Algebraically dependent vs. one element can be expressed as a polynomial of the others

Question

Let $k$ be a subfield of a field $X$, suppose $x_1, \cdots, x_n$ are algebraically dependent, that is there exists a non-zero polynomial with coefficients in $k$ and $p(x_1, \cdots x_n) = 0$.

I guess it does not imply that for some $x_i$ there exists a polynomial with coefficients in $k$ and $x_i = q(x_1, \cdots \tilde x_i \cdots x_n)$.

So is there something analogous to linear dependence of vectors? (inear dependent implies one element is the linear combination of the others)

Answer 1

Your guess (second paragraph) is correct. Algebraic dependence is, however, analogous to linear dependence, if you express the latter in a slightly different way:
Define linear dependence of vectors $\{v_1,\cdots,v_n\}$ to mean that there is a linear relation $\sum_i\lambda_iv_i=0$, with the lambdas being scalars in $k$. Replace “linear relation” with “polynomial relation”, and you get the definition of algebraic dependence.

Answer 2

It is analogous to linear independence if the coefficients are not from a field. For instance, the "vectors" $(4,6)$ and $(6,9)$ are linearly dependent over the integers, but neither is an integer multiple of the other.