1.11 Algebraic and transcendental elements

Let $F$ be a field. An element $\alpha$ of an extension $E$ of $F$ defines a homomorphism

f(X)\mapsto f(\alpha)\colon F[X]\rightarrow E.

There are two possibilities.

Case 1: The kernel of the map is $(0)$ , so that, for $f\in F[X]$ ,

f(\alpha)=0\implies f=0\text{ (in }F[X]\text{).}

In this case, we say that $\alpha$ transcendental over $F$ . The homomorphism $X\mapsto\alpha\colon F[X]\rightarrow F[\alpha]$ is an isomorphism, and it extends to an isomorphism $F(X)\rightarrow F(\alpha)$ of the fields of fractions.

Case 2: The kernel is $\neq(0)$ , so that $g(\alpha)=0$ for some nonzero $g\in F[X]$ . In this case, we say that $\alpha$ is algebraic over $F$ . The polynomials $g$ such that $g(\alpha)=0$ form a nonzero ideal in $F[X]$ , which is generated by the monic polynomial $f$ of least degree such $f(\alpha)=0$ . We call $f$ the minimal (or minimum) polynomial of $\alpha$ over $F$ .⁸⁸ 8 When we order the polynomials by degree, $f$ is a minimal element of the set of polynomials having $\alpha$ as a root. It is also the unique minimal (hence least or minimum) element of the set of monic polynomials having $\alpha$ as a root. See Wikipedia: partially ordered set. It is irreducible, because otherwise there would be two nonzero elements of $E$ whose product is zero. The minimal polynomial is characterized as an element of $F[X]$ by each of the following conditions,

$\diamond$

$f$ is monic, $f(\alpha)=0$ , and $f$ divides every other $g$ in $F[X]$ such that $g(\alpha)=0$ ;
$\diamond$

$f$ is the monic polynomial of least degree such that $f(\alpha)=0;$
$\diamond$

$f$ is monic, irreducible, and $f(\alpha)=0$ .

Note that $g(X)\mapsto g(\alpha)$ defines an isomorphism $F[X]/(f)\rightarrow F[\alpha]$ . Since the first is a field, so also is the second,

F(\alpha)=F[\alpha].

Thus, $F[\alpha]$ is a stem field for $f$ .

Example 1.28

Let $\alpha\in\mathbb{C}$ be such that $\alpha^{3}-3\alpha-1=0$ . Then $X^{3}-3X-1$ is monic, irreducible, and has $\alpha$ as a root, and so it is the minimal polynomial of $\alpha$ over $\mathbb{Q}$ . The set $\{1,\alpha,\alpha^{2}\}$ is a basis for $\mathbb{Q}[\alpha]$ over $\mathbb{Q}$ . The calculations in Example 1.27 show that if $\beta$ is the element $\alpha^{4}+2\alpha^{3}+3$ of $\mathbb{Q}[\alpha]$ , then $\beta=3\alpha^{2}+7\alpha+5$ , and

\textstyle\beta^{-1}=\frac{7}{111}\alpha^{2}-\frac{26}{111}\alpha+\frac{28}{11% 1}.

Remark 1.29

PARI knows how to compute in $\mathbb{Q}[a]$ . For example, factor(X^4+4) returns the factorization

X^{4}+4=(X^{2}-2X+2)(X^{2}+2X+2)

in $\mathbb{Q}[X]$ . Now type F=nfinit(a^2+2*a+2) to define a number field “F” generated over $\mathbb{Q}$ by a root $a$ of $X^{2}+2X+2$ . Then nffactor(F,x^4+4) returns the factorization

X^{4}+4=(X-a-2)(X-a)(X+a))(X+a+2),

in $\mathbb{Q}[a]$ .

A extension $E$ of $F$ is said to be algebraic (and $E$ is said to be algebraic over $F$ ), if all elements of $E$ are algebraic over $F$ ; otherwise it is said to be transcendental (and $E$ is said to be transcendental over $F$ ). Thus, $E/F$ is transcendental if at least one element of $E$ is transcendental over $F$ .

Proposition 1.30

Let $E\supset F$ be fields. If $E/F$ is finite, then $E$ is algebraic and finitely generated (as a field) over $F$ ; conversely, if $E$ is generated over $F$ by a finite set of algebraic elements, then it is finite over $F$ .

Proof.

$\Longrightarrow$ : To say that an element $\alpha$ of $E$ is transcendental over $F$ amounts to saying that its powers $1,\alpha,\alpha^{2},\ldots$ are linearly independent over $F$ . Thus, if $E$ is finite over $F$ , then every element of $E$ is algebraic over $F$ . It remains to show that $E$ is finitely generated over $F$ . If $E=F$ , then it is generated by the empty set. Otherwise, there exists an $\alpha_{1}\in E\smallsetminus F$ . If $E\neq F[\alpha_{1}]$ , then there exists an $\alpha_{2}\in E\smallsetminus F[\alpha_{1}]$ , and so on. Since

[F[\alpha_{1}]\colon F]<[F[\alpha_{1},\alpha_{2}]\colon F]<\cdots<[E\colon F]

this process terminates with $E=F[\alpha_{1},\alpha_{2},\ldots,\alpha_{n}]$ .

$\Longleftarrow$ : Let $E=F(\alpha_{1},...,\alpha_{n})$ with $\alpha_{1},\alpha_{2},\ldots\alpha_{n}$ algebraic over $F$ . The extension $F(\alpha_{1})/F$ is finite because $\alpha_{1}$ is algebraic over $F$ , and the extension $F(\alpha_{1},\alpha_{2})/F(\alpha_{1})$ is finite because $\alpha_{2}$ is algebraic over $F$ and hence over $F(\alpha_{1})$ . Thus, by (1.20), $F(\alpha_{1},\alpha_{2})$ is finite over $F$ . Now repeat the argument. _□

Corollary 1.31

(a) If $E$ is algebraic over $F$ , then every subring $R$ of $E$ containing $F$ is a field.

(b) Consider fields $L\supset E\supset F$ . If $L$ is algebraic over $E$ and $E$ is algebraic over $F$ , then $L$ is algebraic over $F.$

Proof.

(a) If $\alpha\in R$ , then $F[\alpha]\subset R$ . But $F[\alpha]$ is a field because $\alpha$ is algebraic (see p. 1.11), and so $R$ contains $\alpha^{-1}$ .

(b) By assumption, every $\alpha\in L$ is a root of a monic polynomial

X^{m}+a_{m-1}X^{m-1}+\cdots+a_{0}\in E[X].

Each of the extensions

F[a_{0},\ldots,a_{m-1},\alpha]\supset F[a_{0},\ldots,a_{m-1}]\supset F[a_{0},% \ldots,a_{m-2}]\supset\cdots\supset F

is generated by a single algebraic element, and so is finite. Therefore $F[a_{0},\ldots,a_{m-1},\alpha]$ is finite over $F$ (see 1.20), which implies that $\alpha$ is algebraic over $F$ . _□