I'm an extreme beginner with algebraic geometry and am trying to get used to things. Say I have some (algebraically closed) field $k$, in $k^2$ I want to compute the Zariski cotangent space, let's say of something simple, $V(Y-X^2)$, at the origin. The definition I have is that it is the quotient $\mathfrak{m}_{(0,0),V}/\mathfrak{m}_{(0,0),V}^2$ where $\mathfrak{m}_{(0,0),V}$ is the unique maximal ideal (consisting of rational functions which vanish at the origin) of the local ring of $V$ at $(0,0)$, $\mathcal{O}_{(0,0),V}$ (consisting of those rational functions that are defined at the origin). I'd like to do the computation directly using this definition (rather than the multivariable calculus way I've been seeing more often).
So to get this I first think about what $\mathcal{O}_{(0,0),V}$ looks like. This will be the rational functions which are (equivalent to a) quotient of polynomials modulo $(Y-X^2)$, say $g(X,Y) + (Y-X^2)$ and $f(X,Y) + (Y-X^2)$ where the latter is the denominator, such that $f(0,0) + (Y-X^2) \neq 0$ in $k[X,Y]/(Y-X^2)$ which is the same as saying $f(0,0) \neq 0$ in $k$, so we're looking at quotients where the constant term of $f$ is non-zero.
Now for $\mathfrak{m}_{(0,0),V}$. Here by similar reasoning we're looking for quotients where the numerator polynomial has zero constant term and the denominator polynomial has non-zero constant term as before.
I can't see any nice descriptions coming here, nice enough for me to compute the quotient $\mathfrak{m}_{(0,0),V}/\mathfrak{m}_{(0,0),V}^2$. I can see that the coordinate ring $k[X,Y]/(Y-X^2)$ is isomorphic to $k[X]$ and so by passing to there perhaps things would simplify but doing this I get the exact same things as before except now the polynomials are in just the one variable.
Please help.
The Zariski cotangent space is actually very simple conceptually: it is the ideal of germs of functions that vanish at the point, modulo higher derivatives, i.e. modulo the ideal of germs of functions that vanish to second order at the point.
First, let us reduce the problem somewhat by getting rid of the local ring. Let $A$ be a commutative ring and let $\mathfrak{m}$ be a maximal ideal. Then $A_\mathfrak{m}$ is a local ring with maximal ideal $\mathfrak{m}_\mathfrak{m}$. I claim that $\mathfrak{m}_\mathfrak{m} / \mathfrak{m}_\mathfrak{m}^2$ is isomorphic to $\mathfrak{m} / \mathfrak{m}^2$ as an $A / \mathfrak{m}$-vector space. Indeed, there is a field homomorphism $A / \mathfrak{m} \to A_\mathfrak{m} / \mathfrak{m}_\mathfrak{m}$ and it is easily seen to be an isomorphism. Thus, $$\mathfrak{m} / \mathfrak{m}^2 \cong \mathfrak{m} \otimes_A (A / \mathfrak{m}) \cong \mathfrak{m}_\mathfrak{m} \otimes_{A_\mathfrak{m}} (A_\mathfrak{m} / \mathfrak{m}_\mathfrak{m}) \cong \mathfrak{m}_\mathfrak{m} / \mathfrak{m}^2_\mathfrak{m}$$ as claimed.
In particular, for $A = k [x]$ and $\mathfrak{m} = (x)$, we find that $\mathfrak{m}_\mathfrak{m} / \mathfrak{m}_\mathfrak{m}^2$ is just a 1-dimensional $k$-vector space generated by (the residue class of) $x$ (modulo $x^2$).