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Suppose that I have $n$ homogeneous polynomials $f_1, \dots, f_n \in \mathbb{C}[x_1, \dots, x_m]$ and that $n < m$. Is there a well known method or algorithm to determine if these polynomials are algebraically independent?

As far as I know the Jacobian criterion works only for the case where $n=m$.

flag	asked Oct 8 2010 at 20:27 Jon 78●3

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You can consider the equations $f_i=t_i$ with new variables $t_i$, and eliminate the $x$'s using Groebner bases. – Mariano Suárez-Alvarez Oct 8 2010 at 20:57

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(This is explained, if I recall correctly, in Cox's book on Varieties, Ideals and algorithms) – Mariano Suárez-Alvarez Oct 8 2010 at 20:58

jvp · Accepted Answer · 2010-10-08 21:37:08Z UTC

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The polynomials are algebraically independent if and only if $$ df_1 \wedge df_2 \wedge \cdots \wedge df_n $$ is not identically zero. In other words, you have only to check that one of the maximal minors of the matrix $\left( \frac{\partial f_i}{\partial x_j} \right)$ is nonzero.

answered Oct 8 2010 at 21:37

jvp
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Can you give a proof? – Martin Brandenburg Oct 9 2010 at 11:06

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Martin, in char. 0 the map $f:\mathbf{A}^m \rightarrow \mathbf{A}^n$ is smooth on a dense open in the source if and only if it is dominant, and such generic smoothness is equivalent to generic injectivity of the induced map $f^{\ast}(\Omega^1_{\mathbf{A}^n/k}) \rightarrow \Omega^1_{\mathbf{A}^m/k}$. But at the generic point, such injectivity can be checked between $n$th exterior powers. – BCnrd Oct 9 2010 at 12:47

Thanks JVP!, that is what I needed. – Jon Oct 10 2010 at 14:29

AlexR · Answer 2 · 2011-01-05 22:46:12Z UTC

Hi there. A Groebner basis based algorithm that also produces an annihilating polynomial in case the polynomials are algebraically dependent can be found on the Singular site at

http://www.singular.uni-kl.de/Manual/3-0-2/sing_534.htm .

How to show a set of polynomials is algebraically independent?

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