Transcendence degree of Coeff f

Let X={X_1, ..., X_n}, Y={Y_1, ..., Y_n} and f,g \in k[X,Y] polynomials (k is a field of characteristic 0). Let Coeff_Y f be the field generated over k by the coefficients of f seen as function of the Y's.

Prove or disprove:

tr deg_k Coeff_Y f.g \geq tr deg_k Coeff f - 1

where tr deg denotes the transcendence degree of the field extension k/Coeff_Y f.

Note that the -1 is needed, as this example shows:

f = X_1.Y_1 + X_1^2.X_2.Y_2, g = X_2

If g is only a polynomial of the X's, the claim is true. Proof: use the fact that the transcendence degree of k/k(f1, ..., fn) (f1, ..., fn being the coefficients of f) is equal the rank of the matrix, whose columns are given by the gradients of the f's. Then simply apply the product rule of derivation and the property rank(A + B) \geq rank(A) - rank(B).

I have somehow the feeling, that only polynomials g(X,Y) of the form g(X,Y)=g1(X).g2(Y) can indeed decrease the transcendence degree and that for every other polynomial we really have:

tr deg_k Coeff_Y f.g \geq tr deg_k Coeff f

Any ideas about a proof (or, in the worst case, a counterexample)?

Thanks a lot.
.

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