Flatness of schemes I am learning about flatness for the first time and I cannot wrap my head around why the definition with tensor products of a flat module implies geometrically that 1-parameter families of schemes have limits.
I came across these lecture notes by Alexander Ritter which give a very explicit example:

However, I don't understand why $\pi$ flat over $0 \iff$ $X_0$ is $\lim_b X_b$. Any help to understand this particular example or a different intuition of this geometric implication would be very much appreciated.
 A: Note that $\overline{X^*}$ is the scheme-theoretic image of $X^* \to \mathbf A^n_B$: if $X^*$ is reduced, this agrees with the reduced induced structure (see for instance exercise II.3.11 in Hartshorne), and otherwise this is the only sensible definition of the closure of a locally closed subscheme.
In particular, if $X^* \subseteq \mathbf A^n_{B^*}$ is cut out by some ideal $I^*$, then by definition $\overline{X^*} \subseteq \mathbf A^n_B$ is cut out by the contracted ideal $\overline{I^*} = \iota^{-1}(I^*)$, where $\iota \colon B[x_1,\ldots,x_n] \to B^*[x_1,\ldots,x_n]$ is the inclusion. In other words, the factorisation $X^* \to \overline{X^*} \to \mathbf A^n_B$ corresponds to the image factorisation
$$B[x_1,\ldots,x_n] \twoheadrightarrow B[x_1,\ldots,x_n]\big/\ \overline{I^*} \hookrightarrow B^*[x_1,\ldots,x_n]/I^*$$
of rings.
Any module $M$ over the principal ideal domain $B$ is flat over $0$ if and only if it is $t$-torsion-free. Since $B^*[x_1,\ldots,x_n]/I^*$ is a $B^*$-algebra, the element $t$ is invertible there, so $B^*[x_1,\ldots,x_n]/I^*$ is always flat over $0$. Thus, $B[x_1,\ldots,x_n]/\overline{I^*}$ is also flat, since a submodule of a $t$-torsion-free module is $t$-torsion free.
So if $I \subseteq B[x_1,\ldots,x_n]$ is an ideal and $I^*$ its extension to $B^*[x_1,\ldots,x_n]$, then we certainly see that $I = \overline{I^*}$ implies that $B[x_1,\ldots,x_n]/I$ is flat. Conversely, we always have $I \subseteq \overline{I^*}$, and the kernel of
$$B[x_1,\ldots,x_n]/I \twoheadrightarrow B^*[x_1,\ldots,x_n]\big/\ \overline{I^*}$$
is $t$-torsion since this map becomes an isomorphism after inverting $t$. Thus, we see that $B[x_1,\ldots,x_n]/I$ is $t$-torsion-free if and only if $I = \overline{I^*}$.
Of course there is nothing special about $B[x_1,\ldots,x_n]$, and the same argument works for any $B$-algebra $R$ (or $B$-scheme $Y \to \operatorname{Spec} B$).
