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 $\mathrm{Spec}(A)$. (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 $𝔞$, the pre-image of the nilradical of $A$ is an ideal called the radical of $𝔞$ and denoted by $\sqrt{𝔞}$. Explicitly, $x\in \sqrt{𝔞}$ if and only if $x^n\in 𝔞$ for some $n$.

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label=radical prop |claim=Let $𝔦,𝔧◃A$.

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

|proof=Routine.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Exercise |label= |claim=A ring has only one prime ideal if and only if its nilradical is maximal.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Exercise |label= |claim=Every prime ideal in a finite ring is maximal.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label=Ideal prime -> field |claim=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 $x=a{x}^2$. Since $(0)$ is prime, $A$ is a domain. Hence, $1=ax$.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Lemma |label= |claim=Let $𝔭◃A$. Then $𝔭$ is prime if and only if

$𝔭⊊𝔞◃A,𝔭⊊𝔟◃A$ implies $𝔞𝔟\subset ̸𝔭$

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

{{Template:BOOKTEMPLATE/Theorem |name=multiplicative avoidance |label=disjoint maximal ideal |claim=Let $S\subset A$ be a multiplicative system. If $𝔞◃A$ is disjoint from $S$, then there exists a prime ideal $𝔭\supset 𝔞$ that is maximal among ideals disjoint from $S$. |proof=Let $𝔪$ be a maximal element in the set of all ideals disjoint from $S$. Let $𝔞$ and $𝔟$ be ideals strictly larger than $𝔪$. Since $𝔪$ is maximal, we find $a\in 𝔞\cap S$ and $b\in 𝔟\cap S$. By the definition of $S$, $ab\in S$; thus, $𝔞𝔟\subset ̸𝔪$. By the lemma, $𝔪$ is prime then.}}

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

{{Template:BOOKTEMPLATE/Theorem |theorem=Exercise |claim=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 $K=A[s^{-1}]$.

{{Template:BOOKTEMPLATE/Theorem |theorem=Lemma |label= |claim=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 $𝔭$ is a nonzero prime ideal, it contains a nonzero element, say, $s$. Then we can write: $1/s=a/f^n$, or $f^n=as\in 𝔭$; thus, $f\in 𝔭$.}}

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

{{Template:BOOKTEMPLATE/Theorem |label=radical intersection Goldman |claim=Let $A$ be a ring and $𝔞◃A$. Then $\sqrt{𝔞}$ is the intersection of all minimal Goldman ideals of A containing $𝔞$ |proof=By the ideal correspondence, it suffices to prove the case $𝔞=\sqrt{𝔞}=0$. 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 $𝔤$ not containing $f$. It remains to show it is a Goldman ideal. But if $𝔭◃A/𝔤$ is a nonzero prime, then $f\in 𝔭$ since $𝔭$ collapses to zero if it is disjoint from $S$. By Lemma, the field of fractions of $A/𝔤$ is obtained by inverting $f$ and so $𝔤$ is a Goldman ideal. Hence, the intersection of all Goldman ideals reduces to zero.}}

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

{{Template:BOOKTEMPLATE/Theorem |label= |theorem=Lemma |claim=Let $𝔞◃A$. Then $𝔞$ is a Goldman ideal if and only if it is the contraction of a maximal ideal in $A[X]$.}}

{{Template:BOOKTEMPLATE/Theorem |label= |theorem=Theorem |claim=The following are equivalent.

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

|proof=Clear.}}

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

{{Template:BOOKTEMPLATE/Theorem |theorem=Lemma |label= |claim=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.}}

{{Template:BOOKTEMPLATE/Theorem |label=polynomial ring of Jacobson Jacobson |theorem=Theorem |claim=Let $A$ be a Hilbert-Jacobson ring. Then $A[X]$ is a Hilbert-Jacobson ring. |proof=Let $𝔮◃A[X]$ be a Goldman ideal, and $𝔭=𝔮\cap A$. It follows from Lemma something that $A/𝔭$ is a Goldman domain since it is contained in a $A[X]/𝔮$, a Goldman domain. Since $A$ is a Hilbert-Jacobson ring, $𝔭$ is maximal and so $A/𝔭$ is a field and so $A[X]/𝔮$ is a field; that is, $𝔮$ is maximal.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Theorem |label=prime avoidance |name=prime avoidance |claim=Let ${𝔭}_1,...,{𝔭}_r◃A$ be ideals, at most two of which are not prime, and $𝔞◃A$. If $𝔞\subset \bigcup _1^r{𝔭}_i$, then $𝔞\subset {𝔭}_i$ for some $𝔦$. |proof=We shall induct on $r$ to find $a\in 𝔞$ that is in no ${𝔭}_i$. The case $r=1$ being trivial, suppose we find $a\in 𝔞$ such that $a\in ̸{𝔭}_i$ for $i<r$. We assume $a\in {𝔭}_r$; else, we're done. Moreover, if ${𝔭}_i\subset {𝔭}_r$ for some $i<r$, then the theorem applies without ${𝔭}_i$ and so this case is done by by the inductive hypothesis. We thus assume ${𝔭}_i\subset ̸{𝔭}_r$ for all $i<r$. Now, $𝔞{𝔭}_1...{𝔭}_{r-1}\subset ̸{𝔭}_r$; if not, since ${𝔭}_r$ is prime, one of the ideals in the left is contained in ${𝔭}_r$, contradiction. Hence, there is $b$ in the left that is not in ${𝔭}_r$. It follows that $a+b\in ̸{𝔭}_i$ for all $i\le r$. Finally, we remark that the argument works without assuming ${𝔭}_1$ and ${𝔭}_2$ are prime. (TODO: too sketchy.) The proof is thus complete.}}

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:

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label= |claim=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 $p|y$. 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, $p|x$.}}

{{Template:BOOKTEMPLATE/Theorem |label=Chinese remainder |name=Chinese remainder theorem |claim=Let ${𝔞}_1,...,{𝔞}_n◃A$. If ${𝔞}_j+{𝔞}_i=(1)$, then

$\prod {𝔞}_i\to A\to A/{𝔞}_1\times \cdots \times {𝔞}_n\to 0$

is exact.}}

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

{{Template:BOOKTEMPLATE/Theorem |name= |theorem=Proposition |label=Jacobson char |claim=$x\in A$ 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 $𝔪$. But then we have: $1=(1-xy)+xy$, which is a sum of elements in $𝔪$; thus, in $𝔪$, contradiction. Conversely, suppose $x$ is not in the Jacobson radical; that is, it is not in some maximal ideal $𝔪$. Then $(x,𝔪)$ is an ideal containing $𝔪$ but strictly larger. Thus, it contains $1$, and we can write: $1=xy+z$ with $y\in A$ and $z\in 𝔪$. Then $1-xy\in 𝔪$, and $𝔪$ would cease to be proper, unless $1-xy$ is a non-unit. }}

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:

{{Template:BOOKTEMPLATE/Theorem |theorem=Exercise |label= |claim=In $A[X]$, the nilradical and the Jacobson radical coincide.}}

{{Template:BOOKTEMPLATE/Theorem |label=Hopkins |name=Hopkins |claim=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. $\mathrm{Spec}(A)$ is finite and discrete, and $A_{𝔪}$ is noetherian for all maximal ideal $𝔪$.

|proof=(1) $\Rightarrow$ (3): Let $𝔭◃A$ be prime, and $x\in A/𝔭$. Since $A/𝔭$ 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 $A/𝔭$ is a domain, $x$ is a unit then. Hence, $𝔭$ is maximal and so $\mathrm{Spec}(A)$ is discrete. It remains to show that it is finite. Let $S$ be the set of all finite intersections of maximal ideals. Let $𝔦\in S$ be its minimal element, which we have by (1). We write $𝔦={𝔪}_1\cap ...\cap {𝔪}_n$. Let $𝔪$ be an arbitrary maximal ideal. Then $𝔪\cap 𝔦\in S$ and so $𝔪\cap 𝔦=𝔦$ by minimality. Thus, $𝔪={𝔪}_i$ for some i. (3) $\Rightarrow$ (2): We only have to show $A$ is noetherian.}}

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

{{Template:BOOKTEMPLATE/Theorem |label=local equiv. |theorem=Proposition |claim=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 {{Template:BOOKTEMPLATE/ref|Jacobson char}}. (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 $𝔦$ be the set of non-units. If $𝔪◃A$ is maximal, it consists of nonunits; thus, $𝔪\subset 𝔦$ where we have the equality by the maximality of $𝔪$.}}

{{Template:BOOKTEMPLATE/Theorem |label= |theorem=Example |claim=If $p$ is a prime ideal, then $A_p$ is a local ring where $p$ is its unique maximal ideal.}}

{{Template:BOOKTEMPLATE/Theorem |label= |theorem=Example |claim=If $\sqrt{𝔦}$ is maximal, then $A/𝔦$ is a local ring. In particular, $A/{𝔪}^n,(n\ge 1)$is local for any maximal ideal $𝔪$.}}

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

A. Lemma 

• (i) Let $𝔦$ be a proper ideal of $A$. If $M$ is a finite generated $𝔦$-module, then $M=0$.
• (ii) The intersection of all ${𝔪}^k$ 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 $𝔦$,

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 $𝔪$, which contains every non-units, in particular, $a_1$, and thus $1\in 𝔪$, which is nonsense. Thus we find that actually x_2, ..., x_n generates $M$; this contradicts the inductive hypothesis.$◻$

An ideal $𝔮◃A$ is said to be primary if every zero-divisor in $A/𝔮$ is nilpotent. Explicitly, this means that, whenever $xy\in 𝔮$ and $y\in ̸𝔮$, $x\in \sqrt{𝔮}$. In particular, a prime ideal is primary.

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label= |claim=If $𝔮$ is primary, then $\sqrt{𝔮}$ is prime. Conversely, if $\sqrt{𝔮}$ is maximal, then $𝔮$ is primary. |proof=The first part is clear. Conversely, if $\sqrt{𝔮}$ is maximal, then $𝔪=\sqrt{𝔮}/𝔮$ is a maximal ideal in $A/𝔮$. It must be unique and so $A/𝔮$ is local. In particular, a zero-divisor in $A/𝔮$ is nonunit and so is contained in $𝔪$; hence, nilpotent.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Exercise |label= |claim=$\sqrt{𝔮}$ prime $\Rightarrow ̸𝔮$ primary.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Theorem |label=Primary decomposition |name=Primary decomposition |claim=Let $A$ be a noetherian ring. If $𝔦◃A$, then $𝔦$ 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 $𝔦$ be its maximal element. We can write $𝔦$ as an intersection of two ideals strictly larger than $𝔦$. Indeed, since $𝔦$ is not prime by definition in particular, choose $x\in ̸𝔦$ and $y\in ̸𝔦$ such that $xy\in 𝔦$. As in the proof of Theorem {{Template:BOOKTEMPLATE/ref|principal prime pid}}, we can write: $𝔦=𝔧(𝔦+x)$ where $𝔧$ is the set of all $a\in A$ such that $ax\in 𝔦$. By maximality, $𝔧,𝔦+x\in ̸S$. Thus, they are finite intersections of primary ideals, but then so is $𝔦$, contradiction.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label= |claim=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.

{{Template:BOOKTEMPLATE/Theorem |theorem=Lemma |label=Integral equiv |claim=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). }}

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 $A$.

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label= |claim=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 $b$.}}

{{Template:BOOKTEMPLATE/Theorem |name=Noether normalization |label= |claim=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 $k$.

|proof=}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Exercise |name=Artin-Tate |label=Artin-Tate |claim=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.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Exercise |claim=A ring morphism $f:A\to \mathrm{\Omega }$ (where $\mathrm{\Omega }$ is an algebraically closed field) extends to $F:A[b]\to \mathrm{\Omega }$ (Answer: http://www.math.uiuc.edu/~r-ash/ComAlg/)}}

Noetherian rings

{{Template:BOOKTEMPLATE/Theorem |theorem=Exercise |name= |label= |claim=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.

{{Template:BOOKTEMPLATE/Theorem |name=Hilbert basis |label=Hilbert basis |claim=$A$ 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 $I◃A[T]$. Let $L_n$ be the set of all coefficients of polynomials of degree $\le n$ in $I$. Since $L_n◃A$, 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 $f_{1n},f_{2n},...f_{m_{n}n}$ of $I$ whose coefficients $b_{1n},...b_{m_{n}n}$ generate $L_n$. Let $I^{\prime }$ be an ideal generated by $f_{jn}$ for all $j,n$. We claim $I=I^{\prime }$. It is clear that $I\subset I^{\prime }$. 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 $f-g\in I^{\prime }$. Since $g\in I^{\prime }$, $f\in I^{\prime }$ then. If $n\ge d$, then $a\in L_d$ and the same argument shows $f\in I^{\prime }$.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Exercise |label= |claim=Let $A$ be the ring of continuous functions $f:[0,1]\to [0,1]$.$A$ is not noetherian.}}

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

{{Template:BOOKTEMPLATE/Theorem |label= |claim=${\dim }_k(𝔪/{𝔪}^2)\ge \dim A$ |proof=}}

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

{{Template:BOOKTEMPLATE/Theorem |label= |claim=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 $n=1$.}}

{{Template:BOOKTEMPLATE/Theorem |label= |claim=Let $A$ be a finite-dimensional $k$-algebra. If $A$ is a domain with the field of fractions $K$, then $\dim A={\mathrm{trdeg}}_{k}K$. |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, ${\mathrm{trdeg}}_{k}K=n$.}}

{{Template:BOOKTEMPLATE/Theorem |label= |claim=Let $A$ be a domain with (ACCP). Then $A$ is a UFD if and only if every prime ideal $𝔭$ of height 1 is principal. |proof=($\Rightarrow$) By Theorem {{Template:BOOKTEMPLATE/ref|UFD equiv}}, $𝔭$ contains a prime element $x$. Then

$0\subset (x)\subset 𝔭$

where the second inclusion must be equality since $𝔭$ has height 1. ($\Leftarrow$) In light of Theorem {{Template:BOOKTEMPLATE/ref|UFD equiv}}, it suffices to show that $A$ is a GCD domain. (TODO: complete the proof.)}}

{{Template:BOOKTEMPLATE/Theorem |label= |claim=A regular local ring is a UFD.}}

{{Template:BOOKTEMPLATE/Theorem |label=Krull's intersection theorem |name=Krull's intersection theorem |claim=Let $𝔦◃A$ be a proper ideal. If $A$ is either a noetherian domain or a local ring, then $\bigcap _{n\ge 1}{𝔦}^n=0$.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Theorem |label=noetherian power radical |claim=Let $𝔦◃A$. If $A$ is noetherian,

${\sqrt{𝔦}}^n\subset 𝔦\subset \sqrt{𝔦}$ for some $n$.

In particular, the nilradical of $A$ is nilpotent. |proof=It suffices to prove this when $𝔦=0$. 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.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label=noetherian -> completion noetherian |claim=If $A$ is noetherian, then $\hat{A}$ is noetherian. |proof=}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Corollary |label= |claim=If $A$ is noetherian, then $A[[X]]$ is noetherian.}}

Zariski topology

Given $𝔞◃A$, let $\mathrm{V}(𝔞)=\{𝔭\in \mathrm{Spec}(A)|𝔭\supset 𝔞\}$. (Note that $\mathrm{V}(𝔞)=\mathrm{V}(\sqrt{𝔞})$.) It is easy to see

$V(𝔞)\cup V(𝔟)=V(𝔞𝔟)=V(𝔞\cap 𝔟)$, and ${\cap }_{\alpha }V({𝔞}_{\alpha })=V(({𝔞}_{\alpha }|\alpha ))$.

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

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label=Spec quasi-compact |claim=We have:

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

|proof=We have: $X_f\subset \bigcup _{\alpha }{X}_{f_{\alpha }}=X\mathrm{\setminus }V((f_{\alpha }|\alpha ))\iff (f)\subset (f_{\alpha }|\alpha )\iff f\in (f_{{\alpha }_1},...,f_{{\alpha }_n})$.}}

{{Template:BOOKTEMPLATE/Theorem |label= |theorem=Exercise |claim=Let $A$ be a local ring. Then $\mathrm{Spec}(A)$ is connected.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Corollary |label= |claim=$\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ is a closed surjection.}}

{{Template:BOOKTEMPLATE/Theorem |label=locally noetherian -> noetherian |claim=If $A_m$ is noetherian for every maximal ideal $𝔪$ and if $\{𝔪\in \mathrm{Max}(A)|x\in m\}$ is finite for each $x\in A$, then $A$ is noetherian.}}

Integrally closed domain

{{Template:BOOKTEMPLATE/Theorem |theorem=Lemma |label=GCD prop |claim=In a GCD domain, if $(x,y)=1=(x,z)$, then $(x,yz)=1$.}}

{{Template:BOOKTEMPLATE/Theorem |theorem=Proposition |label=In GCD, irreducible -> prime |claim=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 $x|y$.}}

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

{{Template:BOOKTEMPLATE/Theorem |label=ACC |name= |claim=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.}}

{{Template:BOOKTEMPLATE/Theorem |label=UFD equiv |name= |claim=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 $𝔭\in \mathrm{Spec}(A)$. If $𝔭$ is nonzero, it then contains a nonzero element x, which we factor into irreducibles: $x=p_1...p_n$. Then $p_j\in 𝔭$ 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 {{Template:BOOKTEMPLATE/label|disjoint maximal ideal}} (i.e., closed under multiplication). Suppose, on the contrary, there is a nonzero nonunit $x$. It is easy to see that since $x\in ̸S$, $(x)$ and $S$ are disjoint. Thus, by Theorem {{Template:BOOKTEMPLATE/label|disjoint maximal ideal}}, there is a prime ideal $𝔭$ containing $x$ and disjoint from $S$. But, by (2), $𝔭$ contains a prime element $y$; that is, $𝔭$ intersects $S$, contradiction. (1) $\Rightarrow$ (3): By uniqueness of factorization, it is clear that $A$ is a GCD domain. }}

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

{{Template:BOOKTEMPLATE/Theorem |label= |name= |theorem=Corollary |claim=If $A$ is a UFD, then $A[X]$ is a UFD. If A is a principal ideal domain, then $A[[X]]$ is a UFD.}}

{{Template:BOOKTEMPLATE/Theorem |label=Nagata UFD equiv |name=Nagata criterion |claim=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.}}