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:

$0\to 𝔭\to k[t]\to k[x]\to 0$

by letting $x\mapsto t$ and $𝔭$ the kernel of that map. Thus, $x$ is transcendental over $k$ if and only if $𝔭=0$. Since $k[t]$ is a PID, when nonzero, $𝔭$ is generated by a nonzero polynomial called the minimal polynomial of $x$, which must be irreducible since $k[x]$ is a domain and so $𝔭$ 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.

{{Template:BOOKTEMPLATE/Theorem |theorem=Exercise |label= |claim=A complex number is called an algebraic number if it is integral over $𝐐$. 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

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.

{{Template:BOOKTEMPLATE/Theorem |theorem=Theorem |label= |claim=A field $L$ is a separable algebraic over $k$ if and only if every irreducible polynomial has distinct roots (i.e., $f$ and its derivative $f^{\prime }$ have no common root.) |proof=}}

For the remainder of the section, $p$ denotes the characteristic exponent of a field; (i.e., $p=1$ if $\mathrm{char}(k)=0$ and $p=\mathrm{char}(k)$ 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.

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label= |claim=A $k$-algebra $A$ is separable if and only if $A{\otimes }_k{k}_p$ is reduced.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label= |claim=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 $p>1$ 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).}}

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, $p=\mathrm{char}k$.) $L$ 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.)

{{Template:BOOKTEMPLATE/Theorem |theorem=Lemma |label= |claim=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. }}

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label= |claim=A field is perfect if and only if either (i) its characteristic is zero or (ii) $x\mapsto x^p$ is an automorphism of $K$ |proof=First suppose $p=0$. Let $f$ be an irreducible polynomial. If $f$ and $f^{\prime }$ have a common root, then, since $f$ is irreducible, $f$ must divide $f^{\prime }$ and so $f^{\prime }=0$ since $\mathrm{deg}{f}^{\prime }<\mathrm{deg}f$. On the other hand, if $f(t)=a_0+a_{1}t+...+a_n{t}^n$, then

$f^{\prime }(t)=a_1+2a_{2}t+...+n{a}_n{t}^{n-1}\ne 0$.

Thus, a field of characteristic is perfect.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Corollary |label= |claim=A finite field is perfect.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label= |name= |claim=Let $L/K$ be a finite extension. Then $L$ is separable over $K$ if and only if $L$ is separable over $F$ and $F$ is separable over $K$.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label= |claim=Every finite field extension factors to a separable extension followed by a purely inseparable extension. More precisely, }}

{{Template:BOOKTEMPLATE/Theorem |theorem=Exercise |label= |claim=(Clark p. 33) Let $k$ be a field of characteristic 2, $F=k(x,y)$, $u\in F$ a root of $t^2+t+x$, $S=F(u)$ and $K=S(\sqrt{uy})$. Then (i) $K/S$ is purely inseparable and $S/F$ is separable. (ii) There is no nontrivial purely inseparable subextension of K/F.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Theorem |name=Primitive element |label= |claim=Let $L=K[x_1,...,x_n]$ be a finite extension, where $x_2,...,x_n$ (but not necessarily $x_1$) are separable over $K$. Then $L=K[z]$ for some $z\in L$. |proof=It suffices to prove the case $n=2$ (TODO: why?) Let ${\mu }_i$ be the minimal polynomials of $x_i$.}}

{{Template:BOOKTEMPLATE/Theorem |name= |label= |claim=Let $L/K$ be a finitely generated field extension. Then the following are equivalent.

1. $L$ is separable over $K$.
2. $L$ has a separating transcendence basis over $K$.
3. $L{\otimes }_K{K}^{p^{-r}}$ is a domain.}}

Transcendental extensions

{{Template:BOOKTEMPLATE/Theorem |name=Lüroth |label=Lüroth |claim=Any subfield $E$ of $F(X)$ containing $F$ but not equal to $F$ is a pure transcendental extension of $F$.}}

Let $L\supset K$ be a field extension of degree $n<\mathrm{\infty }$. An element$x\in L$ defines a $K$-linear map:

$x_L:L\to L,y\mapsto xy$.

We define

• ${\mathrm{Tr}}_{L/K}(x)=\mathrm{Tr}(x_L).$
• ${\mathrm{Nm}}_{L/K}(x)=\det (x_L).$

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label= |claim=Let $L\supset K\supset F$ be finite field extensions. Then

• (i) ${\mathrm{Tr}}_{L/F}={\mathrm{Tr}}_{L/K}\circ {\mathrm{Tr}}_{K/F}.$
• (ii) ${\mathrm{Nm}}_{L/F}={\mathrm{Nm}}_{L/K}\circ {\mathrm{Nm}}_{K/F}.$}}

{{Template:BOOKTEMPLATE/Theorem |label=Hilbert's Theorem 90 |name=Hilbert 90 |claim=If $L/K$ is a finite Galois extension, then

${\mathrm{H}}^1(\mathrm{Gal}(L/K),L^{\times })=0$.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Corollary |label= |name= |claim=Let $L/K$ is a cyclic extension, and $\sigma$ generate $\mathrm{Gal}(L/K)$. If $a\in L$ such that ${\mathrm{Nm}}_{L/K}(a)=1$, then

$a=\frac{\sigma (b)}{b}$ 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

$K^{\mathrm{Aut}(K/F)}=F.$

Here, we used the notation of invariance:

$K^G=\{x\in K|\sigma (x)=x,\mathrm{\forall }\sigma \in G\}$

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

{{Template:BOOKTEMPLATE/Theorem |label= |theorem=Proposition |claim=GCD domains and valuation domains are integrally closed. |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:

$r^n=-a_{n-1}rs+...+a_{1}r{s}^{n-1}+a_0{s}^n$.

and so $s|r^n$. Since $(r^n,s)=1$ by Lemma {{Template:BOOKTEMPLATE/ref|GCD prop}}, we have that $s$ is a unit in $A$, and thus $r/s\in A$. The case of valuation domains is very similar.}}

{{Template:BOOKTEMPLATE/Theorem |label= |theorem=Proposition |claim="integrally closed" is a local property.}}

{{Template:BOOKTEMPLATE/Theorem |label= |theorem=Proposition |claim=Let $A$ be a domain. The following are equivalent.

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

{{Template:BOOKTEMPLATE/Theorem |label=minimal poly coeff |theorem=Proposition |claim=Let $A$ be an integrally closed domain, and $L$ a finite extension of $A_{(0)}$. Then $x\in L$ is integral over $A$ if and only if its minimal polynomial in $K[X]$ is in $A[X]$.}}

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 $𝔭$ be a prime ideal. We may assume $𝔭$ is nonzero; thus, it contains a nonzero element $x$. We may assume that $x$ is irreducible; thus, prime by unique factorization. If $𝔭$ is prime, then we have $(x)=𝔭$. 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 $A$ 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 $𝔦$. Note that $𝔦$ is not prime; thus, there are $a,b$ such that $ab\in 𝔦$ but $a\in ̸𝔞$ and $b\in ̸𝔦$. Now, $(𝔦+(a))(𝔦+(b))\subset 𝔦$. Since both $𝔦+(a)$ and $𝔦+(b)$ are strictly larger than $𝔦$, which is maximal in $S$, $𝔦+(a)$ and $𝔦+(b)$ are both not in $S$ and both contain products of prime ideals. Hence, $𝔦$ 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.

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