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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
y2 = x2 + x3
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 k2

First of all we have a function f:k1-->C defined by

f(a)=(a2 - 1, a(a2 - 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

(y2 - x2 - x3)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[k1]. Now k[k1]=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 + (t2 - 1)k[t]

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