§ 3. Algebraic extensions

1. Algebraic elements of an algebra

Let A be an algebra over a field K and x an element of A. Two cases are possible:

a) The family of monomials $(x^n)_{n \in \mathbf{N}}$ is free over $K$ . Then we say that $x$ is transcendental over $K$ . There is an isomorphism of the polynomial algebra $K[X]$ onto the subalgebra $K[x]$ of $A$ generated by $x$ , and the latter is of infinite degree over $K$ .

b) There exists an integer $n \geq 1$ such that the monomials $1,\ x,\ \ldots,\ x^{n-1},\ x^n$ are linearly dependent; it comes to the same to say that there exists a polynomial $f \neq 0$ in $K[X]$ such that $f(x) = 0$ . Then we say that $x$ is algebraic over $K$ . The least integer $n \geq 1$ satisfying the above property is called the degree of $x$ over $K$ . If the degree of $x$ over $K$ is $n$ , then the monomials $1,\ x,\ \ldots,\ x^{n-1}$ are linearly independent over $K$ and there exist elements $a_0,\ a_1,\ \ldots,\ a_{n-1}$ of $K$ such that

x^n = a_0 + a_1 x + \cdots + a_{n-1} x^{n-1}

The polynomial

f(X) = X^n - \sum_{k=0}^{n-1} a_k X^k

is the unique monic polynomial of degree

n

K[X]

such that

f(x) = 0

; it is called the minimal polynomial of

x

over

K

Theorem 1

— Let $A$ be an algebra over a field $K$ , $x$ an element of $A$ algebraic over $K$ , $n$ the degree and $f$ the minimal polynomial of $x$ over $K$ .

a) For a polynomial $g \in K[X]$ to satisfy $g(x) = 0$ it is necessary and sufficient that $g$ should be a multiple of $f$ .

b) The mapping $g \mapsto g(x)$ defines by passage to quotients an isomorphism of the quotient algebra $K[X]/(f)$ onto the algebra $K[x]$ , and the elements $1,\ x,\ \ldots,\ x^{n-1}$ form a basis of $K[x]$ over $K$ . In particular, $[K[x]:K] = n$ .

c) Assume that $A$ is an integral domain. Then $K[x]$ is a field and $f$ the unique monic irreducible polynomial in $K[X]$ such that $f(x) = 0$ .

d) For $x$ to be invertible in $A$ it is necessary and sufficient that $f(0) \neq 0$ ; then we have $x^{-1} \in K[x]$ .

There exists a unique algebra homomorphism $\varphi : K[X] \to A$ such that $\varphi(X) = x$ ; we have $\varphi(P) = P(x)$ for every $P \in K[X]$ and the image of $\varphi$ equals $K[x]$ . Let $a$ be the kernel of $\varphi$ : by construction, the minimal polynomial $f$ of $x$ over $K$ belongs to $a$ and it is the monic polynomial of least degree in $a$ . Therefore ( $IV$ , p. 11, Prop. 11) we have $a = (f)$ , whence $a$ . Assertion $b$ ) follows at once from a) and the Cor. of Prop. 10 of $IV$ , p. 11.

Suppose that $A$ is an integral domain. The algebra $K[x]$ is then an integral domain and of finite degree over $K$ , hence it is a field ( $V$ , p. 10, Cor.). The ideal $(f)$ of $K[x]$ is thus maximal, that means that $f$ is irreducible in $K[X]$ ( $IV$ , p. 13). Finally, let $g$ be a monic irreducible polynomial in $K[X]$ such that $g(x) = 0$ ; by $a$ ) it is a multiple of $f$ , hence $g = f$ , and this proves $c$ ).

It remains to prove $d$ ). There exists a polynomial $g \in K[X]$ of degree $n-1$ and an element $a$ of $K$ such that $f(X) = Xg(X) + a$ , whence $f(0) = a$ . If $a = 0$ , we have $xg(x) = f(x) = 0$ , and $g(x) \neq 0$ , so then $x$ is not invertible in $A$ . If on the other hand $a \neq 0$ , then we have $x \cdot [-a^{-1}g(x)] = 1$ , so $x$ is then invertible in $A$ and $x^{-1} = -a^{-1}g(x)$ .

Corollary 1

— Let $A$ be an algebra over a field $K$ . For an element $x$ of $A$ to be algebraic over $K$ it is necessary and sufficient that the subalgebra $K[x]$ of $A$ generated by $x$ should be of finite degree over $K$ . In particular, if $A$ is of finite degree over $K$ , every element of $A$ is algebraic over $K$ .

Corollary 2

— Let $E$ be an extension of $K$ , $A$ an algebra over $E$ and $x$ an element of $A$ algebraic over $K$ . Then $x$ is algebraic over $E$ , the minimal polynomial of $x$ over $E$ divides the minimal polynomial of $x$ over $K$ and the degree of $x$ over $E$ is at most equal to the degree of $x$ over $K$ .

For let $f$ be the minimal polynomial of $x$ over $K$ ; we have $f(x) = 0$ and $f \in E[X]$ , hence $x$ is algebraic over $E$ and $f$ is a multiple of the minimal polynomial of $x$ over $E$ (Th. 1, a).

Remark

Let $E$ be an extension of a field $K$ and $x$ an element of $E$ which is root of a monic irreducible polynomial $f \in K[X]$ . Th. 1, c) then shows that $f$ is the minimal polynomial of $x$ over $K$ .

Example 1

In the field of complex numbers $\mathbf{C}$ , the number $i$ is algebraic of degree 2 over the prime field $\mathbf{Q}$ ; for if $f(X) = X^2 + 1$ , then $f(i) = 0$ , and $x^2 + 1 \neq 0$ for all $x \in \mathbf{Q}$ , hence $i \notin \mathbf{Q}$ . The field $\mathbf{Q}(i)$ is thus an extension of degree 2 of $\mathbf{Q}$ ; it consists of all numbers $a + bi$ , where $a, b$ are rational. Likewise $i$ is algebraic of degree 2 over the field $\mathbf{R}$ of real numbers, and $\mathbf{C}$ is an extension of degree 2 of $\mathbf{R}$ . \

Example 2

Let $K$ be a field and $F$ the field $K(X)$ of rational functions in one indeterminate over $K$ . Let $E$ be the subfield $K(X^3)$ of $F$ ; we have $F = E(X)$ and $X$ is algebraic over $E$ , since it is a root of the polynomial $Y^3 - X^3$ of the ring $E[Y]$ ; this polynomial is irreducible in $E[Y]$ , for in the contrary case it would have at least one factor of the first degree, and there would then exist two non-zero polynomials $u(X)$ , $v(X)$ of $K[X]$ such that $(u(X^3))^3 = X^3(v(X^3))^3$ which is absurd, for if $m$ and $n$ are the degrees of $u$ and $v$ , this would imply $9m = 9n + 3$ , or $3m = 3n + 1$ . The field $F$ is thus of degree 3 over $E$ , and every element of $F$ may be written in just one way as a linear combination $f(X^3) + Xg(X^3) + X^2 h(X^3)$ , where $f, g, h$ are three rational fractions of $K(X)$ .

Example 3

\* In the field $\mathbf{R}$ of real numbers it may be shown $^1$ that the number $\pi$ is transcendental over the prime field $\mathbf{Q}$ . \

2. Algebraic extensions

Definition 1

— An extension $E$ of a field $K$ is said to be algebraic (over $K$ ) if every element of $E$ is algebraic over $K$ . An extension $E$ of $K$ which is not algebraic is called transcendental (over $K$ ).

Proposition 1

— For an extension $E$ of $K$ to be algebraic it is necessary and sufficient that every sub-K-algebra $A$ of $E$ should be a field.

$^1$ Cf. for example D. Hilbert, Gesammelte Abhandlungen, vol. 1, p. 1 (Berlin (Springer), 1932).

The condition is necessary : if $E$ is algebraic over $K$ and $x \neq 0$ is an element of a sub-K-algebra $A$ of $E$ , then $x^{-1} \in K[x] \subset A$ by $V$ , p. 16, Th. 1, d). Therefore $A$ is a field.

The condition is sufficient : if it is satisfied and $x$ is an element $\neq 0$ in $E$ , then the ring $K[x]$ is a field, hence $x^{-1} \in K[x]$ ; in other words, there exists a polynomial $g \in K[X]$ such that $x^{-1} = g(x)$ , or also $xg(x) - 1 = 0$ ; this shows $x$ to be algebraic over $K$ , hence $E$ is an algebraic extension of $K$ .

Proposition 2

— *If an extension E of K is of finite degree n, then it is algebraic and the degree over K of each element of E divides n.

For, given $x \in E$ , $[K(x):K]$ is finite and divides n ( $V$ , p. 10, Cor. 1) and hence $x$ is algebraic over $K$ ( $V$ , p. 17, Cor. 1).

There exist algebraic extensions of infinite degree, for example the algebraic closure of a finite field (V, p. 24, Remark 4).

Theorem 2

— *Let E be a finitely generated extension of K, generated by the elements $a_1, ..., a_m$ , which are algebraic over K; then E is an extension of finite degree of K. If the degree of $a_i$ over $K(a_1, a_2, ..., a_{i-1})$ is $n_i$ (for $1 \leq i \leq m$ ), then the degree of E over K is $n_1 n_2 ... n_m$ and the elements $a_1^{v_1} a_2^{v_2} ... a_m^{v_m}$ ( $0 \leq v_i \leq n_i - 1$ ) form a basis of E over K.

The elements $a_i^{v_i}$ ( $0 \leq v_i \leq n_i - 1$ ) form a basis of $K(a_1, a_2, ..., a_i)$ over $K(a_1, a_2, ..., a_{i-1})$ by Th. 1, b) of $V$ , p. 16; the theorem therefore follows by induction on $m$ from Prop. 25 of II, p. 222.

Corollary 1

— *Let E be an extension of K and A a subset of E consisting of elements that are algebraic over K. Then $K(A)$ is algebraic over K and we have $K[A] = K(A)$ .

For each $x \in K(A)$ belongs to a field $K(F)$ , where F is a finite subset of $A$ ( $V$ , p. 11, Cor.); now $K(F)$ is algebraic over $K$ and equal to $K[F]$ by Th. 2, hence $x$ is algebraic over $K$ and $K(A) = K[A]$ .

Corollary 2

— *Let L be an extension of K and E, F subextensions of L. If F is algebraic over K, the subring $K[E, F]$ of L generated by $E \cup F$ is a field coinciding with $E(F)$ and algebraic over E.

For each element of $F$ , being algebraic over $K$ , is also algebraic over $E$ ( $V$ , p. 17, Cor. 2) hence $E(F)$ is an algebraic extension of $E$ , and we have $E(F) = E[F]$ by Cor. 1.

Remark 1

With the notation of Th. 2, $E = K[a_1, a_2, ..., a_n]$ and hence $E$ is isomorphic to a quotient $K[X_1, X_2, ..., X_m]/\mathfrak{a}$ ; since $E$ is a field, $a$ is a maximal ideal in $K[X_1, ..., X_m]$ .

Remark 2

Let $E$ be an algebraic extension of $K$ , of infinite degree. By Th. 2 there exists an infinite sequence $(a_i)$ , of elements of $E$ such that $a_i \notin K(a_1, a_2, ..., a_{i-1})$ ; Th. 2 shows further that the degree of $K(a_1, a_2, ..., a_n)$ over $K$ takes arbitrarily large values. In other words, if E is an algebraic extension of K such that the degrees [F : K] of the subextensions F of E of finite degree over K are bounded, then E is an extension of finite degree of K.

3. Transitivity of algebraic extensions. Fields that are relatively algebraically closed in an extension field

Proposition 3

— Let E and F be two extension fields of a field K such that K ⊂ E ⊂ F. For F to be algebraic over K it is necessary and sufficient that E should be algebraic over K and F algebraic over E.

The condition is necessary, by V, p. 17, Cor. 2. Let us show that it is sufficient. Let x be any element of F; it is algebraic over E; let g ∈ E[X] be its minimal polynomial over E. If A is the (finite) set of coefficients of g, then g ∈ K(A)[X], hence x is algebraic over K(A) and K(A U {x}) = K(A)(x) is of finite degree over K(A). Now A ⊂ E and E is algebraic over K, hence K(A) is of finite degree over K, by Th. 2. Therefore (V, p. 10, Th. 1), K(A U {x}) is of finite degree over K, which shows x to be algebraic over K (V, p. 18, Prop. 2).

Definition 2

— A subfield K of a field E is said to be relatively algebraically closed in E if every element of E which is algebraic over K, belongs to K.

It comes to the same to say that K is the only algebraic extension of K contained in E. Every field K is relatively algebraically closed in itself. In § 4 we shall study the fields which are relatively algebraically closed in each extension field.

Proposition 4

— Let E be an extension of a field K; the set L of elements of E which are algebraic over K forms a subextension of E which is relatively algebraically closed in E.

For the field K(L) is algebraic over K (Cor. 1 of Th. 2), hence K(L) ⊂ L; it follows that K(L) = L and L is a field. On the other hand, if x ∈ E is algebraic over L, it is so also over K (Prop. 3) and hence belongs to L.

The extension L of K consisting of all the elements of E that are algebraic over K is called the relative algebraic closure of K in E; it is the largest algebraic extension of K contained in E.

Exercises

See the exercises for § 3.