Topics in Abstract Algebra/Field theory
See also: The spectrum of a commutative ring
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.
See also: The spectrum of a commutative ring
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.
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.
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:
Proposition. Let be finite field extensions. Then
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
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.