# 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 $A\neq 0$  be a ring. If every principal ideal in $A$  is prime, then $A$  is a field.
Proof. Let $0\neq x\in A$ . 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 $0\neq f\in A$ . Then $K=A[f^{-1}]$  if and only if every nonzero prime ideal of $A$  contains $f$ .
Proof. ($\Leftarrow$ ) Let $0\neq x\in A$ , and $S=\{f^{n}|n\geq 0\}$ . 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 $0\neq f\in A$ . Let $S=\{f^{n}|n\geq 0\}$ . 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 . We assume $a\in {\mathfrak {p}}_{r}$ ; else, we're done. Moreover, if ${\mathfrak {p}}_{i}\subset {\mathfrak {p}}_{r}$  for some $i , 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 . 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 $i\leq r$ . 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 $(1)\neq (d)\supset (p,x)$ . 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., $x^{n}=ux^{n+1}$  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, $A/{\mathfrak {m}}^{n},(n\geq 1)$ 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 $k\geq 1$  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 $bx_{i}\in M$ , we can write

$bx_{i}=\sum _{j=1}^{n}c_{ij}x_{j}$

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 $0\neq b\in B$ . 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 $a_{0}\neq 0$ . This shows that $bB\cap A\neq 0$ , 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 $\leq n$  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 $0\leq n\leq d$ , choose finitely many elements $f_{1n},f_{2n},...f_{m_{n}n}$  of $I$  whose coefficients $b_{1n},...b_{m_{n}n}$  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 $n\leq d$ , 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 $n\geq d$ , 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. $\dim _{k}({\mathfrak {m}}/{\mathfrak {m}}^{2})\geq \dim A$

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 $\bigcap _{n\geq 1}{\mathfrak {i}}^{n}=0$ .

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.