Let $f:X\to \operatorname{Spec}R$ be a morpism to a spectrum of a discrete valuation ring. Let $\pi \in R$ be a uniformizaer. Let $K:=\operatorname{Frac}R$. Let $x\in X$ and $A:= \mathcal{O}_{X,x}$.
First, Note that Alex Youcis commented below, there is a $R$-algebra structure ( ring homomorphism ) $R \to A$, induced from $\operatorname{Spec}(\mathcal{O}_{X,x}) \to X \to \operatorname{Spec}R$.
Now, my question is, for all prime ideal $\mathfrak{p} \in \operatorname{Spec}(A\otimes_R K)$, $\pi \cdot 1_A \notin
^{a}\iota(\mathfrak{p})$, where $ ^{a}\iota : \operatorname{Spec}(A\otimes_R K) \to \operatorname{Spec}(A)$ is the induced morphism from the canonical morphism $\iota : A \to A \otimes_R K$ ? Or equivalently, does $(\pi \cdot 1_A ) \otimes_R 1_K$ not contained in $\mathfrak{p}$?
This question originates from the Gortz's Algebraic Geometry, proof of the Proposition 14.16 :
Proposition 14.16. Let $R$ be a discrete valuation ring, let $s$ be the special and $\eta$ the generic point of $\operatorname{Spec}R$. Let $f:X\to \operatorname{Spec}R$ be a morphism. Assume that $X$ is locally noetherian. Suppose that all maximal points of the special fiber $f^{-1}(s)$ lie in the closure of $f^{-1}(\eta)$ and that the special fiber $f^{-1}(s)$ is reduced. Then $X$ is flat over $R$.
C.f. In his book p.427, he noted that "Assume moreover that $Y= \operatorname{Spec}R$ for a discrete valuation ring $R$.Then $X$ is flat over $Y$ if and only if for all open affine subschemes $U$ the $R$-module $\Gamma(U,\mathcal{O}_X)$ has no $\pi$-torsion, where $\pi$ is a uniformizing element."
His proof of the Proposition 14.16 is as follows :
Proof. Let $\pi$ be a uniformizing element of $R$. It is enough to show that for all $x\in X$, writing $A=\mathcal{O}_{X,x}$, $\pi$ is a regular element of $A$. Let $a\in A$ with $\pi^{i}a=0$ for some $i>0$. We want to show that $a=0$. Certainly $f^{-1}(\eta) = \operatorname{Spec}(A \otimes_R K)$ is contained in $V(a)$, because its points are prime ideals which do not contain $\pi$. And.. ( omission of next step of proof ; If needed, I'll upload full proof )
( C.f. Possible errata : For the $f^{-1}(\eta)$ in the proof, more correct notation is $f_1^{-1}(\eta)$, where $f_1 : \operatorname{Spec}\mathcal{O}_{X,x} \to X \to \operatorname{Spec}R$ is the composition. )
I don't understand why the bolded statement is true. My rough understanding is as follows :
First, note that the projection map $\operatorname{Spec}(A\otimes_R K) \to \operatorname{Spec}A$ is induced by the canonical morphism $\iota : A \to A\otimes_R K$. In the proof of the Proposition 14.16, we want to show that $^{a}\iota(\operatorname{Spec}A\otimes_R K) \subseteq V(a)$. Let $\mathfrak{p} \in \operatorname{Spec}(A\otimes_R K)$. Then $^{a}\iota(\mathfrak{p})$ is a prime ideal of $A$. Note that $\pi^{i}a= (\pi^{i}\cdot 1_A) \cdot a = 0 \in\ ^{a}\iota(\mathfrak{p})$ as in the proof.
So if $\pi \cdot 1_A \notin\ ^{a}\iota(\mathfrak{p})$, then $ a \in\ ^{a}\iota(\mathfrak{p}) $. So, $^{a}\iota(\mathfrak{p}) \in V(a)$. And is $\pi \cdot 1_A \notin\ ^{a}\iota(\mathfrak{p})$ ture? $\pi \cdot 1_A$ is unit of $A$? If so, why?
Can anyone help?
Best Answer
Yes. $(\pi \cdot 1_A) \otimes_R 1_K$ is not contained in $\mathfrak{p}$ since, by noting that $(\pi \cdot 1_A) \otimes_R 1_K = 1_A \otimes_R ( \pi \cdot 1_K) = 1_A \otimes_R (\frac{\pi}{1})$ and $\frac{\pi}{1}$ is unit in $K:= \operatorname{Frac}R$ ( $\because$ Since the DVR $R$ is not a field by the definition, its maximal ideal is nonzero so that its uniformizaer $\pi$ is also nonzero. So $\frac{\pi}{1}$ is unit in $K:=\operatorname{Frac}R$), so that $1_A \otimes_R (\frac{\pi}{1})$ is unit in $A\otimes_RK$, which should not be contained in $\mathfrak{p}$.