The Reals are not interpretable in the complex numbers Let $L=\{+,\dot{},0,1\}$ be the language of fields. I wish to show that the reals ($N=(\mathbb{R},+,\dot{},0,1)$) are not interpretable in the structure  $M = (\mathbb{C},+,\dot{},0,1)$.
I have the following solution to show that $N$ is not definable in $M$: Suppose that the reals were definable in $M$. Then we can write a formula that defines a linear order inside of $M$ exploiting the fact that $<$ is definable in $N$. This gives a contradiction since $Th(M)$ is $\aleph_{1}$- categorical and hence $\omega$-stable but the above implies its unstable.
I believe the same proof holds with definability replaced with interpretability. 
My questions is: Is there a more elementary proof of this fact? 
Edit1: I saw a lot of comments below. I'm sorry about the typo. It has been corrected but it probably read $Th(N)$ where it should have been $Th(M)$.
 A: OK, here's an argument that avoids stability theory: 
Suppose we could interpret $\mathbb{R}$ in a field $K\models \text{ACF}_0$. Note that $K$ must be uncountable. By elimination of imaginaries in $\text{ACF}_0$, we may assume that the domain of the interpretation is a definable subset of $K$, defined by $\varphi(\overline{x})$, rather than a quotient of a definable set. 
Let $A$ be the finite set of parameters used in the interpretation (i.e. in $\varphi$, as well as in the formulas defining the field operations on the interpreted copy of $\mathbb{R}$), and let $F$ be the algebraic closure in $K$ of the subfield of $K$ generated by $A$. Note that $F$ is countable, and for any $b\in K\setminus F$, there is an automorphism of $K$ which fixes $F$ pointwise but moves $b$ (extend $\{b\}$ to a transcendence basis for $K$ over $F$, and permute the basis).
Now since $\mathbb{R}$ is uncountable, there is a tuple $\overline{a}$ satisfying $\varphi(\overline{x})$ such that one of the entries $a_i$ of the tuple is not in $F$. Let $\sigma\in \text{Aut}(K)$ fix $F$ pointwise but move $a_i$. Then $\sigma$ induces an automorphism of the interpreted copy of $\mathbb{R}$ which moves $\overline{a}$. This contradicts the fact that $\mathbb{R}$ has no nontrivial automorphisms, QED.
Note that this argument does not show that $\text{Th}(\mathbb{R})$ is not interpretable in $\text{Th}(\mathbb{C})$, just that no uncountable rigid model of this theory (e.g. $\mathbb{R}$ or any uncountable subfield) is in the image of any such interpretation. Hence it also shows that $\text{Th}(\mathbb{R})$ and $\text{Th}(\mathbb{C})$ are not bi-interpretable. 
As explained in the question, $\text{Th}(\mathbb{R})$ is not in fact interpretable in $\text{Th}(\mathbb{C})$ - it would be interesting to see a proof of this which avoids stability theory.
