Let $Y = \mathbb{A}^n$ and $P = (a_1, \ldots, a_n)$. The local ring is $\mathcal{O}_{\mathbb{A}^n, P} = k[x_1, \ldots, x_n]_{\mathfrak{m}_P}$ where $\mathfrak{m}_P = (x_1 - a_1, \ldots, x_n - a_n)$. Then

$\mathfrak{m}_P / \mathfrak{m}_P^2 \cong \bigoplus_{i=1}^n k \cdot \overline{(x_i - a_i)},$

so $\dim_k T_P \mathbb{A}^n = n$. The tangent space at every point of $\mathbb{A}^n$ is $n$-dimensional, as expected.

Let $Y = V(f_1, \ldots, f_r) \subseteq \mathbb{A}^n$ and $P \in Y$. For each $f_i$, consider the differential

$d_P f_i = \sum_{j=1}^n \frac{\partial f_i}{\partial x_j}(P) \cdot (x_j - a_j).$

These are linear forms on $k^n$. The Zariski tangent space is

$T_P Y = V(d_P f_1, \ldots, d_P f_r) = \left\{ v \in k^n \;\middle|\; \sum_{j=1}^n \frac{\partial f_i}{\partial x_j}(P) \cdot v_j = 0 \text{ for all } i \right\}.$

In other words, $T_P Y = \ker J(P)$ where $J$ is the Jacobian matrix. We always have $\dim T_P Y \geq \dim Y$, with equality characterizing nonsingularity.

Let $C = V(f) \subseteq \mathbb{A}^2$ be an irreducible plane curve ($\dim C = 1$). Here $n = 2$, $d = 1$, so the Jacobian criterion requires $\operatorname{rank} J(P) = 1$, i.e.,

$\left( \frac{\partial f}{\partial x}(P),\; \frac{\partial f}{\partial y}(P) \right) \neq (0, 0).$

A point $P$ is singular iff $f(P) = 0$ and both partial derivatives vanish at $P$.

Let $C = V(x^2 + y^2 - 1) \subseteq \mathbb{A}^2$ over $k$ with $\operatorname{char}(k) \neq 2$. The Jacobian is

$J = \begin{pmatrix} 2x & 2y \end{pmatrix}.$

At a point $P = (a, b)$ on $C$: $J(P) = (2a,\; 2b)$. This vanishes iff $a = b = 0$, but $(0, 0)$ does not lie on $C$ (since $0 + 0 - 1 \neq 0$). So $C$ is nonsingular at every point.

Let $S = V(x^2 + y^2 + z^2 - 1) \subseteq \mathbb{A}^3$. Here $n = 3$, $d = 2$, so we need $\operatorname{rank} J(P) = 1$. The Jacobian is

$J = \begin{pmatrix} 2x & 2y & 2z \end{pmatrix}.$

This has rank 0 only at the origin, which does not lie on $S$. Hence $S$ is smooth (when $\operatorname{char}(k) \neq 2$).

Let $C = V(y^2 - x^2(x+1)) = V(y^2 - x^3 - x^2) \subseteq \mathbb{A}^2$. Set $f = y^2 - x^3 - x^2$. The partial derivatives are

$\frac{\partial f}{\partial x} = -3x^2 - 2x, \qquad \frac{\partial f}{\partial y} = 2y.$

At the origin $P = (0, 0)$: $f(0,0) = 0$, $\frac{\partial f}{\partial x}(0,0) = 0$, $\frac{\partial f}{\partial y}(0,0) = 0$. Both partial derivatives vanish, so the origin is a singular point (a node).

The tangent space is $T_{(0,0)} C = \ker(0, 0) = k^2$, which is $2$-dimensional -- strictly larger than $\dim C = 1$.

At a generic point, say $P = (0, -1)$: we check $f(0, -1) = 1 - 0 - 0 = 1 \neq 0$, so this point is not on $C$. Try $P = (3, 6)$: $f(3, 6) = 36 - 27 - 9 = 0$ and $\frac{\partial f}{\partial x}(3, 6) = -27 - 6 = -33 \neq 0$. So $(3, 6)$ is a smooth point.

Let $C = V(y^2 - x^3) \subseteq \mathbb{A}^2$. Set $f = y^2 - x^3$.

$\frac{\partial f}{\partial x} = -3x^2, \qquad \frac{\partial f}{\partial y} = 2y.$

At $P = (0, 0)$: $f = 0$, $\frac{\partial f}{\partial x} = 0$, $\frac{\partial f}{\partial y} = 0$. The origin is singular.

At $P = (1, 1)$: $f = 0$, $\frac{\partial f}{\partial x} = -3 \neq 0$. So $(1, 1)$ is smooth. In fact, the origin is the only singular point.

The cusp is "worse" than the node: the local ring $\mathcal{O}_{C, 0}$ has $\mathfrak{m}/\mathfrak{m}^2$ two-dimensional (same as the node), but the tangent cone is different (see below).

Let $C = V(y^2 - x^4) \subseteq \mathbb{A}^2$. Set $f = y^2 - x^4$.

$\frac{\partial f}{\partial x} = -4x^3, \qquad \frac{\partial f}{\partial y} = 2y.$

At $P = (0, 0)$: both vanish, so the origin is singular. The curve factors as $(y - x^2)(y + x^2) = 0$, giving two smooth branches (the parabolas $y = x^2$ and $y = -x^2$) that are tangent to each other at the origin. This is a tacnode: the two branches share the same tangent line $y = 0$ (with contact of order 2).

Compare with the node $y^2 = x^2(x+1)$, where the two branches have distinct tangent directions at the singular point.

Consider $C = V(x^3 + y^3) \subseteq \mathbb{A}^2$ (over a field where $\operatorname{char}(k) \neq 3$). Since $x^3 + y^3 = (x + y)(x + \omega y)(x + \omega^2 y)$ where $\omega$ is a primitive cube root of unity, the curve is actually reducible -- three lines through the origin.

For a genuinely irreducible example, take $C = V(x^4 + y^4 + x^2 y^2 - x^2 - y^2)$. The origin need not be singular here. Instead, consider the curve germ $y^3 = x^3(x+1)$ at the origin. Set $f = y^3 - x^3(x+1)$:

$\frac{\partial f}{\partial x}(0,0) = 0, \qquad \frac{\partial f}{\partial y}(0,0) = 0.$

The origin is a singular point. The lowest-degree terms are $y^3 - x^3 = (y - x)(y - \omega x)(y - \omega^2 x)$, giving three distinct tangent lines. This is an ordinary triple point (or ordinary 3-fold point): a point of multiplicity 3 with 3 distinct tangent directions.

More generally, a singular point of multiplicity $m$ with $m$ distinct tangent directions is called an ordinary $m$-fold point.

For $C = V(y^2 - x^3 - x^2) \subseteq \mathbb{A}^2$, we computed that $\operatorname{Sing}(C)$ is determined by

$f = 0, \quad \frac{\partial f}{\partial x} = -3x^2 - 2x = 0, \quad \frac{\partial f}{\partial y} = 2y = 0.$

From the partials: $y = 0$ and $x(3x + 2) = 0$, so $x = 0$ or $x = -2/3$. Check on $C$: $(0, 0)$ gives $f = 0$ (singular); $(-2/3, 0)$ gives $f = -4/27 + 4/9 = 8/27 \neq 0$ (not on $C$). Thus $\operatorname{Sing}(C) = \{(0,0)\}$, a single point -- a proper closed subset of $C$.

Let $S = V(x^2 - y^2 z) \subseteq \mathbb{A}^3$ (the Whitney umbrella). Set $f = x^2 - y^2 z$:

$\frac{\partial f}{\partial x} = 2x, \quad \frac{\partial f}{\partial y} = -2yz, \quad \frac{\partial f}{\partial z} = -y^2.$

The singular locus is $\{f = 0,\; 2x = 0,\; -2yz = 0,\; -y^2 = 0\}$, which gives $x = 0$, $y = 0$, and then $f = 0$ is automatic. So $\operatorname{Sing}(S) = V(x, y) = \{(0, 0, t) \mid t \in k\}$, the entire $z$-axis.

Here the singular locus is a line sitting inside a surface -- a closed subset of dimension 1 inside a variety of dimension 2.

$\mathbb{A}^n$ is smooth: the local ring at every point is $k[x_1, \ldots, x_n]_{\mathfrak{m}}$, which is a regular local ring of dimension $n$, and $\dim T_P \mathbb{A}^n = n = \dim \mathbb{A}^n$.

$\mathbb{P}^n$ is smooth: it is covered by affine charts $U_i \cong \mathbb{A}^n$, each of which is smooth. Smoothness is a local property, so $\mathbb{P}^n$ is smooth.

Let $Q = V(x_0^2 + x_1^2 + \cdots + x_n^2) \subseteq \mathbb{P}^n$ (assuming $\operatorname{char}(k) \neq 2$). In the affine chart $x_0 = 1$, this is $V(1 + x_1^2 + \cdots + x_n^2)$, i.e., $V(x_1^2 + \cdots + x_n^2 + 1)$. The Jacobian is $(2x_1, \ldots, 2x_n)$, which vanishes only at the origin, but $0 + \cdots + 0 + 1 \neq 0$, so no singular point exists in this chart. By symmetry, the same holds in every chart. So $Q$ is smooth.

More generally, a quadric $V(\sum a_{ij} x_i x_j)$ in $\mathbb{P}^n$ is smooth if and only if the associated symmetric matrix $(a_{ij})$ is nonsingular (i.e., has nonzero determinant).

The Fermat cubic surface $S = V(x_0^3 + x_1^3 + x_2^3 + x_3^3) \subseteq \mathbb{P}^3$ (with $\operatorname{char}(k) \neq 3$). In the chart $x_0 = 1$:

$f = 1 + x_1^3 + x_2^3 + x_3^3, \quad J = (3x_1^2,\; 3x_2^2,\; 3x_3^2).$

$J = 0$ requires $x_1 = x_2 = x_3 = 0$, but then $f = 1 \neq 0$. By symmetry, all charts give the same conclusion: $S$ is smooth.

A classical result: every smooth cubic surface over $\mathbb{C}$ contains exactly 27 lines. This is one of the most famous theorems in 19th-century algebraic geometry (Cayley-Salmon theorem, 1849).

Let $E = V(y^2 - x^3 - ax - b) \subseteq \mathbb{A}^2$ where $4a^3 + 27b^2 \neq 0$ (the discriminant condition). Set $f = y^2 - x^3 - ax - b$:

$\frac{\partial f}{\partial x} = -3x^2 - a, \qquad \frac{\partial f}{\partial y} = 2y.$

A singular point requires $2y = 0$ (so $y = 0$) and $3x^2 + a = 0$ and $x^3 + ax + b = 0$. From $3x^2 = -a$, substitute into $x^3 + ax + b = 0$: one can show this system has a solution iff $4a^3 + 27b^2 = 0$. So the condition $4a^3 + 27b^2 \neq 0$ is precisely what ensures smoothness.

For instance, $y^2 = x^3 - x$ (where $a = -1$, $b = 0$, discriminant $= -4 + 0 = -4 \neq 0$) is smooth. But $y^2 = x^3$ (where $a = 0$, $b = 0$, discriminant $= 0$) has a cusp at the origin.

The Grassmannian $\mathbb{G}(r, n)$ is smooth. This can be seen from its local description: around any point $W \in \mathbb{G}(r, n)$, choose a complementary subspace $W'$ of dimension $n - r$. Then the open set $\{U \in \mathbb{G}(r, n) \mid U \cap W' = 0\}$ is isomorphic to $\mathbb{A}^{r(n-r)}$ (the space of $r \times (n-r)$ matrices, representing $U$ as the graph of a linear map $W \to W'$). Since $\mathbb{A}^{r(n-r)}$ is smooth and smoothness is local, $\mathbb{G}(r, n)$ is smooth.

For $C = V(y^2 - x^3 - x^2)$, the lowest-degree terms at the origin are $y^2 - x^2 = (y - x)(y + x)$. So

$C_{(0,0)} C = V(y^2 - x^2) = V(y - x) \cup V(y + x),$

two distinct lines through the origin. The multiplicity is $\operatorname{mult}_{(0,0)}(C) = 2$.

Geometrically, the two lines $y = x$ and $y = -x$ are the two tangent directions of the two branches of the curve at the node.

For $C = V(y^2 - x^3)$, the lowest-degree terms at the origin are $y^2$ (degree 2). So

$C_{(0,0)} C = V(y^2),$

a "double line" along the $x$-axis. The multiplicity is $\operatorname{mult}_{(0,0)}(C) = 2$.

Although the cusp and the node both have multiplicity 2, their tangent cones differ: the node has two distinct tangent lines, while the cusp has a single tangent line with multiplicity 2. The cusp is "more singular" in this sense.

For $C = V(y^2 - x^4)$, the lowest-degree terms at the origin are $y^2$ (degree 2, since $x^4$ has degree 4). So

$C_{(0,0)} C = V(y^2),$

the same double line as the cusp. The multiplicity is 2. But the tacnode and the cusp are genuinely different singularities (the tacnode is analytically equivalent to two smooth branches tangent to each other; the cusp is an irreducible singular germ). Distinguishing them requires looking beyond the tangent cone, e.g., at higher-order jet spaces or the resolution.

| Curve | Equation | $\operatorname{mult}_{(0,0)}$ | Tangent cone | |---|---|---|---| | Smooth point | $y = x^2$ (at origin: not on curve) | -- | -- | | Node | $y^2 = x^3 + x^2$ | 2 | $y^2 - x^2$: two lines | | Cusp | $y^2 = x^3$ | 2 | $y^2$: double line | | Tacnode | $y^2 = x^4$ | 2 | $y^2$: double line | | Triple point | $y^3 = x^3(x+1)$ | 3 | $y^3 - x^3$: three lines | | $A_4$ singularity | $y^2 = x^5$ | 2 | $y^2$: double line |

The multiplicity determines the "order" of the singularity, while the tangent cone captures the tangent geometry. Two singularities can have the same multiplicity and tangent cone (cusp vs. tacnode) but be analytically different.

For $C = V(y^2 - x^3)$ and $P = (0,0)$:

$\mathcal{O}_{C, P} = (k[x, y]/(y^2 - x^3))_{(x, y)} = k[t^2, t^3]_{(t^2, t^3)}.$

The maximal ideal $\mathfrak{m} = (t^2, t^3)$ satisfies $\mathfrak{m}/\mathfrak{m}^2 = k \cdot \bar{t}^2 \oplus k \cdot \bar{t}^3$ (2-dimensional), but $\dim \mathcal{O}_{C, P} = 1$ (since $C$ is a curve). Since $2 \neq 1$, the local ring is not regular, confirming that the origin is a singular point.

Furthermore, $\mathcal{O}_{C, P}$ is not a UFD: the element $t^2$ is irreducible but not prime (since $t^2 \cdot t^2 = t^4 = t \cdot t^3$, and $t \notin k[t^2, t^3]$). It is also not integrally closed: $t = t^3/t^2$ is integral over $k[t^2, t^3]$ but not in it.

For $C = V(y - x^2) \subseteq \mathbb{A}^2$ and $P = (0, 0)$:

$\mathcal{O}_{C, P} = (k[x, y]/(y - x^2))_{(x, y)} \cong k[x]_{(x)}.$

The maximal ideal is $(x)$, which is principal: $\mathfrak{m}/\mathfrak{m}^2 \cong k$ (1-dimensional), and $\dim \mathcal{O}_{C, P} = 1$. Since $1 = 1$, this is a regular local ring -- a discrete valuation ring (DVR). Every smooth curve has DVRs as its local rings.

Consider $C = V(y^2 - x^2 - x^3) \subseteq \mathbb{A}^2$ with a node at the origin. The blowup of $\mathbb{A}^2$ at the origin replaces $(0,0)$ by a $\mathbb{P}^1$ (the exceptional divisor $E$). In local coordinates:

Chart 1 ($y = tx$): Substituting $y = tx$ into $f = y^2 - x^2 - x^3$:

$t^2 x^2 - x^2 - x^3 = x^2(t^2 - 1 - x).$

The factor $x^2$ corresponds to the exceptional divisor (with multiplicity 2). The strict transform is $\tilde{C}_1 = V(t^2 - 1 - x)$, a smooth curve. On $E$ (where $x = 0$), the strict transform meets $E$ at $t^2 = 1$, i.e., $t = \pm 1$. These are two distinct points, corresponding to the two tangent directions of the node.

Chart 2 ($x = sy$): Substituting $x = sy$ into $f$:

$y^2 - s^2 y^2 - s^3 y^3 = y^2(1 - s^2 - s^3 y).$

The strict transform is $V(1 - s^2 - s^3 y)$, also smooth.

The blowup resolves the node in a single step: the strict transform $\tilde{C}$ is smooth, and $\pi: \tilde{C} \to C$ is an isomorphism away from the origin. Above the origin, $\tilde{C}$ has two points (the two branches are separated).

Consider $C = V(y^2 - x^3) \subseteq \mathbb{A}^2$ with a cusp at the origin.

First blowup (chart $y = tx$): Substitute into $f = y^2 - x^3$:

$t^2 x^2 - x^3 = x^2(t^2 - x).$

The strict transform $\tilde{C}_1 = V(t^2 - x)$ is smooth (it is a parabola in the $(x, t)$-plane). On the exceptional divisor ($x = 0$): $t^2 = 0$, so $t = 0$. The strict transform meets $E$ at a single point with multiplicity 2 (tangency).

But wait -- $\tilde{C}_1 = V(t^2 - x)$ is actually smooth. The intersection with $E$ is transverse after a coordinate change: set $u = t$, then $x = u^2$, and $\tilde{C}_1 \cong \mathbb{A}^1$ via $u \mapsto (u^2, u)$. The cusp is resolved after a single blowup.

However, note that the strict transform meets the exceptional divisor with contact order 2, not transversally. If we want normal crossings (for the purpose of embedded resolution), we would need a second blowup. For the abstract resolution of $C$ alone, one blowup suffices.

| Singularity | Blowups needed | Strict transform meets $E$ at | Behavior | |---|---|---|---| | Node $y^2 = x^2(x+1)$ | 1 | 2 distinct points | Two branches separated | | Cusp $y^2 = x^3$ | 1 | 1 point (tangent to $E$) | Branch smoothed | | Tacnode $y^2 = x^4$ | 2 | After 1st: cusp; after 2nd: smooth | Two tangent branches separated | | $A_4$: $y^2 = x^5$ | 2 | Successive blowups resolve | Higher-order contact resolved iteratively |

For a projective hypersurface $Y = V(F) \subseteq \mathbb{P}^n$ with $F$ homogeneous of degree $d$, the singular points are

$\operatorname{Sing}(Y) = V\!\left(F,\; \frac{\partial F}{\partial x_0},\; \ldots,\; \frac{\partial F}{\partial x_n}\right) \subseteq \mathbb{P}^n.$

By Euler's formula for homogeneous polynomials ($\sum x_i \frac{\partial F}{\partial x_i} = d \cdot F$), the condition $F = 0$ is redundant when $\operatorname{char}(k) \nmid d$: it follows from the vanishing of all partials at a point of $\mathbb{P}^n$.

Concretely: the Klein quartic $V(x^3 y + y^3 z + z^3 x) \subseteq \mathbb{P}^2$ (over $\mathbb{C}$) is smooth. Its partials are:

$\frac{\partial F}{\partial x} = 3x^2 y + z^3, \quad \frac{\partial F}{\partial y} = x^3 + 3y^2 z, \quad \frac{\partial F}{\partial z} = y^3 + 3z^2 x.$

One can verify (e.g., by Grobner basis or direct computation) that the system $F = \frac{\partial F}{\partial x} = \frac{\partial F}{\partial y} = \frac{\partial F}{\partial z} = 0$ has no solution in $\mathbb{P}^2$. The Klein quartic is a smooth curve of genus 3, and it has the largest automorphism group (168 elements) among genus-3 curves.

For "most" choices of coefficients, a degree-$d$ hypersurface in $\mathbb{P}^n$ is smooth. More precisely, the locus of singular hypersurfaces in the parameter space $\mathbb{P}^N$ (where $N = \binom{n+d}{d} - 1$) is a proper closed subset -- the discriminant hypersurface $\Delta$.

For plane curves ($n = 2$):

• Degree 1 (lines): always smooth. $\Delta = \varnothing$.
• Degree 2 (conics): singular iff the $3 \times 3$ matrix has $\det = 0$. $\Delta$ is a degree-3 hypersurface in $\mathbb{P}^5$.
• Degree 3 (cubics): the discriminant $\Delta \subseteq \mathbb{P}^9$ has degree 12.

This is a manifestation of Bertini's theorem: a "general" member of a linear system is smooth.