Facts About Field Extensions

This is a list of definitions and theorems concerning field extensions, that I’m compiling for my learning. Throughout, let $K$ be a field.

As I wrote this, I realized that it was helpful for me to point out theorems that are especially surprising / non-trivial. I will indicate these with a remark immediately following the statement of the result. Conversely, we will omit proofs whose details are immediate. (Many such proofs can be tedious.)

Definition. A field extension of $K$ is a field $L$ such that $K$ is a subfield of $L$ . (We will often just say extension.)

Theorem. Any extension of $K$ can be viewed as a vector space over $K$ .

Definition. The degree of an extension of $K$ is its dimension when viewed as a vector space over $K$ .

Definition. If an extension has finite degree then it is called a finite extension, otherwise it is called an infinite extension.

Definition. Let $L$ be an extension of $K$ . If every element of $L$ is the root of some polynomial in $K\lbrack X\rbrack$ then we say $L$ is an algebraic extension, otherwise we say $L$ is a transcendental extension.

Theorem. Any finite extension is algebraic.

Proof. This is the first answer in abstract algebra – Is any finite-dimensional extension of a field, say $F$, algebraic and finitely generated? – Mathematics Stack Exchange. $\blacksquare$

Corollary. Any transcendental extension is infinite.

Henceforth, let $U$ be an extension of $K$ .

Definition. Let $S \subseteq U$ . By $K(S)$ we mean the “smallest” extension of $K$ containing $S$ , in the following sense: it is the subfield of $U$ containing $K$ and $S$ such that, for any other subfield $L$ of $U$ containing $K$ and $S$ , $K(S) \subseteq L$ . (This subfield always exists, and it is unique as a set.)

Theorem. Either of the following are equivalent to the definition of $K(S)$ :

$K(S)$ is the intersection of all subfields of $U$ that contain $K$ and $S$ .
$K(S) = \left\{ \frac{P(a_{1},\ldots,a_{n})}{Q(b_{1},\ldots,b_{m})}:P \in K\left\lbrack X_{1},\ldots,X_{n} \right\rbrack,Q \in K\left\lbrack X_{1},\ldots,X_{m} \right\rbrack,a_{i} \in S,b_{i} \in S such that Q(b_{1},\ldots,b_{m}) \neq 0 \right\}$ (all quotients of all polynomial expressions in the elements of $S$ with coefficients in $K$ .)

Definition. If $S = \left\{ a_{1},\ldots,a_{n} \right\} \subseteq U$ , then we denote $K\left( a_{1},\ldots,a_{n} \right) = K(S)$ (as a shorthand without the curly braces.) In particular, for a single element $a \in U$ , $K(a)$ is called a simple extension of $K$ .

Theorem. An equivalent definition of $K(a)$ is given by

$\displaystyle K(a) = \left\{ \frac{c_{n}a^{n} + \ldots + c_{1}a^{1} + c_{0}a^{0}}{d_{m}a^{m} + \ldots + d_{1}a^{1} + d_{0}}:n \in \mathbb{Z}_{0},c_{i} \in K,d_{i} \in K\ such\ that\ d_{m}a^{m} + \ldots + d_{1}a^{1} + d_{0} \neq 0 \right\}$

(all quotients of all polynomials in $a$ with coefficients in $K$ .)

Remark. This is exactly the expression of all quotients of all polynomial expressions in the elements of $S$ with coefficients in $K$ from before, written out for $S = \left\{ a \right\}$ .

Theorem. We have $K\left( a_{1},\ldots,a_{n} \right) = K\left( a_{1} \right)\ldots\left( a_{n} \right)$ (a sequence of simple extensions.)

Remark. We will omit the details since they are immediate, but we will give a quick hint: notice that $K\left( a_{1},\ldots,a_{n} \right) \subseteq K\left( a_{1} \right)\ldots\left( a_{n} \right)$ and $K\left( a_{1} \right)\ldots\left( a_{n} \right) \subseteq K\left( a_{1},\ldots,a_{n} \right)$ .

Definition. A sequence of simple extensions $K\left( a_{1} \right)\ldots\left( a_{n} \right)$ for $a_{i} \in U$ is called an iterated extension.

Corollary. Any extension $K(S)$ with $S$ finite is iterated, and any finite extension is iterated.

Theorem. The extension $K(S)$ is “independent” of $U$ up to isomorphism, in the following sense: if $U,U^{'}$ are any two extensions of $K$ containing $S$ , then the two extensions $K(S)$ formed from $U$ and $U^{'}$ are isomorphic to each other. Thus, we may identify elements with each other and speak of $K(S)$ without mentioning $U$ , as long as we know that some extension of $K$ containing $S$ exists.

Theorem. Any sequence of simple algebraic extensions is simple.

Remark. Compared to the previous theorems, this result is much more surprising / non-trivial and less immediate.

Proof. By induction, it suffices to show that, for $a,b \in U$ , there exists $c \in U$ such that $K(a,b) = K(c)$ . A proof of this is provided in Charles Pinter’s A Book of Abstract Algebra, Chapter 31, Theorem 2. $\blacksquare$

Definition. Let $a \in U$ be algebraic over $K$ . The minimal polynomial of $a$ is the monic irreducible polynomial $p \in K\lbrack X\rbrack$ which has $a$ as a root. (This polynomial exists if and only if $a$ is algebraic, and it is unique.)

Theorem. Let $a \in U$ . We have that $K(a)$ is an algebraic extension if and only if $a$ is algebraic over $K$ . Furthermore, if $a$ is algebraic, then the degree of $K(a)$ is equal to the degree of the minimal polynomial of $a$ . (In particular, $K(a)$ is a finite extension.)

Definition. Let $p$ be an irreducible polynomial in $K\lbrack X\rbrack$ . In a way analogous to the construction of the complex numbers from the reals, we may construct from $K$ an extension of $K$ in which $p$ has all roots (if the degree of $p$ is $n$ , then $p$ can be factored as $\left( X - a_{1} \right)\ldots\left( X - a_{n} \right)$ for $a_{i}$ in the extension, where the $a_{i}$ are not necessarily distinct.) Denoting this extension by $L$ , if $a \in L$ is a root of $p$ , then $L$ is isomorphic to $K(a)$ . Hence, we don’t even need existence of $U$ to construct simple algebraic extensions.

Theorem. Let $p,q$ be irreducible polynomials in $K\lbrack X\rbrack$ , and let $a,b$ be roots of $p,q$ respectively. If $q$ is not irreducible over $K(a)$ , then $q$ has all roots in $K(a)$ , and $K(b)$ can be embedded into $K(a)$ .

Definition. We can generalize the previous procedure to any number of polynomials, or even infinitely many polynomials. Let $S$ be a set of irreducible polynomials in $K\lbrack X\rbrack$ . In a way analogous to the construction of the complex numbers from the reals, we may construct from $K$ an extension of $K$ in which every polynomial in $S$ has all roots. Denoting this extension by $L$ , if the set of roots in $L$ of all polynomials in $S$ is denoted by $S^{'}$ , then $L$ is isomorphic to $K\left( S^{'} \right)$ . Hence, we don’t need existence of $U$ to construct any algebraic extension. Furthermore, regardless of existence of $U$ , given $K$ , we can automatically work with roots of any polynomial in $K\lbrack X\rbrack$ .

Definition. If $S$ is the set of *all* irreducible polynomials in $K\lbrack X\rbrack$ , then the extension constructed from the previous definition is called the algebraic closure of $K$ , denoted $\overline{K}$ .

Theorem. There exist non-simple algebraic extensions.

Proof. As a counterexample, take the field of real algebraic numbers as an extension of $\mathbb{Q}$ . This extension is algebraic but has infinite degree, therefore it cannot be simple. $\blacksquare$

Theorem. We have $\overline{\overline{K}} = \overline{K}$ .

Remark. Another interpretation of this statement is that, once we have formed an extension in which any polynomial in $K\lbrack X\rbrack$ has roots, then any polynomial in the potentially much larger set $\overline{K}\lbrack X\rbrack$ will also actually have roots in it. Compared to most of the theorems here, this is quite surprising and non-trivial.

Proof. This is the first answer in abstract algebra – If $K$ is the algebraic closure of $F$, is $K$ also algebraically closed? – Mathematics Stack Exchange. $\blacksquare$

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