# 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.
${\mathfrak {i}},{\mathfrak {j}}\triangleleft A$
* (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.
Exercise.
Proposition A.2.
$A\neq 0$
$A$
$A$
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.
${\mathfrak {p}}\triangleleft A$
${\mathfrak {p}}$
    ${\mathfrak {p}}\subsetneq {\mathfrak {a}}\triangleleft A,{\mathfrak {p}}\subsetneq {\mathfrak {b}}\triangleleft A$ 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).
$S\subset A$
${\mathfrak {a}}\triangleleft A$
$S$
${\mathfrak {p}}\supset {\mathfrak {a}}$
$S$
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 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.
$A$
$K$
$0\neq f\in A$
$K=A[f^{-1}]$
$A$
$f$
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.
$A$
${\mathfrak {a}}\triangleleft A$
${\sqrt {\mathfrak {a}}}$
${\mathfrak {a}}$
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.
${\mathfrak {a}}\triangleleft A$
${\mathfrak {a}}$
$A[X]$
Theorem.
* For any ${\mathfrak {a}}\triangleleft A$, ${\mathfrak {a}}$ is the intersection of all maximal ideals containing ${\mathfrak {a}}$.
* Every Goldman ideal is maximal.
* 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.
$A\subset B$
$B$
$A$
$A$
$B$
Proof. Let $K\subset L$ be the fields of fractions of $A$ and $B$, respectively. $\square$
Theorem A.19.
$A$
$A[X]$
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. $\square$
Theorem A.5 (prime avoidance).
${\mathfrak {p}}_{1},...,{\mathfrak {p}}_{r}\triangleleft A$
${\mathfrak {a}}\triangleleft A$
${\mathfrak {a}}\subset \bigcup _{1}^{r}{\mathfrak {p}}_{i}$
${\mathfrak {a}}\subset {\mathfrak {p}}_{i}$
${\mathfrak {i}}$
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 $i\leq r$. 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.
$x$
$y$
$(x)\cap (y)=(xy)$
$(x,y)$
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).
${\mathfrak {a}}_{1},...,{\mathfrak {a}}_{n}\triangleleft A$
${\mathfrak {a}}_{j}+{\mathfrak {a}}_{i}=(1)$
    $\prod {\mathfrak {a}}_{i}\to A\to A/{\mathfrak {a}}_{1}\times \cdots \times {\mathfrak {a}}_{n}\to 0$
The Jacobson radical of a ring $A$ is the intersection of all maximal ideals.
Proposition A.6.
$x\in A$
$1-xy$
$y\in A$
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.
$A[X]$
Theorem A.17 (Hopkins).
* A is artinian
* A is noetherian and every prime ideal is maximal.
* $\operatorname {Spec} (A)$ 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.
$A$
* $A$ is local.
* For every $x\in A$, either $x$ or $1-x$ is a unit.
* 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.
$p$
$A_{p}$
$p$
Example.
${\sqrt {\mathfrak {i}}}$
$A/{\mathfrak {i}}$
$A/{\mathfrak {m}}^{n},(n\geq 1)$
${\mathfrak {m}}$
Let $(A,{\mathfrak {m}})$ be a local noetherian ring.
* (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.
${\mathfrak {q}}$
${\sqrt {\mathfrak {q}}}$
${\sqrt {\mathfrak {q}}}$
${\mathfrak {q}}$
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.
${\sqrt {\mathfrak {q}}}$
$\not \Rightarrow {\mathfrak {q}}$
Theorem A.8 (Primary decomposition).
$A$
${\mathfrak {i}}\triangleleft A$
${\mathfrak {i}}$
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.
$(0)$

## 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.
$b\in B$
* $b$ is integral over $A$.
* $A[b]$ is finite over $A$.
* $A[b]$ 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.
$f:A\to B$
$B$
* (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).
$A$
$k$
$z_{1},...,z_{d}$
* $A$ is integral over $k[z_{1},...,z_{d}]$.
* $z_{1},...,z_{d}$ are algebraically independent over $k$.
* $z_{1},...,z_{d}$ are a separating transcendence basis of the field of fractions $K$ of $A$ if $K$ is separable over $k$.
Exercise A.10 (Artin-Tate).
$A\subset B\subset C$
$A$
$C$
$A$
$B$
$B$
$A$
Exercise.
$f:A\to \Omega$
$\Omega$
$F:A[b]\to \Omega$

## Noetherian rings

Exercise.
The next theorem furnishes many examples of a noetherian ring.
Theorem A.7 (Hilbert basis).
$A$
$A[T_{1},...T_{n}]$
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.
$A$
$f:[0,1]\to [0,1]$
$A$
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.
$A$
$\dim A[T_{1},...,T_{n}]=n+\dim A$
Proof. By induction, it suffices to prove the case $n=1$. $\square$
Theorem.
$A$
$k$
$A$
$K$
$\dim A=\operatorname {trdeg} _{k}K$
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.
$A$
$A$
${\mathfrak {p}}$
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.
Theorem A.10 (Krull's intersection theorem).
${\mathfrak {i}}\triangleleft A$
$A$
$\bigcap _{n\geq 1}{\mathfrak {i}}^{n}=0$
Theorem A.15.
${\mathfrak {i}}\triangleleft A$
$A$
    ${\sqrt {\mathfrak {i}}}^{n}\subset {\mathfrak {i}}\subset {\sqrt {\mathfrak {i}}}$ for some $n$.
$A$
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.
$A$
${\hat {A}}$
Corollary.
$A$
$A[[X]]$

## 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.
* (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.
$A$
$\operatorname {Spec} (A)$
Corollary.
$\operatorname {Spec} (B)\to \operatorname {Spec} (A)$
Theorem A.12.
$A_{m}$
${\mathfrak {m}}$
$\{{\mathfrak {m}}\in \operatorname {Max} (A)|x\in m\}$
$x\in A$
$A$

## Integrally closed domain

Lemma A.8.
$(x,y)=1=(x,z)$
$(x,yz)=1$
Proposition A.9.
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).
$A$
Theorem A.10.
$A$
* Every nonzero nonunit element is a finite product of prime elements.
* (Kaplansky) Every nonzero prime ideal contains a prime element.
* $A$ 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.
$A$
$A[X]$
$A[[X]]$
Theorem A.13 (Nagata criterion).
$S\subset A$
$A$
$S^{-1}A$