Lemma 69.23.6. Notation and assumptions as in Lemma 69.23.4. If $f$ is flat and of finite presentation, then there exists an $i_3 > i_0$ such that for $i \geq i_3$ we have $f_ i$ is flat, $X_ i = Y_ i \times _{Y_{i_3}} X_{i_3}$, and $X = Y \times _{Y_{i_3}} X_{i_3}$.

**Proof.**
By Lemma 69.7.1 we can choose an $i \geq i_2$ and a morphism $U \to Y_ i$ of finite presentation such that $X = Y \times _{Y_ i} U$ (this is where we use that $f$ is of finite presentation). After increasing $i$ we may assume that $U \to Y_ i$ is flat, see Lemma 69.6.12. As discussed in Remark 69.23.5 we may and do replace the initial diagram used to define the system $(X_ i)_{i \geq i_1}$ by the system corresponding to $X \to U \to B_ i$. Thus $X_{i'}$ for $i' \geq i$ is defined as the scheme theoretic image of $X \to B_{i'} \times _{B_ i} U$.

Because $U \to Y_ i$ is flat (this is where we use that $f$ is flat), because $X = Y \times _{Y_ i} U$, and because the scheme theoretic image of $Y \to Y_ i$ is $Y_ i$, we see that the scheme theoretic image of $X \to U$ is $U$ (Morphisms of Spaces, Lemma 66.30.12). Observe that $Y_{i'} \to B_{i'} \times _{B_ i} Y_ i$ is a closed immersion for $i' \geq i$ by construction of the system of $Y_ j$. Then the same argument as above shows that the scheme theoretic image of $X \to B_{i'} \times _{B_ i} U$ is equal to the closed subspace $Y_{i'} \times _{Y_ i} U$. Thus we see that $X_{i'} = Y_{i'} \times _{Y_ i} U$ for all $i' \geq i$ and hence the lemma holds with $i_3 = i$. $\square$

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