becomes {x i αy i, y i } i N. Let (A) = i j(x i x i) A(y i y ). X is separating if and only if (α) 0.
39 Cardinality of the intersection Proposition. Let F, G be in K[X, Y ] with a common factor, of total degrees at most d. Then V (F, G) has cardinality at most d 2.
40 Cardinality of the intersection Proposition. Let F, G be in K[X, Y ] with a common factor, of total degrees at most d. Then V (F, G) has cardinality at most d 2. Proof. By a generic change of variables, we can suppose that X is a separating element for V = V (F, G) and that the curves are in Noether position.
41 Cardinality of the intersection Proposition. Let F, G be in K[X, Y ] with a common factor, of total degrees at most d. Then V (F, G) has cardinality at most d 2. Proof. By a generic change of variables, we can suppose that X is a separating element for V = V (F, G) and that the curves are in Noether position. This does not change the total degree, or the number of solutions!
42 Cardinality of the intersection Proposition. Let F, G be in K[X, Y ] with a common factor, of total degrees at most d. Then V (F, G) has cardinality at most d 2. Proof. By a generic change of variables, we can suppose that X is a separating element for V = V (F, G) and that the curves are in Noether position. This does not change the total degree, or the number of solutions! Then, then number of solutions of the system is bounded by the degree of R = res Y (F, G).
43 Cardinality of the intersection Proposition. Let F, G be in K[X, Y ] with a common factor, of total degrees at most d. Then V (F, G) has cardinality at most d 2. Proof. By a generic change of variables, we can suppose that X is a separating element for V = V (F, G) and that the curves are in Noether position. This does not change the total degree, or the number of solutions! Then, then number of solutions of the system is bounded by the degree of R = res Y (F, G). The resultant R is the determinant of a matrix of size 2d, with entries of degree at most d. So deg(r) 2d 2.
44 Cardinality of the intersection Proposition. Let F, G be in K[X, Y ] with a common factor, of total degrees at most d. Then V (F, G) has cardinality at most d 2. Proof. By a generic change of variables, we can suppose that X is a separating element for V = V (F, G) and that the curves are in Noether position. This does not change the total degree, or the number of solutions! Then, then number of solutions of the system is bounded by the degree of R = res Y (F, G). The resultant R is the determinant of a matrix of size 2d, with entries of degree at most d. So deg(r) 2d 2. Exercise. Work out the bound d 2 by looking more carefully at the matrix.
45 Some properties of the resultant
46 Properties of the resultant Theorem. Let f, u 1,...,u m and g, v 1,...,v n be new variables and F = f(x u 1 ) (X u m ), G = g(x v 1 ) (X v n ). Then R := res(f, G) = f n g m i (u i v j ). j Proof. Let F i,j = subs(u i = v j, F). Then res(f i,j, G) = 0 but also res(f i,j, G) = subs(u i = v j, R). Hence, R is divisible by all (u i v j ). Working a little bit more (counting degrees and checking the leading terms) finishes the proof.
47 Properties of the resultant Theorem. Let F, G be in K[X] with factorization in K: F = f(x u 1 ) (X u m ), G = g(x v 1 ) (X v n ). Then Proof. Specialization properties. R := res(f, G) = f n g m i (u i v j ). j Corollary res(f, GH) = res(f, G)res(F, H). Proof. True for polynomials with coefficients in a field (cf. above). What about more general cases? It is enough to prove it for polynomials with generic coefficients F = f f m X m, G = g g n X n, H = h h l X l. Because then we can use specialization properties. What we want to prove is that a big polynomial in Z[f i, g j, h k ] is identically zero. Since vanishes for all possible evaluations in e.g. Q, it is zero.
48 Properties of the resultant Theorem. Let F, G be in K[X] with factorization in K: F = f(x u 1 ) (X u m ), G = g(x v 1 ) (X v n ). Then R := res(f, G) = f n g m i (u i v j ). j Proof. Specialization properties. Corollary res(f, GH) = res(f, G)res(F, H). Corollary For F, G with coefficients in a field, R := res(f, G) = f n F(u i )=0 G(u i ).
49 Properties of the resultant Theorem. Let F, G be in K[X] with factorization in K: F = f(x u 1 ) (X u m ), G = g(x v 1 ) (X v n ). Then Proof. Specialization properties. R := res(f, G) = f n g m i (u i v j ). j Corollary res(f, GH) = res(f, G)res(F, H). Corollary For F, G with coefficients in a field, R := res(f, G) = f n F(u i )=0 G(u i ). Corollary For F, G with coefficients in a field, res(f, G) = ( 1) deg(f) deg(g) LeadCoeff(G) deg(f) deg(r) res(g, R), for R such that F = QG + R.
50 Computing resultants
51 Euclid s algorithm Input: F, G in K[Y ]. F 1 := F; F 2 := G; i := 2; while F i <> 0 F i+1 := F i 1 mod F i ; i++; return F i 1 ;
52 Euclid s algorithm Input: F, G in K[Y ]. F 1 := F; F 2 := G; i := 2; while F i <> 0 F i+1 := F i 1 mod F i ; i++; return F i 1 ; Complexity. Euclidean division in degrees n m has cost O((n m)m); With d i = deg F i, Euclid s algorithm has cost (up to big-oh) (d i 1 d i )d i d 1 (d i 1 d i ) d 2 1. i 2 i
53 Euclid s algorithm for resultant in K[Y ] Input: F, G in K[Y ]. F 1 := F; F 2 := G; i := 2; R 1 := 1; while deg(f i ) > 0 F i+1 := F i 1 mod F i ; R i := ( 1) d id i 1 LeadCoeff(F i ) d i 1 d i R i 1 ; i++; if F i <> 0 return R i 1 LeadCoeff(F i ) deg(fi 1). else return 0;
54 Euclid s algorithm for resultant in K[Y ] Correctness As long as deg(f i ) > 0, res(f, G) = R i res(f i, F i+1 ) (by induction). When deg(f i ) 0: If F i = 0, res(f i 1, F i ) = 0 Else, res(f i 1, F i ) is LeadCoeff(F i ) deg(fi 1).
55 Euclid s algorithm for resultant in K[Y ] Correctness As long as deg(f i ) > 0, res(f, G) = R i res(f i, F i+1 ) (by induction). When deg(f i ) 0: If F i = 0, res(f i 1, F i ) = 0 Else, res(f i 1, F i ) is LeadCoeff(F i ) deg(fi 1). Complexity Still O(d 2 ).
56 Resultants of bivariate polynomials Euclid s algorithm in K(Y )[X] creates huge intermediate polynomials. With inputs of total degree d: It takes some work (introducing subresultants) to prove that all intermediate degrees are in O(d 4 ). However, the output we want is a single univariate polynomial of degree d 2.
57 Resultants of bivariate polynomials Euclid s algorithm in K(Y )[X] creates huge intermediate polynomials. With inputs of total degree d: It takes some work (introducing subresultants) to prove that all intermediate degrees are in O(d 4 ). However, the output we want is a single univariate polynomial of degree d 2. Two workarounds: Understand, predict and pre-clean the common factors and denominators; Use a modular algorithm.
58 Modular algorithm, plain version Input: F, G in K[X, Y ] of total degrees d. Compute res(f(c, Y ), G(c, Y )) for d values of c (which do not cancel a leading term); Interpolate the result.
59 Modular algorithm, plain version Input: F, G in K[X, Y ] of total degrees d. Compute res(f(c, Y ), G(c, Y )) for d values of c (which do not cancel a leading term); Interpolate the result. Complexity. O(d 2 d 2 ) + O(Cost interpolation (d 2 )) O(d 4 ) (proof upcoming).
60 Interpolating polynomials Given a 0,...,a l pairwise distinct sample points and values v 0,...,v l, the unique polynomial F such that F(a i ) = v i is l i=0 v i j i (X a j) j i (a i a j )
61 Interpolating polynomials Given a 0,...,a l pairwise distinct sample points and values v 0,...,v l, the unique polynomial F such that F(a i ) = v i is To compute it: Compute M = i (X a i), Deduce all j i (X a j), Deduce all j i (a i a j ), Combine everybody, for a cost of O(l 2 ). l i=0 j i v (X a j) i j i (a i a j )
62 Switching to fast algorithms
63 Speeding things up Let M(d) denote the cost of polynomial multiplication in degree d: M(d) O(d 2 ) for a naive algorithm M(d) O(d log d) using Fast Fourier Transform (if the field has roots of 1) M(d) O(d log d log log d) using Fast Fourier Transform in general. Technically, we ask M(d + d ) M(d) + M(d ).
64 Speeding things up Let M(d) denote the cost of polynomial multiplication in degree d: M(d) O(d 2 ) for a naive algorithm M(d) O(d log d) using Fast Fourier Transform (if the field has roots of 1) M(d) O(d log d log log d) using Fast Fourier Transform in general. Technically, we ask M(d + d ) M(d) + M(d ). Using the fact that Euclidean division can be made in time O(M(d)), both parts can be made faster: Euclid s algorithm: divide-and-conquer and half-gcd techniques, O(d 2 M(d) log(d)) Interpolation using subproduct trees techniques. O(M(d 2 ) log(d))
65 FFT in a nutshell Suppose you want to evaluate F(X) C[X] at all N-roots of 1 with deg(f) < N. 1, exp 2iπ N, exp 4iπ N,...,exp 2(N 1)iπ N,
66 FFT in a nutshell Suppose you want to evaluate F(X) C[X] at all N-roots of 1 with deg(f) < N. 1, exp 2iπ N, exp 4iπ N,...,exp 2(N 1)iπ N, Write F = F even (X 2 ) + XF odd (X 2 ). Then with N = N/2. F(exp 2ikπ N ) = F even(exp 2ikπ ) + 2ikπ N N F odd(exp 2ikπ ), N
67 FFT in a nutshell Suppose you want to evaluate F(X) C[X] at all N-roots of 1 with deg(f) < N. 1, exp 2iπ N, exp 4iπ N,...,exp 2(N 1)iπ N, Write F = F even (X 2 ) + XF odd (X 2 ). Then with N = N/2. So it suffices to F(exp 2ikπ N ) = F even(exp 2ikπ ) + 2ikπ N N F odd(exp 2ikπ ), N F even at all N -roots of 1; F odd at all N -roots of 1; combine the values.
68 FFT in a nutshell Suppose you want to evaluate F(X) C[X] at all N-roots of 1 with deg(f) < N. 1, exp 2iπ N, exp 4iπ N,...,exp 2(N 1)iπ N, Write F = F even (X 2 ) + XF odd (X 2 ). Then with N = N/2. So it suffices to F(exp 2ikπ N ) = F even(exp 2ikπ ) + 2ikπ N N F odd(exp 2ikπ ), N F even at all N -roots of 1; F odd at all N -roots of 1; combine the values. The complexity satisfies T(N) 2T(N/2) + CN so T(N) O(N log N).
69 FFT in a nutshell Proposition The inverse FFT can be performed for the same cost as the direct FFT. Corollary One can multiply F(X), G(X) C[X], both of them having degree < N, in O(N log N) operations Evaluate F and G at 2N-th roots of 1 Multiply the values Do inverse-fft to interpolate the product FG. Extension to any field having roots of unity.
70 Towards a fast Euclidean algorithm We want to make Euclid s algorithm faster, reducing its complexity to O(M(d) log(d)). Warning! Computing all intermediate remainders is too slow (the total number of coefficients is d 2 ). Workaround: Work with the quotients Q i such that F i 1 = Q i F i + F i+1 Main idea: knowing Q i makes it possible to do a single step F i = 0 1 F i 1 1 Q i F i F i+1
71 Towards a fast Euclidean algorithm We want to make Euclid s algorithm faster, reducing its complexity to O(M(d) log(d)). Warning! Computing all intermediate remainders is too slow (the total number of coefficients is d 2 ). Workaround: Work with the quotients Q i such that F i 1 = Q i F i + F i+1 Main idea: knowing Q i makes it possible to do a single step F i = 0 1 F i 1 F i+1 1 Q i F i One can do several steps at once: F j 1 = F i 1 F j 1 Q j 1 Q i F i
72 Towards a fast Euclidean algorithm We want to make Euclid s algorithm faster, reducing its complexity to O(M(d) log(d)). Warning! Computing all intermediate remainders is too slow (the total number of coefficients is d 2 ). Workaround: Work with the quotients Q i such that F i 1 = Q i F i + F i+1 Main idea: knowing Q i makes it possible to do a single step F i = 0 1 F i 1 F i+1 1 Q i F i One can do several steps at once: F j 1 F j = M[i j] F i 1 F i
73 Half GCD: specifications and applications Let F, G K[Y ] with d = deg F > deg G, and let F 1 = F, F 2 = G,...,F N = 0 be the remainders met during Euclid s algorithm.
74 Half GCD: specifications and applications Let F, G K[Y ] with d = deg F > deg G, and let F 1 = F, F 2 = G,...,F N = 0 be the remainders met during Euclid s algorithm. There exists a unique l such that deg(f l 1 ) d/2 > deg(f l ).
75 Half GCD: specifications and applications Let F, G K[Y ] with d = deg F > deg G, and let F 1 = F, F 2 = G,...,F N = 0 be the remainders met during Euclid s algorithm. There exists a unique l such that deg(f l 1 ) d/2 > deg(f l ). The half-gcd algorithm compute the matrix M[2 l], so that F l 1 F l = M[2 l] F 1. F 2
76 Half GCD: specifications and applications Let F, G K[Y ] with d = deg F > deg G, and let F 1 = F, F 2 = G,...,F N = 0 be the remainders met during Euclid s algorithm. There exists a unique l such that deg(f l 1 ) d/2 > deg(f l ). The half-gcd algorithm compute the matrix M[2 l], so that F l 1 F l If F l = 0, F l 1 is the GCD, = M[2 l] F 1. F 2
77 Half GCD: specifications and applications Let F, G K[Y ] with d = deg F > deg G, and let F 1 = F, F 2 = G,...,F N = 0 be the remainders met during Euclid s algorithm. There exists a unique l such that deg(f l 1 ) d/2 > deg(f l ). The half-gcd algorithm compute the matrix M[2 l], so that F l 1 F l If F l = 0, F l 1 is the GCD, = M[2 l] F 1. Else, compute F l+1 (to be sure that all degrees are < d/2), and continue with F l, F l+1. F 2
78 Extension to resultant computation Suppose that the half-gcd algorithm outputs as well: the leading coefficients of F 1,...,F l and their degrees.
79 Extension to resultant computation Suppose that the half-gcd algorithm outputs as well: the leading coefficients of F 1,...,F l and their degrees. Then collecting the leading coefficients and degree of F l+1, we get a recursive algorithm for computing all leading coeffcients and their degrees.
80 Extension to resultant computation Suppose that the half-gcd algorithm outputs as well: the leading coefficients of F 1,...,F l and their degrees. Then collecting the leading coefficients and degree of F l+1, we get a recursive algorithm for computing all leading coeffcients and their degrees. This leads to the resultant (cf. formula in the plain version).
81 Extension to resultant computation Suppose that the half-gcd algorithm outputs as well: the leading coefficients of F 1,...,F l and their degrees. Then collecting the leading coefficients and degree of F l+1, we get a recursive algorithm for computing all leading coeffcients and their degrees. This leads to the resultant (cf. formula in the plain version). Complexity Cost resultant (d) Cost Half gcd (d)+o(m(d))+cost Euclidean division (d)+cost resultant (d/2)
82 Extension to resultant computation Suppose that the half-gcd algorithm outputs as well: the leading coefficients of F 1,...,F l and their degrees. Then collecting the leading coefficients and degree of F l+1, we get a recursive algorithm for computing all leading coeffcients and their degrees. This leads to the resultant (cf. formula in the plain version). Complexity Cost resultant (d) Cost Half gcd (d) + CM(d) + Cost resultant (d/2)
83 Extension to resultant computation Suppose that the half-gcd algorithm outputs as well: the leading coefficients of F 1,...,F l and their degrees. Then collecting the leading coefficients and degree of F l+1, we get a recursive algorithm for computing all leading coeffcients and their degrees. This leads to the resultant (cf. formula in the plain version). Complexity Cost resultant (d) Cost Half gcd (d) + CM(d) + Cost resultant (d/2) so Cost resultant (d) O(Cost Half gcd (d) + M(d))
84 The idea of half-gcd Intuition 1. In the first steps of the Euclidean algorithm, the quotients do not depend on the lower degree terms. X X = (X + 2)(X 999 X ) + R
85 The idea of half-gcd Intuition 1. In the first steps of the Euclidean algorithm, the quotients do not depend on the lower degree terms. X X = (X + 2)(X 999 X ) + R Intuition 2. The degrees of the remainders decrease by one at each step and the quotients have degree 1.
86 The idea of half-gcd Intuition 1. In the first steps of the Euclidean algorithm, the quotients do not depend on the lower degree terms. X X = (X + 2)(X 999 X ) + R Intuition 2. The degrees of the remainders decrease by one at each step and the quotients have degree 1. Hence, a transition matrix of degree l yields remainders of degree d l.
87 The idea of half-gcd Intuition 1. In the first steps of the Euclidean algorithm, the quotients do not depend on the lower degree terms. X X = (X + 2)(X 999 X ) + R Intuition 2. The degrees of the remainders decrease by one at each step and the quotients have degree 1. Hence, a transition matrix of degree l yields remainders of degree d l. Intuition 3. The half-gcd matrix of F 1, F 2 has entries of degrees d/2.
88 The idea of half-gcd Intuition 1. In the first steps of the Euclidean algorithm, the quotients do not depend on the lower degree terms. X X = (X + 2)(X 999 X ) + R Intuition 2. The degrees of the remainders decrease by one at each step and the quotients have degree 1. Hence, a transition matrix of degree l yields remainders of degree d l. Intuition 3. The half-gcd matrix of F 1, F 2 has entries of degrees d/2. Intuition 4. Let m d/2 and G i = F i div X m. Then the half-gcd matrix of G 1, G 2 has entries of degrees m/2. So applying it to F 1, F 2 gives remainders of degrees d m/2 3d/4.
89 The half-gcd (sketch) Input F 1, F 2 with d := deg(f 1 ) > deg(f 2 ) and m := d/2. M := halfgcd(f 1 div X m, F 2 div X m ); [F 0, F 1] t := M[F 1, F 2 ] t ; Q := F 0 quo F 1; F 2 := F 0 QF 1; N = halfgcd(f 1 div X m, F 2 div X m ); return N [ Q] M;
90 The half-gcd (sketch) Input F 1, F 2 with d := deg(f 1 ) > deg(f 2 ) and m := d/2. M := halfgcd(f 1 div X m, F 2 div X m ); [F 0, F 1] t := M[F 1, F 2 ] t ; Q := F 0 quo F 1; F 2 := F 0 QF 1; N = halfgcd(f 1 div X m, F 2 div X m ); return N [ Q] M; Remark. The actual algorithm is slightly more subtle (early exits, more precise choice in the degrees) Complexity. Cost halfgcd (d) 2Cost halfgcd (d/2) + CM(d) so Cost halfgcd (d) O(M(d) log(d)).
91 The half-gcd (sketch) Input F 1, F 2 with d := deg(f 1 ) > deg(f 2 ) and m := d/2. M := halfgcd(f 1 div X m, F 2 div X m ); [F 0, F 1] t := M[F 1, F 2 ] t ; Q := F 0 quo F 1; F 2 := F 0 QF 1; N = halfgcd(f 1 div X m, F 2 div X m ); return N [ Q] M; Remark. The actual algorithm is slightly more subtle (early exits, more precise choice in the degrees) Complexity. Cost halfgcd (d) 2Cost halfgcd (d/2) + CM(d) so Cost halfgcd (d) O(M(d) log(d)) and Cost resultant (d) O(M(d) log(d)).
6.S897 Algebra and Computation February 27, Lecture 6
6.S897 Algebra and Computation February 7, 01 Lecture 6 Lecturer: Madhu Sudan Scribe: Mohmammad Bavarian 1 Overview Last lecture we saw how to use FFT to multiply f, g R[x] in nearly linear time. We also
More information1. Algebra 1.5. Polynomial Rings
1. ALGEBRA 19 1. Algebra 1.5. Polynomial Rings Lemma 1.5.1 Let R and S be rings with identity element. If R > 1 and S > 1, then R S contains zero divisors. Proof. The two elements (1, 0) and (0, 1) are
More informationNotes 6: Polynomials in One Variable
Notes 6: Polynomials in One Variable Definition. Let f(x) = b 0 x n + b x n + + b n be a polynomial of degree n, so b 0 0. The leading term of f is LT (f) = b 0 x n. We begin by analyzing the long division
More informationFast Polynomial Multiplication
Fast Polynomial Multiplication Marc Moreno Maza CS 9652, October 4, 2017 Plan Primitive roots of unity The discrete Fourier transform Convolution of polynomials The fast Fourier transform Fast convolution
More informationFurther linear algebra. Chapter II. Polynomials.
Further linear algebra. Chapter II. Polynomials. Andrei Yafaev 1 Definitions. In this chapter we consider a field k. Recall that examples of felds include Q, R, C, F p where p is prime. A polynomial is
More informationOutline. MSRI-UP 2009 Coding Theory Seminar, Week 2. The definition. Link to polynomials
Outline MSRI-UP 2009 Coding Theory Seminar, Week 2 John B. Little Department of Mathematics and Computer Science College of the Holy Cross Cyclic Codes Polynomial Algebra More on cyclic codes Finite fields
More informationResultants. summary and questions. December 7, 2011
Resultants summary and questions December 7, 2011 1 An exercise about free modules Let A be a unitary commutative integral ring. Let K be the fraction field of A. Let n 1 be an integer. Set V = A n and
More information8 Appendix: Polynomial Rings
8 Appendix: Polynomial Rings Throughout we suppose, unless otherwise specified, that R is a commutative ring. 8.1 (Largely) a reminder about polynomials A polynomial in the indeterminate X with coefficients
More informationChinese Remainder Theorem
Chinese Remainder Theorem Theorem Let R be a Euclidean domain with m 1, m 2,..., m k R. If gcd(m i, m j ) = 1 for 1 i < j k then m = m 1 m 2 m k = lcm(m 1, m 2,..., m k ) and R/m = R/m 1 R/m 2 R/m k ;
More informationLecture 7: Polynomial rings
Lecture 7: Polynomial rings Rajat Mittal IIT Kanpur You have seen polynomials many a times till now. The purpose of this lecture is to give a formal treatment to constructing polynomials and the rules
More informationChapter 3. Rings. The basic commutative rings in mathematics are the integers Z, the. Examples
Chapter 3 Rings Rings are additive abelian groups with a second operation called multiplication. The connection between the two operations is provided by the distributive law. Assuming the results of Chapter
More informationbe any ring homomorphism and let s S be any element of S. Then there is a unique ring homomorphism
21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UFD. Therefore
More information2. THE EUCLIDEAN ALGORITHM More ring essentials
2. THE EUCLIDEAN ALGORITHM More ring essentials In this chapter: rings R commutative with 1. An element b R divides a R, or b is a divisor of a, or a is divisible by b, or a is a multiple of b, if there
More informationThe Sylvester Resultant
Lecture 10 The Sylvester Resultant We want to compute intersections of algebraic curves F and G. Let F and G be the vanishing sets of f(x,y) and g(x, y), respectively. Algebraically, we are interested
More informationSection III.6. Factorization in Polynomial Rings
III.6. Factorization in Polynomial Rings 1 Section III.6. Factorization in Polynomial Rings Note. We push several of the results in Section III.3 (such as divisibility, irreducibility, and unique factorization)
More informationD-MATH Algebra I HS18 Prof. Rahul Pandharipande. Solution 6. Unique Factorization Domains
D-MATH Algebra I HS18 Prof. Rahul Pandharipande Solution 6 Unique Factorization Domains 1. Let R be a UFD. Let that a, b R be coprime elements (that is, gcd(a, b) R ) and c R. Suppose that a c and b c.
More informationChapter 4. Greatest common divisors of polynomials. 4.1 Polynomial remainder sequences
Chapter 4 Greatest common divisors of polynomials 4.1 Polynomial remainder sequences If K is a field, then K[x] is a Euclidean domain, so gcd(f, g) for f, g K[x] can be computed by the Euclidean algorithm.
More information5 Keeping the Data Small: Modular Methods
5 Keeping the Data Small: Modular Methods 5.1 Modular gcd of Polynomials in Z[x] First of all we note the following important fact: Lemma 5.1 (Gauss) For any f, g Z[x] (not both zero) we have cont(fg)
More informationCS 4424 GCD, XGCD
CS 4424 GCD, XGCD eschost@uwo.ca GCD of polynomials First definition Let A and B be in k[x]. k[x] is the ring of polynomials with coefficients in k A Greatest Common Divisor of A and B is a polynomial
More informationReal Solving on Algebraic Systems of Small Dimension
Real Solving on Algebraic Systems of Small Dimension Master s Thesis Presentation Dimitrios I. Diochnos University of Athens March 8, 2007 D. I. Diochnos (Univ. of Athens, µ Q λ ) Real Solving on Bivariate
More informationRINGS: SUMMARY OF MATERIAL
RINGS: SUMMARY OF MATERIAL BRIAN OSSERMAN This is a summary of terms used and main results proved in the subject of rings, from Chapters 11-13 of Artin. Definitions not included here may be considered
More informationThe next sequence of lectures in on the topic of Arithmetic Algorithms. We shall build up to an understanding of the RSA public-key cryptosystem.
CS 70 Discrete Mathematics for CS Fall 2003 Wagner Lecture 10 The next sequence of lectures in on the topic of Arithmetic Algorithms. We shall build up to an understanding of the RSA public-key cryptosystem.
More informationMath 547, Exam 2 Information.
Math 547, Exam 2 Information. 3/19/10, LC 303B, 10:10-11:00. Exam 2 will be based on: Homework and textbook sections covered by lectures 2/3-3/5. (see http://www.math.sc.edu/ boylan/sccourses/547sp10/547.html)
More informationExact Arithmetic on a Computer
Exact Arithmetic on a Computer Symbolic Computation and Computer Algebra William J. Turner Department of Mathematics & Computer Science Wabash College Crawfordsville, IN 47933 Tuesday 21 September 2010
More informationPolynomial evaluation and interpolation on special sets of points
Polynomial evaluation and interpolation on special sets of points Alin Bostan and Éric Schost Laboratoire STIX, École polytechnique, 91128 Palaiseau, France Abstract We give complexity estimates for the
More informationChapter 2: Real solutions to univariate polynomials
Chapter 2: Real solutions to univariate polynomials Before we study the real solutions to systems of multivariate polynomials, we will review some of what is known for univariate polynomials. The strength
More informationLocal properties of plane algebraic curves
Chapter 7 Local properties of plane algebraic curves Throughout this chapter let K be an algebraically closed field of characteristic zero, and as usual let A (K) be embedded into P (K) by identifying
More informationCHAPTER 1. AFFINE ALGEBRAIC VARIETIES
CHAPTER 1. AFFINE ALGEBRAIC VARIETIES During this first part of the course, we will establish a correspondence between various geometric notions and algebraic ones. Some references for this part of the
More informationChapter 1 Divide and Conquer Polynomial Multiplication Algorithm Theory WS 2015/16 Fabian Kuhn
Chapter 1 Divide and Conquer Polynomial Multiplication Algorithm Theory WS 2015/16 Fabian Kuhn Formulation of the D&C principle Divide-and-conquer method for solving a problem instance of size n: 1. Divide
More informationThe Berlekamp algorithm
The Berlekamp algorithm John Kerl University of Arizona Department of Mathematics 29 Integration Workshop August 6, 29 Abstract Integer factorization is a Hard Problem. Some cryptosystems, such as RSA,
More informationFinite Fields. Sophie Huczynska. Semester 2, Academic Year
Finite Fields Sophie Huczynska Semester 2, Academic Year 2005-06 2 Chapter 1. Introduction Finite fields is a branch of mathematics which has come to the fore in the last 50 years due to its numerous applications,
More informationCHAPTER 10: POLYNOMIALS (DRAFT)
CHAPTER 10: POLYNOMIALS (DRAFT) LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN The material in this chapter is fairly informal. Unlike earlier chapters, no attempt is made to rigorously
More informationCDM. Finite Fields. Klaus Sutner Carnegie Mellon University. Fall 2018
CDM Finite Fields Klaus Sutner Carnegie Mellon University Fall 2018 1 Ideals The Structure theorem Where Are We? 3 We know that every finite field carries two apparently separate structures: additive and
More informationLecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman
Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 31, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Symbolic Adjunction of Roots When dealing with subfields of C it is easy to
More informationCDM. Recurrences and Fibonacci
CDM Recurrences and Fibonacci Klaus Sutner Carnegie Mellon University 20-fibonacci 2017/12/15 23:16 1 Recurrence Equations Second Order The Fibonacci Monoid Recurrence Equations 3 We can define a sequence
More informationReal Solving on Bivariate Systems with Sturm Sequences and SLV Maple TM library
Real Solving on Bivariate Systems with Sturm Sequences and SLV Maple TM library Dimitris Diochnos University of Illinois at Chicago Dept. of Mathematics, Statistics, and Computer Science September 27,
More information2. Intersection Multiplicities
2. Intersection Multiplicities 11 2. Intersection Multiplicities Let us start our study of curves by introducing the concept of intersection multiplicity, which will be central throughout these notes.
More informationBetween Sparse and Dense Arithmetic
Between Sparse and Dense Arithmetic Daniel S. Roche Computer Science Department United States Naval Academy NARC Seminar November 28, 2012 The Problem People want to compute with really big numbers and
More informationMATH 431 PART 2: POLYNOMIAL RINGS AND FACTORIZATION
MATH 431 PART 2: POLYNOMIAL RINGS AND FACTORIZATION 1. Polynomial rings (review) Definition 1. A polynomial f(x) with coefficients in a ring R is n f(x) = a i x i = a 0 + a 1 x + a 2 x 2 + + a n x n i=0
More informationCDM. Recurrences and Fibonacci. 20-fibonacci 2017/12/15 23:16. Terminology 4. Recurrence Equations 3. Solution and Asymptotics 6.
CDM Recurrences and Fibonacci 1 Recurrence Equations Klaus Sutner Carnegie Mellon University Second Order 20-fibonacci 2017/12/15 23:16 The Fibonacci Monoid Recurrence Equations 3 Terminology 4 We can
More informationFinite Fields. Sophie Huczynska (with changes by Max Neunhöffer) Semester 2, Academic Year 2012/13
Finite Fields Sophie Huczynska (with changes by Max Neunhöffer) Semester 2, Academic Year 2012/13 Contents 1 Introduction 3 1 Group theory: a brief summary............................ 3 2 Rings and fields....................................
More informationGroebner Bases and Applications
Groebner Bases and Applications Robert Hines December 16, 2014 1 Groebner Bases In this section we define Groebner Bases and discuss some of their basic properties, following the exposition in chapter
More informationφ(xy) = (xy) n = x n y n = φ(x)φ(y)
Groups 1. (Algebra Comp S03) Let A, B and C be normal subgroups of a group G with A B. If A C = B C and AC = BC then prove that A = B. Let b B. Since b = b1 BC = AC, there are a A and c C such that b =
More informationMATH 115, SUMMER 2012 LECTURE 12
MATH 115, SUMMER 2012 LECTURE 12 JAMES MCIVOR - last time - we used hensel s lemma to go from roots of polynomial equations mod p to roots mod p 2, mod p 3, etc. - from there we can use CRT to construct
More informationModular Methods for Solving Nonlinear Polynomial Systems
Modular Methods for Solving Nonlinear Polynomial Systems (Thesis format: Monograph) by Raqeeb Rasheed Graduate Program in Computer Science A thesis submitted in partial fulfillment of the requirements
More information2a 2 4ac), provided there is an element r in our
MTH 310002 Test II Review Spring 2012 Absractions versus examples The purpose of abstraction is to reduce ideas to their essentials, uncluttered by the details of a specific situation Our lectures built
More informationPolynomial Review Problems
Polynomial Review Problems 1. Find polynomial function formulas that could fit each of these graphs. Remember that you will need to determine the value of the leading coefficient. The point (0,-3) is on
More informationIdeals, congruence modulo ideal, factor rings
Ideals, congruence modulo ideal, factor rings Sergei Silvestrov Spring term 2011, Lecture 6 Contents of the lecture Homomorphisms of rings Ideals Factor rings Typeset by FoilTEX Congruence in F[x] and
More informationFactorization in Integral Domains II
Factorization in Integral Domains II 1 Statement of the main theorem Throughout these notes, unless otherwise specified, R is a UFD with field of quotients F. The main examples will be R = Z, F = Q, and
More informationg(x) = 1 1 x = 1 + x + x2 + x 3 + is not a polynomial, since it doesn t have finite degree. g(x) is an example of a power series.
6 Polynomial Rings We introduce a class of rings called the polynomial rings, describing computation, factorization and divisibility in such rings For the case where the coefficients come from an integral
More informationFactorization in Polynomial Rings
Factorization in Polynomial Rings Throughout these notes, F denotes a field. 1 Long division with remainder We begin with some basic definitions. Definition 1.1. Let f, g F [x]. We say that f divides g,
More informationRings. Chapter Homomorphisms and ideals
Chapter 2 Rings This chapter should be at least in part a review of stuff you ve seen before. Roughly it is covered in Rotman chapter 3 and sections 6.1 and 6.2. You should *know* well all the material
More informationMath Introduction to Modern Algebra
Math 343 - Introduction to Modern Algebra Notes Rings and Special Kinds of Rings Let R be a (nonempty) set. R is a ring if there are two binary operations + and such that (A) (R, +) is an abelian group.
More informationCOMPUTER ARITHMETIC. 13/05/2010 cryptography - math background pp. 1 / 162
COMPUTER ARITHMETIC 13/05/2010 cryptography - math background pp. 1 / 162 RECALL OF COMPUTER ARITHMETIC computers implement some types of arithmetic for instance, addition, subtratction, multiplication
More informationRings. Chapter Definitions and Examples
Chapter 5 Rings Nothing proves more clearly that the mind seeks truth, and nothing reflects more glory upon it, than the delight it takes, sometimes in spite of itself, in the driest and thorniest researches
More informationReview all the activities leading to Midterm 3. Review all the problems in the previous online homework sets (8+9+10).
MA109, Activity 34: Review (Sections 3.6+3.7+4.1+4.2+4.3) Date: Objective: Additional Assignments: To prepare for Midterm 3, make sure that you can solve the types of problems listed in Activities 33 and
More informationAbstract Algebra for Polynomial Operations. Maya Mohsin Ahmed
Abstract Algebra for Polynomial Operations Maya Mohsin Ahmed c Maya Mohsin Ahmed 2009 ALL RIGHTS RESERVED To my students As we express our gratitude, we must never forget that the highest appreciation
More informationFactorization in Polynomial Rings
Factorization in Polynomial Rings These notes are a summary of some of the important points on divisibility in polynomial rings from 17 and 18. PIDs Definition 1 A principal ideal domain (PID) is an integral
More information: Error Correcting Codes. November 2017 Lecture 2
03683072: Error Correcting Codes. November 2017 Lecture 2 Polynomial Codes and Cyclic Codes Amnon Ta-Shma and Dean Doron 1 Polynomial Codes Fix a finite field F q. For the purpose of constructing polynomial
More informationCommutative Algebra. Andreas Gathmann. Class Notes TU Kaiserslautern 2013/14
Commutative Algebra Andreas Gathmann Class Notes TU Kaiserslautern 2013/14 Contents 0. Introduction......................... 3 1. Ideals........................... 9 2. Prime and Maximal Ideals.....................
More informationAlgebraic structures I
MTH5100 Assignment 1-10 Algebraic structures I For handing in on various dates January March 2011 1 FUNCTIONS. Say which of the following rules successfully define functions, giving reasons. For each one
More informationNOTES ON FINITE FIELDS
NOTES ON FINITE FIELDS AARON LANDESMAN CONTENTS 1. Introduction to finite fields 2 2. Definition and constructions of fields 3 2.1. The definition of a field 3 2.2. Constructing field extensions by adjoining
More informationMath 121 Homework 3 Solutions
Math 121 Homework 3 Solutions Problem 13.4 #6. Let K 1 and K 2 be finite extensions of F in the field K, and assume that both are splitting fields over F. (a) Prove that their composite K 1 K 2 is a splitting
More informationLecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman
Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 17, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Integral Domains and Fraction Fields 0.1.1 Theorems Now what we are going
More informationDiscrete Mathematics and Probability Theory Fall 2018 Alistair Sinclair and Yun Song Note 6
CS 70 Discrete Mathematics and Probability Theory Fall 2018 Alistair Sinclair and Yun Song Note 6 1 Modular Arithmetic In several settings, such as error-correcting codes and cryptography, we sometimes
More information(Inv) Computing Invariant Factors Math 683L (Summer 2003)
(Inv) Computing Invariant Factors Math 683L (Summer 23) We have two big results (stated in (Can2) and (Can3)) concerning the behaviour of a single linear transformation T of a vector space V In particular,
More informationPolynomial Rings. i=0
Polynomial Rings 4-15-2018 If R is a ring, the ring of polynomials in x with coefficients in R is denoted R[x]. It consists of all formal sums a i x i. Here a i = 0 for all but finitely many values of
More informationSection Properties of Rational Expressions
88 Section. - Properties of Rational Expressions Recall that a rational number is any number that can be written as the ratio of two integers where the integer in the denominator cannot be. Rational Numbers:
More informationCOMMUTATIVE RINGS. Definition 3: A domain is a commutative ring R that satisfies the cancellation law for multiplication:
COMMUTATIVE RINGS Definition 1: A commutative ring R is a set with two operations, addition and multiplication, such that: (i) R is an abelian group under addition; (ii) ab = ba for all a, b R (commutative
More informationA gentle introduction to Elimination Theory. March METU. Zafeirakis Zafeirakopoulos
A gentle introduction to Elimination Theory March 2018 @ METU Zafeirakis Zafeirakopoulos Disclaimer Elimination theory is a very wide area of research. Z.Zafeirakopoulos 2 Disclaimer Elimination theory
More informationRational Functions. Elementary Functions. Algebra with mixed fractions. Algebra with mixed fractions
Rational Functions A rational function f (x) is a function which is the ratio of two polynomials, that is, Part 2, Polynomials Lecture 26a, Rational Functions f (x) = where and are polynomials Dr Ken W
More informationDiscrete Mathematics and Probability Theory Summer 2014 James Cook Note 5
CS 70 Discrete Mathematics and Probability Theory Summer 2014 James Cook Note 5 Modular Arithmetic In several settings, such as error-correcting codes and cryptography, we sometimes wish to work over a
More informationFast Computations in the Lattice of Polynomial Rational Function Fields
Fast Computations in the Lattice of Polynomial Rational Function Fields Franz Binder Abstract By Lüroth s theorem, all intermediate fields of the extension k(x) : k, k an arbitrary field, are simple. Those
More informationMAT 243 Test 2 SOLUTIONS, FORM A
MAT 24 Test 2 SOLUTIONS, FORM A 1. [1 points] Prove the following using Mathematical Induction. L 2 i = L n L n+1 + 2 where L is the Lucas sequence: L 0 = 2 L 1 = 1 L n = L n 1 + L n 2, n 2 Solution: Let
More informationHomework 7 solutions M328K by Mark Lindberg/Marie-Amelie Lawn
Homework 7 solutions M328K by Mark Lindberg/Marie-Amelie Lawn Problem 1: 4.4 # 2:x 3 + 8x 2 x 1 0 (mod 1331). a) x 3 + 8x 2 x 1 0 (mod 11). This does not break down, so trial and error gives: x = 0 : f(0)
More informationAlgorithms for Algebraic Geometry
Chapter 2 Algorithms for Algebraic Geometry Outline: 1. Gröbner basics. 39 47 9 2. Algorithmic applications of Gröbner bases. 48 56 9 3. Resultants and Bézout s Theorem. 57 69 13 4. Solving equations with
More informationAlgebraic Geometry. Contents. Diane Maclagan Notes by Florian Bouyer. Copyright (C) Bouyer 2011.
Algebraic Geometry Diane Maclagan Notes by Florian Bouyer Contents Copyright (C) Bouyer 2011. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation
More informationPolynomials. Chapter 4
Chapter 4 Polynomials In this Chapter we shall see that everything we did with integers in the last Chapter we can also do with polynomials. Fix a field F (e.g. F = Q, R, C or Z/(p) for a prime p). Notation
More informationFast algorithms for polynomials and matrices Part 2: polynomial multiplication
Fast algorithms for polynomials and matrices Part 2: polynomial multiplication by Grégoire Lecerf Computer Science Laboratory & CNRS École polytechnique 91128 Palaiseau Cedex France 1 Notation In this
More informationCALCULUS JIA-MING (FRANK) LIOU
CALCULUS JIA-MING (FRANK) LIOU Abstract. Contents. Power Series.. Polynomials and Formal Power Series.2. Radius of Convergence 2.3. Derivative and Antiderivative of Power Series 4.4. Power Series Expansion
More informationDownloaded from
Question 1: Exercise 2.1 The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case. (i) (ii) (iii) Page 1 of 24 (iv) (v) (v) Page
More informationLECTURE 5, FRIDAY
LECTURE 5, FRIDAY 20.02.04 FRANZ LEMMERMEYER Before we start with the arithmetic of elliptic curves, let us talk a little bit about multiplicities, tangents, and singular points. 1. Tangents How do we
More informationHomework 10 M 373K by Mark Lindberg (mal4549)
Homework 10 M 373K by Mark Lindberg (mal4549) 1. Artin, Chapter 11, Exercise 1.1. Prove that 7 + 3 2 and 3 + 5 are algebraic numbers. To do this, we must provide a polynomial with integer coefficients
More informationLimits at Infinity. Horizontal Asymptotes. Definition (Limits at Infinity) Horizontal Asymptotes
Limits at Infinity If a function f has a domain that is unbounded, that is, one of the endpoints of its domain is ±, we can determine the long term behavior of the function using a it at infinity. Definition
More informationTo hand in: (a) Prove that a group G is abelian (= commutative) if and only if (xy) 2 = x 2 y 2 for all x, y G.
Homework #6. Due Thursday, October 14th Reading: For this homework assignment: Sections 3.3 and 3.4 (up to page 167) Before the class next Thursday: Sections 3.5 and 3.4 (pp. 168-171). Also review the
More informationLecture 2: Gröbner Basis and SAGBI Basis
Lecture 2: Gröbner Basis and SAGBI Basis Mohammed Tessema Suppose we have a graph. Suppose we color the graph s vertices with 3 colors so that if the vertices are adjacent they are not the same colors.
More informationSelected Math 553 Homework Solutions
Selected Math 553 Homework Solutions HW6, 1. Let α and β be rational numbers, with α 1/2, and let m > 0 be an integer such that α 2 mβ 2 = 1 δ where 0 δ < 1. Set ǫ:= 1 if α 0 and 1 if α < 0. Show that
More informationMath 110 Midterm 1 Study Guide October 14, 2013
Name: For more practice exercises, do the study set problems in sections: 3.4 3.7, 4.1, and 4.2. 1. Find the domain of f, and express the solution in interval notation. (a) f(x) = x 6 D = (, ) or D = R
More information3x + 1 (mod 5) x + 2 (mod 5)
Today. Secret Sharing. Polynomials Polynomials. Secret Sharing. Share secret among n people. Secrecy: Any k 1 knows nothing. Roubustness: Any k knows secret. Efficient: minimize storage. A polynomial P(x)
More information15. Polynomial rings Definition-Lemma Let R be a ring and let x be an indeterminate.
15. Polynomial rings Definition-Lemma 15.1. Let R be a ring and let x be an indeterminate. The polynomial ring R[x] is defined to be the set of all formal sums a n x n + a n 1 x n +... a 1 x + a 0 = a
More informationLecture 4.1: Homomorphisms and isomorphisms
Lecture 4.: Homomorphisms and isomorphisms Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4, Modern Algebra M. Macauley (Clemson) Lecture
More informationwhere c R and the content of f is one. 1
9. Gauss Lemma Obviously it would be nice to have some more general methods of proving that a given polynomial is irreducible. The first is rather beautiful and due to Gauss. The basic idea is as follows.
More informationdiv(f ) = D and deg(d) = deg(f ) = d i deg(f i ) (compare this with the definitions for smooth curves). Let:
Algebraic Curves/Fall 015 Aaron Bertram 4. Projective Plane Curves are hypersurfaces in the plane CP. When nonsingular, they are Riemann surfaces, but we will also consider plane curves with singularities.
More informationMath 261 Exercise sheet 5
Math 261 Exercise sheet 5 http://staff.aub.edu.lb/~nm116/teaching/2018/math261/index.html Version: October 24, 2018 Answers are due for Wednesday 24 October, 11AM. The use of calculators is allowed. Exercise
More informationLECTURE 10, MONDAY MARCH 15, 2004
LECTURE 10, MONDAY MARCH 15, 2004 FRANZ LEMMERMEYER 1. Minimal Polynomials Let α and β be algebraic numbers, and let f and g denote their minimal polynomials. Consider the resultant R(X) of the polynomials
More informationFast reversion of power series
Fast reversion of power series Fredrik Johansson November 2011 Overview Fast power series arithmetic Fast composition and reversion (Brent and Kung, 1978) A new algorithm for reversion Implementation results
More informationNumber theory (Chapter 4)
EECS 203 Spring 2016 Lecture 10 Page 1 of 8 Number theory (Chapter 4) Review Questions: 1. Does 5 1? Does 1 5? 2. Does (129+63) mod 10 = (129 mod 10)+(63 mod 10)? 3. Does (129+63) mod 10 = ((129 mod 10)+(63
More informationMathematical Structures Combinations and Permutations
Definitions: Suppose S is a (finite) set and n, k 0 are integers The set C(S, k) of k - combinations consists of all subsets of S that have exactly k elements The set P (S, k) of k - permutations consists
More informationMath 4310 Solutions to homework 7 Due 10/27/16
Math 4310 Solutions to homework 7 Due 10/27/16 1. Find the gcd of x 3 + x 2 + x + 1 and x 5 + 2x 3 + x 2 + x + 1 in Rx. Use the Euclidean algorithm: x 5 + 2x 3 + x 2 + x + 1 = (x 3 + x 2 + x + 1)(x 2 x
More informationSYMMETRY AND SPECIALIZABILITY IN THE CONTINUED FRACTION EXPANSIONS OF SOME INFINITE PRODUCTS
SYMMETRY AND SPECIALIZABILITY IN THE CONTINUED FRACTION EXPANSIONS OF SOME INFINITE PRODUCTS J MC LAUGHLIN Abstract Let fx Z[x] Set f 0x = x and for n 1 define f nx = ff n 1x We describe several infinite
More information