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.

  1.   is separable over  .
  2.   has a separating transcendence basis over  .
  3.   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.

  1. Every finitely generated submodule of a projective  -module is projective.
  2. Every finitely generated nonzero ideal of   is invertible.
  3.   is a valuation domain for every prime ideal  .
  4. Every overring of   is the intersection of localizations of  .
  5. 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