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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 Eisenstein’s 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 = (

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 ID: 22807

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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 Eisenstein’s 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 = (

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