# Uniqueness of deformation family,

Say I want to construct a flat family of affine schemes using deformation theory. To be more specific, suppose I have a local $k$-algebra $A$, and I want to find a flat family $\mathcal{X} \to T$, where $T$ is $\mathrm{Spec}$ of a ring $T$, and the family has $A$ as central fiber.

To do this, I choose among the space of first-order deformations, $T^1_{A/k}$, some subset $\{t_1, \cdots,t_n\}$ of a basis. This solution can be lifted (modulo obstructions, but that's okay) successively, and if we're lucky, the lifting stops (see for example this article for an algorithm), producing a family $\mathcal{X} \to T$ having $(\langle t_1, \cdots,t_n\rangle/\langle t_1,\cdots,t_n\rangle^2)^\lor$ as tangent space at the origin.

The question is the following: what if I didn't choose the $t_i$ as deformation parameters, but some multiple of them, i.e. $\{c_1t_1,\cdots,c_nt_n\}$, where the $c_i$ are non-zero constants, would I get the same family (up to isomorphism)?

It seems obvious that I should, I even suspect the question doesn't have anything to do with deformation theory, but is purely algebraic - but I haven't been able to see an easy solution.

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What do you mean by an isomorphism of deformations? If the isomorphism is supposed to be over $T$, then the answer is negative, because the two families have different Kodaira-Spencer maps. If, on the other hand, you allow automorphisms of $T$, the answer should be positive for trivial reasons. –  Angelo Apr 29 '13 at 19:53
Dear @Angelo, thanks for the answer. In my case I would allow automorphisms of $T$. Could you hint at the trivial reasons why the answer should be true? –  Fredrik Meyer Apr 30 '13 at 8:06
A first order deformation extends to a miniversal deformation if and only if the corresponding Kodaira-Spencer map is an isomorphism. Your modified deformation obviously satisfies this condition. Then you use uniqueness of miniversal deformations. –  Angelo Apr 30 '13 at 9:35