Let $mathbb{k}$ be an algebraically closed field, and let $psubset mathbb{k}[x_1,…,x_n]$ be a prime ideal contained in the ideal $(x_1,…,x_n)$. Suppose $0$ is a nonsingular point of the variety definted by $p$. Let $A=frac{mathbb{k[x_1,…,x_n]}}{p}$ and let $mathbb{K}$ denote the field of fractions of $A$. Suppose $r$ is the transcendence degree of $mathbb{K}/mathbb{k}$. Then, can we find a set $S={i_1,…,i_r}$ consisting of $r$ different elements such that ${x_{i_1},…,x_{i_r}}$ is a transcedence basis of $mathbb{K}$ over $mathbb{k}$, and for each $1leq ileq n$ such that $inotin S$, there exists a polynomials $phi_iinmathbb{k}[x_{i_1},…,x_{i_r}, y]$ such that $phi_i(x_{i_1},…,x_{i_r}, x_i)=0$ in $mathbb{K}$, but $frac{partial phi_i}{partial y}(0)ne0$.

What I am trying to do is to find something of a “basis” of $mathbb{K}/mathbb{k}$, where $x_{i_1},…,x_{i_r},phi_{i_{r+1}},…,phi_{i_n}$ defines $mathbb{K}$ and the transcendence basis, $x_1,…,x_{i_r}$ helps give the tangent space of the variety at the point and the polynomials $phi_{i_{r+1}},…,phi_{i_n}$ helps give a Jacobian matrix that has a nonsingular, diagonal submatrix.

For example: let $p=(x_1+x_2+x_3^4)subset mathbb{k}[x_1,x_2,x_3]$. Now, if I take $S={1,2}$, then we wont be able to find find a $phi_3$ which satisfies out desired conditions. This is because $frac{partial f}{partial x_3}(0)=0$ (here $f=x_1+x_2+x_3^4$). But, if I take $S={1,3}$ and $phi_2=x_1+x_2+x_3^4$, then it will satisfy our requirements.

In the characteristic $0$ situation, I strongly believe it should work. Although I feel it should work in the prime characteristic case, I don’t feel as strongly.

Could someone help me prove or disprove it!!

Thanks in advance.

Let $f_1,cdots,f_s$ be a set of generators of $P$. Let $J:=begin{pmatrix} frac{partial f_i}{partial x_j}(0)end{pmatrix}_{i,j}$ be the Jacobian matrix at the origin. The assumption that $V(P)$ is smooth at the origin means that this matrix is of rank $n-r$, and the tangent space to $V(P)$ at the origin is exactly the kernel of this matrix. Pick a $(n-r)times (n-r)$ minor of this matrix with nonzero determinant, and let $S$ (defined in your question) be the set of indices of the columns not in this minor.

Now I claim that we can find a bijection between rows $i$ in this minor and columns $j$ in this minor so that $frac{partial f_i}{partial x_j}(0)neq 0$: if not, then by the definition of the determinant, we have that the determinant of this minor vanishes, contradiction. By an application of a linear transformation, we may assume that the linear terms of these $f_i$ are exactly $x_i$ and nothing else. This is almost the statement you want – the $f_i$ form the required diagonal minor inside the Jacobian, and I would argue that if this is really your goal, then requiring the $phi_i$ be polynomials in exactly one more indeterminant is unnecessary and in fact not always possible (see the end of the post for a counterexample).

This also shows that the $x_s$ for $sin S$ form a transcendence basis. $K$ is generated as a field over $k$ by the images of all $x_i$, and it is of transcendence degree $r$. The $x_t$ for $tnotin S$ are all algebraic over $k(x_s)$: we can use elimination theory on the $f_i$ to construct the relevant polynomials in $k(x_s)[y]$ which vanish on $x_i$. So our extension $ksubset K$ can be written $ksubset k(x_s)subset K$, and as transcendence degree adds over extension, we see $ksubset k(x_s)$ must have transcendence degree $r$, which means that the $x_s$ form a transcendence basis.

Here’s why you might not be able to get $phi_i$ the way you want. Such a $phi_i$ would be a polynomial vanishing on the projection of $V(P)$ to the hyperplane spanned by $x_{i_1},cdots,x_{i_r}$ and $x_i$. It may happen that $V(P)$ is smooth at the origin, but any such projection is singular at the origin, which could imply that there’s no possible choice of $phi_i$ meeting the condition on the first derivative with respect to $y$.

Consider the curve $X$ in $Bbb A^3$ given parametrically by $$tmapsto(frac12t(t-2)(t-3),frac{-1}{2}t(t-1)(t-3),frac16t(t-1)(t-2)).$$ This is an irreducible closed curve which is smooth at the origin and passes through $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. It’s three projections to the planes $x=0$, $y=0$, and $z=0$ are the curves cut out by the irreducible polynomials $$2y^3+18y^2z-2y^2+54yz^2-21yz=54z^2-54z^3,$$ $$x^3-9x^2z-x^2+27xz^2+6xz=27z^3-27z^2,$$ and $$2x^3+6x^2y-2x^2+6xy^2-5xy=2y^2-2y^3$$ respectively, all of which are singular at the origin. So in this case there are no choices of $phi_i$ which meet both your criteria.