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Why are the following polynomials algebraically independent ?

$f_1=(z+t)(yz+yt+t)(xyz+xyt+xt+t)t^3$

$f_2=(z+t)(yz+yt+t)(xy+x+a)t^5$

$f_3=(z+t)(y+b)(xy+x+a)t^6$

$f_4=(y+b)(xy+x+a)t^7$

where $a,b$ are constants.

I have tried comparing the t-degrees of the monomials but didn't get a clue.

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1 Answer 1

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You can certify the independence of this collection of polynomials by calculating the Jacobian matrix. That is, consider the matrix, $J$, with rows labelled by $f_{1}$, $f_{2}$, $f_{3}$, and $f_{4}$, and columns labelled by $x$, $y$, $z$, and $t$. The entry in row $f_{1}$ and column $x$ is the partial derivative $\partial f_{1}/\partial x$, and the other entries are calculated analogously. Now $J$ is a matrix over the extension field $\mathbb{Q}(x,y,z,t)$, and we can find its determinant in that field. The determinant is non-zero if and only if the collection is algebraically independent. Theorem 6 of this paper has a reference to a proof.

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  • $\begingroup$ More references on the Jacobian criterion for algebraic independence can be found here. $\endgroup$
    – user26857
    Nov 14, 2015 at 22:29
  • $\begingroup$ I tried this in the very beginning but its difficult to find the determinant of the Jacobian matrix. Can't we do this just looking at some particular monomials or say some powers of a variable ? $\endgroup$
    – Jack
    Nov 17, 2015 at 16:20

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