How to verify that one of equations in a polynomial system is redundant? I know that system of polynomial equations
$$
p_1(x_1,\dots,x_n)=0,..., p_N(x_1,\dots,x_n)=0
$$
has infinitely many solutions.
I computed  some of them numerically and notices that they always satisfy one more polynomial equation 
$$
q(x_1,\dots,x_n)=0.
$$
I would like to prove that this is always the case. 
Question:  Does it exist a method to prove that
that $q(x_1,\dots,x_n)=0$ follows from $p_1(x_1,\dots,x_n)=0,..., p_N(x_1,\dots,x_n)=0$?
Extra info:
in my application $p_1, p_N$ contains terms of degrees $3$ and $0$, and $q$  contains terms of degree $6$,$3$, and $0$.
 A: Compute a Groebner basis $G$ for $\{p_1, ..., p_n\}$ with respect to some term order, and divide $q$ by the basis.  If the result is zero then $q$ can be expressed as a polynomial combination of the $p_i$.  In Maple:
P := {p1,p2,...,pn};  # assuming you defined those
G := Groebner[Basis](P, 'tord');  # Maple can choose the term order
r := Groebner[NormalForm](q, G, tord);  # divide q by G

A: Hint: If you compute the row echelon form of the matrix given by the polynomial system, then the all-zero rows will be the redundant ones. 
A: Since the question is not very concrete I will give a broad answer: Given your polynomial equations $p_1,\ldots,p_N$ you can associate to them an ideal 
$$ I:=\left\{\sum_{i=1}^N q_ip_i\vert q_i \in k[x_1,\ldots,x_n]   \right\} $$
In plain english: $I$ contains every linear combination of the $p_i$ with coefficients that are polynomials too. Now you can easily see that every element of $I$ vanishes on the zero locus of the $p_i$. We say that the $p_i$ generate $I$. 
Now your question can be interpreted in 2 ways: you can either ask if there is a algorithm to decide if the $p_i$ are a minimal generating system of $I$. The answer to this is that it's known that in general there is no such algorithm. See for example here. However there my be cases in which you can decide this. But we can say nothing about this until you give us more information.
The second interpretation of your question would be to ask given a polynomial $q$, how to decide if $q^m \in I$ for some $m \in \mathbb{N}$? Again I don't know about a general answer to this question. 
A: When you have a redundant system of $N$ equations of any type for $n$ variables such$$p_i=p_i(x_1,\dots,x_n)=0\qquad i=1,2,\cdots N >n$$ one of the methods for solving consists in the minimization of $$\Phi(x_1,\dots,x_n)=\sum_{i=1}^N p_i^2$$ So, if the result is $0$, then the system is redundant and $\Phi$ contains all the roots.
Now, consider the case of polynomials and see what is $\Phi$. 
A: just sum any number of the given equations and it will yield another redudant equation (since it depends on the summed equations) you can even multiply any of these by a constant, you can also multiply the equations.
$$
q(\vec{x}) = \sum k_j p_j(\vec{x})p_{j'}(\vec{x}) = 0
$$
