Field of rational functions at a non singular point

Let $$\mathbb{k}$$ be an algebraically closed field, and let $$p\subset \mathbb{k}[x_1,...,x_n]$$ 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 $$1\leq i\leq n$$ such that $$i\notin S$$, there exists a polynomials $$\phi_i\in\mathbb{k}[x_{i_1},...,x_{i_r}, y]$$ 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!!

• Your example is wrong: $0$ is a non-regular point for the variety described by $x_1+x_2+x_3^4=0$... Moreover, for the finite-type schemes over algebraically closed fields, a close point is smooth if and only if it is regular; in other words, the regularity (of closed points) can be checked by Jacobian Criterion. Commented May 28, 2020 at 8:07
• @Armandoj18eos It seems to me that the Jacobian matrix of that variety is $(1,1,4x_3^3)$ which is of maximal rank everywhere. Can you explain your claim that the origin isn't regular? Commented May 28, 2020 at 10:18
• I apologize: I red another polynomial, you are right! Commented May 28, 2020 at 10:57

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 $$s\in S$$ 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 $$t\notin S$$ 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 $$k\subset K$$ can be written $$k\subset k(x_s)\subset K$$, and as transcendence degree adds over extension, we see $$k\subset k(x_s)$$ 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 $$t\mapsto(\frac12t(t-2)(t-3),\frac{-1}{2}t(t-1)(t-3),\frac16t(t-1)(t-2)).$$ 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.