抽象代数学习笔记（环论、域论）
Ring Theory
4.8
Definition: A ring is a set together with two binary operation together with "" and "", obeying:
is an Abelian group.
is associative: for .
The distributive laws hold in : and .
The ring is called commutative if multiplication is commutative.
The ring is said to have an identity if there is an element with for .
The additive identity of will always be denoted as and the additive inverse of an element will be denoted as an .
A ring with an identity where is called a division ring (or skew field) if every non-zero element has a multiplicative inverse, i.e., there exists such that . A commutative division ring is called a field.
Examples:
Trivial rings: .
Zero ring: .
The ring of integers: ; rational numbers: ; reals: ; complex numbers: .
The Hamilton Quaternions: elements in the form where and . The Hamilton Quaternions are a non-commutative ring with identity and a division ring.
Ring of functions: Let and be any ring, , , .
Property 1. Let be a ring, then
for .
for .
.
If has identity, then .
Definition. Let be a ring
A non-zero element is called a zero divisor if there exists such that or .
Assume that has identity . An element of is called a unit in if , such that . The set of units in is denoted by .
4.15
Definition. A commutative ring with identity is called an integral domain if it has no zero divisors.
Property 2. Assume , and is not a zero divisor. If , then either or . In particular, if are any elements in an integral domain and , then either or .
Proof. If , then so either or .
Corollary 3. Any finite integral domain is a field.
Proof: Let be a finite integral domain and let be a non-zero element of . By the cancellation law the map is an injective function. Since is finite this map is also surjective. In particular, there is some such that , i.e. is a unit in . Since was an arbitrary non-zero element, is a field.
Definition. A subring of the ring is a subgroup of that is closed under multiplication and addition.
Quadratic Integer Rings
Let be a square-free integer. It is immediately from the addition and multiplication that the subset forms a subring of the quadratic field defined earlier.
If then the slightly larger subset is also a subring.
Polynomial Rings
Let be an indeterminate, be a commutative ring, then with , and each is called a polynomial in with coefficients in . If , the polynomial is said to be of degree , is called the leading term, and is called the leading coefficient. The set of all such polynomials is called the ring of polynomials in the variable in and will be denoted . The polynomial is called monic if .
The ring in which the coefficients are taken makes a substantial difference in the behavior of polynomials. For example, is not a perfect square in , but is a perfect square in .
Property 4. Let be an integral domain and let be non-zero elements of . Then
The degree of equals to the sum of the degree of and the degree of .
The units of are just the units of .
is an integral domain.
Proof: If has no zero divisors then neither does ; if and are polynomials with leading terms and , respectively, then the leading term of is , and . This proves (3) and (1). If is a unit, say in , then the degree of the degree of equals to , so both and are elements of , hence are units in since their product is . This proves (2).
Matrix Rings
Fix an arbitrary ring and let . Let be the set of all matrices with entries from . is a ring under the usual addition and multiplication of matrices.
When , is not commutative even if is commutative.
Group Rings
Fix a commutative ring with identity and let be any finite group with group operation written multiplicatively. Define the group ring, , of with coefficients in to be the set of all formal sums: , , .
The ring is commutative if and only if is commutative.
Ring homomorphisms and Quotient Rings
Definition. Let and be rings:
A ring homomorphism is a map satisfying:
for .
for .
The kernel of the ring homomorphism , .
A bijective ring homomorphism is called an isomorphism.
Example:
The map defined by sending even integers to and odd integers to is a ring homomorphism.
For , the map defined by are not in general a ring homomorphism.
Property 5. Let and and let be a homomorphism.
The image of is a subring of .
The kernel of is a subring of . Furthermore, if then and are inside for .
Proof: Note that both and are empty.
If , then , such that . and . Thus is a subring of .
If , then . Hence , so , i.e., is a subring of . Similarly for , we have , , so .
Definition. Let be a ring, be a subset of and .
and .
A subset of is a left ideal of if
is a subring of
is closed under left multiplication by elements from , i.e., , for .
Similarly, is a right ideal if (i) holds and for .
A subset that is both a left ideal and a right ideal is called an ideal of .
Property 6. Let be a ring and be an ideal of . Then the (additive) quotient group is a ring under the binary operations: and for . Conversely, if is any subring such that the above operation are well-defined, then is an ideal of .
Definition. When is an ideal of the ring with the operations in the previous proposition is called the quotient ring of by .
Theorem 7.
(the First Isomorphism Theorem for Rings) If is a homomorphism of rings, then is an ideal of , the image of is a subring of and is isomorphic as a ring to : .
If is any ideal of , then the map defined by is a surjective ring homomorphism, with kernel . Thus every ideal is the kernel of a ring homomorphism and vice versa.
Theorem 8.
(the Second Isomorphism Theorem for Rings). Let be a subring and be an ideal of , then is a subring of , is an ideal of and .
(the Third Isomorphism Theorem for Rings). Let be ideals of with . Then is an ideal of and .
(the Fourth or Lattice Isomorphism Theorem for rings). Let be an ideal of , the correspondence is an inclusion preserving bijection between the set of subrings of that contain and the set of subrings of . Furthermore, (a subring containing ) is an ideal of if and only if is a subring of .
Definition. Let be ideals of .
Define the sum of and by .
Define the product of and by equals to all finite sums of with .
It is easy to see that is the smallest ideal of containing both and , and that is an ideal contained in .
Definition. Let be a ring with , ,
The ideal generated by : the smallest ideal of containing , denoted by .
.
.
An ideal generated by a single element is called a principal ideal.
An ideal generated by a finite set is called a finitely generated ideal.
Since the intersection of any non-empty collection of ideals of is also an ideal, we have such that is an ideal of .
The left ideal generated by is the intersection of all left ideals of that contain . is precisely the left ideal generated by and is the ideal generated by .
If is commutative then .
Example:
The trivial ideal and the ideal are both principal: and .
In we have for all integers of . These are all the ideals of , so every ideal of is principal. Furthermore, the ideal generated by two nonzero integers and is the principal ideal generated by their greatest common divisor.
Property 9. Let be a ring with and be an ideal of .
if and only if contains a unit.
Assume is commutative. Then is a field if and only if its only ideal are and .
Proof:
If then contains the unit . Conversely, if is a unit in with inverse , then for , . Hence .
is a field if and only if every non-zero element is a unit. If is a field every non-zero ideal contains a unit. If is a field every non-zero ideal contains a unit, so by the first part is the only non-zero ideal. Conversely, if and are the only ideals of , let be any non-zero element of . By hypothesis , so . Thus there exists such that , i.e., is a unit. So is a field.
Corollary 10. If is a field then any non-zero ring homomorphism from to another ring is an injection.
Proof: The kernel of a homomorphism is an ideal. The kernel of a non-zero homomorphism is a proper ideal hence .
4.29
Definition. An ideal in an arbitrary ring is called a maximal ideal if and the only ideals containing are and .
Property 11. In a ring with identity every proper ideal is contained in a maximal ideal.
Property 12. Assume is commutative. The ideal is a maximal ideal if and only if the quotient ring is a field.
Proof: The ideal is maximal if and only if there are no ideals with . By the Lattice Isomorphism Theorem, the ideals of containing correspond bijectively with the ideals of , so is maximal if and only if the only ideals of are and . By Property 9, we see that is maximal if and only if is a field.
Example:
Let be a non-negative integer, the ideal of is a maximal ideal if and only if is a field, if and only if is a prime number.
The ideal is not a maximal ideal in because . In fact, which is not a field.
Definition. Assume is commutative. An ideal is called a prime ideal if and whenever for , at least one of and are inside .
Property 13. Assume is commutative. Then ideal is a prime ideal in if and only if the quotient ring is an integral domain.
Proof: The ideal is prime if and only if and whenever , then either or . Use the bar notation for elements of : . Note that if and only if is zero in . Thus in the terminology of quotients is a prime ideal if and only if and whenever either or , i.e., is an integral domain.
Corollary 14. Let be commutative. Every maximal ideal of is a prime ideal.
Proof: If is a maximal ideal then is a field by Property 12. A field is an integral domain so the corollary follows from Property 13.
Example:
The principal ideals generated by primes in are both prime and maximal ideals. The zero ideal in is prime but not maximal.
The ideal is a prime ideal in since . This ideal is not a maximal ideal. The ideal is a prime ideal in but is not a maximal ideal.
Ring of Fractions.
Theorem 15. Let be a commutative ring. Let be any non-empty subset of that does not contain or any zero divisors and is closed under multiplication. Then there exists a commutative ring with such that contains as a subring and every element of is a unit in . The ring has the following properties:
Every element of is of the form for some and . In particular, if then is a field.
(uniqueness of ) The ring is the "smallest" ring containing in which all elements of become units, in the sense that: any ring containing an isomorphic copy of in which all the elements of become units must also contain an isomorphic copy of .
Definition. Let be as in Theorem 15:
The ring is called the ring of fractions of with respect to and is denoted as .
If is an integral domain and , is called the field of fractions or quotient field of .
The Chinese Remainder Theorem
Definition. The ring direct product of , denoted by , is defined as:
Addition: .
Multiplication: .
Definition. The ideals and are said to be comaximal if .
Theorem 17. (The Chinese Remainder Theorem) Let be ideals in , the map defined by is a ring homomorphism with kernel . If for each with the ideals and are comaximal, then this map is surjective and so .
Proof: We first prove this for , the general case will follow by induction. Let . Consider the map defined by . is a ring homomorphism since it is just the natural projection of into and for the two components. consists of all the elements that are in and , i.e., . When , such that . Thus . If now is an arbitrary element in , then the element maps to this element since . This shows that is surjective. Finally, the ideal is always contained in . If and are comaximal and are as above, then for any , , so .
The general case follows easily by induction from the case of two ideals using and once we show that and are comaximal. By hypothesis, for each there are and such that . Since , it follows that is an element in , this completes the proof.
Corollary 18. Let be a positive integer and let be its factorization into powers of distinct primes. Then . .
Euclidean Domains
Definition: Any function with is called a norm on the integral domain . If for , then is called a positive norm. The integral domain is said to be a Euclidean domain if there exists such that for , such that with or . The element is called the quotient and the remainder.
Euclidean Algorithm: .
Examples:
Fields are trivial examples of Euclidean domains since , , we have .
The integer ring are Euclidean domains with norm given by , the usual absolute value.
If is a field, then is a Euclidean Domain.
Property 1. Every ideal of a Euclidean domain is principal. More precisely, if is any non-zero ideal in the Euclidean domain then , where is any non-zero element of of minimum norm.
Proof: If is zero ideal, then nothing is to prove. So we suppose is not the zero ideal. Let with minimum norm, obviously . Inversely, let and use the division algorithm to write with or . Then . By the minimality of the norm of , we see that must be . Thus , showing .
Example: Let . Since the ideal is not principal, so is not a Euclidean domain for any choice of norm, even though is.
Definition. Let be a commutative ring and , .
is said to be a multiple of if there exists such that . In this case is said to divide or be a divisor of , written .
A greatest common divisor of and is a non-zero element such that , and if , then . A greatest divisor of and will be denoted as , or if is clear in the context.
Since , the properties of greatest common divisor translated into the language of ideals therefore becomes:
If is the ideal generated by and , then is a greatest common divisor of and if:
is contained in .
If is any principal ideal containing then
Property 2. If and are non-zero elements in the commutative ring such that the ideal generated by and is a principal ideal , then .
An integral domain in which every ideal is principal is called a Bezout Domain.
5.6
Property 3. Let be an integral domain. If and generate the same principal ideal, i.e., , then for some unit . In particular, if and are both greatest common divisors of and , then for some unit .
Proof: This is clear if either or is zero, so we may assume and are non-zero. Since , there exists such that . Since , there exists such that . Thus , since , , that is, both are units.
Theorem 4. Let be an Euclidean Domain and let be non-zero elements of and be the last non-zero remainder in the Euclidean Algorithm for and . Then
.
. In particular, such that .
Since the equation is simply another way of stating that is the element of the ideal generated by and . Hence the equation is solvable in integers and if and only if is divisible by .
Principal Ideal Domain (P.I.D)
Definition. A principal ideal domain is an integral domain in which every ideal is principal.
Of course, every Euclidean domain is always a principal ideal domain.
Example:
is a principal ideal domain, but is not.
is not a principal ideal domain, since is not principal.
It is not true that every P.I.D is a Euclidean domain. For example, is a P.I.D, but not a Euclidean domain.
Property 6. Let be a P.I.D and are non-zero elements of . Let be a generator for , then
is a greatest common divisor of and .
, such that .
is unique up to some unit in .
Property 7. Every non-zero prime ideal in a P.I.D is a maximal ideal.
Proof: Let be a non-zero prime ideal in the P.I.D , and be any ideal containing . We must show that or . Since , for some . Since is a prime ideal and , either or must lie in . If then . If , write , then , so , is a unit so .
Corollary 8. If is any commutative ring such that is a P.I.D, then is a field.
Proof: Assume is a P.I.D. Since is a subring of , then must be an integral domain. The ideal is a non-zero prime ideal in because is isomorphic to the integral domain . Thus by Property 7, is a maximal ideal, is a field.
Definition. Let be an integral domain:
Suppose is non-zero and not a unit. Then is called irreducible in whenever with , at least one of must be a unit in . Otherwise, is reducible.
The non-zero element is called prime if is prime.
Two element and of differing by a unit are said to be associate in (i.e. )
Example: In , the irreducible elements are prime numbers, and two integers and are associates of each other if and only if .
Property 10. In a integral domain a prime element is always irreducible.
Proof: Suppose is a prime ideal and , then , so one of , say , is in . Thus for some . So , so , and is a unit. This shows that is irreducible.
However, an irreducible element may not a prime element. Counter example: Let , then is irreducible but not prime since but .
Yet, if is a P.I.D, the notations of prime and irreducible are the same.
Property 11. In a P.I.D, a non-zero element is prime if and only if it is irreducible.
Proof:
Prime->irreducible: Proven in Property 10.
Irreducible->prime: Let be any ideal containing ), then is a principle ideal. Since , for some . But is irreducible, so either or is a unit. So either or . Thus is a maximal ideal, hence prime.
Unique Factorization Domain
Definition. A U.F.D is an integral domain in which every non-zero element , which is not a unit obeys:
, where are irreducible elements inside .
The decomposition is unique up to associates.
Example:
A field is always a U.F.D.
is not a U.F.D since .
Property 12. In a U.F.D, a non-zero element is prime if and only if it is irreducible.
Proof:
Prime->irreducible: Proven in Property 10.
Irreducible->prime: Let be irreducible in and assume for some , we must prove or . Since , for some . Written as product as irreducibles. We see from the decomposition of , must be associate to one of the irreducible occurring either in or . So divides either or .
Theorem 14. A P.I.D is always a U.F.D.
Corollary 15. (Fundamental Theorem of Arithmetic): is a U.F.D.
Fields E.D P.I.D U.F.D Integral domains Rings Groups.
All containments are proper:
is a Euclidean domain, but not a field.
is a P.I.D, but not a Euclidean domain.
is a U.F.D, but not a P.I.D.
is an integral domain, but not a U.F.D.
Polynomial rings
Property 2. Let be an ideal of the ring and , then . In particular, if is a prime ideal of , then is a prime ideal of .
Proof: There is a natural map given by reducing each of the coefficients modulo . Clearly is a surjective homomorphism, and . By the first the first Isomorphism Theorem, . In addition, if is a prime ideal in , then is an integral domain, hence also , is a prime ideal of .
Definition. The polynomial ring in the variables with coefficients in , denoted , is defined inductively by , i.e. a finite sum of non-zero monomial terms: .
Example: consists of all finite sums of monomial terms of the form : for example, are both elements of of degree and .
Polynomial Rings over fields
Theorem 3. Let be a field, then is a Euclidean Domain. Specifically, if and with non-zero, then there exists unique and such that with or .
Proof: If , then take . Thus we may assume and prove the existence of and by induction on . If , then take and . Assume , then write , . Then , is of degree less than . By induction, there exists with with or . Then letting , we have with or . As for uniqueness, suppose and also satisfy , then has degree less than . So , .
Corollary 4. If is a field, then is a P.I.D and U.F.D.
Property 5. (Gauss Lemma) Let be a U.F.D with field of fractions and let . If is reducible in , then is reducible in .
Corollary 6. Let be a U.F.D. be its field of fractions, . Suppose the greatest common divisor of the coefficients of is , then is irreducible in if and only if it is irreducible in .
Proof. By Gauss's Lemma, if is reducible in , then it is reducible in . Conversely, since the greatest common divisor of the coefficients of is , so if is reducible in , then where are not constant polynomial in . Thus the same factorization shows that is reducible in .
Theorem 7. is a U.F.D if and only if is a U.F.D.
Corollary 8. If is a U.F.D, then a polynomial ring in an arbitrary number of variables with coefficients in is also a U.F.D.
5.13
Irreducibility criteria
Property 9: Let be a field and , then has a factor of degree if and only if has a root in , i.e, there exists s.t. .
Proof: If has a factor of degree , i.e., , then . Conversely, suppose , by the Euclidean Algorithm, , where is a constant. Since , .
Property 10. A polynomial of degree or over a field is reducible if and only if it has a root in .
Proof: This follows immediately from Property 9, since a polynomial of degree or is reducible if and only if it has at least one linear factor.
Property 11. Let , if ( are relatively prime) is a root of , then and . In particular, if is a monic polynomial, and for all integers dividing , then has no root in .
Proof. By hypothesis, , so . Thus , since , . Similarly , .
Examples:
in is irreducible.
and in is irreducible.
in is reducible since it has a root .
in is irreducible.
Property 12. Let be a proper ideal in and be a non-constant monic polynomial. If the image of in cannot be factored in into two polynomials of smaller degree, then is irreducible in .
Proof. Suppose cannot be factored in but is reducible in , then there exists monic, non-constant polynomials in such that . Reducing the coefficients modulo gives a factorization in with non-constant factors.
Property 13. (Eisenstein's Criterion) Let be a prime ideal of the integral domain and be a polynomial in () Suppose are all elements of , but , then is irreducible in .
Proof. Assume were reducible, say in , where are non-constant polynomials. Reducing this equation modulo , we have in . Since is a prime ideal, is an integral domain, it follows that both and have zero constant term, , which contradicts with the assumption.
Example:
in is irreducible by Eisenstein's Criterion.
is irreducible in by Eisenstein's Criterion.
Polynomial Rings over Fields
Property 15. The maximal ideals in are the ideals generated by irreducible polynomials . In particular, is a field if and only if is irreducible.
Property 16. Let be a non-constant monic element of and let be its factorization into irreducibles, where the are distinct. Then we have .
Proof: This follows from the Chinese Remainder Theorem, since the ideals and are comaximal if (they are relatively prime in the Euclidean Domain , hence the ideal generated by them is )
Property 17. If has roots in , then has factor. In particular, a polynomial of degree over a field has at most roots.
Proof: The first statement follows easily by induction from Property 9. Since linear factors are irreducible, the second statement follows since is a UFD.
Field Theory
5.13
Definition. The characteristic of a field , denoted by is the smallest positive integer such that if such exist. Otherwise .
is either or a prime number.
Definition. The prime subfield of a field is the subfield of generated by the identity of . It is isomorphic to either (if ) or (if )
Definition. If is a field containing the subfield , then is said to be an extension field (or simply an extension) of , denoted by .
If is any extension fields, then the multiplication in makes into a vector space over . In particular, every field can be considered as a vector field over its prime field.
Definition. The degree (or relative degree of index) of , denoted by , is the dimension of as a vector space over (i.e., . The extension is said to be finite if is finite and infinite otherwise.
Property 2. Let be a homomorphism of fields. Then is either identically or is injective, so that the image of is either or isomorphic to .
Theorem 3. Let be a field and be an irreducible polynomial. Then there exists a field containing an isomorphic copy of in which has a root. Identifying with this isomorphic copy shows that there exists an extension of in which has a root.
Proof: Consider the quotient . Since by assumption is an irreducible polynomial in the P.I.D . The ideal is maximal hence is a field indeed. The canonical projection of to the quotient restricted to gives a homomorphism which is not identically since it maps to . Hence is an isomorphic copy of in . We identify with its isomorphic image in and view as a subfield of . If denote the image of in , then . So does contain a root of the polynomial .
Theorem 4. Let be an irreducible polynomial of degree over the field and let . Let . Then the elements are a basis for as a vector space over . So the degree of the extension is , i.e., . Hence .
Proof: Let , since is a E.D, we have , with . Since , it follows that , which shows that every residue class in is represented by a polynomial of degree less than . Hence the images of in the quotient span the quotient as a vector space over . It remains to see that these elements are linearly independent, so form a basis of . If they are not linearly independent in , then there exists a linear combination in with , not all . This is equivalent to , which is impossible.
5.20
Definition. Let and , then the smallest subfield of containing both and , denoted as is called the field generated by .
If the field is generated by a single element over , , then is said to be a simple extension of and the element is called a primitive element for the extension.
Theorem 6. Let be a field, be an irreducible polynomial. Suppose containing a root of : . Then .
Proof: There exists a natural homomorphism , as . Since , , so we obtain an induced homomorphism from . Since is irreducible, is a field, and is not the zero map, hence is an isomorphism of the field on the left to its image. Since this image is then a subfield of containing and , hence equal to .
Theorem 8. Let be an isomorphism of fields. Let be an irreducible polynomial and be the irreducible polynomial obtained by applying to the coefficients of . Let be a root of , and be a root of . Then there exists an isomorphism from to which maps to and extending , i.e. .
Proof. Clearly, induces a natural isomorphism from to , which maps the maximal ideal to the maximal ideal . Taking quotients by these ideals, we obtain an isomorphism: . By Theorem 6, and . Composing these isomorphisms, we obtain .
Algebraic Extensions: .
Definition. is said to be algebraic over if is a root of some non-zero polynomial . Otherwise is transcendental over . is said to be algebraic if every element of is algebraic over .
Property 9. Let be algebraic over , then there exists a unique monic irreducible polynomial which has as a root. A polynomial has a root if and only if divides in .
Corollary. If and is algebraic over both and , then divides in .
The polynomial is called the minimal polynomial for over . The degree of is called the degree of .
Property 12. The element is algebraic over if and only if the simple extension is finite.
Proof: If is algebraic, then is of finite degree. Conversely, suppose be finite, then the elements, of are linearly dependent over . Say with not all . Proven.
Corollary 13. If the extension is finite, then it is algebraic.
Theorem 14. Let be fields. Then .
Corollary 15. Suppose is a finite extension and let be any subfield of containing , . Then divides .
Example: , , .
Theorem 17. is finite if and only if is generated by a finite number of algebraic element.
Theorem 20. is algebraic over and is algebraic over , then is algebraic over .
Definition. Let be two subfields of a field , then the composite field of and , denoted by , is the smallest subfield of containing both and .
Property 21. Let be two finite extensions of a field containing . Then with equality if and only if an -basis for one of the fields remains linearly independent over the other field. If and are bases for and over respectively, then span over .
Splitting fields
Definition. is called a splitting field for the polynomial if factors completely into linear factors (or splits completely) in , and doesn't factor completely into linearly factors over any proper subfield of containing .
Theorem 25. For any field , if , then there exists an extension of which is a splitting field for .
Example:
The splitting field for over is .
The splitting field for over is .
The splitting field for over is .
The splitting field has degree at most .
Theorem 27. Let be an isomorphism, , . Let be the splitting field for over , and be a splitting field for over . Then extends to an isomorphism
5.27
Definition. The field is called an algebraic closure of if is algebraic over and if every polynomial splits completely over . A field is algebraic closed if every polynomial inside has a root in .
Theorem (Fundamental Theorem of algebra). The field is algebraic closed.
Separable extension
Definition. A polynomial over is called separable if it has no multiple roots. Otherwise it is inseparable.
Property 33. A polynomial has a multiple root if and only if is also a root of . In particular, is separable if and only it is relatively prime to its derivative: .
Example: over is separable.
Corollary 34. Every irreducible polynomial over a field of characteristic (for example, ), is separable. A polynomial over such a field is separable if and only if it is the product of distinct irreducible polynomials.
Property 35. Let be a field of characteristic , then for , , and . Put another way, is an injective field homomorphism from to .
Corollary 36. Suppose is a finite field of characteristic , then every element of is a -th power in .
Property 37. Every irreducible polynomial over a finite field is separable.
Definition. A field of characteristic is called perfect if every element of is a -th power in , i.e. . Any field of characteristic is also called perfect.
Definition. The field is said to be separable (or separable algebraic) over if every element of is the root of a separable polynomial over .
Cyclotomic polynomial & extensions
Definition. The -th cyclotomic polynomial is the polynomial whose roots are the primitive -th roots of unity. .
.
Galois Theory
Definition. An isomorphism of with itself is called an automorphism of . is said to fix an element if .
Definition. Let be an extension of fields. Let be the collection of automorphisms of which fix .
Property 2. Let and be algebraic over . Then for any , is a root of the minimum polynomial for over .
Proof. Suppose is a root of , for any , .
Example:
Let . If , then since these are the two roots of . So , where .
Let , .
Property 3. Let , then the collection of elements of fixed by all elements of is a subfield of .
Proof: Let , then , , .
The subfield of fixed by all the elements of is called the fixed field of .
If , then .
If are two subgroups of automorphisms with associated fixed fields and respectively, then .
Property 5. Let be the splitting field over of the polynomial , then with equality if is separable over .
Definition. Let be a finite extension, the is said to be Galois over and is a Galois extension if . If is Galois the group is called the Galois group of , denoted as .
Corollary. If is the splitting field over of a separable polynomial , then is Galois.
Example:
is Galois but is not.
is Galois.
The splitting field of over is Galois of degree . The roots of are , where . , where and .
The Fundamental Theorem of Galois Theory
Definition. A character of a group with values in a field is a homomorphism from to the . . The characters of are said to be linearly independent over if they are linearly independent as functions of , i.e., there is no non-trivial relation .
Theorem 7. If are distinct characters of with values in then they are linearly independent.
Theorem 9. Let be a subgroup of automorphism of a field and be the fixed field, then .
Corollary 10. Let be any finite extension. Then with equality if and only if is the fixed field of .
6.3
Theorem 13. The extension is Galois if and only if is the splitting field of some separable polynomial over . Furthermore if this is the case then every irreducible polynomial which has a root in is separable and has its all roots in .
Definition. Let be a Galois extension. If , for is called the Galois conjugates of over .
Theorem 14 (Fundamental Theorem of Galois Theory): Let be a Galois extension and , then there exists a bijection between subfields of containing and subgroups of given by the map:
the elements of fixing .
The fixed field of .
which are inverse to each other.
If correspond to respectively, then if and only if .
and .
is always Galois with Galois group .
is Galois over if and only if is a normal subgroup of . If this is the case, then the Galois group is isomorphic to the quotient group . More generally, even if is not necessarily normal in , the isomorphisms of which fix are in one-to-one correspondence with the cosets of in .
If correspond to respectively, then correspond to and correspond to .
Composite extension and simple extension.
Property 19. Suppose is Galois and is any extension. Then is Galois and
Corollary 20. Suppose is Galois and is any finite extension. Then .
Proof:
Property 21. Let be Galois, then
The intersection is Galois over .
is Galois over .
Corollary 22. Let be Galois with , then . Conversely, if is Galois over and , then with .
Definition: Let be indeterminates and , , , .
Theorem 32. The general polynomial over the field is separable with Galois group .
Definition: Define the discriminant of by .
The Fundamental Theorem of Algebra: Every polynomial of degree has precisely roots in .
Proof:
Consider two simple facts:
Every polynomial with real coefficients of odd degree has a root in reals.
Quadratic polynomial have roots in .
Now note that it suffices to prove that has a root in . Suppose is of degree where is odd. We prove that has a root in by induction on . For the statement is trivially true. Suppose now , let be the roots of and set . Then is a Galois extension of containing and the roots of . For any , consider . Any permute the terms, keeping invariant. So , the degree of is . So there exists . There are infinitely many but only finite many values of and , so there exists such that . So , since are roots of the quadratic with coefficients in , hence are elements of .
Definition. is said to be cyclic if it is Galois with a cyclic Galois group.
Property 36. Let be a field of characteristic not dividing which contains the -th roots of unity. Then the extension for is cyclic over of degree not dividing .
Definition. An element which is algebraic over can be expressed by radicals or solved for in terms of radicals if is an element of a field which can be obtained by a succession of simple radical extensions
Where for some . Such a is called a root extension of .
A polynomial can be solved by radicals if all its roots can be solved for in terms of radicals.
Lemma 38. If is contained in a root extension as above, then is contained in a root extension which is Galois over and where each extension is cyclic.
Theorem 39. The polynomial can be solved by radicals if and only if its Galois group is a solvable.
Corollary 40. The general equation of degree can not be solved by radical .
