Let $(X,O_X)$ a locally ringed space, I'd like to define the tangent space $T_x$ for $x \in X$. We can consider the local ring (stalk) $R_x$ in $x$ with maximal ideal $m_x$ and from invertibility we have that $k_x:= R_x/m_x$ is a field. How can I interpreter the notation $m_x/m_x^2$? What is the square of a ideal? How can I prove that $m_x/m_x^2$ is a $\mathbb{R} \simeq R_x/m_x$module?
1 Answers
The square is the product with itsself, $I^2 = I*I$. And you should be familiar with the product of ideals.
Instead of $R_x$ one usually writes $\mathcal{O}_{X,x}$. The residue field is usually denoted by $\kappa(x) = \mathcal{O}_{X,x} / \mathfrak{m}_x$. Of course this doesn't have to be isomorphic to $\mathbb{R}$, it could be any field (namely, every field can be considered as a locally ringed space with one point, and the residue field is the given field). It is isomorphic to $\mathbb{R}$ when we consider a real smooth manifold together with its smooth functions as a locally ringed space.
Observe that $\mathfrak{m}_x/\mathfrak{m}_x^2$ is annihilated by $\mathfrak{m}_x$, hence it is a module over $\mathcal{O}_{X,x}/\mathfrak{m}_x = \kappa(x)$. Here we use the general fact that a module over a quotient ring $A/I$ is the same as a module over $A$ which is annihilated by $I$.
The Zariski tangent space of a locally ringed space $X$ at a point $x$ is defined to be the dual of the $\kappa(x)$vector space $\mathfrak{m}_x/\mathfrak{m}_x^2$. This definition can be applied for example to smooth manifolds (where it gives the usual tangent space) and to schemes.
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*smooth manifolds. What is the tangent space to a nonsmooth manifold :)? – Jun 16 '14 at 23:09

See the last paragraph, it also makes sense for arbitrary locally ringed spaces, and topological manifolds are locally ringed spaces. – Martin Brandenburg Jun 17 '14 at 06:44

I agree but in that case there is no usual tangent space, or is there some notion? I mean the definition for LRS is pretty useless since for topological spaces $m_x/m_x^2=0$. – Jun 17 '14 at 17:32

What does "no usual tangent space" mean? Of course $m_x/m_x^2$ still carries some information (and why does it vanish for topological spaces?), in fact it is part of an even finer invariant, the graded algebra $\bigoplus_{n \geq 0} m_x^n/m_x^{n+1}$. – Martin Brandenburg Jun 17 '14 at 21:41

because $f=\sqrt{f}\cdot\mathrm{sgn}(f)\sqrt{f}$? am I wrong here? – Jun 17 '14 at 22:18