Topics in Abstract Algebra/Field theory

Basic definitions
Let $$L/k$$ be a field extension; i.e., $$k$$ is a subfield of a field $$L$$. Then $$L$$ has a k-algebra structure; in particular, a vector space structure. A transcendental element is an element that is not integral; in other words, $$x$$ is transcendental over $$k$$ if and only if $$k[x]$$ 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 $$L/k$$ and an indeterminate $$t$$, we have the exact sequence:
 * $$0 \to \mathfrak p \to k[t] \to k[x] \to 0$$

by letting $$x \mapsto t$$ and $$\mathfrak p$$ the kernel of that map. Thus, $$x$$ is transcendental over $$k$$ if and only if $$\mathfrak p = 0$$. Since $$k[t]$$ is a PID, when nonzero, $$\mathfrak p$$ is generated by a nonzero polynomial called the minimal polynomial of $$x$$, which must be irreducible since $$k[x]$$ is a domain and so $$\mathfrak p$$ is prime. (Note that if we replace $$k[t]$$ by $$k[t, s]$$, say, then it is no longer a PID; therefore the kernel is no longer principal. So, in general, if a subset $$S \subset L$$ is such that $$k[S]$$ is a polynomial ring where members of $$S$$ are variables, then $$S$$ 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 $$L$$ has finite dimension over $$k$$, the extension is called finite extension. Every finite extension is algebraic. Indeed, if $$x \in L$$ is transcendental over $$k$$, then $$k[x]$$ is a "polynomial ring" and therefore is an infinite-dimensional subspace of $$L$$ and L must be infinite-dimensional as well.

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

For the remainder of the section, $$p$$ denotes the characteristic exponent of a field; (i.e., $$p = 1$$ if $$\operatorname{char}(k) = 0$$ and $$p = \operatorname{char}(k)$$ otherwise.) If the injection
 * $$x \mapsto x^p: k \to k$$

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 $$k_p$$ be the union of $$k$$ adjoined with $$p^e$$-th roots of elements in $$k$$ over all positive integers $$e$$. $$k_p$$ is then called the perfect closure since there is no strictly smaller subfield of $$k_p$$ that is perfect.

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

Separable extensions
Let $$L/K$$ be a field extension, and $$p$$ be the characteristic exponent of $$K$$ (i.e., $$p=1$$ if $$K$$ has characteristic zero; otherwise, $$p = \operatorname{char} k$$.) $$L$$ is said to be separable over $$K$$ if $$L \otimes_K K^{p^{-1}}$$ is a domain. A maximal separable extension $$k$$ is called the separable closure and denoted by $$k_s$$.

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.)

Transcendental extensions
Let $$L \supset K$$ be a field extension of degree $$n < \infty$$. An element$$x \in L$$ defines a $$K$$-linear map:
 * $$x_L : L \to L, y \mapsto xy$$.

We define
 * $$\operatorname{Tr}_{L/K}(x) = \operatorname{Tr}(x_L).$$
 * $$\operatorname{Nm}_{L/K}(x) = \operatorname{det}(x_L).$$

A. Theorem A ﬁeld extension $$L/K$$ is algebraic if and only if it is the direct limit of its ﬁnite subextensions.

A field extension $$K/F$$ is said to be Galois if
 * $$K^{\operatorname{Aut}(K/F)} = F.$$

Here, we used the notation of invariance:
 * $$K^G = \{ x \in K | \sigma(x) = x, \forall \sigma \in G \}$$

(In particular, when $$K/F$$ is a finite extension, $$K/F$$ is a Galois extension if and only if $$|\operatorname{Aut}(K/F)| = [K : F]$$.) When $$K/F$$ is Galois, we set $$\operatorname{Gal}(K/F) = \operatorname{Aut}(K/F)$$, and call $$\operatorname{Gal}(K/F)$$ the Galois group of $$K/F$$.

A. Theorem A field extension $$K/F$$ is Galois if and only if it is normal and separable.

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

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.

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 $$\mathfrak p$$ be a prime ideal. We may assume $$\mathfrak p$$ is nonzero; thus, it contains a nonzero element $$x$$. We may assume that $$x$$ is irreducible; thus, prime by unique factorization. If $$\mathfrak p$$ is prime, then we have $$(x) = \mathfrak p$$. Thus, every prime ideal is principal. $$\square$$

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 $$A$$ be an integral domain. Then $$A$$ is a Dedekind domain if and only if every localization of $$A$$ is a discrete valuation ring.''

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

Proof: Let $$S$$ be the set of all ideals that do not contain a product of nonzero prime ideals. If the lemma is false, $$S$$ is nonempty. Since $$A$$ is noetherian, $$S$$ has a maximal element $$\mathfrak{i}$$. Note that $$\mathfrak{i}$$ is not prime; thus, there are $$a, b$$ such that $$ab \in \mathfrak{i}$$ but $$a \not\in \mathfrak{a}$$ and $$b \not\in \mathfrak{i}$$. Now, $$(\mathfrak i + (a))(\mathfrak i + (b)) \subset \mathfrak{i}$$. Since both $$\mathfrak i + (a)$$ and $$\mathfrak i + (b)$$ are strictly larger than $$\mathfrak{i}$$, which is maximal in $$S$$, $$\mathfrak i + (a)$$ and $$\mathfrak i + (b)$$ are both not in $$S$$ and both contain products of prime ideals. Hence, $$\mathfrak i$$ contains a product of prime ideals. $$\square$$

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