I can show that this is true for your "simple" case.
If g(x,y) ∈ C∞(ℝ2) vanishes on x ≤ 0 then it decomposes as g(x,y) = a(x)G(x,y) where a(x) ∈ C∞(ℝ) vanishes on x ≤ 0 and G(x,y) ∈ C∞(ℝ2).
This can be shown by proving the statements below. They could possibly be standard results, but I've never seen them before.
First, I'll refer to the following sets of functions.
- Let U be thet set of functions f(x) ∈ C∞(ℝ) which vanish on x ≤ 0 and are positive on x > 0.
- Let V be the set of functions f: ℝ+→ ℝ such that x-n f(x) → 0 as x → 0, for each positive integer n.
The statements I need to show the main result are as follows.
Lemma 1: For any f ∈ V, there is a g ∈ U such that f(x)/g(x) → 0 as x → 0.
Proof: Choose any smooth function r: ℝ+→ ℝ+ with r(0) = 1 and r(x) = 0 for x ≥ 1. For example, we can use r(x) = exp(1-1/(1-x)) for x < 1. Then, the idea is to choose a sequence of positive reals αk → 0 satisfying ∑k αk < ∞, and set
$$g(x) = x^{\theta(x)},\ \ \ \theta(x)=\sum_{k=1}^\infty r(x/\alpha_k)$$
for x > 0 and g(x) = 0 for x ≤ 0. Only finitely many terms in the summation will be nonzero outside any neighborhood of 0, so it is a well defined expression, and smooth on x > 0. Clearly, θ(x) → ∞ and, therefore, x-n g(x) → 0 as x → 0. It needs to be shown that all the derivatives of g vanish at 0 so that g ∈ U. As r and all its derivatives are bounded with compact support, r(n)(x) ≤ Knx-n-1 for some constants Kn. The nth derivative of θ is
$$\theta^{(n)}(x)=\sum_k\alpha_k^{-n}r^{(n)}(x/\alpha_k)\le K_nx^{-n-1}\sum_k\alpha_k$$
which has polynomially bounded growth in 1/x. The derivatives of log(g) satisfy
$$\frac{d^n}{dx^n}\log(g(x))=\frac{d^n}{dx^n}\left(\log(x)\theta(x)\right)$$
which also has polynomially bounded growth in 1/x. However, the derivative on the left hand side is g(n)(x)/g(x) plus a polynomial in g(i)(x)/g(x) for i < n. So, induction gives that g(n)(x)/g(x) has polynomially bounded growth in 1/x and, multiplying by g(x), g(n)(x) → 0 as x → 0.
By definition of f ∈ V, there is a decreasing sequence of positive reals εk such that f(x) ≤ xn for x ≤ εn. We just need to make sure that αk ≤ εn+1 for k ≥ n to ensure that g(x) ≥ xn-1 for εn+1 ≤ x ≤ min(εn,1). Then f(x)/g(x) goes to zero at rate x as x → 0.
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Lemma 2: For any sequence f1,f2,... ∈ V there is a g ∈ U such that fk(x)/g(x) → 0 as x → 0 for all k.
Proof: The idea is to apply Lemma 1 to f(x) = Σk λk|fk(x)| for positive reals λk. This works as long as f ∈ V, which is the case if Σk λksupx≤kmin(x,1)-k|fk(x)| is finite, and this condition is easy to ensure.
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Lemma 3: For any sequence f1,f2,... ∈ V there is a g ∈ U such that fk(x)/g(x)n → 0 as x → 0 for all positive integers k,n.
Proof: Apply Lemma 2 to the doubly indexed sequence fk,n = |fk|1/n.
The result follows from applying lemma 3 to the triply indexed sequence fi,j,k(x) = max{|(di+j/dxidyj)g(x,y)|: |y| ≤ k} ∈ V. Then, there is an a ∈ U such that fijk(x)/a(x)n → 0 as x → 0. Set G(x,y) = f(x,y)/a(x) for x > 0 and G(x,y) = 0 for x ≤ 0. On any bounded region for x > 0, the derivatives of G(x,y) to any order are bounded by a sum of terms, each of which is a product of fijk(x,y)/a(x)n with derivatives of a(x), so this vanishes as x → 0. Therefore, G ∈ C∞(ℝ2).
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In fact, using a similar method, the simple case can be generalized to arbitrary submersions.
Let p: M →N be a submersion. If h ∈ C∞(N) and g ∈ C∞(M) satisfy hg = 0 then, g = aG for some G ∈ C∞(M) and a ∈ C∞(N) satisfying ha = 0.
Very Rough Sketch:
If S ⊂ N is the open set {h≠0} then g and all its derivatives vanish on p-1(S).
The idea is to choose a smooth parameter u:N-S →ℝ+ which vanishes linearly with the distance to S. This can be done locally and then extended to the whole of N (I'm assuming manifolds satisfy the second countability property). As all the derivatives of g vanish on p-1(S), u-ng tends to zero at the boundary of S. This uses the fact that p is a submersion, so that u also goes to zero linearly with the distance from p-1(S) in M.
Then, following a similar argument as above, a can be expressed a function of u so that g/a and all its derivatives tend to zero at the boundary of S. Finally, G=0 on the closure of p-1(S) and G=g/a elsewhere.
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I suppose the next question is: does proving the special case above of a single g and h reduce the proof of flatness to algebraic manipulation?
Best Answer
The definition of $f:R\to S$ being faithfully flat that I first saw is that $S\otimes_R-$ is exact and faithful (meaning that $S\otimes_R M=0$ implies $M=0$). I'm not sure exactly what your definition of "faithfully flat" is, but it looks like you're happy with "flat and surjective on spectra." You get flatness for free from étaleness, so I'll show that the extra faithfulness condition implies surjectivity on spectra.
Upon tensoring with $S$, $f$ becomes $f\otimes_R id_S:S\cong S\otimes_R R\to S\otimes_R S$, given by $s\mapsto s\otimes 1$. This is injective since it is a section of multiplication $S\otimes_R S\to S$. By flatness of $S$, this shows that $S\otimes_R \ker f=0$, so $\ker f=0$. So I'll identify $R$ with a subring of $S$.
Let $\mathfrak p\subseteq R$ be a prime ideal. We wish to show that there is a prime $\mathfrak q\subseteq S$ such that $\mathfrak q \cap R=\mathfrak p$. Let $K$ be the kernel of the morphism $R/\mathfrak p\to S/\mathfrak p S$ of $R$-modules. Upon tensoring with $S$, this morphism becomes injective (as before, it's a section of the multiplication map $S/\mathfrak p S\otimes_R S/\mathfrak p S\to S/\mathfrak p S$), so by flatness of $S$, we have $S\otimes_R K=0$, so $K=0$. This shows that $\mathfrak p S \cap R=\mathfrak p$ (if the intersection were any larger, $K$ would be non-zero). So $\mathfrak p$ generates a proper ideal in the localization $(R\setminus \mathfrak p)^{-1}S$. Let $\mathfrak q\subseteq (R\setminus \mathfrak p)^{-1}S$ be a maximal ideal containing $\mathfrak p$. This corresponds to some prime ideal $\mathfrak q$ (slight abuse of notation to use the same letter) of $S$ which contains $\mathfrak p$ but does not intersect $R\setminus \mathfrak p$, so $\mathfrak q\cap R=\mathfrak p$.
See also exercise 16 of Chapter 3 of Atiyah-Macdonald.