# Theorem from Deformation Theory

My question refers to some steps it the proof of Theorem 3.3 part (b) in Christensen's paper treating Deformation theory (see pages 9-11): https://mathematics.stanford.edu/wp.../A.-Christensen-Draft.pdf

(Recently I noticed that the link not work. Fortunatelly it's first google hit for "Christensen Deformation Theory")

Here the excerpt:

As expected firstly (in order to avoid the $$H^1$$ obstruction) the author verifies the affine case.

We start with quasi complact separated scheme $$X$$ and a closed immersion $$i_X:X_0 \subset X$$ where $$X_0$$ is defined by quasicoherent ideal sheaf $$\mathcal{I}_X$$ with $$\mathcal{I}_X^2=0$$.

Futhermore let $$S$$ be some other scheme and $$i_S:S_0 \subset S$$ be a thickering (esp a closed immersion) defined also by quasicoherent ideal sheaf $$\mathcal{J}_S$$ with $$\mathcal{J}_S^2=0$$.

Assume $$X_0$$ is a $$S_0$$-scheme so we have a morphism $$f:X_0 \to S_0$$. Thm 3.3 says that under given conditions we can obtain flat deformation morphism $$X \to S$$ with $$X \times_S S_0=X_0$$.

My problem in the proof is why the obtain morphism is flat?

Back to the local situation:

We set $$SpecB_0:=V_0 \subset S_0, SpecB:=V \subset S$$.

Futhermore $$SpecB_0[t_1,...,t_n]/(f_1,...,f_r) := U_0 \subset X_0$$. And we have smooth morphism $$f:SpecB_0[t_1,...,t_n]/(f_1,...,f_r) \to S_0$$ with $$f_i \in B_0[t_1,...,t_n]$$ and rank $$(\partial g_j/ \partial t_j)_{ij}=r$$ (since $$f$$ smooth!). The author's arguement is that locally we can lift the $$f_i$$ (in the excerpt he calls them $$g_i$$. I guess it's a typo) to elements of $$B[t_1,...,t_n]$$ and this would induce the desired morphism $$U=SpecB[t_1,...,t_n]/(f_1,...,f_r) \to S_0$$.

My question is why is this morphism neccessary flat?

On the ring side it is nothing but $$B \to B[t_1,...,t_n]/(f_1,...,f_r)$$. By our assumption we have $$B_0 =B/I$$ with $$I^2=0$$ and $$B_0 \to B_0[t_1,...,t_n]/(f_1,...,f_r)$$ is smooth.

Why does this imply that $$SpecB[t_1,...,t_n]/(f_1,...,f_r) \to S_0$$ is flat?

Since we want that $$X \times_S S_0=X_0$$ the "base change argument" cannot be applied since here we go "in other direction".

So why is the lifted $$U \to V$$ flat?