Let $f_1,\ldots,f_m \in \mathbb{F}[x_1,\ldots,x_n]$ and suppose $\hat{f}_i = f_i$ $\bmod \langle x_1,\ldots,x_n\rangle^3$ (i.e. the linear and quadratic part of $f_i$). Then if $\hat{f}_1,\ldots,\hat{f}_n$ are algebraically independent would that imply $f_1,\ldots,f_n$ are algebraically independent?
Algebraic Independence of Truncated Polynomials
ag.algebraic-geometryfinite-fields
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If $I(X)$, the ideal of $X$ is empty in degrees less than $d$, then there can be no equations of the secant variety until degree $d+1$, and the ideal in degree $d+1$ consists of all polynomials $P$ such that all partials of $P$ are in $I_d(X)$. There is a similar description for the ideal of the secant variety in higher degrees which I call "multi-prolongation", but one does not know when one has generators for the ideal by this method, and it becomes very difficult to compute. In your case, since you have a hypersurface the termination problem does not arise - your cubic is the unique cubic all of whose partial derivatives are in the ideal of $X$.
Edited: the proof below assumes $k$ is algebraically closed. The proof for the multiplicity inequality has been added.
Given $x \in X := V(f_1, \ldots, f_m)$, let $k$ be the local dimension of $X$ at $x$ (i.e. $k$ is the maximum of the dimension of all irreducible components of $X$ containing $x$).
Claim: $mult_x(X) \leq $ the product of the largest $n-k$ elements of the sequence $\deg(f_1), \dots, \deg(f_n)$.
If $k = 0$, then the claim holds due to Bézout's theorem. This estimate is also sharp: given degrees $d_1, \ldots, d_n$, take generic homogeneous polynomials of the specified degrees in $n$-variables - they intersect only at the origin, and the multiplicity at the origin is precisely $\prod_i d_i$ due to Bézout's theorem.
In the case that $n > k \geq 1$, we will prove the following
Sub-claim 1: For each $j = 1, \ldots, n-k$, there are $g_1, \ldots, g_j$ such that each $g_j$ is a linear combination of the $f_i$, and $V(g_1, \ldots, g_j)$ is a complete intersection on a neighborhood of $x$.
Proof: Proceed by induction on $j$. For $j = 1$, set $g_1 := f_1$. If $k = n - 1$, then stop. Otherwise, for $2 \leq j \leq n - k$, we can assume by induction we found $g_1, \ldots, g_{j-1}$ such that $V(g_1, \ldots, g_{j-1})$ is a complete intersection on a neighborhood of $x$. We claim that there is a linear combination of the $f_1, \ldots, f_n$ which does not identically vanish on any irreducible component of $V(g_1, \ldots, g_{j-1})$. Indeed, otherwise since $k$ is infinite, we can find $m$ linearly independent linear combinations of $f_1, \ldots, f_m$ which vanish identically on one of the irreducible components of $V(g_1, \ldots, g_{j-1})$ containing $x$, which would mean that the local dimension of $X$ at $x$ is $n - j + 1 > k$, a contradiction. Now let $g_j$ be that linear combination of the $f_i$ which does not identically vanish on any irreducible component of $V(g_1, \ldots, g_{j-1})$, and repeat. $\square$
Once you find $g_1, \ldots, g_{n-k}$ as above, let $Y := V(g_1, \ldots, g_{n-k})$. The second observation is that $mult_x(X) \leq mult_x(Y)$, which follows from the following general fact:
Sub-claim 2: Let $X \subseteq Y \subseteq k^n$ be a chain of closed subschemes and $x$ be a closed point of $X$ such that $X$ and $Y$ have the same local dimension at $x$. Then $mult_x(X) \leq mult_x(Y)$.
Proof: For each $q \geq 0$, there is a surjection $$O_{x,Y}/m_x^qO_{x,Y} \to O_{x,X}/m_x^qO_{x,X}$$ where $m_x$ is the ideal of $x$. Treating this surjection as an $O_{x,Y}$-module homomorphism, it follows that $$l(O_{x,Y}/m_x^qO_{x,Y}) \geq l(O_{x,X}/m_x^qO_{x,X})$$ where $l$ is the "length" (see e.g. Atiyah-Macdonald, Proposition 6.9). By the assumption on dimension, for $q \gg 1$, both of these lengths are given by polynomials in $q$ of degree $d$, where $d := \dim_x(X) = \dim_x(Y)$. It follows that the coefficient of $q^d$ in the polynomial providing the values of $l(O_{x,Y}/m_x^qO_{x,Y})$ can not be smaller than the corresponding coefficient of the polynomial for $l(O_{x,X}/m_x^qO_{x,X})$. $\square$
The third observation is that if one of the $f_i$ appears in linear combinations defining more than one $g_j$, say $g_{j_1}, g_{j_2}$, then replacing $g_{j_2}$ by an element of the form $g_{j_2} + ag_{j_1}$ for an appropriate $a \in k$ we may ensure that $f_i$ does not appear in the linear combination defining $g_{j_2}$. In this way we can find appropriate "invertible" linear combinations $g'_1, \ldots, g'_{n-k}$ of $g_1, \ldots, g_{n-k}$ such that
- $V(g_1, \ldots, g_{n-k}) = V(g'_1, \ldots, g'_{n-k})$, and
- there is a reordering of $f_1, \ldots, f_n$ such that $\deg(g'_j) \leq \deg(f_j)$, $j = 1, \ldots, n-k$.
The final observation is that $mult_x(Y) = mult_x(Y \cap H)$ for a generic affine subspace of dimension $k$ containing $x$. On the other hand, if $h_1, \ldots, h_k$ are linear polynomials such that $x$ is an isolated point of $V(g'_1, \ldots, g'_{n-k}, h_1, \ldots, h_k)$, then we are done by the $k = 0$ case.
Best Answer
$x^2 + y$ and $x^4 + 2x^2 y + y^2$ are algebraically dependent, but $x^2 + y$ and $y^2$ are not.