# 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.
Exercise.
$\mathbf {Q}$
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.
Theorem.
$L$
$k$
$f$
$f'$
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.
Proposition.
$k$
$A$
$A\otimes _{k}k_{p}$
Proposition.
* (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 $p>1$ and . Finally, if (iii) is false, then there is an extension $L/k$ such that $L\otimes _{k}k_{p}$ is not reduced. Since $k_{p}$ is algebraic over $k$ by construction, it has a finite extension $F$ such that $L\otimes _{k}F$ is not reduced. This $F$ falsifies (ii). $\square$
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.)
Lemma.
Proof. We may assume that the extension is finite. $\square$
Proposition.
$x\mapsto x^{p}$
$K$
Proof. First suppose $p=0$. Let $f$ be an irreducible polynomial. If $f$ and $f'$ have a common root, then, since $f$ is irreducible, $f$ must divide $f'$ and so $f'=0$ since $\deg f'<\deg f$. On the other hand, if $f(t)=a_{0}+a_{1}t+...+a_{n}t^{n}$, then
    $f'(t)=a_{1}+2a_{2}t+...+na_{n}t^{n-1}\neq 0$.
Thus, a field of characteristic is perfect. $\square$
Corollary.
Proposition.
$L/K$
$L$
$K$
$L$
$F$
$F$
$K$
Proposition.
Exercise.
$k$
$F=k(x,y)$
$u\in F$
$t^{2}+t+x$
$S=F(u)$
$K=S({\sqrt {uy}})$
$K/S$
$S/F$
Theorem (Primitive element).
$L=K[x_{1},...,x_{n}]$
$x_{2},...,x_{n}$
$x_{1}$
$K$
$L=K[z]$
$z\in L$
Proof. It suffices to prove the case $n=2$ (TODO: why?) Let $\mu _{i}$ be the minimal polynomials of $x_{i}$. $\square$
Theorem.
$L/K$
* $L$ is separable over $K$.
* $L$ has a separating transcendence basis over $K$.
* $L\otimes _{K}K^{p^{-r}}$ is a domain.

## Transcendental extensions

Theorem (undefined: Lüroth) (Lüroth).
$E$
$F(X)$
$F$
$F$
$F$
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
* $\operatorname {Tr} _{L/K}(x)=\operatorname {Tr} (x_{L}).$
* $\operatorname {Nm} _{L/K}(x)=\operatorname {det} (x_{L}).$
Proposition.
$L\supset K\supset F$
* (i) $\operatorname {Tr} _{L/F}=\operatorname {Tr} _{L/K}\circ \operatorname {Tr} _{K/F}.$
* (ii) $\operatorname {Nm} _{L/F}=\operatorname {Nm} _{L/K}\circ \operatorname {Nm} _{K/F}.$
Theorem A.8 (Hilbert 90).
$L/K$
    $\operatorname {H} ^{1}(\operatorname {Gal} (L/K),L^{\times })=0$.
Corollary.
$L/K$
$\sigma$
$\operatorname {Gal} (L/K)$
$a\in L$
$\operatorname {Nm} _{L/K}(a)=1$
    $a={\sigma (b) \over b}$ for some $b$.
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.
Proposition.
Proof. Suppose $r/s$ is integral over $A$; i.e.,
    $(r/s)^{n}+a_{n-1}(r/s)^{n-1}+...+a_{1}(r/s)+a_{0}=0$.
We may assume $(r,s)=1$. It follows:
    $r^{n}=-a_{n-1}rs+...+a_{1}rs^{n-1}+a_{0}s^{n}$.
and so $s|r^{n}$. Since $(r^{n},s)=1$ by Lemma A.8, we have that $s$ is a unit in $A$, and thus $r/s\in A$. The case of valuation domains is very similar. $\square$
Proposition.
Proposition.
$A$
* Every finitely generated submodule of a projective $A$-module is projective.
* Every finitely generated nonzero ideal of $A$ is invertible.
* $A_{\mathfrak {p}}$ is a valuation domain for every prime ideal ${\mathfrak {p}}\triangleleft A$.
* Every overring of $A$ is the intersection of localizations of $A$.
* Every overring of $A$ 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.
$A$
$L$
$A_{(0)}$
$x\in L$
$A$
$K[X]$
$A[X]$
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.

## References

* 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