Suppose I take a polynomial $p \in \mathbb{C}[x\_1, \dotsc, x\_n]$ and I consider the zero set $\mathcal{Q}$ of the set of polynomials composed of $p$ and all of the "pure" partials - that is, our set of polynomials consisting of $p$ and all $\frac{\partial\_i p}{\partial x\_k^i}$ for all $i, k$ (obviously, the set is finite, since eventually, the partials become zero). Clearly, if the hypersurface defined by $p$ is nonsingular, then $\mathcal{Q}$ is empty (you only need look at the first derivatives), and for finite characteristic all sort of goofiness may result (e.g., in characteristic 2, only the first derivatives matter, and for $p = x\_1^2+ \dots x\_n^2$ the variety $\mathcal{Q}$ is the same as the zero set of $p.$


Can anything intelligent be said?



A seemingly Groebnerizable problem
**In characteristic zero, a point $a$ is in $\mathcal{Q}$ if and only if $p$ vanishes identically on each line through $a$ parallel to one of the coordinate axes**, i.e., all lines given by fixing all but one of the coordinates of $a$ and allowing the remaining coordinate to vary.


Indeed, the partial derivatives $\partial^i p/\partial x\_k^i$ vanish at $a$ for all orders $i \geq 0$ if and only if $p$ vanishes identically on the line through $a$ parallel to the $k$th coordinate axis. (And $a \in \mathcal{Q}$ if and only if this occurs for each $k$.) This is because, when we restrict $p$ to this line and expand the resulting univariate polynomial in a Taylor series centered at $a$, the Taylor coefficients are given by the partials $\partial^i p/\partial x\_k^i$, evaluated at $a$ (up to some factorial factors). So $p$ vanishes on the line if and only if its restriction to the line vanishes identically, if and only if the Taylor coefficients vanish, if and only if the pure partials of $p$ vanish at $a$.


The set $\mathcal{Q}$ can be nonempty. For example, if $p$ is any linear combination of polynomial multiples of $x\_j x\_k$ for $j \neq k$—that is, if $p$ is any element of the ideal generated by the $x\_j x\_k$, $1 \leq j < k \leq n$, in the polynomial ring—then the origin $0 \in \mathcal{Q}$. One way to see this is that in any pure derivative by, say, $x\_k$ to order $i$, every term of $p$ either is killed (if it has $x\_k$ to a power less than $i$) or else survives, but even then that term still involves some $x\_j$, $j \neq k$. So every pure derivative of $p$ vanishes term-by-term at the origin. Another way to see it, using a touch of algebraic geometry, is that the ideal generated by the products $x\_j x\_k$ is exactly the ideal of polynomials that vanish on the union of the coordinate axes (left to the reader), so $p$ is in the ideal if and only if $p$ vanishes identically on the coordinate axes, which is apparently equivalent to $0 \in \mathcal{Q}$.


Conversely, if $0 \in \mathcal{Q}$, then $p$ can't have any "pure" terms $x\_k^i$ (with a nonzero coefficient), otherwise $\partial^i p/\partial x\_k^i$ would fail to vanish at the origin. This proves that **in characteristic $0$, the origin $0 \in \mathcal{Q}$ if and only if $p$ is a linear combination of polynomials divisible by $x\_j x\_k$ for $j \neq k$**, that is, $p$ is in the ideal generated by the $x\_j x\_k$, $j \neq k$. More generally, $a = (a\_1,\dotsc,a\_n) \in \mathcal{Q}$ if and only if $p$ is in the ideal generated by the products $(x\_j-a\_j)(x\_k-a\_k)$, $j \neq k$.


Unlike the situation in positive characteristic, in characteristic zero, $\mathcal{Q}$ can't coincide with the zero set of $p$, except if the zero set of $p$ is empty or $p=0$. This is because if the zero set of $p$ is nonempty and coincides with $\mathcal{Q}$, then $\mathcal{Q}$ contains some point; then the union of coordinate lines through the point; then the union of all the coordinate lines through those lines, which gives all the coordinate $2$-planes through the original point; then the union of all the coordinate lines through those, which gives all the coordinate $3$-planes through the original point... In the end, the zero set of $p$ is the whole space, so $p=0$.


All of the above statements hold in positive characteristic, too, if the degree of the polynomial $p$ is strictly less than the characteristic.


More generally (briefly) for $m \geq 1$ let $\mathcal{Q}\_m$ be the set of points where $p$ vanishes along with all of $p$'s partial derivatives of all orders that involve at most $m$ variables. The original question and above discussion are for $m=1$. In characteristic $0$, or characteristic strictly greater than the degree of $p$, $a \in \mathcal{Q}\_m$ if and only if $p$ vanishes identically on the union of the $m$-dimensional planes through $a$ that are translates of coordinate $m$-planes, i.e., spanned by standard basis vectors. The origin $0 \in \mathcal{Q}\_m$ if and only if $p$ lies in the ideal generated by products of $m+1$ distinct variables, $x\_{i\_1} \dotsm x\_{i\_{m+1}}$ for $1 \leq i\_1 < \dotsb < i\_{m+1} \leq n$.


