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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 $j_1 \circ i_1 = j_2 \circ i_2$

Since i have not much knowledge about ordered fields would appreciate any help or ideas about this problem.

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  • $\begingroup$ 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? $\endgroup$ – nombre Nov 7 '18 at 20:22
  • $\begingroup$ 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. $\endgroup$ – XYZ Nov 8 '18 at 9:28
  • $\begingroup$ 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? $\endgroup$ – nombre Nov 8 '18 at 15:42
  • $\begingroup$ 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. $\endgroup$ – nombre Nov 12 '18 at 18:48
  • $\begingroup$ 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. $\endgroup$ – XYZ Nov 13 '18 at 18:39
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I am sorry I forgot to actually propose an answer.

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.

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  • $\begingroup$ Just as I thought it takes deeper knowledge to come to the solution. Thank you once again for your effort. $\endgroup$ – XYZ Nov 27 '18 at 15:13

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