if $f$ and $g$ compose with a dominant morphism are the same, show that they agrees on a open dense subset as morphisms of schemes

Question

To be specific about my question, let $A$ be a valuation ring (so local and integral) and $K$ be the fraction field of $A$.

The injection $A\rightarrow K$ induces $i: \mathrm{Spec} \ K\rightarrow\mathrm{Spec}\ A$. Clearly the image (the zero ideal) is dense.

Let $X$ be a scheme over $A$, and $f,g:\mathrm{Spec}\ A\rightarrow X$ be two morphisms of $A$-schemes s.t. $f\circ i=g\circ i$.

We want to show that there exists a non-empty open dense subset $U$ of $\mathrm{Spec}\ A$ s.t. $f|_U=g|_U$ as morphisms of schemes.

Does it hold for more general case, where $i$ is any dominant morphism?

Answer 1

I figured out the proof for my specific case.

Denote $l=f\circ i=g\circ i$. Denote $s$ as the unique point of $\mathrm{Spec} K$.

Let $\mathrm{Spec}\ B$ be an affine open subset of $X$ containing $l(s)$. So $l^{-1}(\mathrm{Spec}\ B)\ne \emptyset$, we can further deduce that $l^{-1}(\mathrm{Spec}\ B)=\mathrm{Spec}\ K$, and similarly $f^{-1}(\mathrm{Spec}\ B)\ne \emptyset$ and $g^{-1}(\mathrm{Spec}\ B)\ne \emptyset$.

Since $A$ is an integral domain, $\mathrm{Spec}\ A$ is integral (i.e. reduced and irreducible). In particular, every non-empty open subset is dense and every pair of non-empty open subset have non-empty intersection.

Apply it to $f^{-1}(\mathrm{Spec}\ B)$ and $g^{-1}(\mathrm{Spec}\ B)$, we have that $\emptyset \ne f^{-1}(\mathrm{Spec}\ B)\cap g^{-1}(\mathrm{Spec}\ B)\subset \mathrm{Spec}\ A$. Since principal open sets form a base, assume $D(t)=\mathrm{Spec}\ A_t\subset f^{-1}(\mathrm{Spec}\ B)\cap g^{-1}(\mathrm{Spec}\ B)$ for some $t\in A\backslash \{0\}$.

This $D(t)$ is our potential candidate of dense open subset. Since $t\notin (0)$, the image (the zero ideal) lies in $D(t)$, hence the map $i:\mathrm{Spec}\ K \rightarrow \mathrm{Spec}\ A$ does factor through $D(t)$.

Note that $l^{-1}(\mathrm{Spec}\ B)=i^{-1}f^{-1}(\mathrm{Spec}\ B)=i^{-1}g^{-1}(\mathrm{Spec}\ B)$

Now $l|_{l^{-1}(\mathrm{Spec}\ B)}=f|_{f^{-1}(\mathrm{Spec}\ B)}\circ i|_{i^{-1}f^{-1}(\mathrm{Spec}\ B)}=g|_{g^{-1}(\mathrm{Spec}\ B)}\circ i|_{i^{-1}g^{-1}(\mathrm{Spec}\ B)}$.

Consider the following diagram (commutes in every square) $\require{AMScd}$ \begin{CD} @.f^{-1}(\mathrm{Spec}\ B)@= f^{-1}(\mathrm{Spec}\ B)@>f|_{f^{-1}(\mathrm{Spec}\ B)} >>\mathrm{Spec}\ B\\ @. @AA inclu_1 A@VV inclu_3 V@VVV\\ \mathrm{Spec}\ K @>i'>> \mathrm{Spec}\ A_t@>inclu_5 >> \mathrm{Spec}\ [email protected]\\ @. @VV inclu_2 V@AA inclu_4 A@AAA\\ @.g^{-1}(\mathrm{Spec}\ B)@= g^{-1}(\mathrm{Spec}\ B)@>g|_{g^{-1}(\mathrm{Spec}\ B)} >>\mathrm{Spec}\ B \end{CD}

We have $i|_{i^{-1}f^{-1}(\mathrm{Spec}\ B)}=inclu_1 \circ i'$ and $i|_{i^{-1}g^{-1}(\mathrm{Spec}\ B)}=inclu_2 \circ i'$. Hence $$l|_{l^{-1}(\mathrm{Spec}\ B)}=f|_{f^{-1}(\mathrm{Spec}\ B)}\circ inclu_1 \circ i'=g|_{g^{-1}(\mathrm{Spec}\ B)}\circ inclu_2 \circ i'$$ Furthermore we have $$f|_{\mathrm{Spec}\ A_t} \circ i'=g|_{\mathrm{Spec}\ A_t} \circ i'$$ Let $\phi_f ,\phi_g:B\rightarrow A_t,\ \phi_{i'}:A_t\hookrightarrow K$ be the corresponding maps on rings.

The above equality implies $\phi_{i'}\circ\phi_f=\phi_{i'}\circ\phi_g$. Since $\phi_{i'}$ is injective, we must have $\phi_f=\phi_g$, which implies $f|_{\mathrm{Spec}\ A_t}=g|_{\mathrm{Spec}\ A_t}$. The result follows.