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AN INTRODUCTION TO THE ZARISKI TOPOLOGY
OSCARMICHEL
Abstract. We give an introduction to the spectrum of a ring and its Zariski
topology, a fundamental tool in algebraic geometry. In addition, we cover the
ring theory and topology necessary for defining and proving basic properties
of the Zariski topology. Finally, we give examples of various ring spectra.
Contents
1. Introduction 1
2. Rings and Ideals 2
3. Topology 8
4. The Spectrum of a Ring 10
Acknowledgments 16
References 16
1. Introduction
Algebraic geometry is the branch of math that studies problems in geometry
that can be solved with algebra, and vice versa. Modern algebraic geometry unfor-
tunately has a reputation for being very difficult and inaccessible to learn. Many
standard algebraic geometry textbooks are written at a graduate level or higher.
The idea that one needs advanced techniques from category theory and commuta-
tive algebra to gain an appreciation for algebraic geometry is far from the truth,
however. It is still possible for the undergraduate student to engage with aspects of
the theory, and it is the goal of this paper to introduce an essential tool of modern
algebraic geometry using only undergraduate ring theory and topology.
In this paper we will study the spectrum of a ring, which gives a way to define a
topological space that can be created from any ring. This topological space, called
the Zariski topology, gives a geometric way to interpret the algebra of a ring using
the language of topology. A quick Google search of “the Zariski topology” is enough
to see its relevance in the theory of modern algebraic geometry, but many sources
will still be saturated with graduate level material. The good news is that there is
still a lot one can learn about the spectrum of a ring without having to know what
a sheaf or a scheme is. We have tried to combine the material that only relies on
basic ring theory and topology into a single source.
This paper should be accessible to second or third year undergraduate math
majors. The paper is divided into three main sections so that readers familiar with
ring theory or topology may skip ahead. Readers who have had a first course in
group theory should have no trouble reading this paper. We will begin with an
overview of ideals in rings, so readers who are unfamiliar with the definition of a
1
2 OSCARMICHEL
ring, a subring, or a product of rings may see [1]. We will assume that all rings are
commutative and with unity. In addition, we assume ring homomorphisms send 1
to 1. No topology background is necessary for reading this paper.
2. Rings and Ideals
We begin our study of ring theory with the definition of an ideal. An ideal is
very similar to a normal subgroup in group theory. This is because ideals allow one
to construct quotient rings similar to the way quotient groups are constructed using
a normal subgroup. As we will see, many theorems of quotient groups reappear in
the form of quotient rings.
Definition 2.1. Let R be a ring and I be a subgroup of R under addition. For
r ∈ R, consider the set rI = {ri | i ∈ I}. If rI ⊆ I for all r ∈ R, then I is said to
be an ideal of R.
Before we continue our study of ideals, we will pause to introduce some notation
for constructing ideals out of ring elements. If A is some subset of a ring R, then
(A)will be the smallest ideal of R containing A. Such an ideal always exists because
Ais contained in the ideal R. We note that if A is finite, A = {a ,a ,...,a }, then
1 2 n
the ideal (A) is also the set of all R-linear combinations of the a . Sometimes we
i
will write the ideal (0) simply as 0.
Onewaytounderstandthedefinitionofanidealistoconsidertheringofintegers
Z. For any n ∈ Z, the set nZ = {nx | x ∈ Z} is an ideal of Z. The ideal 2Z is
exactly the set of even integers. Adding two even numbers together will always give
an even number, and multiplying an even number by any integer will result in an
even number. The key difference between the additive and multiplicative structure
of an ideal is that multiplication by elements outside the ideal must always stay
inside the ideal. This is not true for addition. Going back to our example of the
even integers, an even number will not remain even if an odd number is added to it.
Aswewill see next, the multiplicative structure of an ideal allows for a well-defined
construction of a quotient ring.
Proposition 2.2. Let R be a ring and let I be an ideal of R. Then the additive
quotient group R/I is a ring under addition and multiplication defined by:
(r +I)+(s+I)=(r+s)+I
(r +I)×(s+I)=(rs)+I.
Proof. Since R is an abelian group and I is a normal subgroup, R/I is automatically
an abelian quotient group under addition. We will prove that multiplication in the
quotient group is well defined. The remaining ring axioms should be verified by the
reader. If we chose representatives r,s ∈ R and i,j ∈ I then,
(r +i)(s+j) = rs+rj +is+ij.
Since I is closed under multiplication by elements of R, each of rj, is, and ij is in
I. Furthermore, their sum rj + is + ij is in I. Writing this in terms of cosets we
have the desired result:
(r +I)×(s+I)=(rs)+I.
AN INTRODUCTION TO THE ZARISKI TOPOLOGY 3
In group theory one can understand the structure of a group through group
homomorphisms. The isomorphism theorems for groups establish a relationship
between groups, normal subgroups, quotient groups, and group homomorphisms.
This perspective is very useful for studying rings, too. The following theorems will
prepare us for proving two isomorphism theorems for rings.
Theorem 2.3. Let R and S be rings and let ϕ: R → S be a ring homomorphism.
Then, the image of ϕ is a subring of S, and kerϕ is an ideal of R.
Proof. If s ,s ∈ im(ϕ), then there are r ,r ∈ R such that s = ϕ(r ) and
1 2 1 2 1 1
s =ϕ(r ). From the homomorphism property, we know s +s = ϕ(r )+ϕ(r ) =
2 2 1 2 1 2
ϕ(r +r ) and s s = ϕ(r )ϕ(r ) = ϕ(r r ). Hence, s + s ∈ im(ϕ) and s s ∈
1 2 1 2 1 2 1 2 1 2 1 2
im(ϕ). Finally, 1 ∈ im(ϕ) because ϕ(1) = 1, and this proves that im(ϕ) is a subring
of S.
Next, suppose r ,r ∈ kerϕ. Since ϕ(r ) = ϕ(r ) = 0, it follows again from the
1 2 1 2
homomorphism property that ϕ(r + r ) = 0 which proves r + r ∈ kerϕ. Now
1 2 1 2
let a be any element of R, and let r ∈ kerϕ. Multiplying ϕ(a) and ϕ(r), we see
ϕ(ar) = ϕ(a)ϕ(r) = ϕ(a)0 = 0, and ar ∈ kerϕ.
Theorem 2.4. Let ϕ: R → S be a ring homomorphism. If J is an ideal of S, then
ϕ−1(J) is an ideal of S.
Proof. Suppose r ,r ∈ ϕ−1(J). By definition, ϕ(r ),ϕ(r ) ∈ J, and ϕ(r ) −
1 2 1 2 1
−1
ϕ(r ) = ϕ(r −r ) ∈ J because J is an ideal of S. It follows that r +r ∈ ϕ (J)
2 1 2 1 2
−1 −1
which proves ϕ (J) is closed under addition. Next, suppose a ∈ ϕ (J) and
−1
r ∈ R. Since J is an ideal of S, ϕ(a)ϕ(r) ∈ J. This implies ar ∈ ϕ (J) because
ϕ(ar) = ϕ(a)ϕ(r) ∈ J.
Remark 2.5. It is not true in general that if ϕ: R → S is a ring homomorphism,
then ϕ(J) is an ideal if J is an ideal. However, if ϕ is a surjective homomorphism,
then ϕ(J) is an ideal in S.
We are now ready to prove two ismomorphism theorems for rings. In total,
there are four standard isomorphism theorems for rings, but only two of them will
be presented here. The first isomrophism theorem is a useful tool to prove two
rings are ismorphic, and establishes a relationship between ring homomorphisms
and quotient rings.
Theorem 2.6 (The First Isomorphism Theorem for Rings). If ϕ: R → S is a ring
homomorphism, then R/kerϕ is isomorphic to the image of ϕ. In particular, if ϕ
is surjective, then R/kerϕ ∼ S.
=
Proof. Let I = kerϕ. First we note that R/I is a valid ring because kerϕ is an ideal
by Theorem 2.3. Consider the following map π: R/I → im(ϕ) where r+I 7→ ϕ(r).
First we will prove that this map is well defined. We will use the notation r to denote
the coset r + I. Suppose for some r ,r ∈ R, r = r . Then r − r ∈ I = kerϕ,
1 2 1 2 1 2
which means
π(r ) = ϕ(r ) = ϕ(r +(r −r )) = ϕ(r −r )+ϕ(r ) = 0+ϕ(r ) = π(r ).
1 1 1 2 2 1 2 2 2 2
Next we will prove π is an isomorphism between rings R/kerϕ and im(ϕ). First
note that π is a homomorphism.
π(r r )) = π(r r ) = ϕ(r r ) = ϕ(r )ϕ(r ) = π(r )π(r )
1 2 1 2 1 2 1 2 1 2
4 OSCARMICHEL
π(r +r ) = π(r +r ) = ϕ(r +r ) = ϕ(r )+ϕ(r ) = π(r )+π(r )
1 2 1 2 1 2 1 2 1 2
The map π is surjective. For every ϕ(r) ∈ im(π) we have π(r) = ϕ(r). Finally
suppose π(r ) = π(r ). Then π(r ) − π(r ) = 0 and we get π(r ) − π(r ) =
1 2 1 2 1 2
ϕ(r ) −ϕ(r ) = ϕ(r −r ) = 0. This means r −r ∈ kerϕ = I, so r = r . This
1 2 1 2 1 2 1 2
proves the map is injective, and hence an isomorphism.
If ψ: R → R/kerϕ is the projection map from R to the quotient ring R/kerϕ,
then the diagram below illustrates the proof of Theorem 2.6.
R ϕ S
ψ π
R/kerϕ
To see how the first isomorphism is useful, we will give an example. Recall
the ring of Gaussian integers: Z[i] = {a + bi | a,b ∈ Z}. We will prove that
Z[x]/(x2 +1) ∼ Z[i]. Consider the homomorphism ϕ: Z[x] → Z[i] given by p(x) 7→
=
p(i). This map is surjective because every Gaussian integer a + bi is mapped to
by its corresponding linear polynomial a + bx. Furthermore, the kernel of ϕ is the
2 2 ∼
ideal (x +1), thereby proving Z[x]/(x +1) = Z[i] by Theorem 2.6.
When constructing quotient rings, the ideal structure of the original ring is
preserved. For example, in the ring of integers there are three ideals containing 4Z:
4Z ⊂ 2Z ⊂ Z. In the ring Z/4Z, there are three ideals total: (0), (2), and Z/4Z.
It in not a coincidence that the number of ideals in the quotient ring is the same
as the number of ideals that contain 4Z. This relationship will be made precise in
the next theorem.
Theorem 2.7 (Lattice Isomorphism Theorem). Let I be an ideal of a ring R.
There is an inclusion preserving bijection between the set of ideals of R containing
I and the set of ideals of R/I.
Proof. Let φ: R → R/I be the projection map from R to the quotient ring R/I.
For each ideal J ⊇ I, there is a correspond ideal φ(J) ⊆ R/I since φ is surjective.
−1
Similarly, if K is an ideal of R/I, then φ (K) is an ideal of R. To prove there
is a bijection, we will show φ−1(φ(J)) = J for J ⊇ I. Written explicitly as a set,
φ−1(φ(J)) = {a ∈ R | φ(a) ∈ φ(J)}. When written in this way, it is clear that
J ⊆φ−1(φ(J)). If φ(a) ∈ φ(J), then there is a b ∈ J such that
φ(a) = φ(b)
⇒φ(a−b)=0
⇒a−b∈I
⇒a−b=c, for some c∈I
⇒a=b+c
Since b ∈ J and c ∈ I ⊆ J, we have shown a ∈ J, and thus φ−1(φ(J)) ⊆ J.
The lattice isomorphishm theorem is a fundamental result related to the ideal
structure of a ring. Oftentimes ideals can best be understood by looking at their
containment relationship to other ideals. Next, we will define two special classes of
ideals: prime ideals and maximal ideals.
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