# 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 $n$.

Proposition A.14. Let $\mathfrak i, \mathfrak j \triangleleft A$.

• (i) $\sqrt{\mathfrak i^n} = \sqrt \mathfrak i$
• (ii) $\sqrt{\mathfrak i \mathfrak j} = \sqrt{\mathfrak i \cap \mathfrak j} = \sqrt \mathfrak i \cap \sqrt \mathfrak j$

Proof. Routine. $\square$

Exercise. A ring has only one prime ideal if and only if its nilradical is maximal.

Exercise. Every prime ideal in a finite ring is maximal.

Proposition A.2. Let $A \ne 0$ be a ring. If every principal ideal in $A$ is prime, then $A$ is a field.
Proof. Let $0 \ne x \in A$. Since $x^2$ is in $(x^2)$, which is prime, $x \in (x^2)$. Thus, we can write $x = ax^2$. Since $(0)$ is prime, $A$ is a domain. Hence, $1 = ax$. $\square$

Lemma. Let $\mathfrak p \triangleleft A$. Then $\mathfrak p$ is prime if and only if

$\mathfrak p \subsetneq \mathfrak a \triangleleft A, \mathfrak p \subsetneq \mathfrak b \triangleleft A$ implies $\mathfrak a \mathfrak b \not\subset \mathfrak p$

Proof. ($\Rightarrow$) Clear. ($\Leftarrow$) Let $\overline{x}$ be the image of $x \in A$ in $A / \mathfrak p$. Suppose $\overline{a}$ is a zero-divisor; that is, $\overline{a}\overline{b} = 0$ for some $b \in A \backslash \mathfrak p$. Let $\mathfrak a = (a, \mathfrak p)$, and $\mathfrak b = (b, \mathfrak p)$. Since $\mathfrak a \mathfrak b = ab + \mathfrak p \subset \mathfrak p$, and $\mathfrak b$ is strictly larger than $\mathfrak p$, by the hypothesis, $\mathfrak a \subset \mathfrak p$. That is, $\overline{a} = 0$. $\square$

Theorem A.11 (multiplicative avoidance). Let $S \subset A$ be a multiplicative system. If $\mathfrak a \triangleleft A$ is disjoint from $S$, then there exists a prime ideal $\mathfrak p \supset \mathfrak a$ that is maximal among ideals disjoint from $S$.
Proof. Let $\mathfrak m$ be a maximal element in the set of all ideals disjoint from $S$. Let $\mathfrak a$ and $\mathfrak b$ be ideals strictly larger than $\mathfrak m$. Since $\mathfrak m$ is maximal, we find $a \in \mathfrak a \cap S$ and $b \in \mathfrak b \cap S$. By the definition of $S$, $ab \in S$; thus, $\mathfrak a \mathfrak b \not \subset \mathfrak m$. By the lemma, $\mathfrak m$ is prime then. $\square$

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

Exercise. A domain A is a principal ideal domain if every prime ideal is principal.

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 $K = A[s^{-1}]$.

Lemma. Let $A$ be a domain with the field of fractions $K$, and $0 \ne f \in A$. Then $K = A[f^{-1}]$ if and only if every nonzero prime ideal of $A$ contains $f$.
Proof. ($\Leftarrow$) Let $0 \ne x \in A$, and $S = \{ f^n | n \ge 0 \}$. If $(x)$ is disjoint from $S$, then, by the lemma, there is a prime ideal disjoint from $S$, contradicting the hypothesis. Thus, $(x)$ contains some power of $f$, say, $yx = f^n$. Then $yx$ and so $x$ are invertible in $A[f^{-1}].$ ($\Rightarrow$) If $\mathfrak p$ is a nonzero prime ideal, it contains a nonzero element, say, $s$. Then we can write: $1 / s = a / f^n$, or $f^n = as \in \mathfrak p$; thus, $f \in \mathfrak p$. $\square$

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

Theorem A.21. Let $A$ be a ring and $\mathfrak a \triangleleft A$. Then $\sqrt \mathfrak a$ is the intersection of all minimal Goldman ideals of A containing $\mathfrak a$
Proof. By the ideal correspondence, it suffices to prove the case $\mathfrak a = \sqrt \mathfrak a = 0$. Let $0 \ne f \in A$. Let $S = \{ f^n | n \ge 0 \}$. Since $f$ is not nilpotent (or it will be in $\sqrt{(0)}$), by multiplicative avoidance, there is some prime ideal $\mathfrak g$ not containing $f$. It remains to show it is a Goldman ideal. But if $\mathfrak p \triangleleft A / \mathfrak g$ is a nonzero prime, then $f \in \mathfrak p$ since $\mathfrak p$ collapses to zero if it is disjoint from $S$. By Lemma, the field of fractions of $A / \mathfrak g$ is obtained by inverting $f$ and so $\mathfrak g$ is a Goldman ideal. Hence, the intersection of all Goldman ideals reduces to zero. $\square$

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

Lemma. Let $\mathfrak a \triangleleft A$. Then $\mathfrak a$ is a Goldman ideal if and only if it is the contraction of a maximal ideal in $A[X]$.

Theorem. The following are equivalent.

1. For any $\mathfrak a \triangleleft A$, $\mathfrak a$ is the intersection of all maximal ideals containing $\mathfrak a$.
2. Every Goldman ideal is maximal.
3. Every maximal ideal in $A[X]$ contracts to a maximal ideal in $A$.

Proof. Clear. $\square$

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

Lemma. Let $A \subset B$ be domains such that $B$ is algebraic and of finite type over $A$. Then $A$ is a Goldman domain if and only if $B$ is a Goldman domain.
Proof. Let $K \subset L$ be the fields of fractions of $A$ and $B$, respectively. $\square$

Theorem A.19. Let $A$ be a Hilbert-Jacobson ring. Then $A[X]$ is a Hilbert-Jacobson ring.
Proof. Let $\mathfrak q \triangleleft A[X]$ be a Goldman ideal, and $\mathfrak p = \mathfrak q \cap A$. It follows from Lemma something that $A / \mathfrak p$ is a Goldman domain since it is contained in a $A[X] / \mathfrak q$, a Goldman domain. Since $A$ is a Hilbert-Jacobson ring, $\mathfrak p$ is maximal and so $A / \mathfrak p$ is a field and so $A[X] / \mathfrak q$ is a field; that is, $\mathfrak q$ is maximal. $\square$

 To do: Explain why $A / \mathfrak p$ is a field (or point to a location where it can be understood why it is so...).

Theorem A.5 (prime avoidance). Let $\mathfrak p_1, ..., \mathfrak p_r \triangleleft A$ be ideals, at most two of which are not prime, and $\mathfrak a \triangleleft A$. If $\mathfrak a \subset \bigcup_1^r \mathfrak p_i$, then $\mathfrak a \subset \mathfrak p_i$ for some $\mathfrak i$.
Proof. We shall induct on $r$ to find $a \in \mathfrak a$ that is in no $\mathfrak p_i$. The case $r=1$ being trivial, suppose we find $a \in \mathfrak a$ such that $a \not\in \mathfrak{p}_i$ for $i < r$. We assume $a \in \mathfrak{p}_r$; else, we're done. Moreover, if $\mathfrak{p}_i \subset \mathfrak{p}_r$ for some $i < r$, then the theorem applies without $\mathfrak{p}_i$ and so this case is done by by the inductive hypothesis. We thus assume $\mathfrak p_i \not\subset \mathfrak p_r$ for all $i < r$. Now, $\mathfrak a \mathfrak p_1 ... \mathfrak p_{r-1} \not\subset \mathfrak {p}_r$; if not, since $\mathfrak p_r$ is prime, one of the ideals in the left is contained in $\mathfrak p_r$, contradiction. Hence, there is $b$ in the left that is not in $\mathfrak {p}_r$. It follows that $a + b \not\in \mathfrak p_i$ for all $i \le r$. Finally, we remark that the argument works without assuming $\mathfrak p_1$ and $\mathfrak p_2$ are prime. (TODO: too sketchy.) The proof is thus complete. $\square$

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:

Proposition. Suppose: for every $x$ and $y$, $(x) \cap (y) = (xy)$ whenever (1) is the only principal ideal containing $(x, y)$. Then every irreducible is a prime.
Proof. Let $p$ be an irreducible, and suppose $p | xy$ and $p \not| y$. Since $(p) \cap (x) = (px)$ implies $px | xy$ and $p | y$, there is a $d$ such that $(1) \ne (d) \supset (p, x)$. But then $d | p$ and so $p | d$ ($p$ is an irreducible.) Thus, $p | x$. $\square$

Theorem A.16 (Chinese remainder theorem). Let $\mathfrak a_1, ..., \mathfrak a_n \triangleleft A$. If $\mathfrak a_j + \mathfrak a_i = (1)$, then

$\prod \mathfrak a_i \to A \to A / \mathfrak a_1 \times \cdots \times \mathfrak a_n \to 0$

is exact.

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

Proposition A.6. $x \in A$ is in the Jacobson radical if and only if $1 - xy$ is a unit for every $y \in A$.
Proof. Let $x$ be in the Jacobson radical. If $1-xy$ is not a unit, it is in a maximal ideal $\mathfrak m$. But then we have: $1 = (1 - xy) + xy$, which is a sum of elements in $\mathfrak m$; thus, in $\mathfrak m$, contradiction. Conversely, suppose $x$ is not in the Jacobson radical; that is, it is not in some maximal ideal $\mathfrak m$. Then $(x, \mathfrak m)$ is an ideal containing $\mathfrak m$ but strictly larger. Thus, it contains $1$, and we can write: $1 = xy + z$ with $y \in A$ and $z \in \mathfrak m$. Then $1 - xy \in \mathfrak m$, and $\mathfrak m$ would cease to be proper, unless $1 - xy$ is a non-unit. $\square$

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:

Exercise. In $A[X]$, the nilradical and the Jacobson radical coincide.

Theorem A.17 (Hopkins). Let A be a ring. Then the following are equivalent.

1. A is artinian
2. A is noetherian and every prime ideal is maximal.
3. $\operatorname{Spec}(A)$ is finite and discrete, and $A_\mathfrak m$ is noetherian for all maximal ideal $\mathfrak m$.

Proof. (1) $\Rightarrow$ (3): Let $\mathfrak p \triangleleft A$ be prime, and $x \in A / \mathfrak p$. Since $A / \mathfrak p$ is artinian (consider the short exact sequence), the descending sequence $(x^n)$ stabilizes eventually; i.e., $x^n = u x^{n+1}$ for some unit u. Since $A / \mathfrak p$ is a domain, $x$ is a unit then. Hence, $\mathfrak p$ is maximal and so $\operatorname{Spec}(A)$ is discrete. It remains to show that it is finite. Let $S$ be the set of all finite intersections of maximal ideals. Let $\mathfrak i \in S$ be its minimal element, which we have by (1). We write $\mathfrak i = \mathfrak m_1 \cap ... \cap \mathfrak m_n$. Let $\mathfrak m$ be an arbitrary maximal ideal. Then $\mathfrak m \cap \mathfrak i \in S$ and so $\mathfrak m \cap \mathfrak i = \mathfrak i$ by minimality. Thus, $\mathfrak m = \mathfrak m_i$ for some i. (3) $\Rightarrow$ (2): We only have to show $A$ is noetherian. $\square$

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

Proposition A.17. Let $A$ be a nonzero ring. The following are equivalent.

1. $A$ is local.
2. For every $x \in A$, either $x$ or $1 - x$ is a unit.
3. The set of non-units is an ideal.

Proof. (1) $\Rightarrow$ (2): If $x$ is a non-unit, then $x$ is the Jacobson radical; thus, $1 - x$ is a unit by Proposition A.6. (2) $\Rightarrow$ (3): Let $x, y \in A$, and suppose $x$ is a non-unit. If $xy$ is a unit, then so are $x$ and $y$. Thus, $xy$ is a non-unit. Suppose $x,y$ are non-units; we show that $x+y$ is a non-unit by contradiction. If $x+y$ is a unit, then there exists a unit $a\in A$ such that $1=a(x+y)=ax+ay$. Thus either $ax$ or $1-ax=ay$ is a unit, whence either $x$ or $y$ is a unit, a contradiction. (3) $\Rightarrow$ (1): Let $\mathfrak i$ be the set of non-units. If $\mathfrak m \triangleleft A$ is maximal, it consists of nonunits; thus, $\mathfrak m \subset \mathfrak i$ where we have the equality by the maximality of $\mathfrak m$. $\square$

Example. If $p$ is a prime ideal, then $A_p$ is a local ring where $p$ is its unique maximal ideal.

Example. If $\sqrt \mathfrak i$ is maximal, then $A/\mathfrak i$ is a local ring. In particular, $A/\mathfrak m^n, (n \ge 1)$is local for any maximal ideal $\mathfrak m$.

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.

Proposition. If $\mathfrak q$ is primary, then $\sqrt{\mathfrak q}$ is prime. Conversely, if $\sqrt{\mathfrak q}$ is maximal, then $\mathfrak q$ is primary.
Proof. The first part is clear. Conversely, if $\sqrt \mathfrak q$ is maximal, then $\mathfrak m = \sqrt \mathfrak q / \mathfrak q$ is a maximal ideal in $A/\mathfrak q$. It must be unique and so $A/\mathfrak q$ is local. In particular, a zero-divisor in $A/\mathfrak q$ is nonunit and so is contained in $\mathfrak m$; hence, nilpotent. $\square$

Exercise. $\sqrt{\mathfrak q}$ prime $\not\Rightarrow \mathfrak q$ primary.

Theorem A.8 (Primary decomposition). Let $A$ be a noetherian ring. If $\mathfrak i \triangleleft A$, then $\mathfrak i$ is a finite intersection of primary ideals.
Proof. Let $S$ be the set of all ideals that is not a finite intersection of primary ideals. We want to show $S$ is empty. Suppose not, and let $\mathfrak i$ be its maximal element. We can write $\mathfrak i$ as an intersection of two ideals strictly larger than $\mathfrak i$. Indeed, since $\mathfrak i$ is not prime by definition in particular, choose $x \not\in \mathfrak i$ and $y \not\in \mathfrak i$ such that $xy \in \mathfrak i$. As in the proof of Theorem A.3, we can write: $\mathfrak i = \mathfrak j (\mathfrak i + x)$ where $\mathfrak j$ is the set of all $a \in A$ such that $ax \in \mathfrak i$. By maximality, $\mathfrak j, \mathfrak i + x \not\in S$. Thus, they are finite intersections of primary ideals, but then so is $\mathfrak i$, contradiction. $\square$

Proposition. If $(0)$ is indecomposable, then the set of zero divisors is a union of minimal primes.

## 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.

Lemma A.9. Let $b \in B$. Then the following are equivalent.

1. $b$ is integral over $A$.
2. $A[b]$ is finite over $A$.
3. $A[b]$ is contained in an $A$-submodule of $B$ that is finite over $A$.

Proof. (1) means that we can write:

$b^{n+r} = -(b^{r+n-1} a_{n-1} + ... b^{r+1} a_1 + b^r a_0)$

Thus, $1, b, ..., b^{n-1}$ spans $A[b]$. Hence, (1) $\Rightarrow$ (2). Since (2) $\Rightarrow$ (3) vacuously, it remains to show (3) $\Rightarrow$ (1). Let $M_{/A[b]}$ be generated over $A$ by $x_1, ..., x_n$. Since $b x_i \in M$, we can write

$b x_i = \sum_{j=1}^n c_{ij} x_j$

where $c_{kj} \in A$. Denoting by $C$ the matrix $c_{ij}$, this means that $\det(bI-C)$ annihilates $M$. Hence, $\det(bI-C) = 0$ by (3). Noting $\det(bI-C)$ is a monic polynomial in $b$ we get (1). $\square$

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 $A$.

Proposition. Let $f:A \to B$ be an integral extension where $B$ is a domain. Then

• (i) $A$ is a field if and only if $B$ is a field.
• (ii) Every nonzero ideal of $B$ has nonzero intersection with $A$.

Proof. (i) Suppose $B$ is a field, and let $x \in A$. Since $x^{-1} \in B$ and is integral over $A$, we can write:

$x^{-n} = -(a_{n-1}x^{-(n-1)} + ... + a_1 x^{-1} + a_0)$

Multiplying both sides by $x^{n-1}$ we see $x^{-1} \in A$. For the rest, let $0 \ne b \in B$. We have an integral equation:

$-a_0 = b^n + a_{n-1}b^{n-1} + ... + a_1 b = b(b^{n-1} + a_{n-1}b^{n-2} + ... + a_1)$.

Since $B$ is a domain, if $n$ is the minimal degree of a monic polynomial that annihilates $b$, then it must be that $a_0 \ne 0$. This shows that $bB \cap A \ne 0$, giving us (ii). Also, if $A$ is a field, then $a_0$ is invertible and so is $b$. $\square$

Theorem (Noether normalization). Let $A$ be a finitely generated $k$-algebra. Then we can find $z_1, ..., z_d$ such that

1. $A$ is integral over $k[z_1, ..., z_d]$.
2. $z_1, ..., z_d$ are algebraically independent over $k$.
3. $z_1, ..., z_d$ are a separating transcendence basis of the field of fractions $K$ of $A$ if $K$ is separable over $k$.

Exercise A.10 (Artin-Tate). Let $A \subset B \subset C$ be rings. Suppose $A$ is noetherian. If $C$ is finitely generated as an $A$-algebra and integral over $B$, then $B$ is finitely generated as an $A$-algebra.

Exercise. A ring morphism $f:A \to \Omega$ (where $\Omega$ is an algebraically closed field) extends to $F: A[b] \to \Omega$ (Answer: http://www.math.uiuc.edu/~r-ash/ComAlg/)

## Noetherian rings

Exercise. A ring is noetherian if and only if every prime ideal is finitely generated. (See T. Y. Lam and Manuel L. Reyes, A Prime Ideal Principle in Commutative Algebra for a systematic study of results of this type.)

The next theorem furnishes many examples of a noetherian ring.

Theorem A.7 (Hilbert basis). $A$ is a noetherian ring if and only if $A[T_1, ... T_n]$ is noetherian.
Proof. By induction it suffices to prove $A[T]$ is noetherian. Let $I \triangleleft A[T]$. Let $L_n$ be the set of all coefficients of polynomials of degree $\le n$ in $I$. Since $L_n \triangleleft A$, there exists $d$ such that

$L_0 \subset L_1 \subset L_2, ..., \subset L_d = L_{d+1} = ...$.

For each $0 \le n \le d$, choose finitely many elements $f_{1n}, f_{2n}, ... f_{m_nn}$ of $I$ whose coefficients $b_{1n}, ... b_{m_nn}$ generate $L_n$. Let $I'$ be an ideal generated by $f_{jn}$ for all $j, n$. We claim $I = I'$. It is clear that $I \subset I'$. We prove the opposite inclusion by induction on the degree of polynomials in $I$. Let $f \in I$, $a$ the leading coefficient of $f$ and $n$ the degree of $f$. Then $a \in L_n$. If $n \le d$, then

$a = a_1 b_{1n} + a_2 b_{2n} + ... + a_{m_n} b_{{m_n}n}$

In particular, if $g = a_1 f_{1n} + a_2 f_{2n} + ... + a_{m_n} f_{{m_n}n}$, then $f - g$ has degree strictly less than that of $f$ and so by the inductive hypothesis $f - g \in I'$. Since $g \in I'$, $f \in I'$ then. If $n \ge d$, then $a \in L_d$ and the same argument shows $f \in I'$. $\square$

Exercise. Let $A$ be the ring of continuous functions $f: [0, 1] \to [0, 1]$.$A$ is not noetherian.

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.

Theorem. $\dim_k (\mathfrak m / {\mathfrak m}^2) \ge \dim A$

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

Theorem. Let $A$ be a noetherian ring. Then $\dim A[T_1, ..., T_n] = n + \dim A$.
Proof. By induction, it suffices to prove the case $n = 1$. $\square$

Theorem. Let $A$ be a finite-dimensional $k$-algebra. If $A$ is a domain with the field of fractions $K$, then $\dim A = \operatorname{trdeg}_k K$.
Proof. By the noether normalization lemma, $A$ is integral over $k[x_1, ..., x_n]$ where $x_1, ..., x_n$ are algebraically independent over $k$. Thus, $\dim A = \dim k[x_1, ..., x_n] = n$. On the other hand, $\operatorname{trdeg}_k K = n$. $\square$

Theorem. Let $A$ be a domain with (ACCP). Then $A$ is a UFD if and only if every prime ideal $\mathfrak p$ of height 1 is principal.
Proof. ($\Rightarrow$) By Theorem A.10, $\mathfrak p$ contains a prime element $x$. Then

$0 \subset (x) \subset \mathfrak p$

where the second inclusion must be equality since $\mathfrak p$ has height 1. ($\Leftarrow$) In light of Theorem A.10, it suffices to show that $A$ is a GCD domain. (TODO: complete the proof.) $\square$

Theorem. A regular local ring is a UFD.

Theorem A.10 (Krull's intersection theorem). Let $\mathfrak i \triangleleft A$ be a proper ideal. If $A$ is either a noetherian domain or a local ring, then $\bigcap_{n \ge 1} \mathfrak i^n = 0$.

Theorem A.15. Let $\mathfrak i \triangleleft A$. If $A$ is noetherian,

$\sqrt \mathfrak i^n \subset \mathfrak i \subset \sqrt \mathfrak i$ for some $n$.

In particular, the nilradical of $A$ is nilpotent.
Proof. It suffices to prove this when $\mathfrak i = 0$. Thus, the proof reduces to proving that the nilradical of A is nilpotent. Since $A$ is nilpotent, we have finitely many nilpotent elements $x_1, ..., x_n$ that spans $\sqrt{(0)}$. The power of any linear combination of them is then a sum of terms that contain the high power of some $x_j$ if we take the sufficiently high power. Thus, $\sqrt{(0)}$ is nilpotent. $\square$

Proposition A.8. If $A$ is noetherian, then $\hat A$ is noetherian.

Corollary. If $A$ is noetherian, then $A[[X]]$ is noetherian.

## 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 $X_f = X \backslash V ((f)) = \{ P \in X | P \ni f \}$.

Proposition A.16. We have:

• (i) $X_f$ is quasi-compact.
• (ii) $X_{fg}$ is canonically isomorphic to $\operatorname{Spec}(A[f^{-1}])_g$.

Proof. We have: $X_f \subset \bigcup_\alpha X_{f_\alpha} = X \backslash V((f_\alpha|\alpha)) \Leftrightarrow (f) \subset (f_\alpha|\alpha) \Leftrightarrow f \in (f_{\alpha_1}, ..., f_{\alpha_n})$. $\square$

Exercise. Let $A$ be a local ring. Then $\operatorname{Spec}(A)$ is connected.

Corollary. $\operatorname{Spec}(B) \to \operatorname{Spec}(A)$ is a closed surjection.

Theorem A.12. If $A_m$ is noetherian for every maximal ideal $\mathfrak m$ and if $\{ \mathfrak m \in \operatorname{Max}(A) | x \in m \}$ is finite for each $x \in A$, then $A$ is noetherian.

## Integrally closed domain

Lemma A.8. In a GCD domain, if $(x, y) = 1 = (x, z)$, then $(x, yz) = 1$.

Proposition A.9. In a GCD domain, every irreducible element is prime.
Proof. Let $x$ be an irreducible, and suppose $x | yz$. Then $x | (x, yz)$. If $(x, yz) = 1$, $x$ is a unit, the case we tacitly ignore. Thus, by the lemma, $d = (x, y)$, say, is a nonunit. Since $x$ is irreducible, $x | d$ and so $x | y$. $\square$

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

Theorem (undefined: ACC). Let A be a ring that satisfies the ascending chain conditions on principal ideals (example: noetherian ring). Then every x in $A$ is a finite product of irreducibles.

Theorem A.10. Let A be a domain. The following are equivalent.

1. Every nonzero nonunit element is a finite product of prime elements.
2. (Kaplansky) Every nonzero prime ideal contains a prime element.
3. $A$ is a GCD domain and has (ACC) on principal ideals.

Proof. (3) $\Rightarrow$ (2): Let $\mathfrak{p} \in \operatorname{Spec}(A)$. If $\mathfrak{p}$ is nonzero, it then contains a nonzero element x, which we factor into irreducibles: $x = p_1 ... p_n$. Then $p_j \in \mathfrak{p}$ for some $j$. Finally, irreducibles are prime since $A$ is a GCD domain. (2) $\Rightarrow$ (1): Let $S$ be the set of all products of prime elements. Clearly, $S$ satisfies the hypothesis of Theorem A.11 (i.e., closed under multiplication). Suppose, on the contrary, there is a nonzero nonunit $x$. It is easy to see that since $x \not\in S$, $(x)$ and $S$ are disjoint. Thus, by Theorem A.11, there is a prime ideal $\mathfrak p$ containing $x$ and disjoint from $S$. But, by (2), $\mathfrak p$ contains a prime element $y$; that is, $\mathfrak p$ intersects $S$, contradiction. (1) $\Rightarrow$ (3): By uniqueness of factorization, it is clear that $A$ is a GCD domain. $\square$

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

Corollary. If $A$ is a UFD, then $A[X]$ is a UFD. If A is a principal ideal domain, then $A[[X]]$ is a UFD.

Theorem A.13 (Nagata criterion). Let A be a domain, and $S \subset A$ a multiplicatively closed subset generated by prime elements. Then $A$ is a UFD if and only if $S^{-1}A$ is a UFD.