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

  • $\begingroup$ I'd say the equations $D:xy=0$, $E:xz=0$, $F:yz=0$ are logically dependent, as $D$ implies ($E$ or $F$). I'm not sure that makes the equations $xy=A$, $xz=B$, $yz=C$ dependent, logically or otherwise. And certainly the polynomials $xy$, $xz$, $yz$ are algebraically independent. $\endgroup$ Mar 7, 2013 at 2:47
  • $\begingroup$ Similar question posted to MO, mathoverflow.net/questions/123824/… $\endgroup$ Mar 7, 2013 at 4:30

2 Answers 2


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 $m \neq n$.

  • $\begingroup$ Hm. You raise some interesting points. I take back what I said in my original example - the Jacobian defined by the three equations I have listed is square, so its maximal minor is itself and is indeed zero - so in this case, the polynomials are algebraically dependent and "logically dependent" by your definition. Thank you for the help, your notion of logical independence is exactly what I was attempting to describe. $\endgroup$ Mar 7, 2013 at 23:15
  • $\begingroup$ Can you potentially think of an example of polynomials which are algebraically independent but logically dependent? In my specific case, I am only interested when we are looking at polynomials in $\mathbb{R}[x_{1},x_{2},...,x_{m}]$, so perhaps that makes the notions equivalent, or easier than a general field? Ideally they will be equivalent, as if they are not I might be out of luck for what I am trying to do. $\endgroup$ Mar 7, 2013 at 23:16
  • $\begingroup$ Uh, as I said, $(x, xy)$ is logically dependent but algebraically independent. So is your original example: The Jacobian is $\det \begin{pmatrix} y & x & 0 \\ 0 & z & y \\ z & 0 & x \end{pmatrix} = 2 xyz \neq 0$. $\endgroup$ Mar 8, 2013 at 0:46

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.

  • $\begingroup$ Ah, I just realized I said the polynomials "appear to be linearly independent" in $\mathbb{R}[x, y, z]$. I did not mean to put linearly there. I've now adjusted the original question. $\endgroup$ Mar 7, 2013 at 1:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.