Proof of uniqueness of real-closure of an ordered field

Question

I'm reading the proof of uniqueness of real-closure of an ordered field $F$, that is an algebraic extension $R$ of $F$ such that $R$ is real-closed and the unique order on $R$ extends that of $F$. I understand the proof of existence of such field.

This is theorem 1.3.2 from Real-algebraic geometry by Bochnak, Coste and Roy.

In the proof we have a real-closed field $R'$ and $L\subseteq R$ where $R$ is real-closure of $F$, and $L$ extends $F$ (with order induced by $R$). We have $a\in R\setminus L$ and its minimal polynomial $f$. There is an $F$-homomorphism $\Phi:L\to R'$. We know that $f$ has roots $a_1 < ... < a_n$ in $R$ and $\Phi(f) = g$ has roots $b_1 < ... < b_n$ in $R'$ (those are all roots of those polynomials in those fields). If $a = a_j$, we extend $\Phi$ to an $F$-homomorphism such that $\Phi(a) = b_j$. This is just a standard theorem from field theory that such extension exists so I have no problem with this.

Could someone explain to me why is $\Phi$ order-preserving then?

Edit: I've been looking in Jacobson, and I've been trying to apply his argument here. Take polynomial $h\in L[x]$ that has all of $a_1, ..., a_n$ and $\sqrt{a_{i+1}-a_i}$ for $i = 1, ..., n-1$ as its roots. Then take the extension of $L$ generated by the roots of $h$ in $R$, and primitive element $\theta$ of it. Then extend $\Phi$ to $L(\theta)$ so that $\Phi(\theta) = \theta'$. We can then write $\Phi(\sqrt{a_{i+1}-a_i})^2 = \Phi(a_{i+1})-\Phi(a_i) > 0$. Since $\Phi(a_i)$ must be all roots of $g$ in $R'$, we must have $\Phi(a_i) = b_i$ for $i = 1, ..., n$.

I'm not familiar with primitive elements yet (I'm new to field theory), so it would be nice if someone could double check this argument!

Edit 2: I'm afraid above argument still tells me nothing about being order-preserving, see answer I posted below for a replacement of the argument in the book.

Edit 3: I realize now that in the book by Bochnak, Coste and Roy they do sketch a similar argument to the one in Jacobson's book but they state the needed lemma and proposition inside the proof of the theorem which is why I missed it. Sorry for this goofy mistake, I was genuinely convinced there is an error in the book.

Answer 1

Here's a proof of uniqueness of real-closure of an ordered field based on third volume of Jacobson's Lectures in abstract algebra, chapter 6.

The method is different from the one in Real-algebraic geometry by Bochnak, Coste and Roy, because we're not trying to use Zorn's lemma together with order-preserving maps, and instead proceed by explicitly defining an $F$-isomorphism. It will automatically be order-preserving because it's an isomorphism of real-closed fields, where order is determined algebraically, that is if $R$ is real-closed then $x\geq 0$ iff $x = y^2$ for some $y\in R$.

Let $R, R'$ be real-closures of $F$.

If $a\in R$, let $f\in F[x]$ be the minimal polynomial of $a$. Note that $f$ has the same amount of roots over both $R$ and $R'$ by a straightforward application of Sturm's theorem. If $a_1 < ... < a_n$ are roots of $f$ in $R$, $b_1 < ... < b_n$ are roots of $f$ in $R'$, and $a = a_j$, define $\Phi(a) = b_j$. This defines a map $\Phi:R\to R'$. I claim that $\Phi:R\to R'$ is an $F$-isomorphism. Indeed, $\Phi(s) = s$ for all $s\in F$ is clear from definition. If $a, b\in R$, let $c_1 < ... < c_m$ be roots of minimal polynomials of $a, b, a+b, ab$ in $R$ and let $K = F(c_1, ..., c_m, \sqrt{c_2-c_1}, ..., \sqrt{c_m-c_{m-1}})$, where $\sqrt{c_{i+1}-c_i}$ is understood as the positive solution to $y^2 = c_{i+1}-c_i$, and it exists since $R$ is real-closed. Since $K$ is a finite extension of $F$, and $F$ is of characteristic $0$, there is $\theta\in R$ such that $K = F(\theta)$, this is the primitive element theorem. If $f\in F[x]$ is the minimal polynomial of $\theta$, we can then send $\theta$ to any root $\theta'$ of $f$ in $R'$, and thus obtain an $F$-homomorphism $\Psi:K\to R'$. Since $\Psi(\sqrt{c_{i+1}-c_i})^2 = \Psi(c_{i+1})-\Psi(c_i) > 0$, we see that $\Psi$ preserves the order of roots of minimal polynomials of $a, b, a+b, ab$, and so $\Phi$ and $\Psi$ agree on $a, b, a+b$ and $ab$. Thus $\Phi(a+b) = \Phi(a)+\Phi(b)$ and $\Phi(ab) = \Phi(a)\Phi(b)$, since the same holds for $\Psi$. So $\Phi$ is an $F$-homomorphism. Its image is $\Phi[R]$ is real-closed, and $R'$ is its algebraic extension to a real field. Thus $R' = \Phi[R]$, that is $\Phi$ is an $F$-isomorphism.