## Base change of semi-stable curve still semi-stable

Let $S$ be a scheme.

A smooth curve over $S$ is a smooth projective $S$-scheme of relative dimension $1$ with geometrically connected fibres.

A semi-stable curve over $S$ is a flat projective $S$-scheme of relative dimension $1$ whose geometric fibres are connected, reduced and have only ordinary double singularities.

Let $T\to S$ be a morphism.

Then, if $X\to S$ is a smooth curve over $S$, the base change $X_T\to T$ is a smooth curve over $T$.

Suppose that $X\to S$ is a semi-stable curve over $S$. Why is the base change $X_T\to T$ a semi-stable curve over $T$?

This should be true, but I can't seem to prove it easily. The problem is that I fear the singularities might become worse after a base change.

Let me emphasize that I do not assume the total space $X$ to be non-singular in these definitions. Neither do I assume anything on $S$, besides maybe some finiteness conditions if you'd like.

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What do you mean by a semi-stable curve? – Sasha Jan 7 at 22:01
Flatness is preserved by base-change, as is projectivity and relative dimension (if $T/S$ is locally of finite type). So the only condition left to check is on the geometric fibers. But the geometric fibers of $X_T\to T$ are literally the same as the geometric fibers of $X\to S$. More specifically, if $f: T\to S$ is the morphism and $t$ is a geometric point of $T$, $X_t=X_{f(t)}$ canonically. As $X\to S$ is semi-stable, the geometric fibers satisfy the required conditions so we're done. – Daniel Litt Jan 7 at 22:19
@Daniel Litt: You invoke a fact not obvious to a beginner (as Masse is likely to be) that underlies your use of the phrase "geometric point": if $K/k$ is an extension of algebraically closed fields and $X$ is a $k$-scheme of finite type with pure dimension 1 then $X$ is reduced with $O_{X,x}^{\wedge} = k[[u,v]]/(uv)$ for all non-smooth $x\in X(k)$ if and only if $X_K$ is reduced with $O_{X_K,x}^{\wedge} = K[[u,v]]/(uv)$ for all non-smooth $x\in X_K(K)$. Strictly speaking you use the easier "only if" direction, but one needs "iff" to have a robust notion and proving "if" uses serious technique. – kreck Jan 7 at 23:31
@Masse: It is absolutely not true that $X_T$ is normal as a scheme; consider the case when $T$ is a point! Rethink whatever "exercises in commutative algebra" made you think it is normal (e.g., perhaps you assumed $T$ is normal noetherian and the generic fibers are smooth?). – kreck Jan 7 at 23:33
@Daniel Litt and kreck. Thank you very much. This answers my question fully. – Masse Jan 8 at 7:57