# 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. A complex number is called an algebraic number if it is integral over $\mathbf{Q}$. 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.

Theorem. 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 $f'$ have no common root.)

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. A $k$-algebra $A$ is separable if and only if $A \otimes_k k_p$ 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 $p > 1$ 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. 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. $\square$

Proposition. 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 $f'$ 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

$f'(t) = a_1 + 2 a_2 t + ... + n a_n t^{n-1} \ne 0$.

Thus, a field of characteristic is perfect. $\square$

Corollary. A finite field is perfect.

Proposition. 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 $K$.

Proposition. Every finite field extension factors to a separable extension followed by a purely inseparable extension. More precisely,

Exercise. (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.

Theorem (Primitive element). 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 $x_i$. $\square$

Theorem. 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

Theorem (undefined: Lüroth) (Lüroth). 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 $n < \infty$. 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. Let $L \supset K \supset F$ be finite field extensions. Then

• (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). If $L/K$ is a finite Galois extension, then

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

Corollary. Let $L/K$ is a cyclic extension, and $\sigma$ generate $\operatorname{Gal}(L/K)$. If $a \in L$ such that $\operatorname{Nm}_{L/K}(a) = 1$, then

$a = {\sigma(b) \over b}$ 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. 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:

$r^n = - a_{n-1} r s + ... + a_1 r s^{n-1} + a_0 s^n$.

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. "integrally closed" is a local property.

Proposition. 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. $A_{\mathfrak p}$ is a valuation domain for every prime ideal $\mathfrak p \triangleleft A$.
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.

Proposition A.10. 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 $\mathfrak p$ 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.