Lecture   Separable Extensions I Objectives  Criterion for multiple roots in terms of derivatives  Irreducible polynomials are separable over elds of characteristic zero
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Lecture Separable Extensions I Objectives Criterion for multiple roots in terms of derivatives Irreducible polynomials are separable over elds of characteristic zero

3 Characterization of perfect 64257elds of positive characteristic Key words and phrases Separable polynomial separable element sepa rable extensions derivative of a polynomial perfect 64257elds Let be a 64257eld We have seen that the discriminant o

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Lecture Separable Extensions I Objectives Criterion for multiple roots in terms of derivatives Irreducible polynomials are separable over elds of characteristic zero




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Presentation on theme: "Lecture Separable Extensions I Objectives Criterion for multiple roots in terms of derivatives Irreducible polynomials are separable over elds of characteristic zero"‚ÄĒ Presentation transcript:


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Lecture 9 : Separable Extensions I Objectives (1) Criterion for multiple roots in terms of derivatives (2) Irreducible polynomials are separable over fields of characteristic zero. (3) Characterization of perfect fields of positive characteristic, Key words and phrases: Separable polynomial, separable element, sepa- rable extensions, derivative of a polynomial, perfect fields. Let be a field. We have seen that the discriminant of a polynomial ] vanishes if and only if ) has a repeated root. Calculation of discriminant can be difficult. In this

section we discuss an effective cri- terion in terms of derivatives of polynomials whether certain root of ) is repeated. We will also study fields so that no irreducible polynomial in ] has repeated roots. Let be a splitting field of a monic polynomial ] of degree n. Write in ] the unique factorization of ). ) = ( ∑∑∑ where ,...,r and ,e ,...,e are positive integers. Definition 9.1. The numbers ,e ,...,e are called the multiplicities of ,r ,...,r respectively. If = 1 for some i, then is called a simple root. If then is called a multiple root. A polynomial with no

multiple roots is called a separable polynomial. Proposition 9.2. The numbers of roots and their multiplicities are inde- pendent of a splitting field chosen for over F. Proof. Let and be splitting fields of ) over F. Then there is an isomorphism . This isomorphism gives rise to an isomorphism , 42
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43 Let ) = =1 be the unique factorization of ]. Then )) = =1 )) Since ] is UFD, ,..., ) are the roots of )) = ) with multiplicities ,...,e in respectively. The derivative criterion for multiple roots Let ) = ∑∑∑ ]. We can define derivative of without appealing to

limits. This is preferable since F may not be equipped with a distance function. The derivative of is defined by ) := =0 ia . It is easy to check that the usual formulas for ( )) )) and ( /g )) where = 0 hold for derivatives of polynomials. Theorem 9.3. Let be a monic polynomial. (1) If ) = 0 then every root of is a multiple root. (2) If = 0 then has simple roots if and only if gcd( f,f ) = 1 Proof. (1) Let ) = ( Then 0 = ) = ) + ( Thus ) = ), so is a root of Hence is a multiple root. (2) ( ) Let gcd( f,f ) = 1 and let be a multiple root of ). Then ) = ( ) in some splitting field

of ) over F. Thus ) = ( ) + 2( Hence ) = 0 If ) = gcd( ,f )) ] then ) = ) + for some ,q ]. Hence ) = 0 Therefore, deg 1, so gcd( f,f = 1, which is a contradiction. Therefore ) has only simple roots. Let ,r ,...,r be the roots of ) and assume that they are simple. Then ) = ( )( ∑∑∑ ) and ) = =1 Therefore ( ) does not divide ) any i. Hence and have no common root. Therefore gcd( f,f ) = 1.
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44 Proposition 9.4. (1) Let be an irreducible polynomial. Then is separable if and only if = 0 (2) Irreducible monic polynomials over a field of characteristic zero are sep- arable.

Proof. (1) ( ) If = 0, then every root of ) is a multiple root. ) Suppose is a multiple root of ). Then ) = 0 Since is irreducible, But this is a contradiction since deg deg Therefore ) is separable. (2) If char = 0 and ) is of positive degree, then = 0 Proposition 9.5. Let be a field of positive characteristic . Then is either irreducible in or Proof. Suppose ) = ) where 1 deg m < p Let be a root of ) in a splitting field of Then , so ) = ( Hence is also a root of Thus ) = ( Then F. Since ( p,m ) = 1, there exists x,y such that px my = 1 Hence px my F. Thus Example 9.6. We

construct an irreducible polynomial with a multiple root. Let ) be the quotient field of the polynomial ring Let ) = Then t / Suppose is a th power and Then ip ) = ip Hence = 0 for all . Thus is irreducible. Another way to see that is irreducible is to apply Eisensteinís Criterion with as a prime element. Let be a splitting field of ) over and be a root of ). Then so = ( Hence ) has only one root in E. Proposition 9.7. Let where char p, be an irreducible polynomial. If is not separable then there exists such that ) =
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45 Proof. Since ) = is irreducible and

inseparable, we have ) = ia = 0 Therefore pt for some . Hence ) = pt pt pi Perfect Fields We have seen that irreducible polynomial over fields of characteristic 0 are separable. But over a field of positive characteristic, irreducible polynomial may not be separable. We now discuss a condition on a field of positive characteristic which will ensure that irreducible polynomials in ] are separable. Definition 9.8. Let be a field extension. An algebraic element is a called separable element over if irr α,F is separable. We say K/F is a separable algebraic

extension if each element of is separable. We say is a perfect field if each algebraic extension is separable. Any field of characteristic zero is perfect. By the previous example ) is not perfect. This is basically due to not being a th power in Theorem 9.9. Let be a field of positive characteristic p. Then is perfect if and only if Proof. Suppose Then ] is irreducible and inseparable. Hence is not perfect. ) Let and ] be an irreducible polynomial. If ) is inseparable, then ) = ) = = ( for some F. This contradicts irreducibility of Hence ) is separable. Corollary 9.10. Every

finite field is perfect. Proof. Let By Lagrange theorem applied to the multiplicative group we get = 1 for all Hence for all F. Therefore = (