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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.
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.
