Fix an

algebraically closed base

field *k*
(with

characteristic different from 2) such as
the the

complex numbers.
The nodal cubic curve is the

plane algebraic curve defined
in 2-space by the equation

*y*^{2} = x^{2} + x^{3}

If we sketch the

real points of the curve we get
an alpha-shaped curve as in the diagram.

| /
| /
.-. | /
/ \|/
------- \------------
\ /|\
- | \
| \
| \

Note that the curve crosses over itself at the origin

*(0,0)*. This is a

singularity of the curve. This type
of singualrity is called a

node and makes the curve a
degenerate

elliptic curve.

Let's look at some of the concepts from the writeups
algebraic geometry, Zariski topology, and
the correspondence between closed sets for the Zariski topology
and radical ideals in the polynomial ring
as they are illustrated in this example. Let *C*
denote the set of points of the curve in *k*^{2}

First of all we have a function *f:k*^{1}-->C defined by

*
f(a)=(a*^{2} - 1, a(a^{2} - 1))

This is a surjective polynomial map. It is not injective though
as

*f(1)=f(-1)* (both these points map to the crossing point).

The ideal *I(C)* of the curve is the ideal of the polynomial
ring *k[x,y]* generated by a single polynomial

*
(y*^{2} - x^{2} - x^{3})k[x,y].

Thus the nodal cubic has coordinate ring the

quotient ring
*k[x,y]/I(C)*. Let us give another description
of this ring.
We know that corresponding to

*f* there is a

ring homomorphism
*f*^{*}:k[C]-->k[k^{1}].
Now

*k[k*^{1}]=k[t] is a polynomial
ring in one variable. A quick calculation shows that

*f*^{*} is

injective. Thus the coordinate ring
of

*C* is isomorphic to the

image
which is

*k + (t*^{2} - 1)k[t]