1

All rings are Noetherian and eft. An $A$-algebra $B$ is smooth if it is flat and the fibres are geometrically regular. I want to see some examples of this notion. So I considered the $\mathbb Z$-algebra $\mathbb Z[T]/(T^2+1)$ (Gaussian integers). This is flat (free!) and I computed the fibres as $\mathbb Q[T]/(T^2+1)$ and $(\mathbb Z/p)[T]/(T^2+1)$. Now i think these are geometrically regular as taking tensor products with an algebraic closure gives a product of fields. At the same time I feel the Gaussian integers should not be smooth. Could somebody give me more enlightening examples (apart from the obvious ones: polynomial rings)?

I am particularly interested in the case where $A=\mathbb Z$. Is there something special to say in this case?

user26857
  • 52,094
Dieck
  • 13
  • 1
    Is it correct for $p=2$? Mod 2, $T^2+1=(T+1)^2$. – Mohan Jul 03 '16 at 15:30
  • I wonder what's "eft"? – user26857 Jul 03 '16 at 15:53
  • Why don't you think that $\mathbb Z[i]$ is smooth? – Fredrik Meyer Jul 03 '16 at 16:07
  • @user26857 I should have said that $B$ is essentially of finite type over $A$ (assumptions as in Bourbaki, ALG. COMM, chap. 10). – Dieck Jul 03 '16 at 18:40
  • @Mohan so in that case the fibre is not reduced, in particular not regular. I used to think whether $f^{-1}(p)$ is geometrically regular or not depends on the congruence class of p mod 4, but apparently this is not so. I think that $T^2+1$ is seperable over $p$ for all $p\neq 2$, so the only fibre that is not geometrically regular is for $p=2$? thanks for your help! please post your answer directly as an answer, so I can accept it! – Dieck Jul 03 '16 at 21:14

2 Answers2

1

$\mathbb{Z}\to \mathbb{Z}[T]/(T^2+1)$ has a non-reduced fiber over 2, since $T^2+1\equiv (T+1)^2\bmod 2$.

I have a vague memory that if $A$ is any $\mathbb{Z}$ algebra, which is a domain and integral over $\mathbb{Z}$, then it can not be smooth over $\mathbb{Z}$ (except in the trivial case when $A=\mathbb{Z}$). But, I can neither remember the proof or a suitable reference, so you should take this with appropriate skepticism.

Mohan
  • 17,980
  • So it's well known that every ring of integers of a number field is ramified somewhere over $\mathbb{Z}$. This settles your statement in the case of integrally closed domains which are integral (=finite) over $\mathbb{Z}$. If you have a domain $B$ which is finite over $\mathbb{Z}$ but not integrally closed, then that domain is not normal, and hence not regular. However, regularity ascends along etale morphisms, so this would imply that $B$ cannot be etale over $\mathbb{Z}$. – oxeimon Jul 03 '16 at 22:49
1

If $B$ is a flat $A$-algebra, then all of its fibers have the same dimension. If that dimension is 0, then $B$ being smooth over $A$ is the same as being etale over $A$, which is to say that it is unramified. It's a classical result of algebraic number theory (I believe coming from Minkowski's work on the geometry of numbers) that the only finite etale algebras over $\mathbb{Z}$ are just finite products of $\mathbb{Z}$, indeed any connected finite etale algebra over $\mathbb{Z}$ is just the ring of integers in some number field, which is always ramified over at least 1 (but finitely many) primes of $\mathbb{Z}$.

If you drop the finiteness condition, then there are lots of etale algebras over $\mathbb{Z}$. If $B$ is the ring of integers in some number field, and let $p_1,\ldots,p_n$ be the primes of $\mathbb{Z}$ over which $B$ is ramified, then $B[\frac{1}{p_1p_2\cdots p_n}]$ is etale over $\mathbb{Z}$. Intuitively, by adjoining the inverses of the $p_i$, we've removed the primes at which $B$ is ramified.

If the fibers of $B$ over $A$ have dimension 1, then you're looking at families of smooth curves over $\text{Spec } A$. An easy example is $A[x]$, which can be thought of as a family of genus 0 curves over $\text{Spec }A$. Another interesting result is that there are no elliptic curves over $\mathbb{Z}$ (ie, a family of genus 1 curves over $\text{Spec }\mathbb{Z}$ which admits a section).

oxeimon
  • 12,279