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:
![{\displaystyle 0\to {\mathfrak {p}}\to k[t]\to k[x]\to 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71b36ab7b0729ab6f095298f798179e01448590f)
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
![{\displaystyle \mathbf {Q} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/132d0144479d6f47c30ad82a65d458966ccbe928)
. 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.
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
![{\displaystyle L}](https://wikimedia.org/api/rest_v1/media/math/render/svg/103168b86f781fe6e9a4a87b8ea1cebe0ad4ede8)
is a separable algebraic over
![{\displaystyle k}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3c9a2c7b599b37105512c5d570edc034056dd40)
if and only if every irreducible polynomial has distinct roots (i.e.,
![{\displaystyle f}](https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61)
and its derivative
![{\displaystyle f'}](https://wikimedia.org/api/rest_v1/media/math/render/svg/258eaada38956fb69b8cb1a2eef46bcb97d3126b)
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
![{\displaystyle x\mapsto x^{p}:k\to k}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79c1a58e328ca1ce364d4f5b48c96f4dbba4a142)
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
![{\displaystyle k}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3c9a2c7b599b37105512c5d570edc034056dd40)
-algebra
![{\displaystyle A}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7daff47fa58cdfd29dc333def748ff5fa4c923e3)
is separable if and only if
![{\displaystyle A\otimes _{k}k_{p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/facff26c06c1041c368960756fd5fb598107e6c7)
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)
![{\displaystyle x\mapsto x^{p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f72ef9f7692b1b55c635323f2287d96b1d8cec7)
is an automorphism of
![{\displaystyle K}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b76fce82a62ed5461908f0dc8f037de4e3686b0)
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
![{\displaystyle L/K}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0381945929156997b99bb43e5b7067d18c9a84b)
be a finite extension. Then
![{\displaystyle L}](https://wikimedia.org/api/rest_v1/media/math/render/svg/103168b86f781fe6e9a4a87b8ea1cebe0ad4ede8)
is separable over
![{\displaystyle K}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b76fce82a62ed5461908f0dc8f037de4e3686b0)
if and only if
![{\displaystyle L}](https://wikimedia.org/api/rest_v1/media/math/render/svg/103168b86f781fe6e9a4a87b8ea1cebe0ad4ede8)
is separable over
![{\displaystyle F}](https://wikimedia.org/api/rest_v1/media/math/render/svg/545fd099af8541605f7ee55f08225526be88ce57)
and
![{\displaystyle F}](https://wikimedia.org/api/rest_v1/media/math/render/svg/545fd099af8541605f7ee55f08225526be88ce57)
is separable over
![{\displaystyle K}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b76fce82a62ed5461908f0dc8f037de4e3686b0)
.
Proposition.
Every finite field extension factors to a separable extension followed by a purely inseparable extension. More precisely,
Exercise.
(Clark p. 33) Let
![{\displaystyle k}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3c9a2c7b599b37105512c5d570edc034056dd40)
be a field of characteristic 2,
![{\displaystyle F=k(x,y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/215cfe8e0bc2382fd3baf09bd5b18b234919ce13)
,
![{\displaystyle u\in F}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77b6976322f2bc391667996bad8daf8b31cf4e85)
a root of
![{\displaystyle t^{2}+t+x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bef89f1955a228fa6ba993971ebca2657a86090)
,
![{\displaystyle S=F(u)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edffb5bd164f3caa1664d6c0bfe334d261732ac5)
and
![{\displaystyle K=S({\sqrt {uy}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40392796eeb90980523ab18ca7b89e3c2d50bfa3)
. Then (i)
![{\displaystyle K/S}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83d7e9bb2e5b9a586704e568a2775603bedfbceb)
is purely inseparable and
![{\displaystyle S/F}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4c830dd14ec020f3fd6c80e377d4a9efdab3cd6)
is separable. (ii) There is no nontrivial purely inseparable subextension of K/F.
Theorem (Primitive element).
Let
![{\displaystyle L=K[x_{1},...,x_{n}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c07a462ce9260cb3d7039075dcd369aedcafc633)
be a finite extension, where
![{\displaystyle x_{2},...,x_{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/85bf73d4da3f9f7b2a6ad20f886c13470af76620)
(but not necessarily
![{\displaystyle x_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8788bf85d532fa88d1fb25eff6ae382a601c308)
) are separable over
![{\displaystyle K}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b76fce82a62ed5461908f0dc8f037de4e3686b0)
. Then
![{\displaystyle L=K[z]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48dead19c26a67bfd0c2c778a8c611634c1a05b6)
for some
![{\displaystyle z\in L}](https://wikimedia.org/api/rest_v1/media/math/render/svg/224209d40f76e21870a3a03097b0623c3016ac81)
.
Proof. It suffices to prove the case
(TODO: why?) Let
be the minimal polynomials of
.
Theorem.
Theorem (undefined: Lüroth) (Lüroth).
Any subfield
![{\displaystyle E}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4232c9de2ee3eec0a9c0a19b15ab92daa6223f9b)
of
![{\displaystyle F(X)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00816772e8dff4e6733c478ec77fab0382264a93)
containing
![{\displaystyle F}](https://wikimedia.org/api/rest_v1/media/math/render/svg/545fd099af8541605f7ee55f08225526be88ce57)
but not equal to
![{\displaystyle F}](https://wikimedia.org/api/rest_v1/media/math/render/svg/545fd099af8541605f7ee55f08225526be88ce57)
is a pure transcendental extension of
![{\displaystyle F}](https://wikimedia.org/api/rest_v1/media/math/render/svg/545fd099af8541605f7ee55f08225526be88ce57)
.
Let
be a field extension of degree
. An element
defines a
-linear map:
.
We define
![{\displaystyle \operatorname {Tr} _{L/K}(x)=\operatorname {Tr} (x_{L}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c3d4819ad441ccb8550a9c2176ec4f4006d1140)
![{\displaystyle \operatorname {Nm} _{L/K}(x)=\operatorname {det} (x_{L}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1bba64d3e9b67f21db7d7fec7fe26f09ee93c116)
Proposition.
Let
![{\displaystyle L\supset K\supset F}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52b6ac73a77e621846717a3e301fc5b2000f0930)
be finite field extensions. Then
- (i)
![{\displaystyle \operatorname {Tr} _{L/F}=\operatorname {Tr} _{L/K}\circ \operatorname {Tr} _{K/F}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/efea232adbbe34ba947e5a66fd2b3949d3830c93)
- (ii)
![{\displaystyle \operatorname {Nm} _{L/F}=\operatorname {Nm} _{L/K}\circ \operatorname {Nm} _{K/F}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02bf9b19037b3077f87c2e2584e4519a2e909ef6)
Theorem A.8 (Hilbert 90).
If
![{\displaystyle L/K}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0381945929156997b99bb43e5b7067d18c9a84b)
is a finite Galois extension, then
.
Corollary.
Let
![{\displaystyle L/K}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0381945929156997b99bb43e5b7067d18c9a84b)
is a cyclic extension, and
![{\displaystyle \sigma }](https://wikimedia.org/api/rest_v1/media/math/render/svg/59f59b7c3e6fdb1d0365a494b81fb9a696138c36)
generate
![{\displaystyle \operatorname {Gal} (L/K)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a0ffa68a2cbf93db0330bd5a20bf0e74305d53b1)
. If
![{\displaystyle a\in L}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5299b8fd0430922b749b1febfd9e762300e49ad9)
such that
![{\displaystyle \operatorname {Nm} _{L/K}(a)=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4ee02a93da823922201e16f21c637bbb0de3f4)
, 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
![{\displaystyle K^{\operatorname {Aut} (K/F)}=F.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/74bd1521f1cdd753b83e261192dea1c26bf750e8)
Here, we used the notation of invariance:
![{\displaystyle K^{G}=\{x\in K|\sigma (x)=x,\forall \sigma \in G\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f1711e30655aa4c1fdbaa02d9b971dd82e675b0)
(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.
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.
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
![{\displaystyle A}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7daff47fa58cdfd29dc333def748ff5fa4c923e3)
be an integrally closed domain, and
![{\displaystyle L}](https://wikimedia.org/api/rest_v1/media/math/render/svg/103168b86f781fe6e9a4a87b8ea1cebe0ad4ede8)
a finite extension of
![{\displaystyle A_{(0)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65d3eeaf857ec97b1d03753cfba357640c8845dd)
. Then
![{\displaystyle x\in L}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97fca945ad424639c27ec8dccaf96c0bda408d3d)
is integral over
![{\displaystyle A}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7daff47fa58cdfd29dc333def748ff5fa4c923e3)
if and only if its minimal polynomial in
![{\displaystyle K[X]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bb4d802ca5718a14dc961af8692f35cdfad169b)
is in
![{\displaystyle A[X]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40abd94a808a2369931fac9811cbd1cbdd44497d)
.
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.