Complexity of multi-linear polynomial computing Boolean function Let $f:\{0,1\}^{n}\longmapsto\{0,1\}$ be a Boolean function. As usual,
let $C(f)$ denote circuit complexity of $f$, i.e, the size of the smallest
Boolean circuit computing $f$. 

As we know that every Boolean function
can be computed by polynomials in $\mathbb{F}_{2}[x_{1},x_{2},\ldots,x_{n}]$.
Let $P(f)$ be the size of the smallest arithmetic circuit(over
$\mathbb{F}_{2}$) which computes
a polynomial $P_{f}$ such that $P_{f}$ computes the same function
as $f$ on $\mathbb{F}_{2}^{n}$(correspondingly $\{0,1\}^{n}$). 

It is known that $P(f)\leq\textrm{poly}(C(f))$ and $C(f)\leq\textrm{poly}(P(f))$. 

We also know that there is a unique multi-linear polynomial $M_f\in\mathbb{F}_{2}[x_{1},x_{2},\ldots,x_{n}]$ such that $M_{f}$ computes the same function as $f$ on $\mathbb{F}_{2}^{n}$(correspondingly
$\{0,1\}^{n}$). Let $M(f)$ be the size of the smallest arithmetic circuit(over
$\mathbb{F}_{2}$) computing $M_{f}$. 

It is clear that $C(f)\leq\textrm{poly}(M(f))$.
How about other direction? 


  Can we bound $M(f)$ polynomially in terms of $C(f)$? 


Anything is known about this or something related?
If I understand your question correctly, the answer is no (independently from the field, assuming $\mathsf{VP}\neq\mathsf{VNP}$).