Let $M$ be a (usual, finite dimensional) smooth manifold, and $C^\infty(M)$ the algebra of (real valued) smooth functions on $M$. For each point $a\in M$ and for each subalgebra $A$ in $C^\infty(M)$ (I mean $A$ contains constants and is closed under summing and multiplication) let us denote by $I_a[A]$ the ideal in $A$ consisting of functions which vanish in $a$: $$ f\in I_a[A]\quad \Longleftrightarrow\quad f\in A\quad \&\quad f(a)=0. $$ Let us consider also the ideal $I_a^2[A]$ in $A$ defined by the following rule: $$ f\in I_a^2[A]\quad \Longleftrightarrow\quad \exists g_1,...g_n,h_1,...,h_n\in I_a[A]:\quad f=\sum_{i=1}^ng_i\cdot h_i. $$
From the Hadamard lemma it follows that in the simplest situation, when $A=C^\infty(M)$ the following criterion holds:
Theorem. If $A=C^\infty(M)$, then $f\in I_a^2[A]$ if and only if $f\in I_a[A]$ and $f$ "has zero derivatives in all directions" in $a$ (i.e. $t(f)=0$ for each tangent vector $t\in T_a(M)$).
A question: is it possible that the same is true for any subalgebra $A\subseteq C^\infty(M)$?
One can simplify a little bit this question to the following
Hypothesis 1: For each (unital) subalgebra $A\subseteq C^\infty(M)$ $$ f\in I_a^2[A]\quad \Longleftarrow\quad \Big( f\in I_a[A]\quad\&\quad \forall t\in T_a(M) \quad t(f)=0 \Big). $$
Or to the following
Hypothesis 2: For each (unital) subalgebra $A\subseteq C^\infty(M)$ $$ I_a^2[A]\supseteq A\cap I_a^2[C^\infty(M)]. $$
I am not sure if this is important, but my algebra $A$ differs the points of $M$ $$ a\ne b\in M \quad \Longrightarrow\quad \exists f\in A\quad f(a)\ne f(b), $$ and the tangent vectors in each point $$ 0\ne t\in T_a(M) \quad \Longrightarrow\quad \exists f\in A\quad 0\ne t(f). $$