Parameterfree heir-coheir relation in stable theories

Question

Let $M\models T$, a stable theory. Let $a,b\in U\succ M$. Are the following equivalent?

1. $\varphi(M,b)\neq\varnothing$ for every $\varphi(x,y)\in \color{red}{L(M)}$ such that $\varphi(a,b)$

2. $\varphi(M,b)\neq\varnothing$ for every $\varphi(x,y)\in \color{red}{L}$ such that $\varphi(a,b)$

Answer 1

No, parameters from $M$ can be important for witnessing dependence over $M$. Here's an example.

Let $M$ be an infinite $k$-vector space and $U$ a proper elementary extension. Let $v\in M$ with $v\neq 0$, let $b\in U\setminus M$, and let $a = b+v$. Let's observe two facts:

1. $a\notin M$. Indeed, if $a\in M$, then $b = a - v\in M$, contradicting the choice of $b$.
2. $a$ and $b$ are linearly independent. Indeed, since $b\neq 0$, if $a$ and $b$ are linearly dependent, then $a = cb$ for some $c\in k$. Further, $c\neq 1$, since $v\neq 0$ implies $a\neq b$. But then $cb = b+v$, so $v = (1-c)b$, and hence $b = (1-c)^{-1}v$, so $b\in M$, contradicting the choice of $b$.

Let $\varphi(x,y)$ be the $L(M)$ formula $x = y + v$. Then $\varphi(a,b)\in \text{tp}(a/Mb)$, but is not satisfiable in $M$. Indeed, $a$ is its only realization in $U$, and $a\notin M$.

On the other hand, every $L$-formula in $\text{tp}(a/b)$ is satisfiable in $M$. In fact, every non-zero element of $M$ realizes the whole type! Indeed, for any non-zero $a'\in M$, $a'$ and $b$ are linearly independent (since $b\notin M$), so there is an automorphism of $U$ fixing $b$ and moving $a$ to $a'$.