Let $$K_0$$, $$K_1$$, $$K_2$$ be ordered fields and $$i_k$$:$$K_0$$ $$\rightarrow K_k$$ are ordered inclusions for $$k=1,2$$. Show that there exists ordered field $$K$$ and ordered insclusions $$j_k$$ : $$K_k \rightarrow K$$ for which $$j_1 \circ i_1 = j_2 \circ i_2$$

• This does not seem to be a problem that one could tackle with little knowledge about ordered fields, unless there is a clever method. Can you prove the result if $L_0,L_1,L_2$ are linear orders and we are looking for strictly increasing maps instead of non-decreasing field morphisms? Do you know about Hahn and Kaplansky's embedding theorems? – nombre Nov 7 '18 at 20:22
• Unfortunately I do not have an idea how to approach to this problem at all. No I do not know about those theorems. I will inform myself more about them. Thank you. – XYZ Nov 8 '18 at 9:28
• Actually, trying to write it through I realize my method does not work. I'll try to think about something else. Could you tell me where this problem comes from? – nombre Nov 8 '18 at 15:42
• Have you found a solution yet. There is a way using the adjunction between ordered fields and real-closed fields but this requires to know a little about real closure. – nombre Nov 12 '18 at 18:48
• I think I will give up on searching for solution of this problem because it is too hard for me and requires deeper knowledge which i currently do not have. Thank you so much for your help. – XYZ Nov 13 '18 at 18:39

Let $$F$$ be a $$\kappa$$-saturated ordered field where $$\kappa >|K_1|,|K_2|$$ (for instance a ultrapower of $$\mathbb{R}$$ modulo a free ultrafilter on $$\max(|K_1|,|K_2|)$$).
Thus there is an embedding $$j_1: K_1 \rightarrow F$$. Let $$j_2'$$ be the embedding $$i_2(K) \rightarrow F$$ satisfying $$j_2' \circ i_2= j_1 \circ i_1$$. Since $$F$$ is $$|K_2|^+$$-saturated, the morphism $$j_2'$$ extend to $$K_2$$, giving a solution $$j_2$$ to the almagamation problem.