We use Corollary 3. Let \(x\in \mathbb {Q}\) be non-negative. Then \(x=p/q\) for some integers \(p\geq 0\) and \(q{\gt}0\), so

Consider the sum \(\sum _i x_i\alpha _i^2\) for some \(\alpha _i\in K\) and \(x_i\in F_{\geq 0}\). By \(F\)-linearity, we compute

Since \(\pi (-1)=-1{\lt}0\), we are done by Theorem 5.

Let \(\pi :F(\sqrt{a})\to F\) be the projection induced by the \(F\)-basis \(\{ 1,\sqrt{a}\} \). For \(x,y\in F\), we have \(\pi ((x+y\sqrt{a})^2)=x^2+ay^2\geq 0\). We are done by Lemma 6.

Since \(\operatorname{char}R=0\), we can apply the primitive element theorem. Let \(K=F(\alpha )\) for some \(\alpha \in K\), and let \(f\) be the minimal polynomial of \(\alpha \) over \(K\). Then \(\deg f=[K:F]\) is odd. By Theorem 5, we need to show that the congruence

fails to hold for any non-negative \(a_i\in F\) and polynomials \(g_i\in F[X]\) each of degree at most \(\deg f-1\). Proceed by induction on \(\deg f\); if \(\deg f=1\), then (\(\star \)) reduces to an equality of a non-negative element of \(F\) with a negative one. Otherwise, suppose for a contradiction that (\(\star \)) holds. Without loss of generality, we may assume, for all \(i\), we have \(a_i\neq 0\) and that \(\deg g_i{\lt}\deg f\).

Rearranging (\(\star \)), we have \(\sum _ia_ig_i^2+1=hf\) for some \(h\in F[X]\). Let \(d=\max _i\deg g_i\); note that \(d{\lt}\deg f\) by construction. Since each \(a_i\) is positive, the \(2d\)th coefficient on the left-hand side must be positive. Therefore

Then \(\deg h\) is odd, so \(h\) has an odd-degree irreducible factor \(\tilde h\). We have

but \(\sum _ia_ig_i^2\equiv -1\bmod \tilde h\). We are done by induction.\(_\# \)

For the reverse direction, exactly one of \(1\) and \(-1\) is a square, so \(-1\) is not a square.

For the reverse direction, squares in an ordered field are non-negative.

This is Lemma 2.

Let \(K/R\) be an odd-degree extension of \(R\). By the primitive element theorem, \(K=R(\alpha )\) for some \(\alpha \in K\). Let \(f\) be the minimal polynomial of \(\alpha \) over \(K\). Then \(f\) is irreducible, but \(\deg f=[K:R]\) is odd, so \(f\) has a root in \(R\). Therefore, \([K:R]=\deg f=1\); that is, \(K\cong R\).

Let \(K/R\) be a quadratic extension. Since \(\operatorname{char}R\neq 2\), we have \(K\cong R(\sqrt{a})\) for some \(a\in R\). Since \(a\) cannot be a square in \(R\), we know \(-a\) must be one. Rescaling by \(\sqrt{-a}\), we get \(R(\sqrt{a})\cong R(i)\).

Since \(\operatorname {char}R(i)\neq 2\), it suffices to show that every element of \(R(i)\) is a square. Observe that every \(x\in R\) is a square in \(R(i)\): if \(x\) is not a square in \(R\), then \(-x\) is, and so \(x=(i\sqrt{-x})^2\). Further, since the negation of a sum of squares cannot be a square in \(R\) (otherwise, dividng through and rearranging, \(-1\) is a sum of squares), sums of squares in \(R\) are squares in \(R\). Now, fix \(x=a+bi\in R(i)\) with \(a,b\in R\). If \(b=0\), then \(a=x\) is a square in \(R(i)\). Otherwise, we have \(x=(c+di)^2\), where

Note that \(d\) is well-defined because, if \(c=0\), then \(-a=\sqrt{a^2+b^2}\) and so \(b=0\).

By separability, every finite extension of \(R\) is contained in a finite Galois extension. Since \(R(i)/R\) has no intermediate fields, it suffices to show the result for finite Galois extensions.

Let \(K/R\) be a nontrivial Galois extension of degree \(2^k\cdot a\), where \(k\geq 0\) and \(a\geq 1\) is odd. Applying the Galois correspondence to a Sylow 2-subgroup of \(\operatorname{Gal}(K/R)\) yields an intermediate extension of degree \(a\); by Lemma 14, we have \(a=1\) (and \(k{\gt}0\)). If \(k{\gt}1\), iterating the last construction yields intermediate extensions \(K/L/M/R\) with \([L:M]=[M:R]=2\). By Lemma 15, \(M\cong R(i)\), contradicting Lemma 16.\(_\# \) Therefore \(k=1\) and (by Lemma 15) \(K\cong R(i)\).

An infinite algebraic extension contains finite subextensions of arbitrarily large degree.

Given \(a,b\in F\), find \(c,d\in F\) such that \(a+bi=(c+di)^2\) in \(F(i)\). Then \(a=c^2-d^2\) and \(b=2cd\), so \(a^2+b^2=(c^2+d^2)^2\) is a square in \(F\). By induction, every sum of squares in \(F\) is a square in \(F\).

By Lemma 20, \(R\) is real.

Take a non-square \(a\) in \(R\). Then \(R(\sqrt{a})\cong R(i)\). Suppose \(i\) maps to \(x+y\sqrt{a}\) for some \(x,y\in R\); then \(-1=x^2+ay^2+2xy\sqrt{a}\). Comparing coefficients, we get \(-1=x^2+ay^2\). Since \(-1\) is not a square, \(y\) must be nonzero, and so \(-a=(x/y)^2\) is a square.

Now, fix a nonlinear odd-degree polynomial \(f\in R[X]\). Then \(R[X]/(f)\) cannot be a field since \(R\) has no nontrivial odd-degree extensions, and so \(f\) must be reducible. We are done by induction on the degree.

The field \(R(i)\) is not real since \(-1\) is a square in it. We are done by Corollary 18.

Since ordered fields are real, we are done by Lemma 22.

Let \(f\in R[X]\) be monic and irreducible. The field \(R_f=R[X]/(f)\) is a finite extension of \(R\), so it is classified by Theorem 17. If \(R_f\cong R\), then \(\deg f=1\), so \(f=X-c\) for some \(c\in R\). If \(R_f\cong R(i)\), let the isomorphism be \(\varphi \), and suppose \(\varphi (X+(f))=a+bi\) (\(a,b\in R\)). Note that \(b\neq 0\) since \(\varphi ^{-1}\) is constant on \(R\). Rearranging, we see that \(\varphi ((X-a)^2+b^2+(f))=0\); that is, \((X-a)^2+b^2\in (f)\). Since this polynomial is monic and has the same degree as \(f\), it must in fact be equal to \(f\).

Conversely, linear polynomials over a domain are irreducible by degree, and reducible quadratics have a root. A root of \(f=(X-a)^2+b^2\) with \(a,b\in R\) is an element \(r\in R\) satisfying \((r-a)^2=-b^2\). Since squares are non-negative, if \(b\neq 0\) then \(f\) must be irreducible.

We will prove that, for all \(f\in R[X]\) and all \(a,b\in R\) with \(a\leq b\), if \(f(a)\leq 0\leq f(b)\), then there is some \(c\in [a,b]\) such that \(f(c)=0\).

Fix \(a,b\in R\) with \(a\leq b\), and proceed by induction on \(\deg f\). If \(\deg f=0\), then \(f\) is constant and the result is clear. Otherwise, take a monic irreducible factor \(g\) of \(f\); then \(g\) is classified by Lemma 24.

Observe that, if \(g(a)\) and \(g(b)\) are both positive, then \(f/g\) satisfies the inductive hypothesis, and so has a root in \([a,b]\). If \(g=(X-a)^2+b^2\) with \(a,b\in R\) and \(b\neq 0\), then \(g\) is everywhere positive. If \(g=X-c\) with \(c\in R\), then either \(c\in [a,b]\) and \(g(c)=0\), or \(c\notin [a,b]\) and \(f(a)\) and \(f(b)\) have the same sign. In this last case, either \(f/g\) or \(f/(-g)\) satsifies the inductive hypothesis. In all cases, \(f\) has a root in \([a,b]\).

We use Lemma 12. Let \(a\in R\) be non-negative, and consider the polynomial \(f=X^2-a\). Then \(f(0)=-a\leq 0\), but \(f(a+1)=a^2+a+1{\gt}0\). By the intermediate value property, \(f\) has a root in \(R\), and so \(a\) is a square in \(R\).

Let \(f\) be an odd-degree polynomial over \(R\). Write \(f=a_nX^n+\cdots +a_0\). We will show \(f\) has a root in \(R\). Replacing \(f\) by \(-f\) if necessary, we may assume \(a_n{\gt}0\). For \(x{\gt}1\), we compute

Therefore, when \(x{\gt}\max \{ 1,n\max _i|a_i|/a_n\} \), \(f(x){\gt}0\). A similar calculation shows that \(f(x){\lt}0\) for sufficiently large negative values of \(x\). We are done by the intermediate value property.

We use Lemma 12.

Let \(a\in R\) be non-negative, and suppose \(a\) is not a square. By Corollary 7, there is an ordering making the nontrivial extension \(R(\sqrt{a})/R\) ordered.\(_\# \) Therefore \(a\) is a square in \(R\).

By induction on degree, it suffices to show that irreducible odd-degree polynomials over \(R\) are all linear. Let \(f\in R[X]\) be such a polynomial, and consider the odd-degree field extension \(R_f=R[X]/(f)\). By Lemma 8, there is an ordering making \(R_f/R\) an ordered extension. Therefore \(\deg f=[R_f:R]=1\).

Since ordered fields are real, we are done by Theorem 27.

Let \(C\) be an algebraically closed field of characteristic 0. Observe that the prime subfield \(\mathbb {Q}\) can be ordered. Further, given an ordered subfield \(F\) with \(\bar{F}\neq C\), we can use Lemma 9 to adjoin an element transcendental over \(F\), obtaining a strictly bigger ordered subfield.

Apply Zorn’s lemma to obtain a maximal ordered subfield \(R\subseteq C\); then \(\bar{R}=C\). By Theorem 27, \(R\) must be real closed. By Corollary 19, \(C\cong R(i)\), and so \([C:R]=2\).

Apply Zorn’s lemma to ordered algebraic extensions of \(F\). We are done by Theorem 27.

TODO : decide on the generality of the statement of Sturm’s Theorem

By the primitive element theorem, \(K=F(\alpha )\) for some \(\alpha \in K\). Let \(f\) be the minimal polynomial of \(\alpha \) over \(F\). Since \(F\) has a root in \(K\), it has a root in a real closure of \(K\) (one exists by Lemma 33). By Theorem 34, \(f\) has a root \(\beta \) in \(R\). Therefore define \(\varphi :K\to R\) with \(\varphi (\alpha )=\beta \).

Fix a real closure \(R'\) of \(K\) (one exists by Lemma 33).

By the primitive element theorem, \(K=F(\alpha )\) for some \(\alpha \in K\). Let \(f\) be the minimal polynomial of \(\alpha \) over \(F\), and let \(\alpha _1{\lt}\cdots {\lt}\alpha _m\) be the roots of \(f\) in \(R'\), with \(\alpha =\alpha _k\). By Theorem 34, \(f\) also has \(m\) roots in \(R\); let them be \(\beta _1{\lt}\cdots {\lt}\beta _m\). Since non-negative elements of \(R'\) are squares, and \(x_1,\dots ,x_{m-1}\in R'\) such that \(\alpha _{j+1}-\alpha _j=x_j^2\), and let \(L=K(\alpha _1,\dots ,\alpha _m,r,x_1,\dots ,x_{m-1})\leq R'\). Now, suppose we have a \(K\)-homomorphism \(\psi :L\to R\). Each \(\psi (\alpha _j)\) is equal to a different \(\beta _i\). Then \(\psi (\alpha _{j+1})-\psi (\alpha _j)=\psi (x_j)^2\geq 0\), so \(\psi (\alpha _1){\lt}\cdots {\lt}\psi (\alpha _m)\), and so \(\psi (\alpha _j)=\beta _j\) for all \(j\).

By Lemma 35, there is in fact an \(F\)-homomorphism \(\varphi :L\to R\). We will show that \(\varphi \) is order-preserving. Indeed, fix a non-negative element \(x\in L\). As before, find \(r\in R'\) such that \(x=r^2\), and let \(M=L(r)\leq R'\). Apply Lemma 35 again to obtain an \(L\)-homomorphism \(\psi :M\to R\); then \(\varphi (x)=\psi (x)=\psi (r)^2\geq 0\). Therefore \(\varphi \) maps non-negative elements to non-negative elements, and so is order-preserving. Then \(\varphi |_K\) is the map we want. Note that \(\varphi (\alpha )=\beta _k\).

To see uniqueness, let \(\tilde\varphi :K\to R\) be an order-preserving \(F\)-homomorphism; by existence, \(\tilde\varphi \) extends to a an order-preserving \(K\)-homomorphism \(\tilde\psi :L\to R\). Then \(\tilde\varphi (\alpha )=\tilde\psi (\alpha _k)=\beta _k=\varphi (\alpha )\), and so \(\tilde\varphi =\varphi \).

Let \(R_1\) and \(R_2\) be real closures of \(F\). Applying Zorn’s lemma to the set of ordered extensions intermediate between \(R_1\) and \(F\) having a unique order-preserving \(F\)-embedding into \(R_2\), and using Lemma 36, we obtain an intermediate extension \(R_1/K/F\) with no nontrivial finite ordered extensions and a unique order-preserving \(F\)-embedding \(\varphi :K\to R_2\). If the ordered algebraic extension \(R_2/\varphi (K)\) were nontrivial, then it would contain a nontrivial ordered finite extension, so \(\varphi \) must be surjective (and so an \(F\)-isomorphism). In particular, \(K\subseteq R_1\) is real closed; by maximality (Lemma 22), in fact \(K=R_1\) and so \(\varphi \) is an \(F\)-isomorphism between \(R_1\) and \(R_2\).

Let \(R\) be a real closed field. By Theorem 37, \(R\) has no nontrivial order-preserving automorphisms. Since the ordering on \(R\) is unique (by Lemma 13), every automorphism of \(R\) must be order-preserving.

Fix an ordering on \(K\) extending that on \(F\), and let \(K\) have real closure \(R_K\) (exists by Lemma 33). Then \(R_K/F\) is algebraic, so \(R_K\) is a real closure of \(F\). By Theorem 37, there is an \(F\)-isomorphism to \(R_K\cong R\), and this induces an \(F\)-homomorphism \(K\to R\). Restricting the order on \(R\) to \(K\) via this map recovers the original order on \(K\) by construction.

Moreover, the inverse to order restriction constructed above is unique. Indeed, an inverse \(\varphi :K\to R\) is order-preserving by definition, so it is an order-embedding from \(K\) into its real closure. By Theorem 37, such a map is unique.

By Lemma ??, it suffices to show that \(\bar{R}\cong R(i)\).

Since \(\operatorname{char}\bar{R}\neq 2\), we have \(\bar{R}\cong R(\sqrt{a})\) for some non-square \(a\in R\). Since \(R(\sqrt{a})\) is algebraically closed, \(\sqrt{a}\) is a square in \(R(\sqrt{a})\); find \(x,y\in R\) such that \(\sqrt{a}=(x+y\sqrt{a})^2\). Expanding and comparing coefficients, \(x^2+y^2a=0\) and \(2xy=1\). Rearranging, \(a=-(4x^4)=-1\cdot (2x^2)^2\). Rescaling by \(2x^2\), \(R(i)\cong R(\sqrt{a})=\bar{R}\).

TODO: this needs a lot more preliminaries eg Artin-Schreier theory, Kummer theory

Ordered fields have characteristic 0.

Since \(\mathbb {Q}_{\text{alg}}(i)=\bar{\mathbb {Q}}\), the field \(\mathbb {Q}_{\text{alg}}\) is a finite-index subfield.

By Theorem 41, any finite-index subfield has a unique ordering making it real closed. Let \(R\) be a real closed subfield. Then the ordering on \(R\) restricts to the ordering on \(\mathbb {Q}\) (unique by Corollary 4). Since \(R/\mathbb {Q}\) is algebraic, \(R\) is a real closure of \(\mathbb {Q}\). By Theorem 37, there is a unique \(\mathbb {Q}\)-isomorphism \(\psi :\mathbb {Q}_{\text{alg}}\cong R\). Since \(\mathbb {Q}_{\text{alg}}(i)=R(i)=\bar{\mathbb {Q}}\), \(\psi \) extends uniquely to \(\varphi \in \operatorname{Gal}(\mathbb {Q}_{alg}/\mathbb {Q}(i))\) by mapping \(i\to i\). The subfield \(R\) is then recovered from \(\varphi \in \operatorname{Gal}(\mathbb {Q}_{alg}/\mathbb {Q}(i))\) by taking \(\varphi ^{-1}(\mathbb {Q}_{\text{alg}})\).