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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 $n - m$?


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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
up vote 2 down vote accepted

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.

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The point is that the right notion is not the algebraic independence, but the one of regular sequence. If you start with a regular sequence, then modding out an element lowers the dimension exactly by 1. – Mauro Porta Dec 4 '12 at 19:56

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