Let $A$ be a (commutative) ring and $U$ be an arbitrary subset of $\mathrm{Spec}(A)$. Do there exist a ring $B$ and a ring homomorphism $\varphi\colon A\to B$ such that $\varphi^{-1}(\mathrm{Spec}(B))=U$ ? If $\varphi\colon A\to B$ and $\varphi'\colon A\to B'$ are two ring homomorpisms such that $\varphi^{-1}(\mathrm{Spec}(B))=\varphi'^{-1}(\mathrm{Spec}(B'))=U$, do there exist a ring homomorpism $f\colon B\to B'$ such that $f\varphi=\varphi'$ ?
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2$\begingroup$ Just so you know, there's another use of the term "ring spectrum" in the vernacular. To me, it means a monoid in the category of spectra. Maybe you could change your title to say "subset of Spec(A) realized as inverse image of some Spec(B)" or to say spectrum of a ring rather than ring spectrum? Or not, it's not a big deal. $\endgroup$– David WhiteCommented Oct 28, 2013 at 15:11
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1$\begingroup$ Spec, as a functor taking values in sets or even topological spaces, is not a good functor. In particular it is not faithful (e.g. it can't distinguish between any parallel pair of morphisms with domain a field) so there's no reason to expect anything like your second question to be true. $\endgroup$– Qiaochu YuanCommented Oct 28, 2013 at 19:02
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3 Answers
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For the first question, $\mathrm{Spec}(B)$ is quasi-compact for every ring $B$, so if $U\subset\mathrm{Spec}(A)$ is not quasi-compact, then there cannot exist $f:A\to B$ with $U=f^*(\mathrm{Spec}(B))$. An example of this is $A=k[x_1,x_2,\ldots]$ a polynomial ring over a field $k$ in countably many variables, $$ U=\mathrm{Spec}(A)\backslash\{(x_1,x_2,\ldots)\}. $$ Then $ U_n:=D(x_1,\ldots,x_n) $, $n=1,2,\ldots$, is an open cover of $U$ that has no finite subcover, so $U$ cannot be the the image of any $\varphi^*:\mathrm{Spec}(B)\to\mathrm{Spec}(A)$. I would be interested to know if there is a counterexample with $A$ Noetherian.
For the second question, take $A=\mathbb{Z}$, $B=\mathbb{Z}/(p)$, $B'=\mathbb{Z}/(p^2)$, $p$ a prime, and $\varphi,\varphi'$ the quotient maps. Then $\varphi^*(\mathrm{Spec}(B))=(\varphi')^*(\mathrm{Spec}(B'))=\{(p)\}$, but there don't exist any ring homomorphisms $B\to B'$.
answered Oct 28, 2013 at 15:11
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1$\begingroup$ One can also take $A=\mathbb Q$, $B=\mathbb Q(t)$ and $B'=\overline{\mathbb Q}$, in which case there is no morphism between $B$ and $B'$ from either sides. $\endgroup$– CantlogCommented Oct 30, 2013 at 12:44
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As an addition to Julian Rosen's answer, here is an example with $A$ noetherian and $U$ quasicompact: take $A=\mathbb{Z}$ (or any PID with infinitely many primes), and $U$ = any infinite set of closed points. Assume $B$ is a ring such that the image of $\mathrm{Spec}(B)$ to $\mathrm{Spec}(\mathbb{Z})$ is $U$. Since $U$ does not contain the generic point, we have $\mathbb{Q}\otimes_\mathbb{Z}B=0$. In particular, there is a nonzero integer $n$ such that $n1_B=0$, so $\mathrm{Spec}(B)$ maps to the finite set of divisors of $n$: contradiction.
answered Oct 29, 2013 at 8:03
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There are already two nice counterexamples. Let me just point out for question (1) a characterization of images of $\mathrm{Spec}(B)$ in $X:=\mathrm{Spec}(A)$:
A subset $U$ of $X$ is the image of some $\mathrm{Spec}(B)\to X$ if and only if $U$ is pro-constructible.
See EGA IV.1.9.5. In a noetherian space, $U$ is pro-constructible if it is an arbitrary intersection of constructible subsets (i.e. finite unions of locally closed subsets) of $X$. A constructible subset of $\mathrm{Spec}(\mathbb Z)$ (or any irreducible noetherian scheme of dimension $1$) is either finite or contains the generic point, so a pro-constructible subset is either finite or contains the generic point. This "explains" the example of Laurent.