At a singular point, the maximal ideal of the local ring is too big. A curve has dimension 1, yet at a singular point its maximal ideal needs two generators. This gap between the number of generators and the dimension is what singularity means algebraically.
Below we write out the local rings of two singular points, the node y^2 = x^2(x+1) and the cusp y^2 = x^3. We compute their integral closures, compare their completions and see where the two part ways.
First the standard of comparison. A Noetherian local ring (R, \mathfrak{m}) is regular when the following holds [1].
\dim R = \dim_{R/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^2
The right-hand side is the minimal number of generators of the maximal ideal, called the embedding dimension: by Nakayama's lemma, lifting a basis of \mathfrak{m}/\mathfrak{m}^2 gives generators of \mathfrak{m}. The left-hand side is the Krull dimension, 1 at a point of a curve and 2 at a point of a surface.
At a smooth point the two sides agree. At a smooth point of a curve the local ring is a regular local ring of dimension 1, that is, a discrete valuation ring.
The gap can go only one way. The following inequality always holds.
\dim R \leq \dim_{R/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^2
It follows from Krull's height theorem: if \mathfrak{m} is generated by n elements, its height is at most n. So at a singular point the embedding dimension is strictly larger, and an extra generator is needed.
Geometrically, the tangent space is bigger than the variety. At the node the two tangent lines y = x and y = -x span the whole plane, so a two-dimensional tangent space sits on a one-dimensional curve. At a smooth point there is one tangent line, and the tangent space is one-dimensional like the curve.
Take the origin of the curve y^2 = x^2(x+1). Its coordinate ring and local ring are A = k[x, y]/(y^2 - x^2 - x^3) and \mathcal{O} = A_{\mathfrak{m}} with \mathfrak{m} = (x, y). The curve is irreducible, so A is a one-dimensional domain, and localizing does not lower the dimension.
Here \mathfrak{m}^2 = (x^2, xy, y^2). Both sides of y^2 = x^2 + x^3 lie in \mathfrak{m}^2, and no term of degree 1 appears, so the equation gives no linear relation between x and y.
\dim_k \mathfrak{m}/\mathfrak{m}^2 = 2
The dimension is 1 and the embedding dimension is 2, so the local ring is not regular. In particular \mathfrak{m} is not principal: if \mathfrak{m} = (f), the image of f would span \mathfrak{m}/\mathfrak{m}^2, whose dimension would then be at most 1. This is a one-dimensional local domain that is not a discrete valuation ring.
The same computation works for the cusp y^2 = x^3, with local ring k[x, y]_{(x, y)}/(y^2 - x^3). Both sides of the equation lie in \mathfrak{m}^2, since y^2 has degree 2 and x^3 degree 3. In the figure the neighbourhood of the origin is magnified step by step, and the cusp flattens onto the half-line y = 0, x \geq 0: only the term y^2 of lowest degree is left.
Again no linear relation appears, so the embedding dimension is 2 while the dimension is 1, and the local ring is not regular.
This computation cannot tell the node from the cusp: both have dimension 1 and embedding dimension 2. Other quantities are needed to separate them.
Rewrite the coordinate ring of the cusp with one parameter. Putting x = t^2 and y = t^3 gives y^2 = t^6 = x^3, and this correspondence is an isomorphism.
k[x, y]/(y^2 - x^3) \cong k[t^2, t^3] \subset k[t]
The right-hand side consists of the polynomials in t with no term of degree 1, and the powers 1, t^2, t^3, t^4, \ldots form a basis. Only t is missing, and this one-dimensional gap is the singularity of the cusp.
The fraction field of k[t^2, t^3] is k(t), because t = t^3/t^2. The element t is not in k[t^2, t^3], but it is integral over it: since t^2 \in k[t^2, t^3], it satisfies the monic equation T^2 - t^2 = 0.
So the fraction field contains an integral element outside the ring, and k[t^2, t^3] is not integrally closed.
This is where the two notions meet. A one-dimensional Noetherian local domain is integrally closed exactly when it is a discrete valuation ring [5]. So for points of a curve over an algebraically closed field, being smooth, being regular and being integrally closed are the same thing, and a singular point is a point where the local ring is not integrally closed.
A ring that is not integrally closed can be repaired by taking its integral closure; geometrically this is normalization. For the cusp the integral closure is k[t], a principal ideal domain and hence integrally closed, and it is a finitely generated module over k[t^2, t^3].
The node can be treated the same way. Putting y = tx gives t^2 = x + 1, and hence the following parametrization.
x = t^2 - 1, \quad y = t^3 - t
The element t = y/x of the fraction field satisfies t^2 - x - 1 = 0, so it is integral, and the integral closure is again k[t]. Both singular curves are repaired into the same smooth line.
The difference appears in how many points lie over the origin. Count the values of t that the normalization sends to the origin.
For the node, both t = 1 and t = -1 go to the origin, so two points lie over it. For the cusp, x = t^2 and y = t^3 vanish only at t = 0, so a single point lies over it.
The number of points shows up in the local shape of the integral closure. For the node, the integral closure has two maximal ideals over the origin, and after localizing and completing it splits into a product of two discrete valuation rings, one for each branch through the origin.
For the cusp, the integral closure has a single maximal ideal over the origin, and its completion stays one discrete valuation ring: there is only one branch.
For a reduced ring of finite type over a field, like the coordinate rings here, the integral closure is a finite module over the original ring, so normalization is a finite morphism [4]. Finiteness keeps the points over the origin finite in number, so that they can be counted.
The local rings themselves are both domains, because both curves are irreducible, so at this level the two do not differ yet.
Completion separates them. If the characteristic is not 2, then 1 + x has a square root u in k[[x]], and the equation of the node factors.
k[[x, y]]/(y^2 - x^2(1+x)) = k[[x, y]]/((y - xu)(y + xu))
So the complete local ring of the node is not a domain. For the cusp, y^2 - x^3 stays irreducible in k[[x, y]], and the complete local ring remains a domain. The number of branches is invisible in the local ring and visible after completion.
The number of branches is not enough by itself. To compare the two on the same scale, measure the difference from the integral closure. With \mathcal{O} the local ring at the singular point and \widetilde{\mathcal{O}} its integral closure, put the following.
\delta = \dim_k \widetilde{\mathcal{O}}/\mathcal{O}
For the cusp, \widetilde{\mathcal{O}} = k[[t]] and \mathcal{O} = k[[t^2, t^3]], and the quotient is spanned by t, so \delta = 1. The node also has \delta = 1: different types, the same weight. A higher cusp such as y^2 = x^5 gives k[[t^2, t^5]], which misses t and t^3, so \delta = 2.
This \delta is exactly how much the genus drops [3]. For a plane curve of degree d, the geometric genus is the arithmetic genus (d-1)(d-2)/2 minus the sum of \delta over the singular points.
In the figure the node and the cusp pass through the six quantities from top to bottom, staying together where the values agree and moving apart where they differ. Dimension, embedding dimension and integral closure cannot tell the node from the cusp. The points over the origin and the completion separate them, while \delta is again the same.
So the same word singular hides different information depending on how far down one looks: dimension, embedding dimension, integral closure, completion.
On curves, being singular and failing to be integrally closed were the same. From dimension 2 on they are not. A regular local ring is always integrally closed, but the converse fails: there are integrally closed points that are singular.
The origin of the cone xy = z^2 is an example. Its ring is integrally closed, but the embedding dimension is 3 and the dimension is 2, so it is not regular.
The reason is that normalization removes only singularities in codimension 1, since a normal variety is regular in codimension 1 [2]. On a curve every singular point has codimension 1, which is why normalization alone was enough.
In higher dimensions one needs repeated blow-ups instead. Over a field of characteristic 0 singularities can always be resolved: Heisuke Hironaka proved in 1964 that for every algebraic variety V there is a nonsingular W with a proper birational morphism W \to V [2].
In positive characteristic the problem is solved up to dimension 3 and remains open in general from dimension 4 on. Resolution turns local rings into regular local rings, and normalization of a curve is its simplest case. The figure shows one blow-up separating the two branches of the node.
Some common mistakes are worth listing. A curve without self-intersection need not be smooth, as the cusp shows. A local ring that is a domain need not be regular: the cusp again, even after completion. And the embedding dimension never drops below the dimension.
The number of branches and \delta are different quantities. From dimension 2 on, integrally closed does not imply smooth, since normalization removes only singularities in codimension 1.