Let $(A,\mathfrak{m},k)$ be a (Noetherian) local ring. Let $k[T]/(\tilde{P}(T))$ be a simple field extension of $k$ with $\tilde{P}(T)$ monic and irreducible (and separable). Lift $\tilde{P}(T)$ to monic $P(T)\in A[T]$. Can we show that $A[T]/(P(T))$ is a flat local $A$-algebra?
The claim is used in Qing Liu's book Algebraic Geometry and Arithmetic Curves, Corollary 5.3.17:
A similar claim is in Exercise 4.3.19:
I got it. I mean it indeed seems trivial if you are in the right direction.
Clearly $B:=A[T]/(P(T))$ is a finite free $A$-module with basis $\{1,T,\dots,T^{\mathrm{deg}P-1}\}$. So $A\to B$ is free and thus flat, and in particular, it is a finite ring extension, so $\mathfrak{p}\in \operatorname{Spec}B$ is maximal iff $\mathfrak{p}\cap A$ is maximal, i.e. $\mathfrak{p}\cap A=\mathfrak{m}$. Then the points of $\mathrm{Spec} (B\otimes_A \frac{A}{\mathfrak{m}})=\operatorname{Spec}B \times_{\mathrm{Spec}A}\operatorname{Spec}\kappa(\mathfrak{p})$ are exactly the set of prime ideals of $B$ lying over $\mathfrak{m}$, which must be the set of maximal ideals of $B$. Since $$B\otimes_A \frac{A}{\mathfrak{m}}\cong \frac{A[T]}{(P(T))+\mathfrak{m}[T]}\cong\frac{\frac{A}{\mathfrak{m}}[T]}{(\tilde{P}(T))}=\frac{k[T]}{(\tilde{P}(T))}$$ is a field, it has only one prime ideal. So $B$ must be local.