Topics in Abstract Algebra/Commutative algebra

The set of all prime ideals in a commutative ring $$A$$ is called the spectrum of $$A$$ and denoted by $$\operatorname{Spec}(A)$$. (The motivation for the term comes from the theory of a commutative Banach algebra.)

Spec A
The set of all nilpotent elements in $$A$$ forms an ideal called the nilradical of $$A$$. Given any ideal $$\mathfrak a$$, the pre-image of the nilradical of $$A$$ is an ideal called the radical of $$\mathfrak a$$ and denoted by $$\sqrt{\mathfrak a}$$. Explicitly, $$x \in \sqrt{\mathfrak a}$$ if and only if $$x^n \in \mathfrak a$$ for some $$n$$.

Note that the theorem applies in particular when $$S$$ contains only 1.

A Goldman domain is a domain whose field of fractions $$K$$ is finitely generated as an algebra. When $$A$$ is a Goldman domain, K always has the form $$A[f^{-1}]$$. Indeed, if $$K = A[s_1^{-1}, ..., s_n^{-1}]$$, let $$s = s_1...s_n$$. Then $$K = A[s^{-1}]$$.

A prime ideal $$\mathfrak{p} \in \operatorname{Spec}(A)$$ is called a Goldman ideal if $$A/\mathfrak p$$ is a Goldman domain.

In some rings, Goldman ideals are maximal; this will be discussed in the next section. On the other hand,

A ring satisfying the equivalent conditions in the theorem is called a Hilbert-Jacobson ring.

An element p of a ring is a prime if $$(p)$$ is prime, and is an irreducible if $$p = xy \Rightarrow$$ either $$x$$ or $$y$$ is a unit..

We write $$x | y$$ if $$(x) \ni y$$, and say $$x$$ divides $$y$$. In a domain, a prime element is irreducible. (Suppose $$x = yz$$. Then either $$x | y$$ or $$x | z$$, say, the former. Then $$sx = y$$, and $$sxz = x$$. Canceling $$x$$ out we see $$z$$ is a unit.) The converse is false in general. We have however:

The Jacobson radical of a ring $$A$$ is the intersection of all maximal ideals.

Note that the nilradical is contained in the Jacobson radical, and they coincide in particular if prime ideals are maximal (e.g., the ring is a principal ideal domain). Another instance of this is:

A ring is said to be local if it has only one maximal ideal.

Let $$(A, \mathfrak m)$$ be a local noetherian ring.

A. Lemma
 * (i) Let $$\mathfrak i$$ be a proper ideal of $$A$$. If $$M$$ is a finite generated $$\mathfrak i$$-module, then $$M = 0$$.
 * (ii) The intersection of all $$\mathfrak{m}^k$$ over $$k \ge 1$$ is trivial.

Proof: We prove (i) by the induction on the number of generators. Suppose $$M$$ cannot be generated by strictly less than $$n$$ generators, and suppose we have $$x_1, ... x_n$$ that generates $$M$$. Then, in particular,
 * $$x_1 = a_1 x_1 + a_2 x_2 + ... + a_n x_n$$ where $$a_i$$ are in $$\mathfrak i$$,

and thus
 * $$(1 - a_1) x_1 = a_2 x_2 + ... + a_n x_n$$

Since $$a_1$$ is not a unit, $$1 - a_1$$ is a unit; in fact, if $$1 - a_1$$ is not a unit, it belongs to a unique maximal ideal $$\mathfrak{m}$$, which contains every non-units, in particular, $$a_1$$, and thus $$1 \in \mathfrak{m}$$, which is nonsense. Thus we find that actually x_2, ..., x_n generates $$M$$; this contradicts the inductive hypothesis.$$\square$$

An ideal $$\mathfrak q \triangleleft A$$ is said to be primary if every zero-divisor in $$A/\mathfrak q$$ is nilpotent. Explicitly, this means that, whenever $$xy \in \mathfrak q$$ and $$y \not\in \mathfrak q$$, $$x \in \sqrt{\mathfrak q}$$. In particular, a prime ideal is primary.

Integral extension
Let $$A \subset B$$ be rings. If $$b \in B$$ is a root of a monic polynomial $$f \in A[X]$$, then $$b$$ is said to be integral over $$A$$. If every element of $$B$$ is integral over $$A$$, then we say $$B$$ is integral over $$A$$ or $$B$$ is an integral extension of $$A$$. More generally, we say a ring morphism $$f: A \to B$$ is integral if the image of $$A$$ is integral over $$B$$. By replacing $$A$$ with $$f(A)$$, it suffices to study the case $$A \subset B$$, and that's what we will below do.

The set of all elements in B that are integral over A is called the integral closure of A in B. By the lemma, the integral closure is a subring of $$B$$ containing $$A$$. (Proof: if $$x$$ and $$y$$ are integral elements, then $$A[xy]$$ and $$A[x-y]$$ are contained in $$A[x, y]$$, finite over $$A$$.) It is also clear that integrability is transitive; that is, if $$C$$ is integral over $$B$$ and $$B$$ is integral over $$A$$, then $$C$$ is integral over $$A$$.

Noetherian rings
The next theorem furnishes many examples of a noetherian ring.

Let $$(A, \mathfrak m)$$ be a noetherian local ring with $$k = A/\mathfrak m$$. Let $$\mathfrak i \triangleleft A$$. Then $$\mathfrak i$$ is called an ideal of definition if $$A / \mathfrak i$$ is artinian.

The local ring $$A$$ is said to be regular if the equality holds in the above.

Zariski topology
Given $$\mathfrak a \triangleleft A$$, let $$\operatorname{V}(\mathfrak a) = \{ \mathfrak p \in \operatorname{Spec}(A) | \mathfrak p \supset \mathfrak a \}$$. (Note that $$\operatorname{V}(\mathfrak a) = \operatorname{V}(\sqrt \mathfrak a)$$.) It is easy to see
 * $$V(\mathfrak a) \cup V(\mathfrak b) = V(\mathfrak a \mathfrak b) = V(\mathfrak a \cap \mathfrak b)$$, and $$\cap_\alpha V(\mathfrak a_\alpha) = V((\mathfrak a_\alpha | \alpha))$$.

It follows that the collection of the sets of the form $$\operatorname{V}(\mathfrak a)$$ includes the empty set and $$\operatorname{Spec}(A)$$ and is closed under intersection and finite union. In other words, we can define a topology for $$\operatorname{Spec}(A)$$ by declaring $$\operatorname{Z}(\mathfrak i)$$ to be closed sets. The resulting topology is called the Zariski topology. Let $$X = \operatorname{Spec}(A)$$, and write $$X_f = X \backslash V ((f)) = \{ P \in X | P \ni f \}$$.

Integrally closed domain
In particular, in a polynomial ring that is a GCD domain, every irreducible polynomial is a prime element.

A domain satisfying the equivalent conditions in the theorem is called a unique factorization domain or a UFD for short.