Is $Spec(R_p)$ to $Spec(R)$ is an open immersion?

Question

Let $R$ be a ring. $p$ be a prime ideal of $R$ . Let $R\rightarrow R_p$ be the canonical. Consider the map of schemes $Spec(R_p) \rightarrow Spec(R)$. Is it an open immersion. $Spec(R_p)$ is an open set in $Spec(R)$ is clear.

Answer 1

Look at $R=\mathbb{Z}$ and $\mathfrak{p}=(0)$. (or in fact any other prime ideal.)

Answer 2

Depends on what you really mean.

Assume $R$ is an integral domain for simplicity.

If $R_p$ is the localization at the prime ideal $p$, then $R_p$ is a local ring. The image of Spec $R_p$ in Spec $R$ is then precisely the generic point (i.e., zero ideal of $R$) and the points of Spec $R$ specializing to $p$. This is not open in general (consider for instance a Dedekind domain $R$ with infinitely many prime ideals in which case Spec $R_p$ is just two points if $p$ is non-zero).

If you meant to consider an element $p$ in $R$, and the localization $R_p = R[1/p]$ then Spec $R[1/p] \to $ Spec $R$ is open. It is usually denoted by $D(p)$ and is a so-called principal open of Spec $R$.

Answer 3

As counter-examples have be given, I would like to give a more general explanation. Let $\left( X, \mathscr{O}_X\right)$ be a scheme. You have a canonical morphism $\varphi : \textrm{Spec}\left( \mathscr{O}_{X,x} \right) \rightarrow X$, defined as follows. Let $U$ be an open affine in $X$, containing $x$, and let $A$ be the ring (of global sections of $\mathscr{O}_X$) of (over) $U$. $\mathscr{O}_{X,x}$ is then isomorphic to the localisation $A_{x}$ of $A$ at $x$. To the canonical arrow $A\rightarrow A_x$ corresponds therefore a morphism of schemes $\textrm{Spec}\left( \mathscr{O}_{X,x} \right) \rightarrow \textrm{Spec}\left( A \right) = U$, that composed with the open immersion $U\rightarrow X$ give an arrow $\textrm{Spec}\left( \mathscr{O}_{X,x} \right) \rightarrow X$, depending on $U$. If fact it does not depend on $U$. Indeed, for another $U'$, you can find a third $U''$ such that $U'' \subseteq U\cap U'$, so we can assume that $U\subset U'$, and in this case the obvious diagramm invoving $A$, $A'$ and $\mathscr{O}_{X,x}$ commutes.

Then $\varphi$ induces an homeomorphism from $\textrm{Spec}\left( \mathscr{O}_{X,x} \right)$ to the subset of $X$ consisting in $y\in X$ such that $x\in\overline{\{ y \}}$. There is therefore really intuitively no reason for this set to be open. By the way, the morphisms induced by $\varphi$ on local rings are isomorphism so that $\varphi$ is a monomorphism, even if it is not necessarily an open immersion. On this note, the (set-theoretic) image of $\varphi$ is not generally locally-closed, so it is even not a subscheme.