It's well-known that in a totally transcendental ($\omega$-stable) theory, $p(x)\subseteq q(x)$ is a non-forking extension if and only if $\text{MR}(p) = \text{MR}(q)$. In my answer to [this Math Stackexchange question](https://math.stackexchange.com/questions/3179518/non-forking-extension-is-equivalent-to-sameness-of-morley-rank), I outlined the proof that the same statement is true in a general theory, under the additional assumption that $\text{MR}(p) <\infty$. 


The assumption that $p$ has Morley rank is clearly necessary for one direction of the equivalence: It's easy to come up with examples of $p(x)\subseteq q(x)$ such that $q$ is a forking extension of $p$ but $\text{MR}(p) = \text{MR}(q) = \infty$. But I don't know any examples showing that this assumption is necessary for the converse. So the precise question is:



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> Let $T$ be an arbitrary theory. Suppose $p(x)\in S\_x(A)$ with $\text{MR}(p) = \infty$ and $q(x)\in S\_x(B)$ with $p\subseteq q$. If $q$ does not fork over $A$, is $\text{MR}(q) = \infty$? What if we assume $T$ is (super)stable?
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Morley rank and forking in arbitrary theories
I think the answer is yes: if $p\subseteq q$ is non-forking, then $MR(p)=MR(q)$. To see this, let $p$ be over $A$ and assume that $\phi(x;b)\in q$ has ordinal Morley rank. It follows that $\phi(x;y)\wedge tp\_y(b/A)$ is stable and then by compactness, there is $\psi(y)\in tp(b/A)$ such that $\phi(x;y)\wedge \psi(y)$ is stable. By standard results in (local) stability theory, if we take $(b\_i:i<\omega)$ an independent sequence (in the sense of the stable formula $\phi(x;y)$) in $tp(b/A)$, we have that $p(y)\wedge \bigwedge \neg\phi(x;b\_i)$ is inconsistent. Therefore some finite part of $p(y)$ implies a finite disjunction $\bigvee \phi(x;b\_i)$ and thus has Morley rank at most that of $\phi(x;b)$.


The same argument shows that if $q$ concentrates on a stable definable set, then so does $p$. However, the proof breaks down as soon as we leave stability. I don't know what happens if we replace "stable definable set" by "simple definable set" for instance. (This is essentially asked for NTP$\_2$ theories in Question 6.6 in A. Chernikov, [Theories without the tree property of the second kind](https://arxiv.org/abs/1204.0832) and as far as I know is still open.)


