Proposition 8

— Let K be a commutative field (of arbitrary characteristic) and let $K'$ be a splitting field for the polynomial $X^2 + 1 \in K[X]$ (V, p. 21). Suppose: a) every polynomial in $K[X]$ of odd degree has a root in $K'$;

b) every polynomial in $K'[X]$ of degree 2 has a root in $K'$. Then $K'$ is algebraically closed.

Note first that it is enough to prove that every non-constant polynomial in $K[X]$ has a root in $K'$: this is indeed clear if $K' = K$; if $K' \neq K$ then $[K':K] = 2$; let $a \mapsto a$ denote the unique $K$-automorphism of $K'$ distinct from the identity map; if $f \in K'[X]$ and if $\bar{f}$ denotes the polynomial obtained by applying $a \mapsto \bar{a}$ to the coefficients of $f$, then $f \bar{f} \in K[X]$; if $a \in K'$ is a root of $f \bar{f}$ then $a$ is either a root of $f$ or of $\bar{f}$; thus either $a$ or $a$ is a root of $f$.

Thus let $f$ be a polynomial over $K$ of degree $2^n p$, $p$ odd. We will proceed by induction on $n$, the property being true for $n = 0$ by hypothesis a). Let E be an extension of $K$ in which $f$ splits into linear factors:

Let $b \in K$; put $y_{ij} = a_i + a_j + ba_i a_j \in E$ and

The coefficients of this polynomial are symmetric functions in the $a_i$, with coefficients in $K$; it therefore belongs to $K[X]$ (IV, p. 62, Th. 1); since it has degree $2^n p (2^n p - 1)/2 = 2^{n-1} p'$ ($p'$ odd), it has a root $y_{}$, in $K'$ by inductive hypothesis. If we note that this holds for all $b \in K$, and that $K$ is an infinite field (indeed a finite field, which has monogenous extensions of arbitrarily large odd degree (V, p. 94, Prop. 3), cannot satisfy a)), then we can deduce the existence of at least one pair $(i, j)$ such that

with $b \neq b'$. Then $a_i + a_j$ and $a_i a_j$ are elements of $K'$, hence so are $a_i$ and $a_j$, since they are the roots of the quadratic equation

Q.E.D.

For a generalisation and an alternative proof of Prop. 8, based on Galois theory, see VI, p. 46, Ex. 33.

Let $K$ be an ordered field and let $K' = K(i)$; for every element $z = a + bi$ of $K'$, the norm $z \overline{z} = a^2 + b'$ of $z$ relative to $K$ (III, p. 544, example 1) is a positive element of $K$, which vanishes only for $z = 0$. If every positive element in $K$ is a square (in particular if $K$ is a maximal ordered field), then the positive square root of the norm $z \overline{z}$ is called the absolute value of $z$, and is written $|z|$. Since $|zz'|^2 = |z|^2 |z'|^2$ we have $|zz'| = |z| \cdot |z'|$.

Moreover, the triangle inequality

holds for every pair of elements $z, z'$ of $K'$. Indeed, if $z = a + bi$ and $z' = a' + b'i$, then this inequality is equivalent to

and hence also to

which can be written $(ab' - ba')^2 \geq 0$.

Th. 3 enables us to determine all the irreducible polynomials over a maximal ordered field: