Definition of field Aim lecture: One usually does linear algebra over - PowerPoint PPT Presentation

Definition of field Aim lecture: One usually does linear algebra over some “system of numbers”, or more precisely, a field. We introduce this notion in full generality here. Defn A field consists of an additive (hence abelian) group ( F , +) equipped with a second associative, commutative binary operation × called multiplication such that × has an identity 1 � = 0 called the field or multiplicative identity , 1 there exist inverses for non-zero elements wrt × , that is, given β ∈ F − 0, 2 there is some β − 1 ∈ F with β × β − 1 = 1 = β − 1 × β . the following distributive law holds, for β, β ′ , β ′′ ∈ F we have 3 β × ( β ′ + β ′′ ) = β × β ′ + β × β ′′ . From now on, throughout these lectures, the symbol F will always represent a field of some sort. The addition is always denoted + but the multn will usually be abbreviated to β × β ′ = ββ ′ . Daniel Chan (UNSW) Lecture 4: Fields Semester 2 2012 1 / 10

Examples Daniel Chan (UNSW) Lecture 4: Fields Semester 2 2012 2 / 10

Multiplicative group Prop-Defn Let F be a field as usual. For β, β ′ ∈ F , we have ββ ′ = 0 iff either β = 0 or β ′ = 0. 1 Multn restricts to a binary operation on F × = F − 0. 2 ( F × , × ) is an abelian group called the multiplicative group of F . 3 Proof. Note that 1) = ⇒ 2) whilst 3) follows from field axioms for × and 2) so we only prove 1). 1) ( ⇐ =). 0 + 0 β = 0 β = (0 + 0) β = 0 β + 0 β so cancellation in the additive group ( F , +) gives 0 = 0 β . By commutativity of multn, see also β 0 = 0. ⇒ ) Suppose that ββ ′ = 0 but β � = 0. Picking an inverse β − 1 to β & using 1) (= associativity of multn we see β ′ = β − 1 ββ ′ = β − 1 0 = 0 by the ( ⇐ =) part already proved. This completes the proof of the propn. Rem Since F × is a group, we now know multiplicative inverses are unique & can use other facts about groups. Daniel Chan (UNSW) Lecture 4: Fields Semester 2 2012 3 / 10

Basic properties Note that in any abelian group ( A , +) we can define subtraction by β − β ′ = β + ( − β ′ ). In particular, we can subtract in any field & sim divide by non-zero elements. The field axioms ensure most of the usual arithmetic rules hold Prop Let F be a field as usual & β, β ′ , β ′′ ∈ F . ( − 1) β = − β 1 β ( − β ′ ) = − ( ββ ′ ) 2 β ( β ′ − β ′′ ) = ββ ′ − ββ ′′ 3 Proof. Ex Daniel Chan (UNSW) Lecture 4: Fields Semester 2 2012 4 / 10

Some finite fields This section is not examinable as it depends on you having done MATH1081. It is to give you some interesting examples of fields. Let p be a prime & F p = { 0 , 1 , . . . , p − 1 } . For β, β ′ ∈ F p , we consider the binary operations β + β ′ = ( β + β ′ ) ββ ′ = ( ββ ′ ) mod p , mod p . It is fairly easy to see that these are commutative associative binary operations with identities. Furthermore, ( F p , +) is an abelian group. The existence of multiplicative inverses for β ∈ F − 0 is harder & amounts to the fact that β has an inverse modulo p . Theorem F p is a field with the above addn & multn. Proof. Omitted. In any field, we may let n ∈ Z represent the element n 1. In F p we have p = 0! Daniel Chan (UNSW) Lecture 4: Fields Semester 2 2012 5 / 10

Polynomials A polynomial over (the field) F in the indeterminate x is an expression of the form � p i x i p ( x ) = i ≥ 0 where every p i ∈ F & p i = 0 for i ≫ 0 i.e. there are only finitely many non-zero co-efficients p i . For such a polynomial you can define degree, leading co-efficient, leading term, constant term, term in the usual way. Also, we can write p ( x ) as p ( x ) = p 0 + p 1 x + . . . + p d x d if p i = 0 for i > d . Further, we may omit terms with zero co-efficient. Let F [ x ] denote the set of all polynomials over F . It’s easy to prove Prop ( F [ x ] , +) is an abelian group if we define addition on F [ x ] co-efficient-wise by � � � p i x i ) + ( q i x i ) = ( p i + q i ) x i . ( i i i Note there is no clash in notn with p ( x ) = p 0 + p 1 x + . . . + p d x d . Daniel Chan (UNSW) Lecture 4: Fields Semester 2 2012 6 / 10

Polynomial arithmetic We define multiplication on F [ x ] by � � � � p i x i )( q j x j ) = p i q j ) x k . ( ( i j k i + j = k Prop Multn on F [ x ] is associative & commutative & there is a multiplicative 1 identity, namely, 1. The distributive law holds i.e. for p ( x ) , q ( x ) , r ( x ) ∈ F [ x ] we have 2 p ( x )( q ( x ) + r ( x )) = p ( x ) q ( x ) + p ( x ) r ( x ) . Proof. Long ex. Unfortunately, F [ x ] is never a field. Daniel Chan (UNSW) Lecture 4: Fields Semester 2 2012 7 / 10

Field of rational functions on R & C i p i x i ∈ F [ x ] can be viewed as a Let F = R or C . Then any polynomial p ( x ) = � i p i β i . As you know, in this case, polynomial function p ( x ) : F − → F : β �→ � addn & multn agree with usual pointwise addn & multn of functions. A rational function is a fraction of the form f ( x ) = p ( x ) q ( x ) for some p ( x ) , q ( x ) ∈ F [ x ] with q ( x ) � = 0. We will view this as a partially defined function from F − → F . In particular, we identify fractions if they agree as functions wherever they are both defined. Let F ( x ) denote the set of all such rational functions. Prop R ( x ) and C ( x ) are fields when endowed with pointwise addn & multn. Proof. Easy but long ex. Daniel Chan (UNSW) Lecture 4: Fields Semester 2 2012 8 / 10 Rem In fact there are analogous fields of rational functions F ( x ) for any field, F ,

Matrices Most of the theory of matrices over R you learnt in 1st year carries over to matrices over F for any field F . In particular we have Defn Let ( a ij ) , ( b ij ) ∈ M mn ( F ) , ( c ij ) ∈ M lm ( F ). Matrix addn ( a ij ) + ( b ij ) = ( a ij + b ij ) ij 1 Matrix multn ( c ij ) ij ( a jk ) jk = ( � j c ij a jk ) ik 2 We have the following facts as in 1st year with the same proofs. ( M mn ( F ) , +) is an abelian group. Multn is associative & the identity matrix is a multiplicative identity. The distributive law holds whenever all matrix sums & products are defined. Daniel Chan (UNSW) Lecture 4: Fields Semester 2 2012 9 / 10

Remarks on solving linear equations Fields provide the proper context for solving linear eqns. For example, we may solve the following for y 1 , y 2 ∈ F = R ( x ) by viewing it as a system of linear eqns in y 1 , y 2 with co-effs in F . xy 1 + y 2 = 0 y 1 + xy 2 = x − 1 Point Gaussian elimination works over any field. Daniel Chan (UNSW) Lecture 4: Fields Semester 2 2012 10 / 10