本論文探討交換環上質理想的兩個主要的結構。

Let R be a commutative ring with unity. The spectrum of R,
Spec(R), is the set of all prime ideals of R. There are two important structures on Spec(R), the Zariski topology and the partially ordered set (poset) structure induced by set inclusion.
For each subset E of R, let
V(E)= { P \in Spec(R) ; E \subseteq P }.
Then the sets V(E) satisfy the axioms for closed sets in a topological space. The resulting topology is called the Zariski topology. Under Zariski topology, Spec(R) are well known to have many special
properties: Spec(R) is T_0 and quasi-compact; the quasi-compact open subsets are closed under finite intersection and form an open basis; and every nonempty irreducible closed subset has a generic point. In fact,
these properties do characterize the topological spaces which are homeomorphic to the spectrum of some rings.
A topological space is spectral if it satisfies the properties of Spec(R) we state above. Another important result is that a topological space is spectral if and only if it is a projective limit of finite T_0 spaces.
On the other hand, consider Spec(R) as a poset under set inclusion. I. Kaplansky notes two properties of Spec(R):
(K1) Every totally ordered subset of Spec(R) has a least
upper bound and a greatest lower bound.
(K2) If P \subsetneq Q are distinct prime ideals of R,
then there exists distinct prime ideals P_1 and Q_1
with P \subseteq P_1 \subsetneq Q_1 \subseteq Q such
that there is no prime ideals properly between P_1
and Q_1.
Kaplansky asks whether or not a poset satisfying (K1) and (K2)
must be isomorphic as a poset to Spec(R) for some ring R. In
[L], we see that the answer to Kaplansky's question is no in general, but the answer is yes for two special cases.
One is the case of finite posets (Chapter 4), and the other is the case of trees with (K1), (K2), and finitely many minimal elements (Chapter 5).
(We recall that a poset X is a tree if, for each x in X,
{ y \in X ; y \leq x $ is a totally ordered set.)
Moreover, in [LO], there is a more general result:
A poset X is a tree with (K1) and (K2) if and only if X is order isomorphic to the spectrum of some Bezout ring.
It is interesting to ask that what are the conditions under which a poset X is order isomorphic
to the spectrum of some ring R. The last paragraph of section 2 in [L] says:
The referee of this paper has made the following comments:
It follows easily from [Hochster, Proposition 10] that a
poset X is isomorphic to Spec(R) for some ring R if and
only if X is an inverse limit of finite posets.
In Chapter 7, I give a proof of this statement by using [Hochster, Proposition 10] and some facts in Chapter 6.
In Chapter 6, we discuss some relations between T_0
space structure and poset structure:
Let X be a T_0 topological space. We define an order on X by
$x \leq y \Leftrightarrow y \in \overline{\{x\}}$ (the closure of $\{x\}$),
for $x,y \in X$. Then $(X, \leq)$ becomes a poset and we denote this poset
structure as $X^\P$. If we further assume that $X$ is a spectral space, then
$X^\P$ is a poset satisfying (K1), (K2). On the other hand, let $Y$
be a poset
and, for $x \in Y$, define $cl(x) = \{ y \in Y \mid y \geq x \}$.
Then we can topologize $Y$ by letting $S= \{ \emptyset,cl(x) \mid x \in Y \}$
be a subbasis for closed sets. And then $Y$ is $T_0$ under this topology
and we denote this $T_0$ space structure as $Y^\T$. (Note that $cl(x)$ is
actually the closure of $\{x\}$ in $Y^\T$.)
Observe that (1) for any poset $X$, $X = (X^\T)^\P$ and (2)
for any Noetherian spectral space $Y$, $Y = (Y^\P)^\T$.
In general, it is not always true that, for a spectral space $X$,
$X \cong (X^\P)^\T$. Example 6.4 and 6.8 are two counterexamples.
In fact, we see that
$(X^\P)^\T$ is not even spectral in Example 6.8.
In addition, if $X$ is a finite poset or a tree with (K1), (K2),
and finitely many minimal elements,
then $X^\T$ is also spectral. In both cases, $X \cong Spec(\R)^\P$ for
some ring $\R$ (see Chapter 4 and 5). However, in the later case,
it is possible that $Spec(\R) \ncong (Spec(\R)^\P)^\T$ (see Example 6.4).
So we cannot conclude directly that $X^\T$
is spectral. We have a technical proof of this statement in Theorem 6.3.
Also, in Discussion 6.10, we have some observations about the relations between the continuous maps and the order preserving maps.
In Example 6.4, we use a non-trivial fact in [HW, Theorem 1.1]: if R is the ring of all algebraic integers and if p is a prime number then there are infinitely many maximal ideals of R lying over pZ. For completeness, we include a proof of this fact in appendix.

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