(AC) Every algebraic extension of K is of degree 1 (in other words, K is relatively algebraically closed in every extension field of K).

Let us first prove the equivalence of the properties (AC), (AC') and (AC''). Clearly (AC) implies (AC). Since every non-constant polynomial of K[X] is divisible by an irreducible polynomial (IV, p. 13, Prop. 13) and every polynomial of degree 1 in K[X] clearly admits a root in K, we see that (AC) implies (AC'). The condition (AC') implies by induction on n, that every polynomial of degree n in K[X] is a product of n polynomials of degree 1 (IV, p. 14, Prop. 1), hence (AC') implies (AC).

It remains to see that (AC) and (AC) are equivalent. If (AC'') holds, every element of an extension field L of K which is algebraic over K is of degree 1 (V, p. 16, Th. 1), hence belongs to K, which establishes (AC). Conversely, let f be an irreducible polynomial of degree $n \geq 1$ in K[X]; the quotient algebra K[X]/(f) is of degree n over K and is a field, hence an algebraic extension of degree n of K (V, p. 18, Prop. 2). Now it is clear that (AC) implies (AC'').

Let f be a non-constant polynomial in $\bar{K}[X] \subset \Omega[X]$. Since $\Omega$ is algebraically closed, the polynomial f has at least one root in $\Omega$, and since this root is algebraic over $\bar{K}$, it belongs to $\bar{K}$ (V, p. 19, Prop. 4). Therefore $\bar{K}$ satisfies (AC').

\* Example 2. — By Prop. 2 the set of all complex numbers that are algebraic over Q (often called briefly algebraic numbers) is an algebraically closed field. \*

Let K be a finite field and put $f(X) = 1 + \prod_{a \in K} (X - a)$. The polynomial $f \in K[X]$ is non-constant and $f(a) = 1$ for each $a \in K$. So the field K does not satisfy (AC') and hence is not algebraically closed.

By V, p. 13, Scholium, there exists an extension field $\Omega'$ of $\Omega$ and a K-homomorphism u of E into $\Omega'$. Let $x \in E$; since x is algebraic over K, u(x) is algebraic over u(K) and a fortiori over $\Omega$ (V, p. 17, Cor. 2); since $\Omega$ is algebraically closed, we thus have $u(x) \in O$. It follows that $u$ maps $E$ into $O$.

We may take the polynomials $f_i$ to be monic (Remark 1). Let $i \in I$ and let the degree of $f_i$ be $d_i$. By IV, p. 73, Prop. 5 there exists a commutative algebra $A$, over $K$, not reduced to 0, and elements $\xi_{i,1}, \ldots, \xi_{i,d_i}$ of $A_i$ such that :

a) the algebra $A_i$ is generated by $(\xi_{i,1}, \ldots, \xi_{i,d_i})$;

(1) Take for example $K = \mathbf{Q}$ and $f = X^3 - 2$.

b) we have $f_i(X) = \prod_{k=1}^{d_i} (X - \xi_{i,k})$ in $A_i[X]$.

Let $A$ be the tensor product of the family of algebras $(A_i)_{i \in I}$, and let $\varphi_i$ be the canonical homomorphism of $A_i$ into $A$ (III, p. 470). The algebra $A$ is then commutative and not reduced to 0; by Krull's theorem (I, p. 104), there exists thus a maximal ideal $a$ in $A$ and $E = A/a$ is an extension of the field $K$.

Denote by $\psi$ the canonical homomorphism of $A$ into $E$ and put $x_{i,k} = \psi(\varphi_i(\xi_{i,k}))$ for $i \in I$ and $l \leq k \leq d_i$. Since the algebra $A$ is generated by $\bigcup \varphi_i(A_i)$, the extension $E$ is generated by the family $(x_{i,k})$. Further, we have $i \in I$

Therefore $E$ is a splitting extension of the family $(f_i)_{i \in I}$.

Let $R_i$ be the set of roots of $f_i$ in E and $R = \bigcup_{i \in I} R_i$. Since $f_i$ is a product of polynomials of the first degree lying in $F[X]$, we have $R_i \subset F$. By Def. 2, we have $F = K(R)$; in the same way we find that $F = K(R)$.

This follows from Prop. 5 and V, p. 13, Cor. of Prop. 4.

The condition is necessary by (AC). Conversely, suppose that $\Omega$ is algebraic over $K$ and every non-constant polynomial over $K[X]$ is a product in $\Omega[X]$ of factors of degree 1. Let $\Omega'$ be an algebraic extension of $\Omega$ and let $x \in \Omega'$. Since $x$ is algebraic over $\Omega$ and $\Omega$ is algebraic over $K$, $x$ is algebraic over $K$ (V, p. 19, Prop. 3). Let $f$ be the minimal polynomial of $x$ over $K$. By hypothesis the polynomial $f \in K[X]$ splits in $\Omega[X]$ into a product of factors of degree 1, whence $x \in \Omega$. Thus we have $\Omega' = \Omega$ and $\Omega$ is algebraically closed because it satisfies (AC).

Assertion a) follows from Th. 1 (V, p. 20). Now assume the hypotheses of b) and consider a non-constant polynomial $f \in K[X]$. Let E be a splitting field of $f$ (V, p. 21, Prop. 4); since E is algebraic of finite degree over $K$ (V, p. 18, Th. 2), we may suppose that E is a subextension of $\Omega$. Then the polynomial $f$ is a product of polynomials of degree 1 in $\Omega[X]$ and Prop. 6 shows that $\Omega$ is algebraically closed.

We can now prove the existence and uniqueness (up to isomorphism) of the algebraic closure of a field.

By Prop. 6 an algebraic closure of $K$ is nothing other than a splitting extension for the set of all non-constant polynomials in $K[X]$. Th. 2 therefore follows from V, p. 21, Prop. 4 and V, p. 22, Cor.

It is enough to apply Th. 2 to the algebraic closures $\Omega$ and $(\Omega', u)$ of $K$.