Mutual dependency of polynomial expressions Suppose you are given the values of $m$ polynomial expressions in $n$ variables.
That is we know that $P_1(x_1,x_2,...,x_n)=a_1,P_2(x_1,x_2,...,x_n)=a_2,...,P_m(x_1,x_2,...,x_n)=a_m$ for some $a_1,a_2,...,a_m$.
The values of which polynomial expressions in $x_1,...,x_n$ are now uniquely determined?
For example if we know the value of $ab$ and $a+b$ all symmetric expressions in $a,b$ are now uniquely determined, but all other expressions are not. 
 A: $\def\CC{\mathbb{C}}$The term you are looking for is seminormalization. Specifically, let $S$ be the ring generated by $P_1$, $P_2$, …, $P_m$ and let $Q$ be another polynomial. Then $Q$ is in the seminormalization of $S$ if and only if, whenever we have $P_i(x_1, \ldots, x_n) = P_i(y_1, \ldots, y_n)$ for all $i$, then we have $Q(x_1, \ldots, x_n) = Q(y_1, \ldots, y_n)$.
Throughout this answer, I assume your ground field is the complex numbers.
There is also another condition called "weak normality", which is slightly stricter than "seminormality" in characteristic $p$, but the same in characteristic zero; since I am assuming you are working over the complex numbers, I won't distinguish between them. Here is a good survey on seminormality. 
There is a beautiful though perhaps not very useful description of seminormalization due to Swan. Let $S$ be a subring of $\CC[x_1, \ldots, x_n]$.  The seminormalization of $S$ is the smallest ring $S^{+}$ containing $S$ with the following property: If $Q^2$ and $Q^3$ are in $S$, then so is $Q$. This situation shows the basic example of a case where the seminormalization is more than just $S$: If $P_1=t^2$ and $P_2=t^3$, then $P_1$ and $P_2$ determine the value of $Q = t$.

Here are some more basic rings which might be good enough for your purposes.
The ring generated by the $P_i$ Clearly, any polynomial in the $P_i$ has the property you seek. I'll write $S$ for the ring generated by the $P_i$. In the rest of this answer, I'll consider rings which are larger than $S$. All of these other rings are larger than the seminormalization; they have "too many" functions in them while $S$ has "too few".
The algebraic closure of $S$ in $\mathbb{C}[x_1, \ldots, x_n]$. An element $Q$ of $\CC[x_1, \ldots, x_n]$ is algebraic over $S$ if and only if it obeys a nonzero polynomial relation 
$$A_d Q^d + A_{d-1} Q^{d-1} + \cdots + A_1 Q + A_0 =0$$
with the $A_i$ in $S$. Since the $A_i$ are in $S$, their values will be determined by the values of the $P_i$. As long as the $A_i$ are not all zero, we will then be able to restrict the values of $Q$ to a finite list of length $d$ by finding the roots of the above polynomial. Indeed, if we have chosen a polynomial of minimal degree, then there will generically be $d$ values for $Q$ once we fix the values of the $P_i$. This is the least you should ask for.
There is a very nice test for whether $Q$ is algebraic over the ring generated by the $P_i$: Let $K$ be the field $\mathrm{Frac} \CC[x_1, \ldots, x_n]$. Then $Q$ is algebraic over the ring generated by the $P_i$ if and only if $\nabla Q := (\partial Q/\partial x_1, \partial Q/\partial x_2, \dots, \partial Q/\partial X_n)$ is in the $K$ linear span of the vectors 
$\nabla P_1$, $\nabla P_2$, …., $\nabla P_m$.
The intersection $\mathrm{Frac} S \cap \CC[x_1, \ldots, x_n]$. One might want the number $d$ above to be $1$, so that there will generically be one value of $Q$ once the values of the $P_1$, ..., $P_m$ are fixed. In other words, $Q$ is in this ring of we can find $A$ and $B$ in $S$ so that $AQ+B=0$. 
The integral closure of $S$ in $\mathbb{C}[x_1, \ldots, x_n]$. Recall above that there are at most $d$ values for $Q$ unless all of the $A_i$ are zero. We can insist that this doesn't happen by forcing one of them to be one. Specifically: An element $Q$ of $\CC[x_1, \ldots, x_n]$ is integral over $S$ if it obeys a polynomial relation
$$ Q^d + A_{d-1} Q^{d-1} + \cdots + A_1 Q + A_0 =0$$
with $A_{d-1}$, ..., $A_0$ in $S$. So the values of $Q$ are always given by the roots of a degree $d$ polynomial. That doesn't mean there are always $d$ values; the polynomial might have multiple roots. However, one can show the following nice property: If $Q$ is integral over the ring generated by the $P_i$, and $(x_1, \ldots, x_n)$ ranges over a region such that the values of the $P_i$ remain bounded, then the value of $Q$ remains bounded.
The normalization of $S$ This is the intersection of the previous two examples. $Q$ is in the normalization of $S$ if both $Q$ satisfies a polynomial $$ Q^d + A_{d-1} Q^{d-1} + \cdots + A_1 Q + A_0 =0$$
with $A_{d-1}$, ..., $A_0$ in $S$ and $AQ+B=0$, with $A$ and $B$ in $S$. For example, if $(P_1, P_2)$ are $(t^2, t^3)$, then $t$ is in the normalization because we have both $t^2-P_1=0$ and $P_1 t - P_2=0$.
So, when $Q$ is in the normalization, there will generically be one value of $Q$ given the values of the $P_i$ and, if the values of the $P_i$ are small, then the values of $Q$ are small. 

Here are some contrasting examples to distinguish these concepts.
Let $P_1 = t^2$ and consider $Q=t$. Then $Q^2 = P_1$, but $Q$ cannot be written as a ratio of polynomials in $P_1$. Given a value for $t^2$, there are at most $2$ values for $t$ and, if $t^2$ is bounded then so is $t$, but we generically cannot determine $t$ from $t^2$. So $t$ is in the integral closure of $S$, but it not in $\mathrm{Frac}(S)$.
Conversely, let $P_1 = x$, $P_2 = xy$ and $Q = y$. Then $Q = P_2/P_1$ so, generically, the values of $P_1$ and $P_2$ determine $Q$. However, if $P_1$ and $P_2$ are less than (say) $1$, we can deduce no bound for $Q$. This reflects that $y$ is not integral over $\CC[x,xy]$.
Finally, let's see why normalization isn't the same as seminormalization. Take $P_1 = t(t-1)$,  $P_2=t^2(t-1)$ and $Q = t$. Then $Q=P_2/P_1$, so $Q$ is in $\mathrm{Frac}(S)$, and $Q^2-Q = P_1$ so $Q$ is integral over $S$. So $Q$ is in the normalization of $S$. Indeed, for generic values of $t$, the value of $Q$ can be recovered from $(P_1, P_2)$ and, if $(P_1, P_2)$ are bounded then $t$ is bounded. However, when $t=0$ and when $t=1$, we have $(P_1,P_2)=(0,0)$. So $P_1$ and $P_2$ cannot distinguish the points $t=0$ and $t=1$, while $Q$ can, so $Q$ is not in the seminormalization of $S$.
