Is it possible to find totally ordered fields $K$ and $L$, and a map $f: K \to L$ that is an isomorphism of the ordered additive group structures such that $f(1)=1$, but which is not an isomorphism of fields?
I know that this is impossible if $K$ or $L$ is a subfield of the real numbers. My guess is that it is probably possible for non-Archimedean fields, but I don't know for sure.
My interest in this question is in relation to the foundations of plane geometry.
Thank you for any ideas.
Consider $K$ the field given by the Laurent series of $t$ over $\mathbb{Q}$ which have only finitely many terms with $t$ up to a negative exponent, i.e. $$K=\left\{\sum_{k\geq n}a_kt^k\,|\,n\in\mathbb{Z}\text{, for all }k\geq n\text{, }a_k\in\mathbb{Q}\right\}$$ with the usual addition and multiplication, i.e., $$\sum_k a_kt^k+\sum_kb_kt^k=\sum_k(a_k+b_k)t^k$$ and $$\left(\sum_{k\geq n} a_kt^k\right)\left(\sum_{k\geq m} b_kt^k\right)=\sum_{k\geq n+m}\left(\sum_{\substack{i+j=k\\i\geq n,j\geq m}}a_ib_j\right)t^k\text{.}$$ This field can be made an ordered field by defining the set of positive elements to be $$K^+=\left\{\sum_{k\geq n}a_kt^k\in K\,|\,a_n>0\right\}\text{.}$$ One easily verifies that $K^+$ is closed under addition and multiplication and that every element of $K$ is in $K^+$, $-K^+$ or $\{0\}$. Therefore $x\geq y$ given by $$x-y\in K^+\cup\{0\}$$ defines a total order compatible with the field structure of $K$.
Here, the counterexamples is given by the function $$f:K\rightarrow K$$ defined by $$f\left(\sum_{k\geq n} a_kt^k\right)=\sum_{k\geq n} (1+\delta^1_k)a_kt^k\text{,}$$ where $\delta^1_k$ is the Kornecker's delta -equal to one if $1=k$, zero otherwise-, which is monotonous since it takes positives to positives. (In a more simple way, $f$ multiplies by $2$ the coefficient of $t$.) Clearly, $f$ is an isomorphism of the ordered additive groups such that $f(1)=1$, but it is not compatible with multiplication since $$t^2=f(t^2)\neq f(t)^2=4t^2\text{.}$$
(The problem is that since $Q$ is not dense in $K$ with respect the order topology, we cannot guarantee using continuity that the fact that $f$ preserves the multiplication on $\mathbb{Q}$ implies that it preserves it on $K$.)
EDIT: For a counterexample with an ordered field whose positive elements are the squares, consider a field $L$ whose elements are the formal series given by $$\sum_{k\geq n}a_kt^{k/2^m}$$ with $a_k\in\mathbb{R}$ and $n,m\in\mathbb{Z}$. Define the sum, multiplication and order in an identical way as we did in the previous field. For concreteness, this means that in $L$: $$\sum_{\gamma}a_\gamma t^{\gamma}+\sum_{\gamma}b_\gamma t^{\gamma}=\sum_\gamma (a_\gamma+b_\gamma)t^\gamma\text{,}$$ $$\left(\sum_{k\geq n_0}a_kt^{k/2^{m_0}}\right)\left(\sum_{k\geq n_1}a_kt^{k/2^{m_1}}\right)=\sum_{k\geq n_02^{m_1}+n_12^{m_1}}\left(\sum_{\substack{i2^{m_1}+j2^{m_0}=k\\i\geq n_0,j\geq n_1}}a_ib_j\right)t^{k/2^{m_0+m_1}} \text{,}$$ and $$L^+=\left\{\sum_{k\geq n}a_kt^{k/2^m}\in L\,|\,a_n>0\right\}\text{.}$$ In this way the previous counterexample works in exactly the same way, we consider $$f:L\rightarrow L$$ that multiplies the coefficient of $t$ by $2$. And ones easily check that is $f$ is also in this case an isomorphism of ordered additive groups, such that it is not compatible with the multiplicative structure.
We only have to check that every element has an square root, for that we will use a direct computation. If $$f=\sum_{k\geq n}a_kt^{k/2^m}\in L^+\text{,}$$ with $a_n>0$, for $$g=\sum_{k\geq n}b_kt^{k/2^{m+1}}$$ we plan the equation $$g^2=f$$ which translates into $$ (2-\delta_{k-n}^n)b_{k-n}b_n+\sum_{\substack{i+j=k\\i,j> n}}b_ib_j=\sum_{\substack{i+j=k\\i,j\geq n}}b_ib_j=\begin{cases} a_{k/2}&\text{ if }k\text{ even}\\ 0&\text{ if }k\text{ odd} \end{cases}\text{,}$$ for $k\geq 2n$. By solving -which it is only possible when $a_n\geq 0$, otherwise $b_{n}^2=a_n$ has not solution-, we have that $$g^2=f$$ when $$ b_k=\begin{cases} \sqrt{a_n} & \text{if }k=n\\ (2\sqrt{a})^{-1}\left(a_{k+n}-\sum_{\substack{i+j=n+k\\i,j>n}}b_ib_j\right) &\text{if }k+n\neq 2n\text{ even}\\ -(2\sqrt{a})^{-1}\sum_{\substack{i+j=n+k\\i,j>n}}b_ib_j&\text{if }k+n\text{ odd} \end{cases}\text{.}$$ In other words, $g$ is an square root of $f$.