Commutative local Artinian ring without a section in characteristic $0$

Question

Let $R$ be a commutative local Artinian ring, with unique maximal ideal $\mathfrak m$ and residue field $k = R/\mathfrak m$.

If $R$ is a finitely generated algebra over a perfect field $F$, then the Jordan-Chevalley decomposition applied to the regular representation of $R$ seems to imply that that the surjection $R \twoheadrightarrow k$ necessarily has a (unique!) section $k \hookrightarrow R$ of rings. I'm pretty sure my reasoning there is solid (you have to apply the JC decomposition to multiple commuting operators, but I'm pretty sure you can do that).

An example of $R$ where the residue field has no section is $R = \mathbb Z/p^n \mathbb Z$, where $p$ is a prime integer and $n > 1$.

It seems like this failure of the existence of a section occurs naturally in an "arithmetic setting", whereas it never seems to happen in a "geometric setting", if these phrases can be given any precise meaning.

For the sake of a precise question to address my intuition here: Is it possible to find an example for which the section $k \hookrightarrow R$ does not exist in characteristic $0$?

I'm not sure whether it's fair to categorize non-perfect fields as an "arithmetic not geometric" setting; feel free to pitch in your opinion on that point as well if you think such a discussion would be productive.

Answer 1

The Cohen structure theorem answers your question. First, recall the following definition.

Definition: Let $(R,\mathfrak{m})$ be a complete local ring. A subring $\Lambda\subseteq R$ is called a coefficient ring if the following conditions hold:

• $\Lambda$ is a complete local ring with maximal ideal $\Lambda\cap\mathfrak{m},$
• the residue field of $\Lambda$ maps isomorphically to the residue field of $R,$ and
• $\Lambda\cap\mathfrak{m} = p\Lambda,$ where $p$ is the characteristic of the residue field of $R.$

In particular, if $(R,\mathfrak{m})$ is a complete local ring with residue field $k$ of characteristic $0$ and $\Lambda$ is a coefficient ring, then $\Lambda$ is the residue field of $R$, as it has unique maximal ideal $\Lambda\cap\mathfrak{m} = (0).$ Thus $\Lambda$ is a field, and the second condition above forces $\Lambda = k.$

Notice also that any Artinian local ring $(R,\mathfrak{m})$ is a complete local ring, as $\mathfrak{m}^n = 0$ for some $n.$

Theorem: Let $(R,\mathfrak{m})$ be a complete local ring. Then

1. $R$ has a coefficient ring, and
2. if $\mathfrak{m}$ is a finitely generated ideal, then $R$ is isomorphic to a quotient $\Lambda[\![x_1,x_2,\dots, x_r]\!]/I$ where $\Lambda$ is either a field or a Cohen ring.

You of course do not need the full power of this theorem for your result. Let us prove directly that there is a section in your case.

Claim: If $(R,\mathfrak{m})$ is a complete local ring with residue field $k$ of characteristic $0,$ then there is a section $k\to R$ of the quotient map $R\to R/\mathfrak{m}.$ (In particular, a section exists if $R$ is Artinian.)

Proof: We will show inductively that a section exists to the canonical quotient map $R/\mathfrak{m}^n\to k.$ The base case is obvious. Then suppose that a section of $R/\mathfrak{m}^r\to k$ exists for $r = n-1.$ Since $\operatorname{char}k = 0,$ $k/\Bbb{Q}$ is separable, hence formally smooth (see here), and thus a morphism $k\to R/\mathfrak{m}^n$ exists making the diagram $$ \require{AMScd} \begin{CD} R/\mathfrak{m}^n @>>> R/\mathfrak{m}^{n-1}\\ @AAA @AAA\\ \Bbb{Q} @>>> k\\ \end{CD} $$ commutative (with the arrow $k\to R/\mathfrak{m}^n$ included, of course). This completes the proof, as $R\cong\varprojlim_n R/\mathfrak{m}^n.$ $\square$

Final remark. The map $\Bbb{Q}\to R/\mathfrak{m}^n$ exists because we have a map $\Bbb{Q}\to R.$ As $\operatorname{char}k = 0$ it follows that for any prime $p\in\Bbb{Z},$ $p\not\in\mathfrak{m}.$ Thus, $p\in R^\times,$ so the canonical map $\Bbb{Z}\to R$ extends to a map $\Bbb{Q}\to R$ by the universal property of localization.