The output consists of lists of the form [ the location of the singularity, the multiplicity, the delta invariant, the number of local branches ]. The location is given as a list of three homogeneous coordinates, (x, y, z). The points (x, y, 1) are points in the affine plane $C^2$, where $C$ is the field of constants. The points (x, y, 0) are on the line at infinity. (In this example, there are no singularities at infinity.)

A point is singular if and only if the multiplicity of that point is > 1, and also if and only if the delta invariant is > 0. In this example, all of the singularities are double points. A double point has multiplicity 2 and delta invariant 1.

The genus of an algebraic curve equals $\frac{\left(d-1\right)\left(d-2\right)}{2}$  minus the sum of the delta invariants. Because these have already been determined by the previous command, computing the genus is easy now:

The genus only depends on the algebraic function field of the curve. This field does not change if we apply birational transformations, so the genus is invariant under such transformations. This means, for example, that $\mathrm{subs}\left(y\=y-x^2\,f\right)$ must have genus 0 as well:

An irreducible algebraic curve allows a parametrization if and only if the genus is 0. A parametrization is a birational map from the projective line (= the field of constants union {infinity}) to the algebraic curve. This map is 1-1, except for a finite number of points. It is a 1-1 map between the places of the projective line and the places of the algebraic curve $f$.

This parametrization algorithm computes a parametrization over an algebraic extension of the constants of degree <= 2 if degree(f,{x,y}) is even, and a rational (that is, no field extension) parametrization if the degree of the curve is odd, as in this example.

If we substitute an arbitrary number for $t$ (avoiding roots of the denominators to avoid "division by zero" messages), we get a point on the curve.

Verify if this is indeed a point on the curve:

The function field of an irreducible algebraic curve f can be identified with the field C(x)[y]/(f). This is an algebraic extension of C(x) of degree degree(f,y). In some applications (integration of algebraic functions and the method that algcurves[parametrization] uses), one must be able to recognize the poles of elements in the function field. For this purpose, one can compute a basis (as a C[x] module) for the ring of functions in C(x)[y]/(f) that have only poles on the line at infinity. This basis is computed as follows:

Note: This did not require computation time because it has already been determined for use in the parametrization algorithm. The map( normal, b ) command makes the output look somewhat smaller.

The integral basis has a factor $k$ in the denominator if and only if there is a singularity on the line x=RootOf(k). This can only happen if $k^2$ divides the discriminant discrim(f, y). The integral basis contains information about the singularities in a form that is useful for computations. The advantage of this form is that it is rational--one requires no algebraic extensions of the field of constants to denote the integral basis, whereas we do need algebraic numbers to denote the Puiseux expansions.

Suppose that we are only interested in the singularities on the line x=0. Then we can compute a local integral basis for the factor $x$. A local integral basis for a factor $x-\mathrm{\alpha }$ is a basis for all elements in the function field that are integral over C[[$x-\mathrm{\alpha }$]]. An element of the function field is integral over C[[$x-\mathrm{\alpha }$]] if it has no pole at the places on the line $x\&equals;\mathrm{\alpha }$.

An example of the kind of information that the integral basis contains is the sum of the multiplicities of the factor $x-\mathrm{\alpha }$ in the denominators. This sum equals the sum of the delta invariants of the points on the line $x\&equals;\mathrm{\alpha }$. So this local integral basis for the set of factors {x} tells us that the sum of the delta invariants on the line x=0 is 2.

Until now an algebraic curve was represented by a polynomial in two variables, x and y. An algebraic curve is normally not viewed as lying in the affine plane $C^2$ (where C is the field of constants), but in the projective plane $P^2$(C). The notation as a nonhomogeneous polynomial in two variables is convenient if we want to study the affine part of the curve (for example, in the integral basis computation), but not if we are interested in the part of the curve on the line at infinity. Often (for example, for computing the genus), the part of the curve at infinity is needed as well. The nonhomogeneous notation in two variables can be converted to the homogeneous notation as follows:

This can be converted again to f with

Now the line at infinity is the line z=0 on homogeneous(f,x,y,z).

By switching x and z we can move the line x=0 to infinity.

We see that now there are two singularities at infinity, namely (1,0,0) and (-1,1,0). This may look different in the output of singularities, because in homogeneous coordinates, the points (x, y, z) and ($cx\&comma;cy\&comma;cz$) are the same for nonzero $c$.

This curve is given as a homogeneous polynomial; however, the input for the algorithms in this package must be the curve in its nonhomogeneous representation:

This polynomial is a curve of degree 10 having a maximal number of cusps according to the Plucker formulas. It was found by Rob Koelman. It has 26 cusps and no other singularities.

Now check if these points are indeed cusps. The multiplicities are 2 and the delta invariants are 1, so that part is correct. To decide if these points are cusps, we can use Puiseux expansions. Take one of these points:

Now compute the Puiseux expansions at the line x = <the x coordinate of this point> :

To obtain the y coordinates of the points on the line $x\&equals;{\mathrm{Point}}_1$ from this, we need only substitute $x\&equals;{\mathrm{Point}}_1$.

We see that there are eight different points on this line. The $\mathrm{RootOf}\left(.. {\mathrm{\_Z}}^6\&plus;..\right)$ stands for six conjugated points (namely the roots of the polynomial inside the $\mathrm{RootOf}$). However, the expression $3\cdot \mathrm{RootOf}\left(196{\mathrm{\_Z}}^2\&plus;1\right)$ is only one point, because our field of definition is not $Q$ anymore, but $\mathrm{Q}\left(\mathrm{RootOf}\left(196{\mathrm{\_Z}}^2\&plus;1\right)\right)$. This is because we needed to extend the field $Q$ to be able to "look" on the line x=$\mathrm{RootOf}\left(196{\mathrm{\_Z}}^2\&plus;1\right)$.

The Puiseux series in this set $\mathrm{p0}$ (which have only been determined up to minimal accuracy) are series (with fractional powers) in ($x-{\mathrm{Point}}_1$). Substitute the following to get series in x instead of in ($x-{\mathrm{Point}}_1$). That makes it somewhat easier to read.

For determining the type of the singularity, the coefficients here are not relevant. We have an expansion of the form

$\mathrm{constant}\cdot x^{\frac{3}{2}}\&plus;\mathrm{constant}\cdot x\&plus;\mathrm{constant}\&plus;\mathrm{higher\_order\_terms}$

The higher order terms (which have not yet been determined) have no influence on the type of the singularity, nor do the precise values of these constants. These expansions show that there are six regular points on this line and two cusps.
One can easily get more terms of the Puiseux expansions, although that is not necessary for determining the type of the singularities.

We see that if we compute more terms, the results can get bigger quickly.

A different way to show information about a singularity is the plot_knot command. The input of this procedure is a polynomial f in $x$ and $y$, for which the singularity that we are interested in is located at $x\&equals;0\&comma;y\&equals;0$. For example, the curve $f$ on the top of this worksheet has a singularity at 0, a double point.

The curve $f$ is irreducible, and so it consists of only one component. But locally, around the point 0, it has two components. Information on these components and their intersection multiplicities can be given in the form of Puiseux pairs, obtained by computing the Puiseux expansions. A different way of representing this information is as follows: By identifying $C^2$ with $R^4$, the curve can be viewed as a two-dimensional surface over the real numbers. Now we can draw a small sphere inside $R^4$ around the point 0. The surface of the sphere has dimension 3 over R. The intersection with the curve (which has dimension 1 over the complex numbers, so dimension 2 over the real numbers) consists of a number of closed curves over the real numbers, inside a space (the sphere surface) of dimension 3. After applying a projection from the sphere surface to $R^3$, these curves can be plotted. (See also: E. Brieskorn, H. Knorrer: Ebene Algebraische Kurven, Birkhauser 1981.)

In this plot, each component will correspond to one of the local components. Furthermore, the winding number in the plot equals the intersection multiplicity of the two branches of the curve. In this example this number is 1. Of course, we want to see more complicated 3-D plots. For this, we need only make the singularity more complicated, and the intersection multiplicities of the branches higher. Because we are interested in the curve only locally, it does not matter if the curve is irreducible. However, the input of plot_knot must be square-free.

We see that a cusp gives a 2-3 torus knot. More generally, if $\mathrm{igcd}\left(p\&comma;g\right)\&equals;1$, then $\mathrm{plot\_knot}\left(x^p-y^q\&comma;x\&comma;y\right)$ gives a p-q torus knot.
It gets more interesting when we have plots consisting of more components. For this, we need only have a singularity consisting of more components. In this example, we start with a 2-3 torus knot using $y^3-x^2$. To obtain a high intersection multiplicity, we add a high power of $x$, and multiply these two components. Then we get:

Getting good plots sometimes requires tweaking with the various options (see plot_knot), or changing some of the coefficients (for example, the coefficient of $x^5$). Plot options can be experimented with interactively by clicking the plot and using the plot menus or the context panel for the plot.  A useful option is Lighting Schemes available using the Color menu (or specified as the lightmodel option to the plot_knot call).

For curves with genus $0$, one can compute a parametrization--a bijection between the curve and a projective line. One can view this projective line as a normal form for curves with genus $0$. For curves with genus $1$, we can also compute a normal form, the Weierstrass normal form. In this form the curve is written as  F=$y^2$- (polynomial in $x$ of degree $3$). To avoid ambiguity, we will denote the Weierstrass normal form with the variables $\mathrm{x0}$ and $\mathrm{y0}$ instead of $x$ and $y$.

Now the curves f and F are birationally equivalent. The Weierstrass form algorithm computes such an equivalence in two directions, [w[2] , w[3]] is a morphism from f to F, and [w[4] , w[5]] is the inverse morphism. Check this for the point (-2,2,1) on $f$.

Now check if this is on F:

Now try the inverse, and see if we get the point (-2,2,1):

The Weierstrassform procedure handles hyperelliptic curves as well.  A curve f is called hyperelliptic if and only if the genus is >1 and f is birational to a curve F of the form $F\&equals;Y^2-P\left(X\right)$ where P is a polynomial in X. This means that the algebraic function field C(x)[y]/(f) is isomorphic to C(X)[Y]/(F). So this is similar to the elliptic case, the only difference is that the degree of F is higher.

The procedure is_hyperelliptic tests if a curve f is hyperelliptic.