# Algebraic independence and dimension of a variety

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!

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I leave the algebraic geometers to say more, but to me the answer to your question is no. For example, if $R=\mathbb{C}[X_1,X_2]$, $f_1=X_1+X_2$ and $f_2=X_1^2-X_2^2$, then the ideal $I=(f_1,f_2)$ is contained in $(X_1+X_2),$ hence its height is one. –  user26857 Dec 4 '12 at 18:54
What is true, however – modulo various niceness conditions – is that the dimension of a variety with $m$ algebraically independent coordinate functions is $m$. –  Zhen Lin Dec 4 '12 at 19:03
Finally, as a partial converse, we can say that if a prime ideal $\mathfrak p$ is of height $m$, then we can find $n-m$ algebraically independent elements defining functions over $k[X_1,\ldots,X_n]/\mathfrak p$. This is possible thanks to the refined Noether's normalization Lemma proved for example in Eisenbud's Commutative Algebra with a view towards Algebraic Geometry, Theorem 13.3 page 283. –  Mauro Porta Dec 4 '12 at 21:22