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Let ($A,m$) be a local, Noetherian ring. If $n$ is the minimal number of generators of the unique maximal ideal $m$, then by Krull's Hauptidealsatz and Nakayama's Lemma, we have the following inequality: $\dim(A) \leq n = \dim_{A/m}(m/m^{2})$. So, one way to determine $n$ is to compute $\dim_{A/m}(m/m^{2})$. And this might be a trivial question, but in general how difficult is it to determine the dimension of $m/m^{2}$ as an $A/m$ vector space? Also, what other techniques can one use to determine $n$? Also, in my notation $\dim_{A/m}(m/m^{2})$ is the dimension of a vector space, and $\dim(A)$ is the Krull dimension of the ring $A$.

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How is your ring given? The difficulty usually depends on the data... –  Mariano Suárez-Alvarez Oct 4 '11 at 5:15
Well, I do not have a specific ring in mind. I wanted to see if there are general statements one can prove about $dim_{A/m}(m/m^{2})$, but I guess my question is too broad. –  Rankeya Oct 4 '11 at 5:23
The one estatement one can make is: it can be arbitrarily difficult to compute that dimension! –  Mariano Suárez-Alvarez Oct 4 '11 at 5:26
Are there other way to compute the minimal number of generators? –  Rankeya Oct 4 '11 at 5:28
That number is called the embedding dimension of the local ring. Googling for that should provide useful information. –  Mariano Suárez-Alvarez Oct 4 '11 at 5:32

1 Answer 1

up vote 4 down vote accepted

An easy case is the geometric one.
Suppose you have an algebraic variety $V$ over a field $k$ defined by the polynomials $F_1,F_2,...,F_r\in k[X_1,...,X_n]$, that is $V=V(F_1,F_2,...,F_r)\subset \mathbb A^n_k$ and the local ring you are interested in is $A=\mathcal O_{V,P}$ for some rational point $P\in V$ with coordinates $(a_1,a_2,...,a_n)\in k^n$.
You can consider the jacobian matrix $Jac(P)=(\frac {\partial F_i}{\partial X_j}(P))_{i,j} \in k^{r\times n}$
The number you are interested in is then characterized by the rank over $k$ of this matrix and given by the formula

$$ dim_k (m_{V,P} / m^2_{V,P})=n-rank(Jac(P)) $$

An unintuitive example If $V\subset \mathbb A^n_k$ is the curve parametrically given by $x_1=t^n,x_2=t^{n+1},...,x_n=t^{2n}$, all polynomials vanishing on the curve will have their partial derivatives zero at at the origin $O=(0,...,0)$ and so $Jac(O)$ is the zero matrix. Hence $ dim_k (m_{V,O} / m^2_{V,O})=n-rank(Jac(O)) =n $.
This reflects that the curve is very singular at the origin, which is what the invariant you are interested in is meant to detect (it would be $1$ instead of $n$ for a smooth curve).

A nice application You can prove with the above formula that in $\mathbb A^3_k$ the union of three coplanar lines through the origin is not isomorphic to the union of the three coordinate axes, not even locally at the origin: the singularities are different (which is not so easy to see with the naked eye...)

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+1 A very nice application indeed! I'll remember it. –  Zev Chonoles Oct 4 '11 at 14:25
Thank you for the kind words, @Zev. –  Georges Elencwajg Oct 4 '11 at 14:34
This is really illuminating. Thank you for taking the time to reply to my question. –  Rankeya Oct 4 '11 at 23:34

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