Suppose $(A,\mathfrak m)$ is a Neotherian local $k$-algebra with residue field $k$. Then, we define (the coordinate ring of) its algebraic tangent cone to be the $k$-algebra $A_c = \sum_{i\ge 0} \mathfrak m^i/\mathfrak m^{i+1}$.

On the other hand, we also have the Zariski tangent space $(\mathfrak m/\mathfrak m^2)^\vee$, whose coordinate ring is the symmetric algebra $A_t = \sum_{i\ge 0} \text{Sym}^i(\mathfrak m/\mathfrak m^2)$. There is an evident surjective map of $k$-algebras $A_t\to A_c$, which gives an embedding of the algebraic tangent cone into the Zariski tangent space.

Griffiths-Harris, in their book, say that the tangent cone to a variety (over $\mathbb C$) at a point is the collection of tangent vectors obtainable as velocities of analytic arcs through that point lying on the variety. A naïve attempt at formalizing this in this algebraic setting could be the following.

A $k$-algebra map $A\to k[t]/t^2$ is the same as a Zariski tangent vector, and we can ask if this can be lifted (non-uniquely) to a $k$-algebra map $A\to k[[t]]$ (roughly this corresponds to going from a vector to a "formal path" tangent to it). We could then ask if the image of the embedding $\text{Spec } A_c\to\text{Spec }A_t = (\mathfrak m/\mathfrak m^2)^\vee$ corresponds exactly to those algebra maps to $k[t]/t^2$ which admit lifts to $k[[t]]$.

This is not the correct characterization as the example of $\mathbb C[[x,y]]/(y^2-x^3)$ shows, since any two formal power series $f,g$ in the variable $t$ with $g^2 = f^3$ must be divisible by $t^2,t^3$ respectively. Is there a way then to modify the above question so that the answer provides a nice characterization of the image of the tangent cone inside the Zariski tangent space?

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    $\begingroup$ If $(A,\mathfrak{m})$ is the germ of a $k$-scheme $(X,x)$, then form the product $X\times \text{Spec} \ k[[t]]$ with its projection to the DVR $\text{Spec}\ k[[t]]$. Next, blow this up at the point $x$ in the closed fiber, $\nu:\mathcal{X} \to X\times \text{Spec}\ k[[t]]$. Every formal arc of $X$ centered at $x$ gives a section of the projection, which then lifts uniquely to $\mathcal{X}$ by the valuative criterion applied to $\nu$. The fiber of $\nu$ over $x$ is the (projective completion) of the algebraic tangent cone, cf. "deformation to the normal cone". $\endgroup$ Commented Mar 15, 2018 at 18:44

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You may wish to read the discussion on pages 106-108 of Eisenbud/Harris "The Geometry of Schemes", where they explicitly compute the tangent cone of $(y^2-x^3)$ as a degeneration of the tangent cones in the family $(y^2 - tx^2 - x^3)_{t \in \mathbb{A}^1}$.

In a direction close to what @JasonStarr said, though one requiring a background in jet schemes, you could look at Example 5.13 in Ein/Mustață "Jet Schemes and Singularities" where they describe some features of the arc space of the cusp that you seem to be curious about.


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