MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to understand the structure of monomorphisms/epimorphisms in the category of local rings (with local homomorphisms), or dually in the category of local schemes. Let $LR$ denote this category.

  • Every monomorphism is injective.

Proof: Let $R \to S$ be a monomorphism and $a,b \in R$ mapped to the same element. Regard $a,b$ as ring maps $R[t] \to R$. Let $\mathfrak{p},\mathfrak{q}$ the preimages of the maximal ideal $\mathfrak{m}$. Then we get two morphisms of local rings $R[t]_{\mathfrak{p}} \times_S R[t]_{\mathfrak{q}} \to R$ which agree when composed with $R \to S$, thus they are equal. Evaluating at $(t,t)$ gives $a=b$. There is a shorter proof which uses $R[t]_{(t)+\mathfrak{m}}$.

  1. Is every monomorpism regular? (i.e. the equalizer of some morphism pair)

  2. Is every monomorphism extremal? (i.e. $i=ep$ with $p$ epi implies $p$ iso)

  3. Is every epimorphism, which is also injective, and isomorphism?

  4. Is every epimorphism surjective?

Clearly we have 4 => 3 <=> 2 and 1 => 2. Since in $LR$ every morphism can be factorized as $ep$ where $p$ is surjective and $e$ is injective, we also have 3 => 4.

In the category of rings, 1,2,3,4 are false (see this discussion). However, the counterexamples don't carry over to $LR$. In the "category theory bible" (joy of cats!) you can find nothing about local rings. The questions are partially motivated by Anton's problem about coequalizers in the category of (local) schemes.

In Lazards article about flat epimorphisms (pdf link), there is an example (1.6) in which 4 fails. I have not understood it yet. What is $Z$ there? Perhaps $B$ is a localization of $k[T,Z]$ and not $k[T]$? But I don't believe that $f$ is well-defined.

If the answers turn out to be "no":

  • Is there a nice description for the regular monomorphisms? What about field extensions?

If $L/K$ is a finite galois extension, then $K \to L$ is a regular monomorphism iff $L/K$ is cyclic (see Brians comments).

share|cite|improve this question
1 fails for quadratic Galois ext'n of fields $K/k$ (due to non-locality of $K \otimes_k K$ and how its isom with $K \times K$ is defined). In category of local noetherian rings, map from ring to completion is an epimorphism (due to Krull intersection theorem, hence the noetherian condition), and in non-complete case provides counterexamples to 2, 3, 4. Doesn't show 2,3,4 fail in $LR$, but kills a reason to care, since affirmative results then seem useless. Category theory is like point-set topology: can get wrapped up in "weird examples"; not necessarily a good use of time. – BCnrd May 9 '10 at 16:44
If $K/k$ is a cyclic galois extension with generator $\sigma$ of the galois group, then $k \to K$ is regular since it is the equalizer of $id,\sigma : K \to K$. – Martin Brandenburg May 9 '10 at 17:25
@Martin: sorry, for #1 I got disoriented by Spec and was thinking of co-equalizers. But how about a Galois extension whose Galois group doesn't admit a pair of generators? By restricting to local rings can you find an equalizer pair description for that? – BCnrd May 9 '10 at 17:45
I'm thinking about that, but it's pretty difficult to start. Even for nice small extensions ... – Martin Brandenburg May 9 '10 at 18:36
@Martin: here's proof of non-existence when Gal. group G for K/k is non-cyclic. Suppose $s,t:K \rightrightarrows A$ is equalizer pair for $k \rightarrow K$ in category $LR$. Then $A$ is naturally a $K \otimes_k K$-algebra, and by locality of $A$ and structure of $K \otimes_k K$ (using $K/k$ is Galois), the algebra structure factors through projection to a factor field. Thus, $t=s \circ g$ for some $g \in G$. Hence, by univ. property of $A$ and inj. of $s$, any $f:B \rightarrow K$ s.t. $g \circ f=f$ factors through $k$. Thus, $K^g=k$, so $G=\langle g \rangle$. Contradiction. – BCnrd May 9 '10 at 19:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.