[proofplan] We prove the equivalence by isolating the algebraic content of real closedness. The central implication is that if $F(i)$ is algebraically closed and $F$ is not, then the nonzero squares of $F$ form the positive cone of an ordering; conjugation in $F(i)$ then forces odd-degree polynomials over $F$ to have roots in $F$. Conversely, if an ordered field has square roots for all positive elements and roots for all odd-degree polynomials, then $F(i)$ is quadratically closed and has no nontrivial finite extension, by a finite Galois-theoretic reduction to odd-degree extensions and quadratic extensions. The maximal ordered-extension formulation is equivalent to the same condition by the standard algebraic [extension theorem](/theorems/59) for orderings. [/proofplan] [step:Reduce the terminology of real closed ordered fields to the square-root and odd-degree conditions] In this theorem, an ordered field $(F,<)$ is called real closed when every positive element of $F$ is a square in $F$ and every polynomial in $F[X]$ of odd degree has a root in $F$. With this convention, conditions $1$ and $3$ are identical. The remaining work is therefore to connect this square-root and odd-degree formulation with the algebraic condition that $F(i)$ is algebraically closed and with maximality among algebraic ordered field extensions. The maximality equivalence is proved directly in the final step, using the standard existence theorem for real closures of ordered fields. [/step] [step:Construct the ordering from the algebraic closedness of $F(i)$] Assume that $F$ is not algebraically closed and that $K:=F(i)$ is algebraically closed. Since $K$ is algebraically closed but $F$ is not, we have $i \notin F$, so $X^2+1$ is irreducible over $F$ and $\operatorname{char} F \neq 2$. Let $F^\times := F \setminus \{0\}$ denote the multiplicative group of nonzero elements of $F$. Define \begin{align*} P := (F^\times)^2 = \{a^2 : a \in F^\times\}. \end{align*} We prove that $P$ is the positive cone of an ordering on $F$. First, $P \cdot P \subset P$ is immediate. Also $P \cap (-P)=\varnothing$, because if $a^2=-b^2$ for nonzero $a,b \in F$, then $(a/b)^2=-1$, hence $i \in F$, contradicting $i \notin F$. Next let $x \in F^\times$. Since $K$ is algebraically closed, there exists $z \in K$ with $z^2=x$. Write $z=a+bi$ with $a,b \in F$. Then \begin{align*} x=z^2=(a^2-b^2)+2ab\,i. \end{align*} Because $x \in F$ and $1,i$ are $F$-linearly independent, $2ab=0$. Since $\operatorname{char} F \neq 2$, either $a=0$ or $b=0$. If $b=0$, then $x=a^2 \in P$; if $a=0$, then $x=-b^2 \in -P$. Hence \begin{align*} F^\times = P \sqcup (-P). \end{align*} It remains to prove $P+P \subset P$. Let $u^2,v^2 \in P$ with $u,v \in F^\times$. Then \begin{align*} u^2+v^2 = u^2\left(1+\left(\frac{v}{u}\right)^2\right). \end{align*} Thus it suffices to show that $1+c^2 \in P$ for each $c \in F$. Since $K$ is algebraically closed, choose $z \in K$ satisfying \begin{align*} z^2=1+ci. \end{align*} Write $z=a+bi$ with $a,b \in F$. Define the norm map $N_{K/F}:K \to F$ by \begin{align*} N_{K/F}(x+yi)=(x+yi)(x-yi)=x^2+y^2 \end{align*} for $x,y \in F$. Applying $N_{K/F}$ to $z^2=1+ci$ gives \begin{align*} 1+c^2=N_{K/F}(1+ci)=N_{K/F}(z^2)=N_{K/F}(z)^2=(a^2+b^2)^2. \end{align*} The element $1+c^2$ is nonzero, because $1+c^2=0$ would imply $c^2=-1$ and hence $i \in F$. Therefore $a^2+b^2 \neq 0$, and $1+c^2 \in P$. Hence $P+P \subset P$. Define an ordering $<$ on $F$ by \begin{align*} x<y \quad \Longleftrightarrow \quad y-x \in P. \end{align*} The preceding properties of $P$ are exactly the positive-cone axioms for an ordered field. By construction, the positive elements for this ordering are precisely the nonzero squares of $F$. [guided] The goal is to recover an order using only the algebraic hypothesis that $F(i)$ is algebraically closed. Let $F^\times := F \setminus \{0\}$ denote the multiplicative group of nonzero elements of $F$. The natural candidate is the set of nonzero squares: \begin{align*} P := (F^\times)^2. \end{align*} For $P$ to define an ordering, we must prove three facts: every nonzero element is exactly one of positive or negative, products of positives are positive, and sums of positives are positive. The product condition is built into the definition of squares. The separation condition $P \cap (-P)=\varnothing$ follows from the fact that $i \notin F$. Indeed, if $a^2=-b^2$ with $a,b \neq 0$, then $(a/b)^2=-1$, so $F$ already contains a square root of $-1$, contrary to $i \notin F$. Now take any $x \in F^\times$. Since $F(i)$ is algebraically closed, $x$ has a square root in $F(i)$. Write that square root as $a+bi$ with $a,b \in F$. Squaring gives \begin{align*} x=(a+bi)^2=(a^2-b^2)+2ab\,i. \end{align*} The element $x$ lies in $F$, so its $i$-coefficient is $0$. Since $\operatorname{char}F \neq 2$, we get $ab=0$. If $b=0$, then $x=a^2$ is a square; if $a=0$, then $x=-b^2$ is the negative of a square. Thus every nonzero element lies in exactly one of $P$ and $-P$. The only subtle point is closure under addition. Let $u^2,v^2 \in P$. Since $u \in F^\times$, we can factor \begin{align*} u^2+v^2=u^2\left(1+\left(\frac{v}{u}\right)^2\right). \end{align*} Thus it is enough to show that every element of the form $1+c^2$ is a square in $F$. The direct attempt to take a square root of $1+c^2$ in $F(i)$ does not determine whether that square root lies in $F$ or in $Fi$. Instead we take a square root of $1+ci$ and then apply the quadratic norm. Since $F(i)$ is algebraically closed, choose $z \in F(i)$ such that \begin{align*} z^2=1+ci. \end{align*} Write $z=a+bi$ with $a,b \in F$. Define the norm map $N_{K/F}:K \to F$ by \begin{align*} N_{K/F}(x+yi)=(x+yi)(x-yi)=x^2+y^2 \end{align*} for $x,y \in F$. This norm is multiplicative because it is multiplication by the conjugate: $N_{K/F}(ww')=N_{K/F}(w)N_{K/F}(w')$ for $w,w' \in K$. Applying $N_{K/F}$ to $z^2=1+ci$ gives \begin{align*} 1+c^2=N_{K/F}(1+ci)=N_{K/F}(z^2)=N_{K/F}(z)^2=(a^2+b^2)^2. \end{align*} The element $1+c^2$ is nonzero, because $1+c^2=0$ would imply $c^2=-1$ and hence $i \in F$, contrary to the present case. Therefore $a^2+b^2 \neq 0$, and $1+c^2$ is the square of the nonzero element $a^2+b^2 \in F$. Hence $1+c^2 \in P$, so $P+P \subset P$. Therefore $P$ is a positive cone, and the relation \begin{align*} x<y \quad \Longleftrightarrow \quad y-x \in P \end{align*} defines an ordering of $F$. Its positive elements are exactly the nonzero squares. [/guided] [/step] [step:Derive roots of odd-degree polynomials from conjugation over $F(i)$] Continue assuming condition $2$, and use the ordering just constructed. Let $f \in F[X]$ have odd degree $n$. Since $K=F(i)$ is algebraically closed, $f$ splits over $K$: \begin{align*} f(X)=c\prod_{j=1}^{n}(X-\alpha_j), \end{align*} where $c \in F^\times$ and $\alpha_j \in K$ for $1 \leq j \leq n$. Let $\sigma:K \to K$ be the $F$-automorphism defined by \begin{align*} \sigma(a+bi)=a-bi \end{align*} for $a,b \in F$. Since the coefficients of $f$ lie in $F$, the roots of $f$ in $K$ are stable under $\sigma$. Every root not lying in $F$ has a two-element orbit $\{\alpha,\sigma(\alpha)\}$. Because the total number of roots counted with multiplicity is odd, at least one root is fixed by $\sigma$. The fixed field of $\sigma$ is $F$, so this root lies in $F$. Thus every odd-degree polynomial over $F$ has a root in $F$. Together with the previous step, condition $2$ implies condition $3$. [/step] [step:Show that the square-root and odd-degree conditions force $F(i)$ to be algebraically closed] Assume condition $3$, and let $<$ be such an ordering on $F$. Since $-1<0$, the element $-1$ is not a square in $F$, so $F$ is not algebraically closed and $K:=F(i)$ is a quadratic extension of $F$. First we prove that every element of $K$ has a square root in $K$. Let $z=a+bi \in K^\times$ with $a,b \in F$. Define \begin{align*} r \in F \end{align*} to be the positive square root of $a^2+b^2$, which exists because $a^2+b^2>0$. The elements \begin{align*} \frac{r+a}{2}, \qquad \frac{r-a}{2} \end{align*} are nonnegative in $F$, since $r^2-a^2=b^2 \geq 0$ and $r \geq |a|$ in the ordered field. Hence they have square roots in $F$. Choose $c,d \in F$ satisfying \begin{align*} c^2=\frac{r+a}{2}, \qquad d^2=\frac{r-a}{2}, \end{align*} and choose the sign of $d$ so that $2cd=b$ when $b \neq 0$; if $b=0$, take $d=0$ when $a\geq 0$ and $c=0$ when $a<0$. Then \begin{align*} (c+di)^2=(c^2-d^2)+2cd\,i=a+bi=z. \end{align*} Thus every element of $K$ has a square root in $K$. Since $K$ contains $i$, we have $\operatorname{char} K \neq 2$. For any quadratic polynomial $X^2+pX+q \in K[X]$, completing the square gives \begin{align*} X^2+pX+q=\left(X+\frac{p}{2}\right)^2-\left(\frac{p^2}{4}-q\right), \end{align*} and the element $p^2/4-q \in K$ has a square root in $K$. Hence every quadratic polynomial over $K$ splits in $K$, so $K$ has no quadratic extension. Now suppose, for contradiction, that $K$ is not algebraically closed. Then there exists an element algebraic over $K$ whose [minimal polynomial](/page/Minimal%20Polynomial) over $K$ has degree greater than $1$; adjoining one such element gives a finite extension $E/K$ with $E \neq K$. Since $F$ is ordered, $\operatorname{char} F=0$, so every algebraic extension of $F$ is separable. Therefore the normal closure of the finite extension $E/F$ is a finite Galois extension. Choose such a finite Galois extension $L/F$ containing $E$, and define \begin{align*} G:=\operatorname{Gal}(L/F), \qquad H:=\operatorname{Gal}(L/K). \end{align*} By the [Fundamental Theorem of Galois Theory](/page/Fundamental%20Theorem%20of%20Galois%20Theory), the equality $[K:F]=2$ gives \begin{align*} [G:H]=[K:F]=2. \end{align*} We use two finite group facts: the [Sylow Theorems](/page/Sylow%20Theorems), which imply that a Sylow $2$-subgroup $S \leq G$ has odd index $[G:S]$, and the elementary fact that every nontrivial finite $2$-group has a subgroup of index $2$. Let $S \leq G$ be a Sylow $2$-subgroup. By the [Fundamental Theorem of Galois Theory](/page/Fundamental%20Theorem%20of%20Galois%20Theory), its fixed field $L^S$ satisfies \begin{align*} [L^S:F]=[G:S], \end{align*} which is odd. We now prove that $L^S=F$. The extension $L^S/F$ is finite and separable because it is a subextension of the finite Galois extension $L/F$. By the [Primitive Element Theorem](/page/Primitive%20Element%20Theorem), there exists $\theta \in L^S$ such that $L^S=F(\theta)$. Hence the minimal polynomial $m_\theta \in F[X]$ has degree \begin{align*} \deg m_\theta=[L^S:F], \end{align*} which is odd. If $[L^S:F]>1$, then $m_\theta$ is an irreducible polynomial of odd degree greater than $1$. By the hypothesis on odd-degree polynomials, $m_\theta$ has a root in $F$, contradicting irreducibility unless $\deg m_\theta=1$. Therefore $[L^S:F]=1$, so $L^S=F$, and [the Galois correspondence](/theorems/1898) gives $S=G$. Thus $G$ is a finite $2$-group. Since $E \neq K$ and $E \subset L$, the subgroup $H=\operatorname{Gal}(L/K)$ is nontrivial. As a nontrivial finite $2$-group, $H$ has a subgroup $H_0 \leq H$ of index $2$. By the [Fundamental Theorem of Galois Theory](/page/Fundamental%20Theorem%20of%20Galois%20Theory), the fixed field $L^{H_0}$ is a quadratic extension of $K=L^H$, contradicting the fact that $K$ has no quadratic extension. Therefore no such $E$ exists, and $K=F(i)$ is algebraically closed. [guided] The contradiction argument reduces algebraic closedness of $K=F(i)$ to finite Galois theory. Suppose $K$ is not algebraically closed. Then some polynomial over $K$ has an irreducible factor of degree greater than $1$, and adjoining one root gives a finite extension $E/K$ with $E \neq K$. We need a finite Galois extension over $F$ in order to use fixed fields. This is legitimate because $F$ is ordered, so $\operatorname{char} F=0$, and hence all finite extensions of $F$ are separable. Taking the normal closure of the finite extension $E/F$ gives a finite Galois extension $L/F$ containing $E$. Define \begin{align*} G:=\operatorname{Gal}(L/F), \qquad H:=\operatorname{Gal}(L/K). \end{align*} The extension $K/F$ has degree $2$ because $K=F(i)$ and $i \notin F$. Therefore the [Fundamental Theorem of Galois Theory](/page/Fundamental%20Theorem%20of%20Galois%20Theory) gives \begin{align*} [G:H]=[K:F]=2. \end{align*} Let $S \leq G$ be a Sylow $2$-subgroup. The [Sylow Theorems](/page/Sylow%20Theorems) say that the index $[G:S]$ is odd, since $S$ contains the full power of $2$ dividing $|G|$. By the [Galois correspondence](/page/Galois%20Correspondence), the fixed field $L^S$ is a finite extension of $F$ and \begin{align*} [L^S:F]=[G:S]. \end{align*} Thus $L^S/F$ has odd degree. Now we use the odd-degree root hypothesis correctly. It is not enough to say that every element of an odd-degree extension has odd degree over $F$; that assertion is false. Instead, because $L^S/F$ is finite and separable, the [Primitive Element Theorem](/page/Primitive%20Element%20Theorem) gives an element $\theta \in L^S$ with $L^S=F(\theta)$. The minimal polynomial $m_\theta \in F[X]$ then satisfies \begin{align*} \deg m_\theta=[L^S:F], \end{align*} so its degree is odd. If this degree were greater than $1$, then $m_\theta$ would be an irreducible odd-degree polynomial over $F$ with a root in $F$ by hypothesis, which is impossible for an irreducible polynomial of degree greater than $1$. Hence $\deg m_\theta=1$, so $L^S=F$. The equality $L^S=F$ means, by the Galois correspondence, that $S=G$. Hence $G$ is a finite $2$-group. Since $E \neq K$ and $E \subset L$, the subgroup $H=\operatorname{Gal}(L/K)$ is nontrivial. Every nontrivial finite $2$-group has a subgroup $H_0 \leq H$ of index $2$. Applying the Galois correspondence once more, $L^{H_0}/L^H$ has degree $2$. But $L^H=K$, so this gives a quadratic extension of $K$, contradicting the previous conclusion that every quadratic polynomial over $K$ splits in $K$. Therefore $K$ has no nontrivial finite extension and is algebraically closed. [/guided] Hence condition $3$ implies condition $2$. [/step] [step:Relate real closedness to maximality among algebraic ordered extensions] Assume condition $3$. By the previous step, $F(i)$ is algebraically closed. Let $E/F$ be an algebraic [field extension](/page/Field%20Extension) with an ordering extending the given ordering on $F$. If $E \neq F$, choose $\alpha \in E \setminus F$, and let $m_\alpha \in F[X]$ denote the minimal polynomial of $\alpha$ over $F$. Since every odd-degree polynomial over $F$ has a root in $F$, an irreducible polynomial over $F$ of odd degree must have degree $1$. Hence $\deg m_\alpha$ is even. Because $F(i)$ is algebraically closed, the extension $F(i)/F$ is an [algebraic closure](/page/Algebraic%20Closure) of $F$. The standard embedding theorem for algebraic extensions into an algebraic closure gives an $F$-embedding \begin{align*} \varphi:F(\alpha) \to F(i). \end{align*} The image $\varphi(F(\alpha))$ is an intermediate field between $F$ and $F(i)$. Since $[F(i):F]=2$, the only intermediate fields are $F$ and $F(i)$; because $\alpha \notin F$, the image is not $F$. Therefore $F(\alpha) \cong_F F(i)$. In particular, $F(\alpha)$ contains an element $\beta$ with $\beta^2=-1$. But an ordered field cannot contain such an element, since every square is nonnegative while $-1<0$. This contradicts the assumption that the ordering on $E$ extends the ordering on $F$. Therefore condition $3$ implies condition $4$. Conversely, assume condition $4$, and let $<$ be an ordering on $F$ with no proper algebraic ordered extension. We invoke the [Existence of Real Closures](/page/Real%20Closure): every ordered field has an algebraic ordered extension that is real closed, and the given ordering extends to that real closure. Applying this theorem to $(F,<)$ gives an algebraic ordered extension $R/F$ with $R$ real closed. By maximality, $R=F$. Therefore $(F,<)$ is real closed, so every positive element of $F$ is a square and every odd-degree polynomial over $F$ has a root in $F$. Hence condition $4$ implies condition $3$. [guided] The implication from condition $3$ to condition $4$ uses the algebraic closedness of $F(i)$ to control all algebraic extensions of $F$. Suppose an algebraic ordered extension $E/F$ extending the order were proper, and choose $\alpha \in E \setminus F$. Let $m_\alpha \in F[X]$ be the minimal polynomial of $\alpha$ over $F$. If $\deg m_\alpha$ were odd, then the odd-degree root hypothesis would give a root of $m_\alpha$ in $F$. Since $m_\alpha$ is irreducible, that would force $\deg m_\alpha=1$, contradicting $\alpha \notin F$. Thus $\deg m_\alpha$ is even. The important field-theoretic point is that $E$ need not literally be a subfield of $F(i)$. Instead, because $F(i)$ is algebraically closed and algebraic over $F$, it is an algebraic closure of $F$. The embedding theorem for algebraic extensions into an algebraic closure gives an $F$-embedding \begin{align*} \varphi:F(\alpha) \to F(i). \end{align*} The image is an intermediate field between $F$ and the quadratic extension $F(i)$. A degree-$2$ extension has no intermediate fields other than its endpoints, so $\varphi(F(\alpha))$ is either $F$ or $F(i)$. Since $\alpha \notin F$, the image is not $F$, and hence $F(\alpha) \cong_F F(i)$. Therefore $F(\alpha)$ contains an element $\beta$ with $\beta^2=-1$. This cannot occur in an ordered field: squares are nonnegative, while $-1$ is negative. Hence no proper algebraic ordered extension exists. For the converse, condition $4$ says that the given ordering on $F$ admits no proper algebraic ordered extension. The [Existence of Real Closures](/page/Real%20Closure) says that every ordered field admits an algebraic ordered extension $R/F$ that is real closed and whose order extends the original order. Applying this to $(F,<)$, maximality forces $R=F$. Thus $(F,<)$ is already real closed, meaning every positive element is a square and every odd-degree polynomial over $F$ has a root in $F$. [/guided] Combining the previous steps gives \begin{align*} 1 \iff 3 \iff 2 \iff 4. \end{align*} This proves the equivalence of all four conditions. [/step]