Field of rational functions at a non singular point

Question

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!!

Thanks in advance.

Answer 1

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.

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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.