The results in this writeup are a sharpening of

Hilbert's Nullstellensatz
so you might want to read that first. Also take a look at

Zariski topology.
Let

*k* be an

algebraically closed field (for example

**C**).
We are going to show there is a bijective correspondence between closed
subsets of

*k*^{n} (for the Zariski topology)
and

radical ideals in the

polynomial
ring

*k[x*_{1},...,x_{n}].
Thus we have a correspondence between geometric objects (the closed sets)
and algebraic objects (the ideals) and we can thus study geometry
using algebra.

Recall that if *I* is an ideal of
*k[x*_{1},...,x_{n}] there is an associated
closed subset of *k*^{n}
(for the Zariski topology) denoted by *Z(I)*.

On the other hand if *X* is a subset of *k*^{n}
then write

*I(X) = { f* in *k[x*_{1},...,x_{n}] : *f(x)=0*, for all *x* in *X }.
*

Note that

*I(X)* is an ideal.

**Theorem** There is a bijective correspondence between radical ideals
of *k[x*_{1},...,x_{n}] and closed subsets
of *k*^{n} for the Zariski topology. In detail, if *I*
is a radical ideal then *I=I(Z(I))* and if *X* is a closed set
then *X=Z(I(X))*.

**Proof:**
First the easy part.
*I(X)* vanishes on *X* so *X<=Z(I(X))*. On the other hand
suppose that *X=Z({f*_{1},...,f_{m}}). Clearly
then *f*_{1},...,f_{m} are in *I(X)*
and so *X=Z({f*_{1},...,f_{m}})>=Z(I(X)).
This shows that the map *X|--> I(X)* is an injective map from
closed sets to ideals; in fact, to radical ideals.
For if *f*^{m} is in *I(X)* then this means
that *f*^{m} vanishes at all points of *X*. Obviously
then so does *f* and so *f* is in *I(X)*.

To complete the proof we now have to show that for a radical ideal we have
*I(Z(I))=I*. In fact we prove the slightly more general fact

**Lemma**
For any ideal of *k[x*_{1},...,x_{n}]
we have *I(Z(I))=*rad(*I*).

**Proof:**
Observe that we do at least have rad(*I*)*<=I(Z(I))*.
Since if *f*^{m} is in *I* then *f* will
kill any point that is killed by *f*^{m}.

The remainder of the proof uses a trick of Rabinowitsch. Let *f*
be a nonzero polynomial in *I(Z(I))*. we want to show that it is in the radical
of *I*. Introduce a new variable *y* and think about the ideal
*J* of the polynomial ring *R=k[x*_{1},...,x_{n},y]
that is generated by *I* and
*fy-1*. I claim that that *J=R*. Think about a point
*p=(a*_{1},...,a_{n},b) in *k*^{n+1}.
If this point is a zero of *J* then *(a*_{1},...,a_{n})
is a zero of *I*. Thus, by assumption, *f(a*_{1},...,a_{n})=0.
It follows that *(fy-1)(p)=-1* This is absurd since *p* is
supposed to be a zero of *fy-1*. This contradiction shows that
*J* has no zeroes. If *J* is proper then it
is contained in a maximal ideal. But, by Hilbert's Nullstellensatz,
such a maximal ideal has a zero. it follows that *J=R*.
In particular there exist
*f*_{i} in *I* and polynomials *h*_{i},*h*
in *R* such that

*
1 = f*_{1}h_{1} + ... + f_{t}h_{t} + h(fy-1)

Now define a ring homomorphism
*F:R-->k(x*_{1},...,x_{n})
by mapping each *x*_{i} to itself and *y* to *1/f*.
Here *k(x*_{1},...,x_{n}) denotes the field
of rational functions, i.e. the field of fractions of
*k[x*_{1},...,x_{n}].
Applying *F* to the equation above we get

*1 = f*_{1}F(h_{1}) + ... + f_{t}F(h_{t})

Of course each

*F(h*_{1}) has the form

*p/f*^{r}
with

*p* in

*k[x*_{1},...,x_{n}].
Clearing the denominators we see that a power of

*f* is in the ideal

*I*. The result is proven.