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Let f: F_p[[X_1,...,X_d]] --->> R be a surjection from a power series ring. Also assume that there is another surjection g: F_p[[Y_1,...,Y_d]] --->> R to the same local ring R.

Question: Is it possible to find an isomorphism h: F_p[[X_1,...,X_d]] \cong F_p[[Y_1,...,Y_d]] such that f = g \circ h ?

Namely we make the morphism commutative with respect to given morphisms f and g.

Please teach me. Many thanks.


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I think this is possible (over any base field $k$, not just $\mathbb{F}_p$). You can rephrase your question to ask: if an isomorphism $\phi: k[[x_1, \dots, x_d]]/I \to k[[y_1, \dots, y_d]]/J$ is specified, when can it be lifted to an isomorphism $k[[x_1, \dots, x_d]] \to k[y_1 , \dots, y_d]]$? Or even further, if you're given a surjection $$\pi: A = k[[x_1, \dots, x_d]] \to k[[y_1, \dots, y_d]]/J = B,$$ when does it come from an isomorphism $$\widetilde{\pi} : A \to C = k[[y_1, \dots, y_d]]?$$ Perhaps this follows from some "formal" properties, but in any case it's not a difficult exercise. You work with the map on tangent spaces $$ \overline{\pi} : {\frak{m}}_A/{\frak{m}}_A^2 \to {\frak{m}}_B/ {\frak{m}}_B^2 \cong {\frak{m}}_C/ (J \cap {\frak{m}}_C + {\frak{m}}_C^2) $$ which must be surjective. So you just have to extend it to an isomorphism of the $d$-dimensional vector spaces, by choosing an isomorphism of $K = \mathrm{ker}(\overline{\pi})$ with $(J \cap {\frak{m}}_C + {\frak{m}}_C^2)/ {\frak{m}}_C^2$, and then it will automatically induce an isomorphism $\widetilde{\pi}$.

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Dear Kumar, thanks a lot! You helped me greatly. Sincerely, Pierre MATSUMI – Pierre MATSUMI Jun 20 '13 at 5:48

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