Algebraic Independence of Equations vs Polynomials I am considering the difference between algebraic independence of a system of equations and polynomials. Are these two notions equivalent? For example, for $x, y, z$ real,
$xy = A$
$yz = B$
$xz = C$
Seems to be algebraically dependent, because if we set $A=0$, then either $x=0$ or $y=0$, which then implies either $C=0$ or $B=0$. Admittedly, I am not sure of a polynomial $f$ such that $f(xy, yz, xz) = 0$.
However, if we were to view these as polynomials in $x, y, z$, i.e. elements of $\mathbb{R}[x, y, z]$, they seem to be algebraically independent i.e., all the maximal minors of $(\frac{\partial f}{\partial x_{i}})$ are nonzero. Am I correct in saying these two notions are not the same?
EDIT: It is also possible I am confusing the notion of coupling and algebraic dependence - i.e., maybe the suggested equations are algebraically independent, but are coupled, which is why specifying the solution to two sets the solution of the third. Clarification on this point would be appreciated as well
 A: Your notion of algebraically independent equations is not one that I am familiar with. Using your example, I would guess that the definition is that $f_1(x_1, \ldots, x_n)$, $f_2(x_1, \ldots, x_n)$, ..., $f_m(x_1, \ldots, x_n)$ are "algebraically independent as equations" if, given the knowledge of the values of any $m-1$ of the $f_i$, the remaining $f$ is still free to assume any value. I'm going to term this "logical independence". I have not seen this condition before, but I'll try to make some useful remarks about it.
Claim The polynomials $f_1$, $f_2$, ..., $f_m$ are logically independent iff the map 
$$F:(x_1, \ldots, x_n) \mapsto (f_1(x), \ldots, f_m(x))$$
from $k^n \to k^m$ is surjective.  
Proof: It is clear that, if $F$ is surjective, then the $f_i$ are algebraically independent. Conversely, suppose that the $f_i$ are logically independent. Let $(x_1, \ldots, x_m)$ be an arbitrary point in $k^m$.
We will show, by induction on $i$, that there is a point $(x_1, x_2, \ldots, x_i, y_{i+1}, \ldots, y_m)$ in the image of $F$. The base case, $i=0$, is vacuous. Suppose that $(x_1, \ldots, x_i, y_{i+1}, \ldots, y_m)$ is in the image of $F$. Since the $f_i$ are logically independent, the equations $f_1=x_1$, $f_2=x_2$, ..., $f_i=x_i$ does not let us deduce anything about $f_{i+1}$. In particular, it does not let us deduce $f_{i+1} \neq x_{i+1}$. So there is some $z$ with $f_1(z) = x_1$, ..., $f_i(z) = x_i$ and $f_{i+1}(z)=x_{i+1}$. This completes the induction. $\square$.
As your example demonstrates, logical independence is stronger than algebraic independence: $f_1$, ..., $f_m$ are algebraically independent (over an infinite field) if and only if there is a nonzero polynomial $P$ which vanishes on the image of $F$. A simpler example is
$$x \ \mbox{and} \ xy.$$
When $k$ is not algebraically closed, logical independence seems to be a very hard notion to work with. For example, 
$$x^{37}+y^{37}+z^{37} \ \mbox{and} \ xyz$$
are logically dependent over $\mathbb{Q}$ by the work of Taylor and Wiles. It would probably be easy to find equations where it is still an open problem whether they are logically independent.
When $k$ is algebraically closed, this is a reasonable notion, but I've never seen a more concise name for it than "the map $(f_1, \ldots, f_m)$ is surjective".
When $k$ is algebraically closed and $m=n$, we have the following result: If $k[x_1, \ldots, x_n]$ is a finite $k[f_1, \ldots, f_n]$-module, this implies that the $f_i$ are logically independent. The adjective for this is "the map $F$ is finite". "Finite" is strictly stronger than surjective. For example,  $(x,y) \mapsto (x+xy, x^2 y)$ is surjective over $\mathbb{C}$ (exercise!) but I claim that $\mathbb{C}[x+xy, x^2 y]$ is not a finite $\mathbb{C}[x,y]$ module. (Harder exercise. Hint:  Show that $\mathbb{C}[x,xy]$ IS finite over $\mathbb{C}[x+xy,xy^2]$.)
I don't know of a good generalization of "finite" to the case $m \neq n$.
A: EDIT: OP has edited reference to linear independence out of the question, so what follows is no longer relevant. 
Linear independence and algebraic independence are very different things. For example, the polynomials $x$ and $x^2$ are linearly independent, since there are no real numbers $a$ and $b$ such that $ax+bx^2$ is the zero polynomial, but they are algebraically dependent, since if $f(u,v)=u^2-v$ then $f(x,x^2)$ is identically zero. 
Another example: $1$ and $\sqrt2$ are linearly, but not algebraically, independent over the rationals. 
