TITLE: Algebraic independence and dimension of a variety
QUESTION [7 upvotes]: A set of polynomials $\{f_1,\ldots,f_m\}$ in $k[x_1,\ldots,x_n]$ are algebraically independent over $k$ iff for all polynomials $p \in k[y_1,\ldots,y_m]$, $p(f_1,\ldots,f_m) = 0$ implies that $p = 0$. 
In linear algebra, the dimension of a subspace of $k^n$ defined by $m$ linearly independent equations is $n - m$. Is the analogous statement in algebraic geometry true: that the dimension of a variety in $k^n$ defined by $m$ algebraically independent polynomials is $n - m$?
Thanks!

REPLY [5 votes]: One definition of dimension for an algebraic variety is the transcendence degree of its function field (which, for an affine variety, is the fraction field of its coordinate ring).  Intuitively, each polynomial you use to define a variety in an n-dimensional ambient space reduces dimension by 1, but as YACP's comment above illustrates, this is not always the case even if the polynomials are algebraically independent.