Fields Rings with extra properties: 2. multiplication is commutative, 3. every non-zero element has an inverse. Here 22 / 151 1. there is a multiplicative identity that is difgerent from 0, xy = 0 ⇒ x = 0 or y = 0 . Proof: Suppose xy = 0 but x ̸ = 0. Then x − 1 exists. Thus 0 = x − 1 0 = x − 1 ( xy ) = ( x − 1 x ) y = 1 y = y
Similar thing happens any fjnite fjeld . Can also happen in infjnite fjelds. Examples of fjelds: Suggested Exercise: 4.2. 23 / 151 ‘Strange’ things still possible: Z 2 is a fjeld but 1 + 1 = 0 . 1. Q , R , C all with usual operations. 2. Z p when p is a prime. Note: Z n is not a fjeld if n = 1 or a composite number.
Intermediate Structures difgerent from 0, s.t. Note: Every fjeld is an ID (but not conversely). elements (cf prime numbers) and unique decomposition of elements like integer case. In UFD’s greatest common divisors are guaranteed to exist. 24 / 151 ▶ Integral domain: ( ID ) commutative ring with identity, xy = 0 ⇒ x = 0 or y = 0 . Consequence: If ax = ay and a ̸ = 0 then x = y . ▶ Unique factorization domain: ( UFD ) notion of irreducible
1. d is a common divisor of a , b and Note: Division is of no interest in fjelds. Conversely if d is a gcd of a , b and u is an invertible element of R then ud is also a gcd of a , b . 25 / 151 Defjnition: Let a , b ∈ R , where R is a ring. We say that a divides b , written as a | b , if and only if b = ac for some c ∈ R . Defjnition: Let a , b ∈ D , where D is an integral domain. Then ▶ d is a common divisor of a , b if d | a and d | b . ▶ d is a greatest common divisor (gcd) of a , b if 2. for all common divisors c of a , b we have c | d . Note: If a ̸ = 0 or b ̸ = 0 then necessarily d ̸ = 0. Also 0 is the only gcd of 0, 0. Fact: If d 1 , d 2 are two gcd’s of a , b then there is an invertible element u of R s.t. d 1 = ud 2 .
Canonical and Normal Representations A representation is: representations. notion of 0 and subtraction.) 26 / 151 ▶ Canonical if equality of objects is same as equality of ▶ Each object has exactly one representation. ▶ Normal if 0 has only one representation. (In a system with a ▶ This means that we can test objects for equality. a = b ⇐ ⇒ a − b = 0 ⇐ ⇒ R ( a − b ) ≡ R ( 0 ) .
Integers and Rationals Integers Use a large base B which 1. fjts into a word (usually leave a bit for carries), 2. is usually a power of 2 or 10 and is largest power (of 2 or 10) Representation: hold digits in a linked list or an array. 27 / 151 s.t. B 2 representable in host machine arithmetic.
Karatsuba’s Algorithm No improvement. But Pays ofg for integers of suffjciently many digits. Two integers of length n in base B : Leads to time 28 / 151 (adjust appropriately for n odd). Now x = aB n / 2 + b , y = cB n / 2 + d , xy = acB n + ( bc + ad ) B n / 2 + bd . bc + ad = ( a + b )( c + d ) − ac − bd . { k 1 , if n = 1; t ( n ) = 3 t ( n / 2 ) + k 2 n , if n > 1. ( k 1 , k 2 constants). Solution: t ( n ) = Θ( n log 2 3 ) , (log 2 3 ≈ 1 . 67 ) .
Fractions Defjnition: a , b integers not both 0. Greatest common divisor , Gives canonical form. Representation: any structure that can hold a pair of integers. 29 / 151 gcd( a , b ) , is largest integer d dividing both a and b . Always represent a / b as p / q with q ≥ 1 and gcd( p , q ) = 1. So can convert to an integer type if and only if q = 1.
Rational arithmetic Much better: Division same. Justifjed because Required canonical form is: 30 / 151 a a / b , c / d in canonical form. bd = ac / gcd( ac , bd ) b × c d = ac bd / gcd( ac , bd ) d 1 = gcd( a , d ) , d 2 = gcd( b , c ) . ( a / d 1 )( c / d 2 ) ( b / d 2 )( d / d 1 ) . gcd( a , b ) = gcd( c , d ) = 1 = ⇒ gcd( ac , bd ) = gcd( a , d ) gcd( b , c ) .
For addition/subtraction put: d Suggested Exercise: 4.5 Now: bd a b Compute r.h.s. in canonical form. 31 / 151 b ± c d = p q , p ′ = a gcd( b , d ) + c gcd( b , d ) q ′ = gcd( b , d ) p = p ′ / gcd( p ′ , q ′ ) , q = q ′ / gcd( p ′ , q ′ ) .
Euclid’s Algorithm for the Integers Simple properties of gcd’s: fj 32 / 151 1. gcd( a , b ) = gcd( b , a ) . 2. gcd( a , b ) = gcd( | a | , | b | ) . 3. gcd( 0 , b ) = | b | . 4. gcd( a , b ) = gcd( a − b , b ) . Simple (ineffjcient) algorithm ( a , b ≥ 0): if a = 0 then b elif a < b then gcd( b , a ) else gcd( a − b , b )
Improved version (Euclid’s Algorithm) . and where . . 33 / 151 q is quotient of a , b and r remainder . Have Assume a ≥ 0, b > 0 and put a = qb + r , 0 ≤ r < b , q ∈ Z . gcd( a , b ) = gcd( b , r ) . Algorithm: Put r 0 = a , r 1 = b : r 0 = q 1 r 1 + r 2 r 1 = q 2 r 2 + r 3 r 2 = q 3 r 3 + r 4 r s − 2 = q s − 1 r s − 1 + r s r s − 1 = q s r s + r s + 1 0 ≤ r i < r i − 1 , for 1 ≤ i ≤ s + 1 . r s + 1 = 0
Extended version Rewrite last step as so Process can be continued until where u , v are integers . Can compute u , v by ‘forwards’ Euclid’s algorithm. Suggested Exercise: 4.9 34 / 151 r s = r s − 2 − q s − 1 r s − 1 . Remainder r s − 1 can be written as r s − 1 = r s − 3 − q s − 2 r s − 2 r s = − q s − 1 r s − 3 + ( 1 + q s − 1 q s − 2 ) r s − 2 . r s = ur 0 + vr 1 Conclusion: If d = gcd( a , b ) then there are integers u , v s.t. d = ua + vb . Lemma: Z n is a fjeld if and only if n is a prime.
but x very useful. Polynomials 35 / 151 ▶ R a commutative ring with 1. ▶ x a brand new symbol—called an indeterminate over R . ▶ Polynomials in indeterminate x with coeffjcients from R : a 0 + a 1 x + a 2 x 2 + · · · + a n x n + · · · where a i ∈ R and all all but fjnitely many are 0 ▶ Could just as well write ( a 0 , a 1 , a 2 , . . . ) ▶ a i is coeffjcient of x i . ▶ a 0 is constant term . ▶ Set of all such polynomials denoted by R [ x ] .
Convenient abbreviation: b i x i instead of Sensible convention: write ifg 36 / 151 Equality: ∞ ∑ a i x i = a 0 + a 1 x + a 2 x 2 + · · · + a n x n + · · · i = 0 ∞ ∞ ∑ ∑ a i x i = i = 0 i = 0 a 0 = b 0 , a 1 = b 1 , a 2 = b 2 , . . . 2 + 5 x 3 − 3 x 5 2 + 0 x + 0 x 2 + 5 x 3 + 0 x 4 − 3 x 5 + 0 x 6 + · · ·
37 / 151 whenever both sides defjned. Turning R [ x ] into a Ring ▶ Defjne + , ∗ on polynomials in the usual way. ▶ Makes R [ x ] into commutative ring with 1. Further defjnitions: For p ∈ R [ x ] defjne: ▶ Degree , deg( p ) ; undefjned for zero polynomial. ▶ Leading coeffjcient , lc ( p ) ; undefjned for zero polynomial. ▶ Basic facts: ( ) deg( p ± q ) ≤ max deg( p ) , deg( q ) , deg( pq ) ≤ deg( p ) + deg( q ) , deg( pq ) = deg( p ) + deg( q ) , if lc ( p ) lc ( q ) ̸ = 0 ,
Polynomial Functions Given Defjne corresponding function p very difgerent objects. Consider equality of polynomials v. equality of polynomial functions. Fact: Two notions of equality coincide if R an infjnite integral domain. 38 / 151 p = a 0 + a 1 x + · · · + a n x n ˆ p : R → R , ˆ p ( α ) = a 0 + a 1 α + · · · + a n α n . Note: p , ˆ
but For R fjnite two notions very difgerent: 39 / 151 R = { r 1 , r 2 , . . . , r n } , Z ( x ) = ( x − r 1 )( x − r 2 ) · · · ( x − r n ) . Suppose R is not the zero ring (so 1 ̸ = 0). Now Z ( x ) ̸ = 0 , in R [ x ] , ˆ Z ( x ) = 0 .
Polynomials in Several Indeterminates New indeterminate y . Polynomials in y , coeffjcients are polynomials in x . 40 / 151 R [ x ] a ring. Get ring R [ x ][ y ] . Essentially same ring as R [ y ][ x ] . (N.B. used xy = yx .) Denote by R [ x , y ] . Elements look like ∞ ∑ r ij x i y j , i , j = 0 where r ij ∈ R . Distinguish between total degree , deg( p ) , degree in x , deg x ( p ) , and degree in y , deg y ( p ) .
Power products: expressions x i 1 n 41 / 151 Can do same for indeterminates x 1 , x 2 , . . . , x n . 1 · · · x i n Degree of this is i 1 + i 2 + · · · + i n . Notion of degree for polynomials in R [ x 1 , x 2 , . . . , x n ] . Coeffjcient of a power product t in a polynomial p : coefg ( t , p ) . Convention: if X = { x 1 , x 2 , . . . , x n } write R [ X ] instead of R [ x 1 , x 2 , . . . , x n ] .
Factorization and Greatest Common Divisors to express p as: Answer: Yes if you are careful about what you mean by ‘unique’. Question: Is this factorization unique? R a UFD. Then Split h , k likewise. Eventually get to 42 / 151 is non-invertible) and p has no non-invertible constant factors. Try deg( pq ) = deg( p ) + deg( q ) , for all p , q ∈ R [ x ] . Given non-zero f ∈ R [ x ] put f = ap where a is constant (either 1 or p = hk where deg( h ) < deg( p ) , deg( k ) < deg( p ) . p = p e 1 2 · · · p e s s , 1 p e 2 where each p i can’t be split up, i.e. it is irreducible . Consequence: : R a UFD ⇒ R [ x 1 , x 2 , . . . , x n ] a UFD.
maximum possible degree over all common divisors of f , g . Question: Given f , g as above with h a gcd. Suppose p is a common divisor of maximum degree how does p relate to h ? Thus p is a gcd except for possibly missing a constant factor. factor of highest degree then h is a gcd of f , g . Can make it unique by insisting it is monic . 43 / 151 Fact: If R is a UFD then gcd’s exist in R [ x ] . Note: if p | q in R [ x ] then deg( p ) ≤ deg( q ) . Fact: Assume f ̸ = 0 or g ̸ = 0. Any gcd h of f , g ∈ R [ x ] has • If p is a common factor, p | h so deg( p ) ≤ deg( h ) . Answer: By choice of p we have deg( h ) ≤ deg( p ) . By above fact deg( p ) ≤ deg( h ) , i.e., deg( p ) = deg( h ) . Since h is a gcd and p a common factor, p | h . Thus h = ap and so deg( a ) = 0, i.e. a ∈ R . Fact: Let k be a fjeld and f , g ∈ k [ x ] . Suppose h is a common Standard abuse of notation: gcd( f , g ) stands for a gcd of f , g .
Euclid’s Algorithm for Univariate Polynomials . . . 44 / 151 Suggested Exercise: Prove that q , r are unique. q is quotient , r is remainder . Assume coeffjcients are from a fjeld and g ̸ = 0. Can put f = qg + r , r = 0 or deg( r ) < deg( g ) . Algorithm: Put r 0 = f , r 1 = g : r 0 = q 1 r 1 + r 2 r 1 = q 2 r 2 + r 3 r 2 = q 3 r 3 + r 4 r s − 2 = q s − 1 r s − 1 + r s r s − 1 = q s r s + r s + 1 where r s + 1 = 0 and deg( r i ) < deg( r i − 1 ) , 1 ≤ i ≤ s . Must eventually have r i = 0 since deg( r 0 ) > deg( r 1 ) > . . . > deg( r i ) > . . . ≥ 0 .
Rational Coeffjcients Problem: Coeffjcients blow up exponentially. 45 / 151 ▶ Working with fractions ⇒ many integer gcd computations. ▶ Can slow things down. ▶ Try to use only integer arithmetic. Fact: If f , g ∈ Z [ x ] , deg( f ) > deg( g ) then can fjnd q , r ∈ Z [ x ] s.t. lc ( g ) deg( f ) − deg( g )+ 1 f = qg + r , where r = 0 or deg( r ) < deg( g ) .
Well Known Example The sequence of remainders obtained by applying the modifjed algorithm is Possible way out: Take out gcd of coeffjcients at each Above board method: Sub-resultant polynomial remainder sequences. OK but compicated. 46 / 151 f = x 8 + x 6 − 3 x 4 − 3 x 3 + 8 x 2 + 2 x − 5 , g = 3 x 6 + 5 x 4 − 4 x 2 − 9 x + 21 . − 15 x 4 + 3 x 2 − 9 , 15795 x 2 + 30375 x − 59535 , 1254542875143750 x − 1654608338437500 , 12593338795500743100931141992187500 . stage—errm . . .
Extended Euclidean Algorithm for Polynomials Just like integer case get polys u , v s.t. Moreover can ensure: or or 47 / 151 uf + vg = gcd( f , g ) . u = 0 deg( u ) < deg( g ) v = 0 deg( v ) < deg( f )
Rational Expressions Gives us a fjeld. Caution: Again distinguish between functions and elements of 48 / 151 ▶ k a fjeld. k ( x 1 , . . . , x n ) = { p / q | p , q ∈ k [ x 1 , . . . , x n ] & q ̸ = 0 } . ▶ Equality: p / q = p ′ / q ′ ⇔ pq ′ − p ′ q = 0 , in k [ x 1 , . . . , x n ] . ▶ Defjne + , ∗ by: ( p / q ) + ( p ′ / q ′ ) = ( pq ′ + p ′ q ) / qq ′ , ( p / q )( p ′ / q ′ ) = pp ′ / qq ′ . k ( x 1 , . . . , x n ) .
Representation of Polynomials and Rational Expressions Basic types: Dense Sparse Recursive Distributed 49 / 151
Recursive Representation An expression of the isomorphism Example: represented as y is main indeterminate. Generally: Use c i x i n 50 / 151 R [ x 1 , . . . , x n ] ∼ = R [ x 1 , . . . , x n − 1 ][ x n ] . Regard x n as the main indeterminate. 3 xy 2 + 2 y 2 − 4 x 2 y + y − 1 ( 3 x + 2 ) y 2 + ( − 4 x 2 + 1 ) y + ( − 1 ) y 0 , ∑ each c i a polynomial represented similarly.
Distributive Representation 1 x 0 c t t t Can now write Consider power products in given indeterminates e.g. n ) is least, 1 x 3 x 7 Pick a total order on power products s.t. x 2 51 / 151 5 . 2 · · · x 0 ▶ 1 (i.e. x 0 ▶ each power product has only fjnitely many others less than it. ∑ p ( x 1 , . . . , x n ) = t ≤ ¯ where c t ∈ R for each t .
Example suitable ordering: Total degree then lexicographic. 1. sort according to degree, 2. within each degree use lexicographic ordering: order indeterminates, e.g. then x i 1 n 52 / 151 x 1 > L x 2 > L · · · > L x n 1 · · · x i n n > L x j 1 1 · · · x j n if and only if there is a k such that i l = j l for 1 ≤ l < k and i k > j k .
Dense Representations Example: Distributed representation Problem: Can lead to a great deal of wasted space, t Record all coeffjcients up to highest degree main indeterminate or 53 / 151 Example: Recursive representation m highest power product. ∑ c i x i ← → ( c 0 , . . . , c m ) . i = 0 ∑ c t t ← → ( c 1 , c t 1 , . . . , c ¯ t ) , t ≤ ¯ where ( . . . ) denotes a list or array. Consider x 1000 + 1 or x 4 y 7 + x + 1.
Sparse Representations power product. Example: In second example x e 1 n represented by 54 / 151 ▶ Drop all zero coeffjcients. ▶ With each non-zero coeffjcient record corresponding degree or x 1000 + 1 ← → (( 1 , 1000 ) , ( 1 , 0 )) , x 4 y 7 + 2 x + 1 ← → (( 1 , ( 4 , 7 )) , ( 2 , ( 1 , 0 )) , ( 1 , ( 0 , 0 ))) . 1 · · · x e n ( e 1 , . . . , e n ) .
Rational Expressions Consider: Maple does not. can lead to problems. 55 / 151 ▶ Pair of polynomials ⟨ f,g ⟩ ▶ Numerator in normal form ⇒ ⟨ f,g ⟩ in normal form. ▶ Dangerous temptation: Remove gcd( f , g ) . 1 − x n 1 − x = 1 + x + · · · + x n − 1 . Take e.g. n = 2 20 . ▶ L.h.s. needs less than 10 bytes. ▶ R.h.s. needs well over a 1,000,000 bytes! ▶ Nevertheless Axiom does remove gcd( f , g ) automatically, ▶ Maple uses sum of products representation; very compact but
Intermediate Expression Swell x 2 . 1 2 n Basic algebra shows: Consider: Vandermonde determinant 1 1 1 1 1 1 x 1 x 2 . . 2 1 . . . . . 1 . . . x 2 2 x 2 . 56 / 151 . n 1 1 1 x 1 x 2 x n . 1 x 2 2 x 2 x 2 . . . . � � . . . � � � � . . . � � � � . . . � � V ( x 1 , x 2 , . . . , x n ) = . � � � � � � � � x n − 1 x n − 1 x n − 1 . . . � � ∏ ( x j − x i ) . V ( x 1 , x 2 , . . . , x n ) = 1 ≤ i < j ≤ n � � . . . � � � � . . . � � � � . . . � � x n + 1 � � Z ( x 1 , x 2 , . . . , x n + 1 ) = . . . . � � n + 1 � � � � � � � � x n − 1 x n − 1 x n − 1 . . . � � n + 1
Obviously: Perfectly decent sum of products representation. 57 / 151 But expanding along fjrst row: Z ( x 1 , x 2 , . . . , x n + 1 ) = 0 . n + 1 ∑ ( − 1 ) i + 1 V ( x 1 , . . . , ˆ Z ( x 1 , x 2 , . . . , x n + 1 ) = x i , . . . , x n + 1 ) i = 1 n + 1 ∑ ∏ ( − 1 ) i + 1 = ( x k − x j ) , i = 1 1 ≤ j < k ≤ n + 1 j , k ̸ = i Expansion leads to n ! terms before any cancellation.
Keeping the Data Small: Modular Methods defjne its content & primitive part by: Conclusion: Can restrict attention to primitive polynomials—gcd also primitive. 58 / 151 Gcd of Polynomials in Z [ x ] Defjnition: For f ∈ Z [ x ] , f = a m x m + a m − 1 x m − 1 + · · · + a 0 cont ( f ) = gcd( a m , a m − 1 , . . . , a 0 ) , pp ( f ) = f / cont ( f ) . Lemma: (Gauss) For any f , g ∈ Z [ x ] we have cont ( fg ) = cont ( f ) cont ( g ) and pp ( fg ) = pp ( f ) pp ( g ) . Corollary: : For f , g ∈ Z [ x ] cont (gcd( f , g )) = gcd( cont ( f ) , cont ( g )) , pp (gcd( f , g )) = gcd( pp ( f ) , pp ( g )) .
59 / 151 Defjnition: Put Suggested Exercise: Let f , g ∈ Z [ x ] and h be their gcd in Z [ x ] . Prove that h is also a gcd of f , g in Q [ x ] . Useful fact: lc (gcd( f , g )) | gcd( lc ( f ) , lc ( g )) . Equivalantly: If a ̸ | lc ( f ) or a ̸ | lc ( g ) then a ̸ | lc (gcd( f , g )) . ( f mod p ) = ( a m mod p ) x m + ( a m − 1 mod p ) x m − 1 + · · · + ( a 0 mod p ) . Abbreviate ( f mod p ) to f p . Gives us a function ϕ : Z [ x ] → Z p [ x ] f �→ f p . ϕ ( 1 ) = 1, ϕ ( f + g ) = ϕ ( f ) + ϕ ( g ) , ϕ ( fg ) = ϕ ( f ) ϕ ( g ) . ▶ Example of a ring homomorphism .
60 / 151 Thus Put A = x 8 + x 6 − 3 x 4 − 3 x 3 + 8 x 2 + 2 x − 5 , B = 3 x 6 + 5 x 4 − 4 x 2 − 9 x + 21 . A = PH , B = QH , in Z [ x ] , where H = gcd( A , B ) . Consider modulo 5; A 5 = P 5 H 5 , B 5 = Q 5 H 5 , in Z 5 [ x ] . Direct computation in Z 5 [ x ] shows: gcd( A 5 , B 5 ) = 1 . So H 5 = 1, more accurately H 5 is a constant. Now [ ] 5 ̸ | lc ( A ) & 5 ̸ | lc ( B ) ⇒ 5 ̸ | lc ( H ) ⇒ deg( H ) = deg( H 5 ) ≤ deg(gcd( A 5 , B 5 )) = 0 ⇒ deg( H ) = 0 ⇒ H is a constant . gcd( A , B ) = 1 .
General Strategy Input Output mod p 1 mod p 2 ... mod p s Combine using CRA 61 / 151
Problems to Address be? How do we recover them? (Use symmetric representation of remainders.) 3. Which primes should we choose? Are there any that should be avoided? 62 / 151 1. How do we combine the various results in the Z p i [ x ] into a single result in Z [ x ] ? 2. Given A , B ∈ Z [ x ] how big can the coeffjcients of gcd( A , B )
Detailed Example Observations: 1. A , B primitive so H primitive. Note: Full algorithm does not do this step, only done here to keep number of coeffjcients down to 4. 63 / 151 A = 3 x 4 + 4 x 3 − 6 x 2 − 3 x + 2 , B = 9 x 5 + 21 x 4 + 6 x 3 + x 2 + x − 2 , H = gcd( A , B ) . 2. deg( H ) ≤ min(deg( A ) , deg( B )) = 4. 3. Easy computation shows A ̸ | B so deg( H ) < 4. Can put H = h 3 x 3 + h 2 x 2 + h 1 x + h 0 .
Aim: Work modulo p for p a prime (maybe use several p ). Compute Sensible to ensure so that which means 64 / 151 F p = gcd( A p , B p ) using Euclid’s algorithm in Z p [ x ] . Hope: F p = H p . Not guaranteed. p ̸ | lc ( A ) or p ̸ | lc ( B ) , deg( F p ) ≥ deg( H p ) = deg( H ) . Note: Even if p ̸ | lc ( A ) or p ̸ | lc ( B ) might get deg(gcd( A p , B p )) > 3 gcd( A p , B p ) ̸ = H p .
Get Conclusion: Must be something wrong with 2 as a modulus. No sign of trouble—carry on with hopeful heart. 65 / 151 First modulus p = 2: A 2 = x 4 + x , B 2 = x 5 + x 4 + x 2 + x , Euclid’s algorithm in Z 2 [ x ] gives: gcd( A 2 , B 2 ) = x 4 + x . Second modulus p = 3: No good—divides lc ( A ) and lc ( B ) . Third modulus p = 5: A 5 = 3 x 4 + 4 x 3 + 4 x 2 + 2 x + 2 , B 5 = 4 x 5 + x 4 + x 3 + x 2 + x + 3 , F 5 = gcd( A 5 , B 5 ) = x 3 + 4 x 2 + 2 x + 1 .
Test fails: So 5 might be a bad choice or need more work to recover coeffjcients of H completely. (At least one of them has been ‘collapsed’ by taking it modulo 5.) 66 / 151 Test: View F 5 as an element of Z [ x ] . See if F 5 | A & F 5 | B . Fourth modulus p = 7: F 7 = gcd( A 7 , B 7 ) = x 3 + 5 x + 4 , and F 7 ̸ | A .
Assumption: Both 5 and 7 are good moduli. h 2 h 0 h 0 h 1 h 1 Yields: Four pairs of simultaneous congruences: 67 / 151 h 2 h 3 h 3 ≡ 1 (mod 5 ) , ≡ 1 (mod 7 ) , ≡ 4 (mod 5 ) , ≡ 0 (mod 7 ) , ≡ 2 (mod 5 ) , ≡ 5 (mod 7 ) , ≡ 1 (mod 5 ) , ≡ 4 (mod 7 ) .
Example: Find all solutions to Substitute into second congruence: i.e. So: Now 68 / 151 First congruence gives: h 0 ≡ 1 (mod 5 ) , h 0 ≡ 4 (mod 7 ) . h 0 = 1 + 5 q , for q ∈ Z . 5 q ≡ 3 (mod 7 ) . 3 · 5 − 2 · 7 = 1 ⇒ 3 · 5 ≡ 1 (mod 7 ) q ≡ 3 · 3 (mod 7 ) , ≡ 2 (mod 7 ) . For simultaneous solution take q = 2 + 7 q ′ in 1 + 5 q to get for q ′ ∈ Z h 0 = 11 + 35 q ′ , h 0 ≡ 11 (mod 35 ) .
Solve other pairs of congruences to get: as candidate for H 35 . Note: Never did any work modulo 35. Assumption: Coeffjcients of H all in range Conclusion: Already have H , not just H 35 . Simple calculation shows: Give up?—never! 69 / 151 F 35 = x 3 + 14 x 2 + 12 x + 11 − 17 < h ≤ 18 . F 35 ̸ | A .
monic results. but monic is best. Much better: Know that where 70 / 151 Crucial observation: When fjnding gcd’s in Z p [ x ] we returned ▶ In fact any non-zero constant multiple would do just as well ▶ Assuming p is a good prime, H p = lc ( H ) gcd( A p , B p ) in Z p [ x ] . Desperate way out: Find lc ( H ) and multiply monic gcd’s by it. lc ( H ) | c c = gcd( lc ( A ) , lc ( B )) = 3 . Take, in Z 5 [ x ] and Z 7 [ x ] : 5 = 3 F 5 = 3 x 3 + 2 x 2 + x + 3 , F ∗ 7 = 3 F 7 = 3 x 3 + x + 5 . F ∗
7 : Make it primitive—OK already. Now easy to see so 71 / 151 Candidate from F ∗ 5 , F ∗ 35 = 3 x 3 + 7 x 2 + x − 2 . F ∗ F ∗ 35 | A & F ∗ 35 | B , in Z [ x ] , gcd( A , B ) = F ∗ 35 , in Z [ x ] .
The Chinese Remainder Problem D a Euclidean domain—i.e. integral domain in which a version of Euclidean Algorithm works. Given: 72 / 151 1. Remainders r 1 , . . . , r n ∈ D . 2. Moduli m 1 , . . . , m n ∈ D − { 0 } which are pairwise coprime, i.e. gcd( m i , m j ) = 1 for i ̸ = j . Problem: Find r ∈ D such that r ≡ r i (mod m i ) for 1 ≤ i ≤ n .
Direct Solution . if and only if . . 73 / 151 Then x is a solution to the system Let M i = m 1 m 2 · · · m i − 1 m i + 1 · · · m n for 1 ≤ i ≤ n . Find b 1 , b 2 , . . . , b n such that b i M i ≡ 1 (mod m i ) , for 1 ≤ i ≤ n (the b i exist because gcd( M i , m i ) = 1). x ≡ r 1 (mod m 1 ) x ≡ r 2 (mod m 2 ) x ≡ r n (mod m n ) x ≡ r 1 b 1 M 1 + r 2 b 2 M 2 + · · · + r n b n M n (mod M ) , where M = m 1 m 2 · · · m n .
74 / 151 Solutions of (1) have form: Thus Use Extended Euclidean Algorithm to fjnd c : Base Case n = 2 r ≡ r 1 (mod m 1 ) ( 1 ) r ≡ r 2 (mod m 2 ) ( 2 ) r 1 + σ m 1 . So have to fjnd σ such that: r 1 + σ m 1 ≡ r 2 (mod m 2 ) . cm 1 ≡ 1 (mod m 2 ) . σ = c ( r 2 − r 1 ) (mod m 2 ) . r 1 + σ m 1 ≡ r 1 + c ( r 2 − r 1 ) m 1 ≡ r 1 + r 2 − r 1 (mod m 2 ) .
75 / 151 General problem now reduces to: if and only if Again have: . . hold for x if and only if . congruences Observation: Solution r = r 1 + σ m 1 is such that the simultaneous x ≡ r 1 (mod m 1 ) x ≡ r 2 (mod m 2 ) x ≡ r (mod m 1 m 2 ) . General case: Solve fjrst two congruences to obtain r 12 as answer. x ≡ r 12 (mod m 1 m 2 ) x ≡ r 3 (mod m 3 ) x ≡ r i (mod m i ) , 1 ≤ i ≤ n , x ≡ r (mod m 1 m 2 · · · m n ) .
Conclusion bounded as follows Moreover there is exactly one such r . Suggested Exercise: Prove the claim in the preceding Theorem. 76 / 151 Can work with conveniently sized moduli m 1 , . . . , m n and then construct result for single large modulus m 1 m 2 · · · m n . Theorem: For the case D = Z the solution r computed by CRA n or 0 ≤ r < m 1 m 2 · · · m n . Theorem: For the case D = k [ x ] the solution r ( x ) computed by CRA n is is either 0 or bounded in degree as follows deg( r ) < deg( m 1 ) + · · · + deg( m n ) . Moreover there is exactly one such r ( x ) .
Chinese Remainder Theorem for the Integers To sum up, stated purely as a theorem we have: . . . 77 / 151 that Theorem: Assume r 1 , r 2 . . . , r n ∈ Z and m 1 , m 2 , . . . , m n ∈ Z where m i > 1, for 1 ≤ i ≤ n , and m i , m j are comprime (i.e., gcd( m i , m j ) = 1) for 1 ≤ i < j ≤ n . Then there is an integer x such x ≡ r 1 (mod m 1 ) x ≡ r 2 (mod m 2 ) x ≡ r n (mod m n ) . Moreover setting M = m 1 m 2 · · · m n we have that x + qM is also a solution for all q ∈ Z and all solutions are of this form.
Integer Case for all d , But want possibly negative integers. 78 / 151 applies. Choose moduli m 1 , . . . , m n to be distinct primes: ▶ Automatically coprime. ▶ Z p a fjeld so in Z p [ x ] gcd’s exists and Euclidean Algorithm ▶ This is critical . ▶ p not a prime means Z p is not an ID, gcd’s need not exist in Z p [ x ] . ▶ Example: in Z 6 [ x ] we have 3 x d + 1 | 2 x 3 x d + 1 | 4 x & since 2 x = ( 3 x d + 1 ) 2 x and 4 x = ( 3 x d + 1 ) 4 x . ▶ Use of CRT gives coeffjcients in range: 0 ≤ r < M = m 1 m 2 · · · m n .
Shift CRA results to range: where Symmetric representation of remainders. Conclusion: If trying to recover R with then choose moduli so that 79 / 151 − M / 2 < r ′ ≤ M / 2 , { if r ≤ M / 2 ; r , r ′ = r − M , if r > M / 2 . Can recover R uniquely if − M / 2 < R ≤ M / 2. | R | ≤ B M > 2 B .
Bound on Coeffjcients of gcd m Bit of a shame really. —FALSE— value than the largest absolute value of the coeffjcients of A or B . i b 2 n a 2 m 1 a 2 80 / 151 n Theorem: (Landau-Mignotte Inequality) Let A = ∑ m i = 0 a i x i and B = ∑ n i = 0 b i x i in Z [ x ] and suppose that B is a factor of A . Then � � | b i | ≤ 2 n | b n | ∑ ∑ � i . � | a m | i = 0 i = 0 Corollary: Let A , B ∈ Z [ x ] . The absolute value of each coeffjcient of gcd( A , B ) is bounded by � � � � � ∑ � ∑ 2 min( m , n ) gcd( a m , b n ) min . i , 1 � � | a m | | b n | i = 0 i = 0 Conjecture: Coeffjcients of gcd( A , B ) are no larger in absolute
Choosing Good Primes so invertible constant multiple. Note: We interpret equalities between gcds as being up to an 81 / 151 Put ▶ A , B ∈ Z [ x ] , G = gcd( A , B ) . ▶ Choose a prime p s.t. p ̸ | lc ( A ) or p ̸ | lc ( B ) so p ̸ | lc ( G ) . A = PG , B = QG , A p = P p G p , B p = Q p G p . Problem: G p might not be gcd( A p , B p ) in Z p [ x ] . Example: A = x − 3, B = x + 2, p = 5. gcd( A , B ) = 1 , in Z [ x ] , gcd( A 5 , B 5 ) = x + 2 , in Z 5 [ x ] .
Call a prime p which doesn’t work unlucky , i.e. Same as for some constant c . primes. Question: How many unlucky primes are there? 82 / 151 Lemma: Let A , B ∈ Z [ x ] and p a prime which does not divide both lc ( A ) , lc ( B ) . Then deg(gcd( A p , B p )) ≥ deg(gcd( A , B )) . deg(gcd( A p , B p )) > deg(gcd( A , B )) . gcd( A p , B p ) ̸ = c gcd( A , B ) p Note: Could have gcd( A p , B p ) = c gcd( A , B ) p for p dividing both lc ( A ) , lc ( B ) . But then we have no reliable way of detecting bad
Very useful tool: introduced by J. Sylvester 19th century. a m Have n rows of a -entries, m rows of b -entries, blank spaces 0. b 0 b n b 0 b n b 0 b n a 0 a m a 0 a m a 0 83 / 151 The resultant of A , B is A = a m x m + a m − 1 x m − 1 + · · · + a 0 , B = b n x n + b n − 1 x n − 1 + · · · + b 0 , both non-zero. Could have a m = 0 or b n = 0. . . . � � a m − 1 � � . . . � � a m − 1 � � · · � � � � · · � � � � · · � � � � · · � � � � . . . Res ( A , B ) = � � a m − 1 � � . . . � � b n − 1 � � . . . � � b n − 1 � · · � � � � · · � � � � · · � � � � · · � � � � . . . � b n − 1 Note: Strictly speaking should wrote Res m , n ( A , B ) .
Proof: First Claim: A , B have non-constant common factor ifg Simple proof based on unique factorization. Now put 84 / 151 Theorem: Suppose that a m ̸ = 0 or b n ̸ = 0. Then A and B have a non-constant common factor if and only if Res ( A , B ) = 0. ψ A = ϕ B for some non-zero ϕ and ψ , with deg( ϕ ) < m & deg( ψ ) < n . ϕ = α m x m − 1 + · · · + α 1 , ψ = β n x n − 1 + · · · + β 1 . When can ψ A = ϕ B ?
Equivalent to: . . . Use determinant condition for existence of non-trivial solution to 85 / 151 a 0 β 1 = b 0 α 1 , a 1 β 1 + a 0 β 2 = b 1 α 1 + b 0 α 2 , a m β n = b n α m . View as set of homogeneous equations in m + n unknowns: α 1 , . . . , α m , β 1 , . . . , β n . MX = 0 .
86 / 151 So 1 Example: 0 3 1 0 Thus 0 1 Lemma: Let A , B , p , A p , B p be as above and put G = gcd( A , B ) . Assume that A p ̸ = 0 and B p ̸ = 0. If p ̸ | Res ( A / G , B / G ) then gcd( A p , B p ) = G p . A = 3 x 4 + 4 x 3 − 6 x 2 − 3 x + 2 , B = 9 x 5 + 21 x 4 + 6 x 3 + x 2 + x − 2 , G = gcd( A , B ) = 3 x 3 + 7 x 2 + x − 2 . A / G = x − 1 , B / G = 3 x 2 + 1 . � � − 1 � � � � Res ( A / G , B / G ) = − 1 = 4 � � � � � �
87 / 151 5. goto 2 (all the primes were unlucky) 6. 2. od 3. fj (all previous primes were unlucky) 4. 1. MODGCD ( A , B ) �→ G g := gcd( lc ( A ) , lc ( B )) ; M := 2 g Landau_Mignote_Bound ( A , B ) ; p := new prime not dividing g ; C p := gcd( A p , B p ) computed in Z p [ x ] ; (ensure lc ( C p ) = 1) G p := ( g mod p ) C p in Z p [ x ] if deg( G p ) = 0 then return 1 fj ; P := p ; G := G p ; while P ≤ M do p := new prime not dividing g ; C p := gcd( A p , B p ) ; (ensure lc ( C p ) = 1) G p := ( g mod p ) C p ; if deg( G p ) < deg( G ) then goto 4 fj ; if deg( G p ) = deg( G ) then G := CRA ( G , G p , P , p ) ; P := pP H := pp ( G ) ; if H | A and H | B then return H fj ;
Let This yields 88 / 151 A = ( x − 2 )( x + 1 )( x 3 + 2 x − 1 ) = x 5 − x 4 − 3 x 2 − 3 x + 2 , B = ( x − 2 ) 2 ( x + 1 ) 2 = x 4 − 2 x 3 − 3 x 2 + 4 x + 4 . g = 1 , √ √ M = 2 · 1 · 2 4 · 1 · min( 24 , 46 ) ≤ 160 .
Trace of algorithm: 89 / 151 G 2 = x 3 + x , p = 2 : P = 2, G = x 3 + x , G 3 = x 2 − x + 1, so 2 was unlucky; p = 3 : P = 3, G = x 2 − x + 1 G 5 = x 2 − x − 2, p = 5 : G = x 2 − x − 2, this is gcd( A , B ) . Note: Algorithm would do 2 more steps to ensure P > 160
Polynomial Simplifjcation Basics of Algebraic Geometry Defjnition: The Variety corresponding to the polynomials is the set of their common zeros: 90 / 151 ▶ k a fjeld, ▶ X = { x 1 , . . . , x n } indeterminates over k , ▶ p 1 ( x 1 , . . . , x n ) , . . . , p m ( x 1 , . . . , x n ) ∈ k [ X ] . V ( p 1 , . . . , p m ) = { ( a 1 , . . . , a n ) ∈ k n | p i ( a 1 , . . . , a n ) = 0 , for 1 ≤ i ≤ n } . ▶ Subset of k n (variety depends on k and n ). ▶ Defjnition makes sense for arbitrary S ⊆ k [ X ] : V ( S ) = { ( a 1 , . . . , a n ) ∈ k n | p ( a 1 , . . . , a n ) = 0 , for all p ∈ S } .
Ideals Take: Put Obviously Thus Can add any set of polynomials like q to S without changing the variety. 91 / 151 p 1 , . . . , p s ∈ S , q 1 , . . . , q s ∈ k [ X ] . q = q 1 p 1 + · · · + q s p s . q ( a 1 , . . . , a n ) = 0 , for all ( a 1 , . . . , a n ) ∈ V ( S ) . V ( S ∪ { q } ) = V ( S ) .
92 / 151 Note: Bases not unique. Abstract defjnition: I is an ideal if and only if Have V commutative rings. S Note: Exactly the same defjnition of ideal applies to arbitrary Defjnition: The ideal of k [ X ] generated by S , denoted by ( S ) , is: ( S ) = { q 1 p 1 + · · · + q s p s | s ≥ 1 , q i ∈ k [ X ] , p i ∈ S , for 1 ≤ i ≤ s } . ( ) ( ) = V ( S ) . Say that S is a basis of ideal I if I = ( S ) . ( I is generated by S .) 1. I ̸ = ∅ , 2. p 1 , p 2 ∈ I ⇒ p 1 q , p 1 − p 2 ∈ I for all q ∈ k [ X ] . Fact: If S 1 ⊆ S 2 then ( S 1 ) ⊆ ( S 2 ) . Fact: If I is an ideal and p 1 , . . . , p s ∈ I , q 1 , . . . , q s ∈ k [ X ] then q 1 p 1 + · · · + q s p s ∈ I . Fact: If I is an ideal and S ⊆ I then ( S ) ⊆ I .
and infjnitely more. 93 / 151 S ⊆ k [ x , y ] with elements p 1 = x 2 y + x − 1 , p 2 = xy 2 + y − 1 . Then ( S ) contains ( 2 x + 3 y 2 ) p 1 = 3 x 2 y 3 + 2 x 3 y + 3 xy 2 + 2 x 2 − 3 y 2 − 2 , yp 1 − xp 2 = x − y ,
Consider Therefore Final set of equations is in triangular form so very easy to solve. Thus Thus 94 / 151 p 1 = x + y − 2 z − 1 , p 2 = 2 x − 3 y − z + 2 , p 3 = x − y + z , from Q [ x , y , z ] and let I = ( p 1 , p 2 , p 3 ) . Now p 4 = p 2 − 2 p 1 = − 5 y + 3 z + 4 ∈ I p 5 = p 3 − p 1 − 2 / 5 p 4 = 9 / 5 z − 3 / 5 ∈ I ( p 1 , p 4 , p 5 ) ⊆ I . Easily p 2 , p 3 ∈ ( p 1 , p 4 , p 5 ) so I = ( p 1 , p 4 , p 5 ) . V ( I ) = V ( p 1 , p 4 , p 5 ) = V ( x + y − 2 z − 1 , − 5 y + 3 z + 4 , 9 / 5 z − 3 / 5 ) .
Major Problem Question: Does every ideal have a fjnite basis?. Geometric signifjcance: Given fjgures in n dimensional space defjned by infjnitely many polynomial equations. Are there fjnitely many equations that defjne precisely the same fjgures? of the invariants’). proof! 95 / 151 | X | = 1: Yes—easy (follows from Euclidean Algorithm). | X | = 2: Yes—long & complicated proof by Gordan (the ‘King | X | arbitrary: Yes—Hilbert’s Basis Theorem (1888) very short
a fjnite basis. Method of proof: non-constructive. Gordan’s reaction: ‘Das ist nicht Mathematik. Das ist Theologie’. Not just sour grapes—fairly typical at the time. Later on: Hilbert produced constructive proof based on earlier non-constructive one. 96 / 151 Theorem: [Hilbert’s Basis Theorem, (1888)] Every ideal of k [ X ] has
Can view V as a function Have obvious function I in opposite direction: assigns to variety V the ideal Questions: 97 / 151 Ideals → Varieties . Varieties → Ideals I ( V ) = { p | p ∈ k [ X ] & p ( a 1 , . . . , a n ) = 0 , for all ( a 1 , . . . , a n ) ∈ V } . 1. is I = I V ( I ) for an arbitrary ideal I of k [ X ] ? 2. is V = V I ( V ) for an arbitrary variety V of k n ?
Easily: In fact always have But can have 98 / 151 1. I ⊆ I V ( I ) for all ideals I of k [ X ] , 2. V ⊆ V I ( V ) for all varieties V of k n , V = V I ( V ) I ̸ = I V ( I ) , e.g. take V = V ( p ( x ) 2 ) , p ( x ) non-constant.
has a root in k Assumption: from now on k is algebraically closed. Theorem: [Hilbert’s Nullstellensatz, (1893)] Let I be an ideal of that 99 / 151 Defjnition: k is algebraically closed if every non-constant p ∈ k [ x ] Example: C , fjeld of complex numbers. k [ X ] and q a polynomial of k [ X ] which is zero at all points of V ( I ) , i.e. q ∈ I V ( I ) . Then q s ∈ I for some integer s > 0. Concrete form: If q , p 1 , . . . , p m ∈ k [ X ] and q vanishes whenever p 1 , . . . , p m do then there exist s > 0 and q 1 , . . . , q m ∈ k [ X ] such q s = q 1 p 1 + · · · + q m p m .
Concrete form: A simultaneous system of polynomial equations: . . . does not have a simultaneous solution if and only if Note: Nullstellensatz defjnitely false if k not algebraically closed: 100 / 151 Equivalent form: V ( I ) = ∅ if and only if 1 ∈ I (i.e. I = k [ X ] ). p 1 ( x 1 , . . . , x n ) = 0 p 2 ( x 1 , . . . , x n ) = 0 p m ( x 1 , . . . , x n ) = 0 1 = q 1 p 1 + · · · + q m p m for some q 1 , . . . , q m ∈ k [ X ] . consider p = x 2 + 1 ∈ R [ x ] .
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