Topics in Abstract Algebra/Commutative algebra

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

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

Proposition A.14.

Let ${\displaystyle {\mathfrak {i}},{\mathfrak {j}}\triangleleft A}$.

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

Proof. Routine. ${\displaystyle \square }$

Exercise.

Exercise.

Proposition A.2.

Let ${\displaystyle A\neq 0}$ be a ring. If every principal ideal in ${\displaystyle A}$ is prime, then ${\displaystyle A}$ is a field.

Proof. Let ${\displaystyle 0\neq x\in A}$. Since ${\displaystyle x^{2}}$ is in ${\displaystyle (x^{2})}$, which is prime, ${\displaystyle x\in (x^{2})}$. Thus, we can write ${\displaystyle x=ax^{2}}$. Since ${\displaystyle (0)}$ is prime, ${\displaystyle A}$ is a domain. Hence, ${\displaystyle 1=ax}$. ${\displaystyle \square }$

Lemma.

Let ${\displaystyle {\mathfrak {p}}\triangleleft A}$. Then ${\displaystyle {\mathfrak {p}}}$ is prime if and only if

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

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

Theorem A.11 (multiplicative avoidance).

Let ${\displaystyle S\subset A}$ be a multiplicative system. If ${\displaystyle {\mathfrak {a}}\triangleleft A}$ is disjoint from ${\displaystyle S}$, then there exists a prime ideal ${\displaystyle {\mathfrak {p}}\supset {\mathfrak {a}}}$ that is maximal among ideals disjoint from ${\displaystyle S}$.

Proof. Let ${\displaystyle {\mathfrak {m}}}$ be a maximal element in the set of all ideals disjoint from ${\displaystyle S}$. Let ${\displaystyle {\mathfrak {a}}}$ and ${\displaystyle {\mathfrak {b}}}$ be ideals strictly larger than ${\displaystyle {\mathfrak {m}}}$. Since ${\displaystyle {\mathfrak {m}}}$ is maximal, we find ${\displaystyle a\in {\mathfrak {a}}\cap S}$ and ${\displaystyle b\in {\mathfrak {b}}\cap S}$. By the definition of ${\displaystyle S}$, ${\displaystyle ab\in S}$; thus, ${\displaystyle {\mathfrak {a}}{\mathfrak {b}}\not \subset {\mathfrak {m}}}$. By the lemma, ${\displaystyle {\mathfrak {m}}}$ is prime then. ${\displaystyle \square }$

Note that the theorem applies in particular when ${\displaystyle S}$ contains only 1.

Exercise.

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

Lemma.

Let ${\displaystyle A}$ be a domain with the field of fractions ${\displaystyle K}$, and ${\displaystyle 0\neq f\in A}$. Then ${\displaystyle K=A[f^{-1}]}$ if and only if every nonzero prime ideal of ${\displaystyle A}$ contains ${\displaystyle f}$.

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

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

Theorem A.21.

Let ${\displaystyle A}$ be a ring and ${\displaystyle {\mathfrak {a}}\triangleleft A}$. Then ${\displaystyle {\sqrt {\mathfrak {a}}}}$ is the intersection of all minimal Goldman ideals of A containing ${\displaystyle {\mathfrak {a}}}$

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

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

Lemma.

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

Theorem.

Proof. Clear. ${\displaystyle \square }$

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

Lemma.

Let ${\displaystyle A\subset B}$ be domains such that ${\displaystyle B}$ is algebraic and of finite type over ${\displaystyle A}$. Then ${\displaystyle A}$ is a Goldman domain if and only if ${\displaystyle B}$ is a Goldman domain.

Proof. Let ${\displaystyle K\subset L}$ be the fields of fractions of ${\displaystyle A}$ and ${\displaystyle B}$, respectively. ${\displaystyle \square }$

Theorem A.19.

Let ${\displaystyle A}$ be a Hilbert-Jacobson ring. Then ${\displaystyle A[X]}$ is a Hilbert-Jacobson ring.

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

Theorem A.5 (prime avoidance).

Let ${\displaystyle {\mathfrak {p}}_{1},...,{\mathfrak {p}}_{r}\triangleleft A}$ be ideals, at most two of which are not prime, and ${\displaystyle {\mathfrak {a}}\triangleleft A}$. If ${\displaystyle {\mathfrak {a}}\subset \bigcup _{1}^{r}{\mathfrak {p}}_{i}}$, then ${\displaystyle {\mathfrak {a}}\subset {\mathfrak {p}}_{i}}$ for some ${\displaystyle {\mathfrak {i}}}$.

Proof. We shall induct on ${\displaystyle r}$ to find ${\displaystyle a\in {\mathfrak {a}}}$ that is in no ${\displaystyle {\mathfrak {p}}_{i}}$. The case ${\displaystyle r=1}$ being trivial, suppose we find ${\displaystyle a\in {\mathfrak {a}}}$ such that ${\displaystyle a\not \in {\mathfrak {p}}_{i}}$ for ${\displaystyle i<r}$. We assume ${\displaystyle a\in {\mathfrak {p}}_{r}}$; else, we're done. Moreover, if ${\displaystyle {\mathfrak {p}}_{i}\subset {\mathfrak {p}}_{r}}$ for some ${\displaystyle i<r}$, then the theorem applies without ${\displaystyle {\mathfrak {p}}_{i}}$ and so this case is done by by the inductive hypothesis. We thus assume ${\displaystyle {\mathfrak {p}}_{i}\not \subset {\mathfrak {p}}_{r}}$ for all ${\displaystyle i<r}$. Now, ${\displaystyle {\mathfrak {a}}{\mathfrak {p}}_{1}...{\mathfrak {p}}_{r-1}\not \subset {\mathfrak {p}}_{r}}$; if not, since ${\displaystyle {\mathfrak {p}}_{r}}$ is prime, one of the ideals in the left is contained in ${\displaystyle {\mathfrak {p}}_{r}}$, contradiction. Hence, there is ${\displaystyle b}$ in the left that is not in ${\displaystyle {\mathfrak {p}}_{r}}$. It follows that ${\displaystyle a+b\not \in {\mathfrak {p}}_{i}}$ for all ${\displaystyle i\leq r}$. Finally, we remark that the argument works without assuming ${\displaystyle {\mathfrak {p}}_{1}}$ and ${\displaystyle {\mathfrak {p}}_{2}}$ are prime. (TODO: too sketchy.) The proof is thus complete. ${\displaystyle \square }$

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

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

Proposition.

Suppose: for every ${\displaystyle x}$ and ${\displaystyle y}$, ${\displaystyle (x)\cap (y)=(xy)}$ whenever (1) is the only principal ideal containing ${\displaystyle (x,y)}$. Then every irreducible is a prime.

Proof. Let ${\displaystyle p}$ be an irreducible, and suppose ${\displaystyle p|xy}$ and ${\displaystyle p\not |y}$. Since ${\displaystyle (p)\cap (x)=(px)}$ implies ${\displaystyle px|xy}$ and ${\displaystyle p|y}$, there is a ${\displaystyle d}$ such that ${\displaystyle (1)\neq (d)\supset (p,x)}$. But then ${\displaystyle d|p}$ and so ${\displaystyle p|d}$ (${\displaystyle p}$ is an irreducible.) Thus, ${\displaystyle p|x}$. ${\displaystyle \square }$

Theorem A.16 (Chinese remainder theorem).

Let ${\displaystyle {\mathfrak {a}}_{1},...,{\mathfrak {a}}_{n}\triangleleft A}$. If ${\displaystyle {\mathfrak {a}}_{j}+{\mathfrak {a}}_{i}=(1)}$, then

${\displaystyle \prod {\mathfrak {a}}_{i}\to A\to A/{\mathfrak {a}}_{1}\times \cdots \times {\mathfrak {a}}_{n}\to 0}$

is exact.

The Jacobson radical of a ring ${\displaystyle A}$ is the intersection of all maximal ideals.

Proposition A.6.

${\displaystyle x\in A}$ is in the Jacobson radical if and only if ${\displaystyle 1-xy}$ is a unit for every ${\displaystyle y\in A}$.

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

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 ${\displaystyle A[X]}$, the nilradical and the Jacobson radical coincide.

Theorem A.17 (Hopkins).

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

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

Proposition A.17.

Let ${\displaystyle A}$ be a nonzero ring. The following are equivalent.

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

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

Example.

If ${\displaystyle p}$ is a prime ideal, then ${\displaystyle A_{p}}$ is a local ring where ${\displaystyle p}$ is its unique maximal ideal.

Example.

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

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

A. Lemma 

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

Proof: We prove (i) by the induction on the number of generators. Suppose ${\displaystyle M}$ cannot be generated by strictly less than ${\displaystyle n}$ generators, and suppose we have ${\displaystyle x_{1},...x_{n}}$ that generates ${\displaystyle M}$. Then, in particular,

${\displaystyle x_{1}=a_{1}x_{1}+a_{2}x_{2}+...+a_{n}x_{n}}$ where ${\displaystyle a_{i}}$ are in ${\displaystyle {\mathfrak {i}}}$,

and thus

${\displaystyle (1-a_{1})x_{1}=a_{2}x_{2}+...+a_{n}x_{n}}$

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

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

Proposition.

If ${\displaystyle {\mathfrak {q}}}$ is primary, then ${\displaystyle {\sqrt {\mathfrak {q}}}}$ is prime. Conversely, if ${\displaystyle {\sqrt {\mathfrak {q}}}}$ is maximal, then ${\displaystyle {\mathfrak {q}}}$ is primary.

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

Exercise.

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

Theorem A.8 (Primary decomposition).

Let ${\displaystyle A}$ be a noetherian ring. If ${\displaystyle {\mathfrak {i}}\triangleleft A}$, then ${\displaystyle {\mathfrak {i}}}$ is a finite intersection of primary ideals.

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

Proposition.

If ${\displaystyle (0)}$ is indecomposable, then the set of zero divisors is a union of minimal primes.

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

Lemma A.9.

Let ${\displaystyle b\in B}$. Then the following are equivalent.

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

Proof. (1) means that we can write:

${\displaystyle b^{n+r}=-(b^{r+n-1}a_{n-1}+...b^{r+1}a_{1}+b^{r}a_{0})}$

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

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

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

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 ${\displaystyle B}$ containing ${\displaystyle A}$. (Proof: if ${\displaystyle x}$ and ${\displaystyle y}$ are integral elements, then ${\displaystyle A[xy]}$ and ${\displaystyle A[x-y]}$ are contained in ${\displaystyle A[x,y]}$, finite over ${\displaystyle A}$.) It is also clear that integrability is transitive; that is, if ${\displaystyle C}$ is integral over ${\displaystyle B}$ and ${\displaystyle B}$ is integral over ${\displaystyle A}$, then ${\displaystyle C}$ is integral over ${\displaystyle A}$.

Proposition.

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

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

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

${\displaystyle x^{-n}=-(a_{n-1}x^{-(n-1)}+...+a_{1}x^{-1}+a_{0})}$

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

${\displaystyle -a_{0}=b^{n}+a_{n-1}b^{n-1}+...+a_{1}b=b(b^{n-1}+a_{n-1}b^{n-2}+...+a_{1})}$.

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

Theorem (Noether normalization).

Let ${\displaystyle A}$ be a finitely generated ${\displaystyle k}$-algebra. Then we can find ${\displaystyle z_{1},...,z_{d}}$ such that

1. ${\displaystyle A}$ is integral over ${\displaystyle k[z_{1},...,z_{d}]}$.
2. ${\displaystyle z_{1},...,z_{d}}$ are algebraically independent over ${\displaystyle k}$.
3. ${\displaystyle z_{1},...,z_{d}}$ are a separating transcendence basis of the field of fractions ${\displaystyle K}$ of ${\displaystyle A}$ if ${\displaystyle K}$ is separable over ${\displaystyle k}$.

Exercise A.10 (Artin-Tate).

Let ${\displaystyle A\subset B\subset C}$ be rings. Suppose ${\displaystyle A}$ is noetherian. If ${\displaystyle C}$ is finitely generated as an ${\displaystyle A}$-algebra and integral over ${\displaystyle B}$, then ${\displaystyle B}$ is finitely generated as an ${\displaystyle A}$-algebra.

Exercise.

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

Exercise.

The next theorem furnishes many examples of a noetherian ring.

Theorem A.7 (Hilbert basis).

${\displaystyle A}$ is a noetherian ring if and only if ${\displaystyle A[T_{1},...T_{n}]}$ is noetherian.

Proof. By induction it suffices to prove ${\displaystyle A[T]}$ is noetherian. Let ${\displaystyle I\triangleleft A[T]}$. Let ${\displaystyle L_{n}}$ be the set of all coefficients of polynomials of degree ${\displaystyle \leq n}$ in ${\displaystyle I}$. Since ${\displaystyle L_{n}\triangleleft A}$, there exists ${\displaystyle d}$ such that

${\displaystyle L_{0}\subset L_{1}\subset L_{2},...,\subset L_{d}=L_{d+1}=...}$.

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

${\displaystyle a=a_{1}b_{1n}+a_{2}b_{2n}+...+a_{m_{n}}b_{{m_{n}}n}}$

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

Exercise.

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

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

Theorem.

${\displaystyle \dim _{k}({\mathfrak {m}}/{\mathfrak {m}}^{2})\geq \dim A}$

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

Theorem.

Let ${\displaystyle A}$ be a noetherian ring. Then ${\displaystyle \dim A[T_{1},...,T_{n}]=n+\dim A}$.

Proof. By induction, it suffices to prove the case ${\displaystyle n=1}$. ${\displaystyle \square }$

Theorem.

Let ${\displaystyle A}$ be a finite-dimensional ${\displaystyle k}$-algebra. If ${\displaystyle A}$ is a domain with the field of fractions ${\displaystyle K}$, then ${\displaystyle \dim A=\operatorname {trdeg} _{k}K}$.

Proof. By the noether normalization lemma, ${\displaystyle A}$ is integral over ${\displaystyle k[x_{1},...,x_{n}]}$ where ${\displaystyle x_{1},...,x_{n}}$ are algebraically independent over ${\displaystyle k}$. Thus, ${\displaystyle \dim A=\dim k[x_{1},...,x_{n}]=n}$. On the other hand, ${\displaystyle \operatorname {trdeg} _{k}K=n}$. ${\displaystyle \square }$

Theorem.

Let ${\displaystyle A}$ be a domain with (ACCP). Then ${\displaystyle A}$ is a UFD if and only if every prime ideal ${\displaystyle {\mathfrak {p}}}$ of height 1 is principal.

Proof. (${\displaystyle \Rightarrow }$) By Theorem A.10, ${\displaystyle {\mathfrak {p}}}$ contains a prime element ${\displaystyle x}$. Then

${\displaystyle 0\subset (x)\subset {\mathfrak {p}}}$

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

Theorem.

Theorem A.10 (Krull's intersection theorem).

Let ${\displaystyle {\mathfrak {i}}\triangleleft A}$ be a proper ideal. If ${\displaystyle A}$ is either a noetherian domain or a local ring, then ${\displaystyle \bigcap _{n\geq 1}{\mathfrak {i}}^{n}=0}$.

Theorem A.15.

Let ${\displaystyle {\mathfrak {i}}\triangleleft A}$. If ${\displaystyle A}$ is noetherian,

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

In particular, the nilradical of ${\displaystyle A}$ is nilpotent.

Proof. It suffices to prove this when ${\displaystyle {\mathfrak {i}}=0}$. Thus, the proof reduces to proving that the nilradical of A is nilpotent. Since ${\displaystyle A}$ is nilpotent, we have finitely many nilpotent elements ${\displaystyle x_{1},...,x_{n}}$ that spans ${\displaystyle {\sqrt {(0)}}}$. The power of any linear combination of them is then a sum of terms that contain the high power of some ${\displaystyle x_{j}}$ if we take the sufficiently high power. Thus, ${\displaystyle {\sqrt {(0)}}}$ is nilpotent. ${\displaystyle \square }$

Proposition A.8.

If ${\displaystyle A}$ is noetherian, then ${\displaystyle {\hat {A}}}$ is noetherian.

Corollary.

If ${\displaystyle A}$ is noetherian, then ${\displaystyle A[[X]]}$ is noetherian.

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

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

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

Proposition A.16.