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 ​.