Geometric interpretation of $M/M^2$ and $M^n /M^{n+1}$, for $M$ maximal ideal of a DVR

Question

Consider a number field $K$ and its ring of integers $\mathcal{O}_K$. It is well known that it is a discrete valuation ring, so that, for $M$ maximal ideal, one has $\mathcal{O}_K/M \simeq M/M^2$. My question is if someone can tell me how this can be seen from a geometric perspective. How can one see that $M^n/M^{n+1}\simeq \mathcal{O}_K / M$? Thanks in advance!

Answer 1

Discrete valuation rings are regular local rings of dimension one. Regular means that the tangent space at each point has the correct dimension - in this case, this is one. So the tangent space at the unique closed point is a one-dimensional vector space, which as a module over the residue field is isomorphic to the residue field.

EDIT (question changed to also ask about $M^n/M^{n+1}$):

To treat the case of $M^n/M^{n+1}$, consider the tangent cone at the unique closed point, given by $\operatorname{Spec}$ of $\operatorname{gr}_M \mathcal{O}_{X,x}\cong \bigoplus_{n\geq 0} M^n/M^{n+1}$. At any regular point of any variety, the tangent cone is just isomorphic to the tangent space: in this case, that means $\Bbb A^1$, which implies that $\operatorname{gr}_M \mathcal{O}_{X,x}$ is a polynomial ring in one variable, which has all nonnegatively graded pieces isomorphic as modules over the $0$th graded piece.