non-forking extension is equivalent to sameness of morley rank given monster model of $\mathcal{M}\models T$ in $\mathcal{L}$, $p\in S_n(A)$ and $p\subseteq q\in S_n(B)$ where $A\subseteq B$, 
 $q$ does not fork over $A$ iff $MR(p)=MR(q)$(MR : morley rank).
$\leftarrow$ is easy to prove by the statement "if $q$ forks over A then $MR(q)< MR(p)$." I saw proof of $\rightarrow$ which premises that both concepts(nonforking extension and same morley rank) satisfy some unique properties like invariance under automorphism, local character, transitivity, etc. I just want to know whether some simple direct proof exists. What I tried was zorn's lemma for $P=\{q'\mid p\subseteq q'\text{(partial) type in $\mathcal{L_B}$}, MR(p)=MR(q'), q' \text{ does not fork over A}\}$ but failed to show maximal element is complete type in $S_n(B)$.
 A: The $\Leftarrow$ direction is only true if we assume that $\text{MR}(p)<\infty$. It's easy to come up with examples where $q$ is a forking extension of $p$, but $\text{MR}(q) = \text{MR}(p) = \infty$.
So you'll usually find this statement proven under the assumption that $T$ is totally transcendental ($\omega$-stable), so every type has an ordinal-valued Morley rank. But it's interesting to ask to what extent it's true in a general theory.
There's a very simple argument for the $\Rightarrow$ direction, assuming $p$ is stationary and has ordinal-valued Morley rank.
Claim: Let $p(x)\in S_x(A)$ be a stationary type with $\text{MR}(p) < \infty$, and suppose $p(x)\subseteq q(x)\in S_x(B)$ and $q$ does not fork over $A$. Then $\text{MR}(p) = \text{MR}(q)$.
Proof: Let $q'(x)\in S_x(B)$ be such that $p(x)\subseteq q'(x)$ and $\text{MR}(q') = \text{MR}(p)$. By the $\Leftarrow$ direction (this is where we use $\text{MR}(p)<\infty$), $q'$ does not fork over $A$. But since $p$ is stationary, it has a unique non-forking extension to a type over $B$. It follows that $q' = q$ and $\text{MR}(p) = \text{MR}(q)$.

Ok, but we want the statement for all types, not just stationary ones. That is, we want to prove
Theorem: Let $p(x)\in S_x(A)$ be a type with $\text{MR}(p)<\infty$, and suppose $q(x)$ is a non-forking extension of $p(x)$. Then $\text{MR}(p) = \text{MR}(q)$.
Here's a plan for proving this:

*

*Show that we can reduce to the case $A = \text{acl}^{\text{eq}}(A)$: Work in $T^{\text{eq}}$. Suppose $p(x)\in S_x(A)$ with $\text{MR}(p)<\infty$, and suppose $q(x)\in S_x(B)$ is a non-forking extension of $p(x)$. Let $A' = \text{acl}^{\text{eq}}(A)$, and let $B' = B\cup A'$. Let $q'(x)\in S_x(B')$ be an extension of $q(x)$ which doesn't fork over $A$, and let $p'(x) = q'(x)|_{A'}\in S_x(A')$. We also have $p(x) = p'(x)|_A$, so $\text{MR}(p') = \text{MR}(p)$ (this is Exercise 6.2.8 in Tent & Ziegler). Now if we know the theorem is true for types over  $\text{acl}^{\text{eq}}$-closed sets, we can conclude that $\text{MR}(q') = \text{MR}(p') = \text{MR}(p)$, since $q'$ doesn't fork over $A\subseteq A'$. And $p\subseteq q\subseteq q'$, so also $\text{MR}(q) = \text{MR}(p)$.

*If you're happy to assume that $T$ is totally transcendental (or even just stable), then we're done by the Claim, since any type over an $\text{acl}^{\text{eq}}$-closed set is stationary.

*On the other hand, to prove the theorem in an arbitrary theory $T$, we need to do more work. The idea is to prove that any type $p(x)$ with ordinal-valued Morley rank is a stable type (meaning that there is no formula $\varphi(x;y)$ which has the order property witnessed by $(a_i)_{i\in \omega}$ and $(b_j)_{j\in \omega}$ such that all of the $a_i$ realize $p(x)$), and any stable type over an $\text{acl}^{\text{eq}}$-closed set is stationary. This puts together Exercise 8.3.5(1) and Exercise 8.5.4(2) in Tent & Ziegler. In general, Tent & Ziegler is a great reference for this kind of material, and there are good solutions/hints for the exercises in the back of the book. The proof that I've outlined above follows the suggested solution to Exercise 8.5.5(2).

Finally, observe that the proof above really uses our assumption that $p$ has ordinal-valued Morley rank. This raises the following questions, to which I don't know the answer:

Does there exist a theory $T$ and types $p\subseteq q$ such that $q$ is a non-forking extension of $p$, $\text{MR}(p) = \infty$, and $\text{MR}(q) < \infty$? Is there an example when $T$ is superstable?

Edit: Pierre Simon has given a negative answer to this question on MathOveflow.
