In my reading I recently encountered the following problem:
Let $A$ be a DVR and let $Y=\mathrm{Spec}\,A$. Let $K$ be the fraction field of $A$ and let $k$ be the residue field. The canonical homomorphisms $i:A \rightarrow K$ and $\pi:A \rightarrow k$ yield a ring homomorphism $\phi: A \rightarrow K \times k$ given by $a \mapsto (i(a),\pi(a))$. Let $X=\mathrm{Spec}(K \times k)$ and let $f:X \rightarrow Y$ be the corresponding scheme morphism. Show that $K \times k$ is a finitely generated $A$-algebra, that $f$ is bijective, and that $\mathrm{dim}\,X=0$ and $\mathrm{dim}\,Y=1$.
How might one prove this? Any help would be appreciated! Thanks!
We have $Y=\operatorname {Spec}(A)=\{\eta,M\}$, consisting of the zero ideal $\eta=(0)$ and the maximal ideal $M=(\pi) $, where $\pi$ is a uniformizer.
On the other hand, $X=\operatorname {Spec}(K)\sqcup \operatorname {Spec}(k)=\{y\}\sqcup \{m\}$ and since the morphism $f:X\to Y$ satisfies $f(y)=\eta, f(m)=M$, $f$ is bijective.
The scheme $X$ has dimension zero because it is discrete, and $X$ has dimension $1$ since its only two primes are connected by the inclusion relation $(0) \subsetneq M$.
Finally $K\times k$ is finitely generated over $A$ as the product of the finitely generated algebras $K=A[\frac 1\pi]$ and $k=A/M$.