1

Today, while reading some articles, I had this doubt trying to justify a passage:

Hypothesis

Suppose $O_K$ is some complete discrete valuation ring (it is the ring of integers of some complete field in my case, but it shouldn't be important). Let $B$ be an $O_K$-algebra, finite as an $O_K$-module, and complete intersection. Moreover you know that there's an ideal $\mathfrak a$ in $B$ such that $\Omega_{B/O_K}$ is a $B/\mathfrak a$-free module.

Question

Suppose that you get $x_1,...,x_n$ such that $d(x_1),...,d(x_n)$ is an $A/\mathfrak a$-basis for $\Omega_{B/A}$. Consider the only continuous homomorphism $O_K[[X_1,...,X_n]] \to B$ sending $X_i \to x_i$. Why is this surjective and how the fact that $B$ is complete intersection implies that being surjective there need to be $n$ power series that generates the kernel (that is same number of equation and unknown)?

I think i can prove surjectivity directly (but constructing step by step a counterimage for any element using the fact that at the linear stage it is a set of generator). But what about such a simple structure of the kernel: from the definition i have of complete intersection i just know that $B$ is expressible in such a form, but how can i prove it also for different presentation?(maybe i have the wrong definitions).

Is it true something more general like: if i take $x_1,...,x_n$ with $d(x_1),...,d(x_n)$ a minimal set of generators for the differentials, then the power series in them are all B and they induce a complete intersection presentation for B?

user26857
  • 52,094
Warner
  • 11

1 Answers1

1

It seems that you are asking if being a complete intersection is an intrinsic property of a ring. It is, in the following sense:

Proposition: Let $R$ be a quotient of a regular local ring $S$, i.e. $R = S/I$ for some $S$-ideal $I$. Then $R$ is a complete intersection iff $I$ is generated by a regular sequence.

This is proved in Bruns-Herzog, Cohen-Macaulay Rings, Theorem 2.3.3. In your case, $\mathcal{O}_K[[X_1, \ldots, X_n]]$ is regular local, and $\dim B = \dim \mathcal{O}_K = 1$, so if $B \cong \mathcal{O}_K[X_1, \ldots, X_n]]/I$, then $\text{ht}(I) = n$, so if $B$ is a complete intersection, then $I$ is generated by a regular sequence of length $n$.

zcn
  • 15,640