# Can the dimension of the Zariski tangent space of a complex curve at a singular point be arbitrarily big?

Can the dimension of the Zariski tangent space of a complex curve at a singular point be arbitrarily big ?

Is there a formula relating the dimension of the Zariski tangent space and the order of the singularity?

What are the basic references (books) for learning about classification of complex curve singularities (not necessarily plane)?

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One way to increase the dimension of the Zariski tangent space, which is probably not the answer you are looking, is to add fuzz at the origin by making the coordinate ring non-reduced. So if you take the scheme Spec $k[x_{1}, \ldots, x_{n}]/(x_{1}^{4}, \ldots, x_{n-1}^{4})$, this gives a non-reduced curve whose Zariski tangent space at the origin has basis $\{x_{1}, \ldots, x_{n-1}, x_{n}\}$ and is hence dimension $n$. I'll try to think of a proper example (i.e. a reduced curve). – Siddharth Venkatesh Apr 18 '14 at 0:13
The union of the axis in the affine space $\mathbb C^n$ is reduced with tangent space at the origin of dimension $n$, but is reducible. The equations $x_1^2-x_2^3, x_1^2-x_3^5, \dots, x_1^2-x_n^{2n-1}$ define a curve with tangent space at the origin of dimension $n$, and it is probably irreducible. – user143488 Apr 18 '14 at 0:31
The equation $x^n-y^{2n+1}$ defines a singular plane curve, and whatever is the definition of the "order", the latter should be big when $n$ is big. – user143488 Apr 18 '14 at 0:33
Yeah, the union of the n axes was the simplest example I could think of that was reduced. – Siddharth Venkatesh Apr 18 '14 at 0:35

Another way to make a curve with many branches through a point is to start with $\mathbb A^1$ over $\mathbb C$ (say) and then glue together the points $t = 0, 1, \ldots, n$ (for some the value of $n$).
That is, we let $A$ be the $\mathbb C$-subalgebra of $\mathbb C[t]$ consisting of polynomials $f$ such that $f(0) = f(1) = \ldots = f(n).$ This a finitely generated $\mathbb C$-algebra corresponding to an (irreducible) curve with a singularity through which there are $n+1$ branches. A computation shows that the tangent space at the origin has dimension $n+1$.
Consider the curve singularity at the origin of the image $C$ of the map $z \mapsto (z^e, z^{e + 1} \ldots, z^{e + n})$ where $e$ is a large integer (say larger than $n + 1$). Any polynomial equation for $C$ must have vanishing constant and linear terms. Hence the Zariski tangent space of $C$ at $0$ has dimension $n + 1$.