# Topics in Abstract Algebra/Field theory

## Basic definitions edit

Let be a field extension; i.e., is a subfield of a field . Then has a *k*-algebra structure; in particular, a vector space structure. A *transcendental element* is an element that is not integral; in other words, is *transcendental* over if and only if is (isomorphic to) the polynomial ring in one variable. The situation can be phrased more abstract as follows. Given an element *x* in an extension and an indeterminate , we have the exact sequence:

by letting and the kernel of that map. Thus, is transcendental over if and only if . Since is a PID, when nonzero, is generated by a nonzero polynomial called the *minimal polynomial* of , which must be irreducible since is a domain and so is prime. (Note that if we replace by , say, then it is no longer a PID; therefore the kernel is no longer principal. So, in general, if a subset is such that is a polynomial ring where members of are variables, then is said to be *algebraically independent*; By convention, the empty set is algebraically independent, just as it is linearly independent.) Finally, as a custom, we call an integral field extension an *algebraic extension*.

When has finite dimension over , the extension is called *finite extension*. Every finite extension is algebraic. Indeed, if is transcendental over , then is a "polynomial ring" and therefore is an infinite-dimensional subspace of and *L* must be infinite-dimensional as well.

**Exercise.** *A complex number is called an algebraic number if it is integral over . The set of all algebraic numbers is countable.*

A field is called *algebraically closed* if it admits no nontrivial algebraic field extension. (A field is always an algebraic extension of itself, a trivial extension.) More concretely, a field is algebraically closed if every root of a polynomial over that field is already in that field. It follows from the Axiom of Choice (actually equivalent to it) that every field is a subfield of some algebraically closed field.

## Separable extensions edit

A field extension is said to be separable if it is separable as *k*-algebra; i.e., is reduced for all field extension . The next theorem assures that this is equivalent to the classical definition.

**Theorem.** *A field is a separable algebraic over if and only if every irreducible polynomial has distinct roots (i.e., and its derivative have no common root.)*

For the remainder of the section, denotes the characteristic exponent of a field; (i.e., if and otherwise.) If the injection

is actually surjective (therefore, an automorphism), then a field is called *perfect*. Examples: Fields of characteristic zero and finite fields are perfect. Imperfect fields are therefore rather rare; they appear in algebraic geometry, a topic in later chapters. We let be the union of adjoined with -th roots of elements in over all positive integers . is then called the *perfect closure* since there is no strictly smaller subfield of that is perfect.

**Proposition.** *A -algebra is separable if and only if is reduced.*

**Proposition.** *The following are equivalent.*

- (i) A field is perfect.
- (ii) Every finite extension is separable.
- (iii) Every extension is separable.

*Proof.* Suppose (ii) is false; it is then necessary that and . Finally, if (iii) is false, then there is an extension such that is not reduced. Since is algebraic over by construction, it has a finite extension such that is not reduced. This falsifies (ii).

In particular, any extension of a perfect field is perfect.

## Separable extensions edit

Let be a field extension, and be the characteristic exponent of (i.e., if has characteristic zero; otherwise, .) is said to be *separable* over if is a domain. A maximal separable extension is called the *separable closure* and denoted by .

A field is said to be *perfect* if its separable closure is algebraically closed. A field is said to be *purely inseparable* if it equals its separable closure. (As the reader would notice, the terminology so far is quite confusing; but it is historical.)

**Lemma.** *An algebraic extension is separable if and only if the minimal polynomial of any element has no multiple root.**Proof.* We may assume that the extension is finite.

**Proposition.** *A field is perfect if and only if either (i) its characteristic is zero or (ii) is an automorphism of **Proof.* First suppose . Let be an irreducible polynomial. If and have a common root, then, since is irreducible, must divide and so since . On the other hand, if , then

- .

Thus, a field of characteristic is perfect.

**Corollary.** *A finite field is perfect.*

**Proposition.** *Let be a finite extension. Then is separable over if and only if is separable over and is separable over .*

**Proposition.** *Every finite field extension factors to a separable extension followed by a purely inseparable extension. More precisely,*

**Exercise.** *(Clark p. 33) Let be a field of characteristic 2, , a root of , and . Then (i) is purely inseparable and is separable. (ii) There is no nontrivial purely inseparable subextension of K/F.*

**Theorem (Primitive element).** *Let be a finite extension, where (but not necessarily ) are separable over . Then for some .**Proof.* It suffices to prove the case (TODO: why?) Let be the minimal polynomials of .

**Theorem.** *Let be a finitely generated field extension. Then the following are equivalent.*

- is separable over .
- has a separating transcendence basis over .
- is a domain.

## Transcendental extensions edit

**Theorem (undefined: Lüroth) (Lüroth).** *Any subfield of containing but not equal to is a pure transcendental extension of .*

Let be a field extension of degree . An element defines a -linear map:

- .

We define

**Proposition.** *Let be finite field extensions. Then*

- (i)
- (ii)

**Theorem A.8 (Hilbert 90).** *If is a finite Galois extension, then*

*.*

**Corollary.** *Let is a cyclic extension, and generate . If such that , then*

*for some .*

**A. Theorem** *A ﬁeld extension is algebraic if and only if it is the direct limit of its ﬁnite subextensions.*

A field extension is said to be *Galois* if

Here, we used the notation of invariance:

(In particular, when is a finite extension, is a Galois extension if and only if .) When is Galois, we set , and call the *Galois group* of .

**A. Theorem** *A field extension is Galois if and only if it is normal and separable.*

## Integrally closed domain edit

A domain is said to be *integrally closed* if equals the integral closure of in the field of fractions.

**Proposition.** *GCD domains and valuation domains are integrally closed.**Proof.* Suppose is integral over ; i.e.,

- .

We may assume . It follows:

- .

and so . Since by Lemma A.8, we have that is a unit in , and thus . The case of valuation domains is very similar.

**Proposition.** *"integrally closed" is a local property.*

**Proposition.** *Let be a domain. The following are equivalent.*

- Every finitely generated submodule of a projective -module is projective.
- Every finitely generated nonzero ideal of is invertible.
- is a valuation domain for every prime ideal .
- Every overring of is the intersection of localizations of .
- Every overring of is integrally closed.

A domain satisfying any/all of the equivalent conditions in the proposition is called the *Prüfer domain*. A notherian Prüfer domain is called a *Dedekind domain*.

**Proposition A.10.** *Let be an integrally closed domain, and a finite extension of . Then is integral over if and only if its minimal polynomial in is in .*

A *Dedekind domain* is a domain whose proper ideals are products of prime ideals.

**A. Theorem** *Every UFD that is a Dedekind domain is a principal ideal domain.*

Proof: Let be a prime ideal. We may assume is nonzero; thus, it contains a nonzero element . We may assume that is irreducible; thus, prime by unique factorization. If is prime, then we have . Thus, every prime ideal is principal.

**Theorem** *Let *A* be an integral domain. Then *A* is a Dedekind domain if and only if:*

*(i) A is integrally closed.**(ii) A is noetherian, and**(iii) Every prime ideal is maximal.*

**A. Theorem** *Let *A* be a Dedekind domain with fraction field *K*. Let L be a finite degree field extension of K and denote by S the integral closure of R in L. Then S is itself a Dedekind domain.*

**A Lemma** *Let be an integral domain. Then is a Dedekind domain if and only if every localization of is a discrete valuation ring.*

**Lemma** *Let be a noetherian ring. Then every ideal contains a product of nonzero prime ideals.*

Proof: Let be the set of all ideals that do not contain a product of nonzero prime ideals. If the lemma is false, is nonempty. Since is noetherian, has a maximal element . Note that is not prime; thus, there are such that but and . Now, . Since both and are strictly larger than , which is maximal in , and are both not in and both contain products of prime ideals. Hence, contains a product of prime ideals.

A local principal ideal domain is called a *discrete valuation ring*. A typical example is a localization of a Dedekind domain.

## Henselian rings edit

## References edit

- Pete L. Clark. Commutative Algebra
- Pete L. Clark. Field Theory
- J.S. Milne. A Primer of Commutative Algebra
- J.S. Milne. Fields and Galois Theory
- Matsumura, Commutative ring theory