If $f(m) = 0, df(m) = 0$, then $f(x) = g(x)h(x)$ locally, with $g(m) = 0 = h(m)$?

Thumbing through "Foundations of Differentiable Manifolds and Lie Groups" by Frank Warner today, I saw the following

Lemma: $\;$ $T_{m}M$ is naturally isomorphic to $(F_m/{F_m}^2)^\ast$.

Here $M$ is a smooth manifold, $T_mM$ is the tangent space at $m \in M$ and $F_m$ is the ideal of all germs of smooth functions from $M$ into $\mathbb R$ which vanish at $m$. The isomorphism is given by $$T_mM \to (F_m/{F_m}^2)^\ast, \quad X \mapsto \left( [f] \mapsto X(f) \right)$$ where $f \in C^\infty(U)$ represents a germ on some neighbourhood $U$ of $m$.

I observed a corollary of this: Since both vector spaces are finite dimensional, we obtain an isomorphism between the dual spaces $T_m^\ast M \simeq (F_m/{F_m}^2)^{\ast \ast} = F_m/{F_m}^2$, given by $[g] \mapsto {\mathrm dg}_m$.

But this then implies the following:

If $f\in C^\infty(\mathbb R^n)$ such that $f(0) = 0$ (i.e. $[f]\in F_m$ with $m=0$) and ${\mathrm df}_0 = 0$, then, there exists a neighbourhood $U$ of $0$ in $\mathbb R^n$ and functions $g,h \in C^\infty(U)$, such that $g(0) = 0, h(0) = 0$ and $$f(x) = g(x) h(x) \quad \forall x\in U$$

I was wondering whether one could obtain the above result directly?

For the one-dimensional case this is not too difficult: Just use Taylorexpansion to get $$f(x) = f(0) + df(0)x + r(x)x^2 = r(x)x^2$$

And define $h(x) = x\cdot r(x),\; g(x) = x$.

But already in dimension $2$, I don't know how one would factor something like $x^2 + y^2$. So here are my

Questions:

• Do you see any flaws my derivation of the corollary?
• Do you see a way of factoring $x^2 + y^2$ in $F_0$ (as described above)?.
• Is there a general way of doing this / a direct proof of this "factorization-property"?

I'm looking forward to your thoughts!

-

The problem is that $F_m^2$ is not the set of all products of pairs of things in $F_m$, it's sums of such things.

Hence, your conclusion should be that if $f(0) = 0$ and $d_0 f = 0$, then $f(x) = \sum_i g_i(x)h_i(x)$ where $g_i(0) = h_i(0) = 0$.

This conclusion is actually correct - see Little Bézout theorem for smooth functions and comments. In fact, you can take $g_i(x) = x_i$.

In your example of $x^2+y^2$, you get $x^2 + y^2 = x*x + y*y$, i.e., $g_1(x,y) = h_1(x,y) = x$ and $g_2(x,y) = h_2(x,y) = y$.

-
This makes sense (I'm lacking algebra, it seems), it also matches my mental image of differentiable functions much better. Thanks a lot for reading through it and clearing this up! =) – Sam Oct 5 '11 at 19:57