How "separable" (not in that sense) is a polynomial?

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Since "separable" is used for different meaning in separable polynomial and separation of variable, I am having trouble searching for anything related to my question. So I hope someone can help with this.

Consider a multivariable polynomial $P(x_{1},\ldots,x_{n})$. For convenience, we will instead write $P(x_{1},\ldots,x_{n})$ as $P(A)$ where $A=\{x_{1},\ldots,x_{n}\}$. We look at all the way $P(A)$ can be written as $\prod\limits_{i=1}^{k}Q_{i}(B_{i})$ where $B_{i}\subseteq A$ and $Q_{i}$ are (possibly multivariable) polynomial in term of all elements of $B_{i}$. For each such product, we associate with it a number $S=\max\limits_{i}|B_{i}|$. Looking at all the way the polynomial can be written as such a product, we search for an $S_{\min}=\min S$.

The question is about finding a good upperbound, or an exact formula, of $S_{\min}$, dependent on $P$.

Just to make sure thing are clear, this is some example (I will use name $x,y,z$ for variable):

Example 1: $P=y^{2}+xy^{2}+x+1$. We can write $P=(y^{2}+1)(x+1)$. So $S$ for this decomposition is $1$. It is clear that no decomposition can give smaller $S$. So $S_{\min}=1$.

Example 2: $P=zx^{2}+zy^{2}$. We can write $P=z(x^{2}+y^{2})$. So $S$ for this decomposition is $2$. There is no decomposition such that each polynomial in the product only use $1$ variable. Hence $S_{\min}=2$.

Example 3: $P=x+y+z$. The only possible decomposition is the trivial one $P=P$ (not counting some trivial variants) so $S_{\min}=3$.

I would appreciate any significant helps in the matter. An answer to any one of the following would be fine:

-A full exact formula.

-A good upperbound.

-A good necessary or sufficient condition for $S_{\min}=1$.

-A good necessary or sufficient condition for $S_{\min}<n$.

-Some references on (Internet accessible) literature on the issue.

Note: the ring of coefficient is assumed to be a field of characteristic $0$.

Thank you for your help.

My "attempt":

Ring of polynomial of several variable is UFD. So it all boil down to $S$ of the prime factorization. The subproblem on conditions for $S_{\min}=1$ when the field is algebraically closed is easy.