Commutative Algebra

Seminar Note

Author

Liyve

Published

June 25, 2025

Modified

June 25, 2025

Abstract

Note about [1, Introduction to Commutative Algebra].

1 Rings and Ideals

1.1 Rings and Ring Homomorphisms

Definition 1 (Ring) A ring $A$ is a set with two binary operations, usually called addition and multiplication, such that:

$(A, +)$ is an abelian group,
$(A, \cdot)$ is a semigroup,
Multiplication is distributive over addition: for all $a, b, c \in A$, $a \cdot (b + c) = a \cdot b + a \cdot c$ and $(a + b) \cdot c = a \cdot c + b \cdot c$.
Multiplication is commutative: for all $a, b \in A$, $a \cdot b = b \cdot a$.
There exists a multiplicative identity $1 \in A$ such that for all $a \in A$, $a \cdot 1 = 1 \cdot a = a$.

Definition 2 (Ring Homomorphism) A ring homomorphism is a mapping $f : A \to B$ between rings $A$ and $B$ such that for all $a, a' \in A$:

$f(a + a') = f(a) + f(a')$,
$f(a \cdot a') = f(a) \cdot f(a')$,
$f(1_A) = 1_B$.

1.2 Ideals and Quotient Rings

Definition 3 (Ideal) An ideal $\mathfrak{a}$ of a ring $A$ is a subset $\mathfrak{a}\subseteq A$ such that:

$(\mathfrak{a}, +)$ is a subgroup of $(A, +)$,
For all $a \in \mathfrak{a}$ and $r \in A$, both $r a$ and $a r$ are in $\mathfrak{a}$ (i.e., $\mathfrak{a}$ is closed under multiplication by elements of $A$).

Definition 4 (Quotient Ring) The quotient ring $A/\mathfrak{a}$ is defined as follows: Let $A$ be a ring and $\mathfrak{a}$ an ideal of $A$. The set of cosets

\[ A/\mathfrak{a}= \{ a + \mathfrak{a}\mid a \in A \} \]

forms a ring with operations defined by

\[ (a + \mathfrak{a}) + (b + \mathfrak{a}) = (a + b) + \mathfrak{a}, \quad (a + \mathfrak{a}) \cdot (b + \mathfrak{a}) = (a b) + \mathfrak{a}. \]

The natural projection $\pi: A \to A/\mathfrak{a}$ given by $\pi(a) = a + \mathfrak{a}$ is a surjective ring homomorphism with kernel $\mathfrak{a}$.

Proposition 1 (Correspondence of Ideals) Let $A$ be a ring and $\mathfrak{a}\vartriangleleft A$ an ideal. There is a bijective correspondence between the set of ideals of $A$ containing $\mathfrak{a}$ and the set of ideals of the quotient ring $A/\mathfrak{a}$.

Explicitly, for each ideal $\mathfrak{b}$ of $A$ with $\mathfrak{a}\subseteq \mathfrak{b}$, the image $\bar{\mathfrak{b}} = \mathfrak{b}/\mathfrak{a}$ is an ideal of $A/\mathfrak{a}$. Conversely, for each ideal $\bar{\mathfrak{b}}$ of $A/\mathfrak{a}$, its preimage under the natural projection $\pi: A \to A/\mathfrak{a}$ is an ideal of $A$ containing $\mathfrak{a}$.

This correspondence preserves inclusion, sums, intersections, and properties such as being prime or maximal (with suitable conditions). \[ \{\mathfrak{b}\vartriangleleft A \mid \mathfrak{a}\subseteq \mathfrak{b}\} \leftrightarrow \{\bar{\mathfrak{b}}\vartriangleleft A/\mathfrak{a}\} \]

Definition 5 (Kernel) Let $f: A \to B$ be a ring homomorphism. The kernel of $f$, denoted $\ker f$, is the set

\[ \ker f = \{ a \in A \mid f(a) = 0_B \} \] where $0_B$ is the additive identity in $B$. The kernel $\ker f$ is an ideal of $A$.

Definition 6 (Image) Let $f: A \to B$ be a ring homomorphism. The image of $f$, denoted $\operatorname{Im}f$, is the set

\[ \operatorname{Im}f = \{ f(a) \mid a \in A \} \] which is a subring of $B$.

1.3 Zero-Divisors, Nilpotent Elements and Units

Definition 7 (Zero Divisor) Let $A$ be a ring. An element $a \in A$, $a \neq 0$, is called a zero-divisor if there exists a nonzero $b \in A$ such that $a b = 0$ or $b a = 0$.

Definition 8 (Integral Domain) A ring $A$ is called an integral domain if $A \neq \{0\}$ and $A$ has no zero-divisors; that is, for all $a, b \in A$, if $a b = 0$, then either $a = 0$ or $b = 0$.

Definition 9 (Nilpotent) Let $A$ be a ring. An element $a \in A$ is called nilpotent if there exists a positive integer $n$ such that $a^n = 0$.

Definition 10 (Unit) An element $u \in A$ of a ring $A$ is called a unit if there exists $v \in A$ such that $u v = v u = 1$, where $1$ is the multiplicative identity in $A$. The set of all units in $A$ is denoted by $A^\times$.

Definition 11 (Principal Ideal) An ideal $\mathfrak{a}$ of a ring $A$ is called a principal ideal if there exists an element $a \in A$ such that \[ \mathfrak{a}= (a) = \{ r a \mid r \in A \}. \] That is, $\mathfrak{a}$ is generated by a single element $a$.

Proposition 2 Let $A\neq 0$ , then TFAE:

A is a field
the only ideals in $A$ are $(0)$ and $A(=(1))$.
$\forall f:A \to B\neq 0$ is injective.

Proof

Proof.

$(1)\implies (2)$ :: Let $\mathfrak{a}\vartriangleleft A$. If $\mathfrak{a}\neq 0$, then $\exists x$ is a unit ,$x\in \mathfrak{a}$
$(2)\implies (3)$ :: The kernel $\ker f$ is either $\{0\}$ or $A$. If $\ker f = A$, then $f$ is the zero map, so $\operatorname{Im}f = \{0\}$, contradicting $B \neq 0$. Thus, $\ker f = \{0\}$, so $f$ is injective.
$(3)\implies (1)$ :: Let $x$ be not a unit. $(x)\neq (1)$. Let $B=A/(x)$, $f(x)=0\implies x=0$.

1.4 Prime Ideals and Maximal Ideals

Definition 12 (Prime Ideal) An ideal $\mathfrak{p}$ in $A$ is prime if $\mathfrak{p}\neq(1)$ and if $xy\in \mathfrak{p}\implies x\in \mathfrak{p}\text{ or } y\in \mathfrak{p}$.

Definition 13 (Maximal Ideal) An ideal $\mathfrak{m}$ in $A$ is maximal if $\mathfrak{m}\neq(1)$ and there is no ideal $\mathfrak{a}$ s.t. $\mathfrak{m}\subsetneq\mathfrak{a}\subsetneq(1)$.

Proposition 3

$\mathfrak{p}$ is prime ideal $\Leftrightarrow$ $A/\mathfrak{p}$ is integral domain.
$\mathfrak{m}$ is maximal ideal $\Leftrightarrow$ $A/\mathfrak{m}$ is field. Hence maximal ideals are prime.
Let $f:A\to B$ is ring homomorphism. $\mathfrak{p}$ is a prime ideal in $B$, then $f^{-1}(\mathfrak{p})$ is prime in $A$.

Proof

Proof.

$(1)(2)$ :: Omitted. cf.[2, Ch. 3, Sec.4, p.110, thm.7, thm.8]
$(3)$ :: You can consider the preimage $f^{-1}(\mathfrak{p}) = \{ a \in A \mid f(a) \in \mathfrak{p}\}$. If $xy \in f^{-1}(\mathfrak{p})$, then $f(xy) = f(x)f(y) \in \mathfrak{p}$. Since $\mathfrak{p}$ is prime, $f(x) \in \mathfrak{p}$ or $f(y) \in \mathfrak{p}$, so $x \in f^{-1}(\mathfrak{p})$ or $y \in f^{-1}(\mathfrak{p})$.

In particular, you can consider $A/f^{-1}(\mathfrak{p})\cong B/\mathfrak{p}$.

Remark

Remark. Note that if $\mathfrak{m}\vartriangleleft B$ is maximal, then $f^{-1}(\mathfrak{m})$ is a maximal ideal of $A$ if $f$ is surjective. In general, the preimage of a maximal ideal under a ring homomorphism need not be maximal unless the map is surjective.

Let $f: \mathbb{Z} \to \mathbb{Q}$ be the natural embedding, $\mathfrak{m}=(0)$. $\mathbb{Q}$ is a field, $\mathfrak{m}$ is maximal, but its preimage $f^{-1}(\mathfrak{m})=(0)$ in $\mathbb{Z}$ is properly contained in $(p)$, for any $p\in \mathbb{N}$.

Lemma 1 (Zorn’s lemma) Let $S$ be a non-empty partially ordered set such that every chain (i.e., totally ordered subset) in $S$ has an upper bound in $S$. Then $S$ contains at least one maximal element; that is, there exists $m \in S$ such that if $m \leq s$ for some $s \in S$, then $m = s$.

Theorem 1 (Existence of Maximal Ideals) Every nonzero ring $A$ with $1$ has at least one maximal ideal.

Proof

Proof. Let $S$ be the set of all proper ideals of $A$, partially ordered by inclusion. $S$ is nonempty since $(0)$ is a proper ideal (as $A \neq 0$). Any chain of ideals in $S$ has an upper bound given by the union of the chain, which is again a proper ideal. By Zorn’s Lemma, $S$ has a maximal element, which is a maximal ideal of $A$.

Corollary 1 (Every Ideal is Contained in a Maximal Ideal) If $\mathfrak{a}$ be a proper ideal of $A$, then $\exists \mathfrak{m}$ is maximal, s.t. $\mathfrak{a}\subseteq \mathfrak{m}$.

Proof

Proof. Let $\mathfrak{a}$ be a proper ideal of $A$ (i.e., $\mathfrak{a}\neq (1)$). Consider the quotient ring $A/\mathfrak{a}$. By the existence of maximal ideals, $A/\mathfrak{a}$ has a maximal ideal $\bar{\mathfrak{m}}$. The preimage $\mathfrak{m}= \pi^{-1}(\bar{\mathfrak{m}})$ under the natural projection $\pi: A \to A/\mathfrak{a}$ is a maximal ideal of $A$ containing $\mathfrak{a}$.

Corollary 2 (Every Non-Unit is Contained in a Maximal Ideal) Every non-unit element of $A$ is contained in some maximal ideal of $A$. Let $a \in A$ be a non-unit. Then the ideal $(a)$ generated by $a$ is a proper ideal, i.e., $(a) \neq (1)$. By the previous corollary, there exists a maximal ideal $\mathfrak{m}$ such that $(a) \subseteq \mathfrak{m}$. Thus, $a \in \mathfrak{m}$.

Proof

Definition 14 (Local Ring) A ring $A$ is called a local ring if it has a unique maximal ideal $\mathfrak{m}$. That is, there exists exactly one maximal ideal in $A$.

Definition 15 (Residue Field) Let $A$ be a local ring with unique maximal ideal $\mathfrak{m}$. The residue field of $A$ is the quotient ring \[ k = A/\mathfrak{m} \] which is a field. The natural projection $A \to k$ is called the residue map.

Proposition 4

Let $A$ be a ring and $\mathfrak{m}\neq(1)$, s.t. $\forall x \in A\setminus\mathfrak{m}$ is a unit. Then $A$ is a local ring, and $\mathfrak{m}$ is maximal.
Let $A$ be a ring and $\mathfrak{m}$ maximal ideal of $A$, s.t. $1+\mathfrak{m}$ is a unit of $A$. Then $A$ is a local ring.

Proof

Proof.

$(1)$ :: Every non-unit is contained in $\mathfrak{m}$. Hence $\mathfrak{m}$ is the only maximal ideal.
$(2)$ :: $\forall \mathfrak{n}\vartriangleleft A$. If $\mathfrak{n}\not\subseteq\mathfrak{m}$, take $x\in \mathfrak{n}\setminus\mathfrak{m}$. $(x)+\mathfrak{m}=(1)$. $\exists y\in A, m\in \mathfrak{m}, xy+m=1 \implies xy=1-m$ is a unit. Then $\mathfrak{n}=(1)$. Contradiction!

Definition 16 (Semi-local Ring) A ring $A$ is called semi-local if $A$ has only finitely many maximal ideals.

Definition 17 (PID) An integral domain $A$ is called a principal ideal domain (PID) if every ideal of $A$ is principal; that is, for every ideal $\mathfrak{a}\subseteq A$, there exists $a \in A$ such that $\mathfrak{a}= (a) = \{ r a \mid r \in A \}$.

Proposition 5 In PID, $\mathfrak{a}$ is prime $\Leftrightarrow$ $\mathfrak{a}$ is maximal.

Proof

Proof. If $(x)\neq(1)$ is prime. Let $(x)\subsetneq(y)$. Then $x\in(y)\implies \exists z$ s.t. $x=yz$. $y\not\in(x)\implies z\in (x)\implies \exists t,\ \text{s.t. } z=xt.$$

1.5 Nilradical and Jacobson Radical

Proposition 6

The set $\mathfrak{N}$ of all nilpotent elements of $A$ is an ideal. \[ \mathfrak{N}= \{ a \in A \mid a \text{ is nilpotent} \} \]
And $A/\mathfrak{N}$ has no non-zero nilpotent element.

Proof

Proof.

$(1)$ :: If $x\in\mathfrak{N}$, then $ax\in \mathfrak{N},\ \text{for}\ \forall\ a\in A$. $\forall x,\ y\in\mathfrak{N},\ \exists m,\ n,\ x^m=y^n=0$, then \[ (x+y)^{m+n-1}=0\implies x+y\in \mathfrak{N}. \]
$(2)$ :: If $\bar{x}^n=0$, $x^n\in \mathfrak{N}\implies \exists k,\ x^{nk}=0\implies x\in\mathfrak{N}\implies\bar{x}=0$.

Definition 18 (Nilradical) The set $\mathfrak{N}$ is called Nilradical of $A$.

Proposition 7 The nilradical $\mathfrak{N}$ of a ring $A$ is equal to the intersection of all prime ideals of $A$

That is, an element $a \in A$ is nilpotent if and only if $a$ belongs to every prime ideal of $A$.

Let \[ \mathfrak{N}' = \bigcap_{\mathfrak{p}\text{ prime}} \mathfrak{p} \] We need to show $\mathfrak{N}=\mathfrak{N}'$

Proof

Proof.

$(\mathfrak{N}\subseteq \mathfrak{N}')$ :: If $x\in \mathfrak{N}$, then $x^n=0\in \mathfrak{p}$ for any $\mathfrak{p}$. It implies $x\in \mathfrak{p}$ for any $\mathfrak{p}$.
$(\mathfrak{N}' \subseteq \mathfrak{N})$ :: Suppose $\forall n>0$, $x^n\neq0$. Let \[ \Sigma=\{\mathfrak{a}\vartriangleleft A \mid x^n\not\in\mathfrak{a}, \forall n>0\}. \] Let $T$ be a totally ordered chain in $\Sigma$. Consider $\mathfrak{a}_T = \bigcup_{\mathfrak{a}\in T} \mathfrak{a}$. We claim that $\mathfrak{a}_T \in \Sigma$.

$\mathfrak{a}_T$ is an ideal: Since $T$ is a chain, the union of the ideals in $T$ is again an ideal.
For all $n > 0$, if $x^n \in \mathfrak{a}_T$, then $x^n \in \mathfrak{a}$ for some $\mathfrak{a}\in T$, contradicting the definition of $\Sigma$.

Thus, every chain in $\Sigma$ has an upper bound, so by Zorn’s Lemma, $\Sigma$ has a maximal element, say $\mathfrak{p}$. We claim that $\mathfrak{p}$ is a prime ideal.

Suppose $a, b \notin \mathfrak{p}$. Then the ideals $\mathfrak{a}_1 = \mathfrak{p}+ (a)$ and $\mathfrak{a}_2 = \mathfrak{p}+ (b)$ strictly contain $\mathfrak{p}$, so by maximality, there exist $n_1, n_2 > 0$ such that $x^{n_1} \in \mathfrak{a}_1$ and $x^{n_2} \in \mathfrak{a}_2$. Thus, \[ x^{n_1} = y_1 + a z_1, \quad x^{n_2} = y_2 + b z_2 \] for some $y_1, y_2 \in \mathfrak{p}$, $z_1, z_2 \in A$. Then \[ x^{n_1 + n_2} = (x^{n_1})(x^{n_2}) = (y_1 + a z_1)(y_2 + b z_2) \] Expanding and using that $\mathfrak{p}$ is an ideal, all terms except $a b z_1 z_2$ are in $\mathfrak{p}$, so \[ x^{n_1 + n_2} - a b z_1 z_2 \in \mathfrak{p}\implies x^{n_1 + n_2} \in \mathfrak{p}+ (a b) \] Thus, $x^{n_1 + n_2} \in \mathfrak{p}+ (a b)$, so by maximality, $x^m \in \mathfrak{p}+ (a b)$ for some $m > 0$, but $x^m \notin \mathfrak{p}$ by construction, so $a b \notin \mathfrak{p}$.

Therefore, $\mathfrak{p}$ is a prime ideal not containing any power of $x$, contradicting $x \in \bigcap_{\mathfrak{p}\text{ prime}} \mathfrak{p}$. Thus, $\mathfrak{N}= \mathfrak{N}'$.

Definition 19 (Jacobson Radical) Let $\mathfrak{R}$ be the intersection of all maximal ideals of $A$: \[ \mathfrak{R}= \bigcap_{\mathfrak{m}\text{ maximal}} \mathfrak{m} \] This ideal is called the Jacobson radical of $A$.

Proposition 8 $x\in\mathfrak{R}\Leftrightarrow 1-xy$ is a unit in $A$ for all $y\in A$

Proof

Proof. $(\implies)$: Suppose $x \in \mathfrak{R}$, but $1 - x y$ is not a unit for some $y \in A$. Then the ideal $(1 - x y)$ is proper, so it is contained in some maximal ideal $\mathfrak{m}$. Thus, $1 - x y \in \mathfrak{m}$. But $x \in \mathfrak{R}\subseteq \mathfrak{m}$, so $x y \in \mathfrak{m}$, hence $1 = (1 - x y) + x y \in \mathfrak{m}$, which is impossible since $\mathfrak{m}$ is proper. Therefore, $1 - x y$ must be a unit for all $y \in A$.

$(\Longleftarrow)$ :: Suppose $x \notin \mathfrak{m}$ for some maximal ideal $\mathfrak{m}$. Then the ideal generated by $x$ and $\mathfrak{m}$ is the whole ring: $(x) + \mathfrak{m}= (1)$. This means there exist $y \in A$ and $t \in \mathfrak{m}$ such that $x y + t = 1$, or equivalently, $1 - x y = t \in \mathfrak{m}$. Since $\mathfrak{m}$ is maximal, $1 - x y$ is not a unit only if it lies in some maximal ideal, but by assumption $x \notin \mathfrak{m}$, so $1 - x y$ cannot be non-invertible. Therefore, if $1 - x y$ is a unit for all $y \in A$, then $x$ must be contained in every maximal ideal, i.e., $x \in \mathfrak{R}$.

1.6 Operations on Arbitrary Families of Ideals

Let $\{\mathfrak{a}_i\}_{i\in I}$ be a family of ideals in a ring $A$.

Definition 20 (Sum of Ideals) The sum $\sum_{i\in I} \mathfrak{a}_i$ is defined as: \[ \sum_{i\in I} \mathfrak{a}_i = \left\{ a_1 + a_2 + \cdots + a_n \mid a_k \in \mathfrak{a}_{i_k},\; i_k \in I,\; n \geq 1 \right\} \]

Definition 21 (Intersection of Ideals) The product $\prod_{i\in I} \mathfrak{a}_i$ is defined as: \[ \prod_{i\in I} \mathfrak{a}_i = \left\{ \sum_{k=1}^m a_{1,k} \cdots a_{n,k} \mid a_{j,k} \in \mathfrak{a}_j,\; m \geq 1 \right\} \] (For infinite families, the product is usually defined only for finite subfamilies.)

Definition 22 (Product of Ideals) The intersection $\bigcap_{i\in I} \mathfrak{a}_i$ is defined as: \[ \bigcap_{i\in I} \mathfrak{a}_i = \{ a \in A \mid a \in \mathfrak{a}_i \text{ for all } i \in I \} \]

Distributive law: \[ \mathfrak{a}(\mathfrak{b}+i\mathfrak{c})=\mathfrak{a}\mathfrak{b}+\mathfrak{a}\mathfrak{c} \]
Modular law: \[ \mathfrak{a}\cap(\mathfrak{b}+\mathfrak{c})=\mathfrak{a}\cap\mathfrak{b}+\mathfrak{a}\cap\mathfrak{c}, \text{if } \mathfrak{a}\supseteq \mathfrak{b}\text{or } \mathfrak{a}\supseteq \mathfrak{c} \] In general, we have ${\mathfrak{a}+\mathfrak{b}}(\mathfrak{a}\cap\mathfrak{b})\subseteq\mathfrak{a}\mathfrak{b}$. Clearly,$\mathfrak{a}\mathfrak{b}\subseteq\mathfrak{a}\cap\mathfrak{b}$, hence $\mathfrak{a}\cap\mathfrak{b}=\mathfrak{a}\mathfrak{b}$ provided $\mathfrak{a}+\mathfrak{b}=(1)$.

Definition 23 (Coprime) Let $\mathfrak{a}, \mathfrak{b}$ be ideals of $A$. We call $\mathfrak{a}, \mathfrak{b}$ are coprime, when $\mathfrak{a}+\mathfrak{b}=A$.

Definition 24 (Direct Product of Rings) Let $\{A_i\}_{i \in I}$ be a family of rings. The direct product $\prod_{i \in I} A_i$ is defined as \[ \prod_{i \in I} A_i := \left\{ (x_i)_{i \in I} \mid x_i \in A_i \text{ for all } i \in I \right\} \] with addition and multiplication defined componentwise: \[ (x_i) + (y_i) = (x_i + y_i), \quad (x_i) \cdot (y_i) = (x_i y_i) \] for all $(x_i), (y_i) \in \prod_{i \in I} A_i$.

Let $A_i$ be rings, and let $p_i : \prod_{j \in I} A_j \to A_i$ be the projection onto the $i$-th component, defined by $p_i((x_j)_{j \in I}) = x_i$.

Definition 25 (Chinese Remainder Map) Let $\{\mathfrak{a}_i\}_{i \in I}$ be a family of ideals of $A$. Define the canonical ring homomorphism \[ \Phi: A \to \prod_{i \in I} A/\mathfrak{a}_i, \quad a \mapsto (a + \mathfrak{a}_i)_{i \in I} \] where each component is the natural projection $\phi_i : A \to A/\mathfrak{a}_i$, $a \mapsto a + \mathfrak{a}_i$.

This map $\Phi$ is a ring homomorphism, called the Chinese Remainder map associated to the family $\{\mathfrak{a}_i\}$.

Proposition 9 Let $\{\mathfrak{a}_i\}_{i=1}^n$ be a family of ideals of $A$.

$\forall i\neq j$, $\mathfrak{a}_i, \mathfrak{a}_j$ are coprime, then $\prod_{i=1}^n\mathfrak{a}_i=\bigcap_{i=1}^n\mathfrak{a}_i$.
$\phi$ is surjective $\Leftrightarrow$ $\mathfrak{a}_i, \mathfrak{a}_j$ are coprime.
$\phi$ is injective $\Leftrightarrow$ $\bigcap_{i=1}^n\mathfrak{a}_i=0$.

Proof

Proof. Omitted. cf.[1, Ch. 1, sec.6, p.7, prop.1.10].

Theorem 2 (Chinese Remainder Theorem) Let $A$ be a ring and $\mathfrak{a}_1, \ldots, \mathfrak{a}_n$ be ideals of $A$ such that $\mathfrak{a}_i + \mathfrak{a}_j = (1)$ for all $i \neq j$ (i.e., the ideals are pairwise coprime). Then the canonical map \[ \Phi: A \to \prod_{i=1}^n A/\mathfrak{a}_i, \quad a \mapsto (a + \mathfrak{a}_1, \ldots, a + \mathfrak{a}_n) \] is surjective, and its kernel is $\bigcap_{i=1}^n \mathfrak{a}_i$. Thus, \[ A / \left( \bigcap_{i=1}^n \mathfrak{a}_i \right) \cong \prod_{i=1}^n A/\mathfrak{a}_i \] as rings.

In particular, if $A = \mathbb{Z}$ and the $\mathfrak{a}_i = (n_i)$ with $\gcd(n_i, n_j) = 1$ for $i \neq j$, then \[ \mathbb{Z}/(n_1 n_2 \cdots n_k) \cong \mathbb{Z}/n_1 \times \cdots \times \mathbb{Z}/n_k. \]

Proof

Proof. Let $\Phi: A \to \prod_{i=1}^n A/\mathfrak{a}_i$ be the canonical map, $a \mapsto (a+\mathfrak{a}_1, \ldots, a+\mathfrak{a}_n)$.

Kernel: $\ker \Phi = \bigcap_{i=1}^n \mathfrak{a}_i$, since $a \in \ker \Phi$ iff $a \in \mathfrak{a}_i$ for all $i$.
Surjectivity: For any $(b_1+\mathfrak{a}_1, \ldots, b_n+\mathfrak{a}_n) \in \prod_{i=1}^n A/\mathfrak{a}_i$, we want $a \in A$ such that $a \equiv b_i \pmod{\mathfrak{a}_i}$ for all $i$.

Since the ideals are pairwise coprime, for each $i$ there exists $e_i \in A$ such that $e_i \equiv 1 \pmod{\mathfrak{a}_i}$ and $e_i \equiv 0 \pmod{\mathfrak{a}_j}$ for $j \neq i$. (This follows from the Chinese Remainder construction: for each $i$, let $J_i = \bigcap_{j \neq i} \mathfrak{a}_j$, then $J_i + \mathfrak{a}_i = (1)$, so $1 = x_i + y_i$ with $x_i \in J_i$, $y_i \in \mathfrak{a}_i$; set $e_i = x_i$.)

Then set $a = \sum_{i=1}^n b_i e_i$. For each $k$, $a \equiv b_k e_k \equiv b_k \pmod{\mathfrak{a}_k}$, since $e_k \equiv 1 \pmod{\mathfrak{a}_k}$ and $e_i \equiv 0 \pmod{\mathfrak{a}_k}$ for $i \neq k$.

Thus, $\Phi$ is surjective.

Isomorphism: By the First Isomorphism Theorem, $A/\ker\Phi \cong \operatorname{Im}\Phi = \prod_{i=1}^n A/\mathfrak{a}_i$.

Therefore, \[ A / \left( \bigcap_{i=1}^n \mathfrak{a}_i \right) \cong \prod_{i=1}^n A/\mathfrak{a}_i. \]

Remark

Remark. The union of ideals is not necessarily an ideal unless one contain the others.

In general, the union $\mathfrak{a}\cup \mathfrak{b}$ fails to be closed under addition. For example, in $\mathbb{Z}$, the ideals $(2)$ and $(3)$ have union $\{ \ldots, -6, -4, -3, -2, 0, 2, 3, 4, 6, \ldots \}$, but $2 \in (2)$ and $3 \in (3)$, yet $2 + 3 = 5 \notin (2) \cup (3)$.

Proposition 10

Let $\mathfrak{p}_1, \ldots, \mathfrak{p}_n$ be prime ideals and let $a$ be an ideal contained in $\bigcup_{i=1}^n \mathfrak{p}_i$. Then $a \subseteq \mathfrak{p}_i$ for some $i$.
Let $a_1, \ldots, a_n$ be ideals and let $\mathfrak{p}$ be a prime ideal containing $\bigcap_{i=1}^n a_i$. Then $\mathfrak{p} \supseteq a_i$ for some $i$. If $\mathfrak{p} = \bigcap a_i$, then $\mathfrak{p} = a_i$ for some $i$.

Proof

Proof. Omitted. cf.[1, Ch. 1, sec.6, p.8, prop.1.11].

Definition 26 (Quotion of Ideals) The set $(\mathfrak{a}:\mathfrak{b})=\{x\in A\mid x\mathfrak{b}\subseteq\mathfrak{a}\}$ is quotien of $\mathfrak{a}$ and $\mathfrak{b}$. This set is an ideal of $A$.

If $\mathfrak{b}=(x)$ is a principal ideal of $A$, then $(\mathfrak{a}:\mathfrak{b})$ is denoted by $(\mathfrak{a}:x)$.

Definition 27 (Annihilator) The set $(0:\mathfrak{b})=\{x\in A\mid x\mathfrak{b}=0\}$ is called the annihilator of $\mathfrak{b}$. It is denoted by $\operatorname{Ann}(\mathfrak{b})$.

Proposition 11 (Zero-Divisors) The set of zero-divisors of a ring $A$ is the set \[ D = \{ a \in A \mid \exists\, b \in A,\, b \neq 0,\ ab = 0 \text{ or } ba = 0 \}. \] This set is not necessarily an ideal, but it is a union of ideals of $A$. \[ D = \bigcup_{x\neq0}\operatorname{Ann}(x), \] Moreover, it is a union of prime ideals of $A$. \[ D = \bigcup_{\mathfrak{p}\ \text{prime}} \mathfrak{p}, \] where the union is taken over all prime ideals of $A$.

In particular, every zero-divisor lies in some prime ideal.

Definition 28 (Radical of an Ideal) Let $\mathfrak{a}$ be an ideal of a ring $A$. The radical of $\mathfrak{a}$, denoted $\sqrt{\mathfrak{a}}$ or $\operatorname{r}(\mathfrak{a})$, is defined as \[ \sqrt{\mathfrak{a}} = \{ x \in A \mid \exists n > 0,\ x^n \in \mathfrak{a}\} \] That is, $x$ is in the radical of $\mathfrak{a}$ if some power of $x$ lies in $\mathfrak{a}$. The radical $\sqrt{\mathfrak{a}}$ is itself an ideal of $A$.

If $\mathfrak{a}= (0)$, then $\sqrt{(0)}$ is the set of all nilpotent elements, i.e., the nilradical of $A$.

Proposition 12

$\operatorname{r}(\mathfrak{a})\supseteq\mathfrak{a}$.
$\operatorname{r}(\operatorname{r}(\mathfrak{a}))=\operatorname{r}(\mathfrak{a})$.
$\operatorname{r}(\mathfrak{a}\mathfrak{b})=\operatorname{r}(\mathfrak{a}\cap\mathfrak{b})=\operatorname{r}(\mathfrak{a})\cap\operatorname{r}(\mathfrak{b})$.
$\operatorname{r}(\mathfrak{a})=(1)\Leftrightarrow\mathfrak{a}=(1)$.
$\operatorname{r}(\mathfrak{a}+\mathfrak{b})=\operatorname{r}(\operatorname{r}(\mathfrak{a})+\operatorname{r}(\mathfrak{b}))$.
If $\mathfrak{p}$ is prime, $\operatorname{r}(\mathfrak{p}^n)=\mathfrak{p}$ for all $n>0$.

Proof

Proof. Left to the reader. (Easy to check)

Proposition 13 \[\operatorname{r}(\mathfrak{a})=\bigcap_{\mathfrak{p}\supseteq\mathfrak{a}\ \text{prime}}\mathfrak{p}\].

Hint: Consider nilradical of the quotient ring $A/\mathfrak{a}$, and the corresponding of ideals.

Proof (Proof.1)

Proof (Proof.1). Let $\pi: A \to A/\mathfrak{a}$ be the canonical projection. By the Correspondence Theorem, there is a bijection between the set of prime ideals of $A$ containing $\mathfrak{a}$ and the set of prime ideals of $A/\mathfrak{a}$.

The nilradical of $A/\mathfrak{a}$, denoted $\mathfrak{N}(A/\mathfrak{a})$, is the intersection of all prime ideals of $A/\mathfrak{a}$: \[ \mathfrak{N}(A/\mathfrak{a}) = \bigcap_{\bar{\mathfrak{p}} \text{ prime in } A/\mathfrak{a}} \bar{\mathfrak{p}} \] The preimage of this intersection under $\pi$ is the intersection of all prime ideals of $A$ containing $\mathfrak{a}$: \[ \pi^{-1}(\mathfrak{N}(A/\mathfrak{a})) = \bigcap_{\mathfrak{p}\supseteq \mathfrak{a},\ \mathfrak{p}\text{ prime}} \mathfrak{p} \]

On the other hand, $\mathfrak{N}(A/\mathfrak{a})$ consists of all elements $\bar{x} = x + \mathfrak{a}$ such that $(x + \mathfrak{a})^n = \mathfrak{a}$ for some $n \geq 1$, i.e., $x^n \in \mathfrak{a}$. Thus, \[ \pi^{-1}(\mathfrak{N}(A/\mathfrak{a})) = \{ x \in A \mid x^n \in \mathfrak{a}\text{ for some } n \geq 1 \} = \operatorname{r}(\mathfrak{a}) \]

Therefore, \[ \operatorname{r}(\mathfrak{a}) = \bigcap_{\mathfrak{p}\supseteq \mathfrak{a},\ \mathfrak{p}\text{ prime}} \mathfrak{p} \]

Proof (Proof.2)

Proof (Proof.2). Let $x \in \operatorname{r}(\mathfrak{a})$. Then $x^n \in \mathfrak{a}$ for some $n > 0$. For any prime ideal $\mathfrak{p}\supseteq \mathfrak{a}$, since $\mathfrak{p}$ is prime and $x^n \in \mathfrak{p}$, it follows that $x \in \mathfrak{p}$. Thus, $x$ is in every prime ideal containing $\mathfrak{a}$, so $x \in \bigcap_{\mathfrak{p}\supseteq \mathfrak{a}\ \text{prime}} \mathfrak{p}$.

Conversely, suppose $x \notin \operatorname{r}(\mathfrak{a})$. Then $x^n \notin \mathfrak{a}$ for all $n > 0$. Consider the quotient ring $A/\mathfrak{a}$ and the image $\bar{x}$ of $x$. Then $\bar{x}^n \neq 0$ for all $n > 0$. By the proof of the nilradical as intersection of primes, there exists a prime ideal $\bar{\mathfrak{p}}$ of $A/\mathfrak{a}$ not containing any power of $\bar{x}$. The preimage $\mathfrak{p}$ of $\bar{\mathfrak{p}}$ under the projection $A \to A/\mathfrak{a}$ is a prime ideal of $A$ containing $\mathfrak{a}$ but not $x$. Thus, $x \notin \bigcap_{\mathfrak{p}\supseteq \mathfrak{a}\ \text{prime}} \mathfrak{p}$.

Therefore, $\operatorname{r}(\mathfrak{a}) = \bigcap_{\mathfrak{p}\supseteq \mathfrak{a}\ \text{prime}} \mathfrak{p}$.

Definition 29 (Radical of a Subset) More general, let $S \subseteq A$ be any subset of a ring $A$. The radical of $S$, denoted $\sqrt{S}$ or $\operatorname{r}(S)$, is defined as the intersection of all prime ideals of $A$ containing $S$: \[ \sqrt{S} = \bigcap_{\mathfrak{p}\supseteq S,\ \mathfrak{p}\ \text{prime}} \mathfrak{p} \]

Proposition 14

$\operatorname{r}(\bigcap_{\alpha}E_\alpha)=\bigcap_{\alpha}\operatorname{r}(E_\alpha).$
$D= \bigcap_{x\neq0}\operatorname{r}(\operatorname{Ann}(x)).$
$\operatorname{r}(\mathfrak{a}),\ \operatorname{r}(\mathfrak{b})\text{ are coprime }\implies \mathfrak{a},\ \mathfrak{b}\text{ are coprime}.$

1.7 Extension and Contraction of Ideals

Let $f: A \to B$ be a ring homomorphism.

Definition 30 (Extension) Given an ideal $\mathfrak{a}\subseteq A$, the extension of $\mathfrak{a}$ to $B$ is the ideal \[ \mathfrak{a}^e = f(\mathfrak{a})B = \left\{ \sum_{i=1}^n f(a_i) b_i \mid a_i \in \mathfrak{a},\, b_i \in B,\, n \geq 1 \right\} \] That is, $\mathfrak{a}^e$ is the ideal of $B$ generated by the image of $\mathfrak{a}$.

Definition 31 (Contraction) Given an ideal $\mathfrak{b}\subseteq B$, the contraction of $\mathfrak{b}$ to $A$ is the ideal \[ \mathfrak{b}^c = f^{-1}(\mathfrak{b}) = \{ a \in A \mid f(a) \in \mathfrak{b}\} \]

Proposition 15

The extension of an ideal is always an ideal; the contraction of an ideal is always an ideal.
If $\mathfrak{a}\subseteq A$, then $\mathfrak{a}\subseteq (\mathfrak{a}^e)^c$.
If $\mathfrak{b}\subseteq B$, then $(\mathfrak{b}^c)^e \subseteq \mathfrak{b}$.
The set $C=\{\mathfrak{a}^e\mid \mathfrak{a}\vartriangleleft A\}$, and $E=\{\mathfrak{b}^c\mid \mathfrak{b}\vartriangleleft B\}$, then $C=\{\mathfrak{a}\mid \mathfrak{a}^{ec}=\mathfrak{a}\}$, and $E=\{\mathfrak{b}\mid \mathfrak{b}^{ce}=\mathfrak{b}\}$.
There is a correspondence between ideals of $A$ and ideals of $B$ that are stable under extension and contraction, i.e., there is a bijective between $E$ and $C$.
If $f$ is surjective, then every ideal of $B$ is the extension of its contraction.
The contraction of a prime ideal of $B$ is a prime ideal of $A$.
The extension of a prime ideal of $A$ need not be prime in $B$.

Proof

Proof. Left to the reader. (Easy to check) cf.[1] {{ch.1, sec.7, p.10, prop.1.17}}

1.8 Spectrum and Zariski Topology

This section all of proofs will be omitted, since we have discussed in seminar

Definition 32 (Spectrum of a Ring) The spectrum of a ring $A$, denoted $\operatorname{Spec}A$, is the set of all prime ideals of $A$: \[ \operatorname{Spec}A = \{\, \mathfrak{p}\subseteq A \mid \mathfrak{p}\text{ is a prime ideal} \,\} \]

Proposition 16 (Toplogy Structure of Spectum) Let $A$ be a ring and let $X$ be the set of all prime ideals of $A$. For each subset $E$ of $A$, let $V(E) = \{\, \mathfrak{p}\in \operatorname{Spec}A \mid E \subseteq \mathfrak{p}\,\}$. Then we have: - If $\mathfrak{a}$ is the ideal generated by $E$, $V(E)=V(\mathfrak{a})=V(\operatorname{r}(\mathfrak{a}))$. - $V(0) = \operatorname{Spec}A$; $V(1) = \varnothing$. - $V\left(\bigcup_{\alpha} \mathfrak{a}_\alpha\right) = \bigcap_{\alpha} V(\mathfrak{a}_\alpha)$. - $V(\mathfrak{a}) \cup V(\mathfrak{b}) = V(\mathfrak{a}\cap \mathfrak{b}) = V(\mathfrak{a}\mathfrak{b})$.

Definition 33 (Zariski Topology) The spectrum $\operatorname{Spec}A$ is equipped with the Zariski topology, where the closed sets are of the form \[ V(E) = \{\, \mathfrak{p}\in \operatorname{Spec}A \mid E \subseteq \mathfrak{p}\,\} \] for some subset $E \subseteq A$.

In particular, for an ideal $\mathfrak{a}\subseteq A$, $V(\mathfrak{a}) = \{\, \mathfrak{p}\mid \mathfrak{a}\subseteq \mathfrak{p}\,\}$.

Proposition 17 (Open set of Spectum) For each $f\in A$, let $X_f$ denote complement of $V(f)$ in $X=\operatorname{Spec}A$. 0. The basic open sets are complements of $V(f)$ for $f \in A$: $X_f$. The basic open sets is a basis of Zariski topology.

$X_f\cap X_g=X_{fg}$.
$X_f=\emptyset\Leftrightarrow f$ is nilpotent.
$X_f=X\Leftrightarrow f$ is a unit.
$X_f=X_g\Leftrightarrow\operatorname{r}((f))=\operatorname{r}((g))$.
Each $X_f$ is quasi-compact.
An open subset of $X$ is quasi-compact if and only if it is a finite union of basic open sets $X_{f_1}, \ldots, X_{f_n}$ for some $f_1, \ldots, f_n \in A$.

Proposition 18 (Closures of Spectum) Denote a prime ideal of $A$ by a letter $x$ or $y$ when thinking of it as a point of $X=\operatorname{Spec}A$. When thinking of $x$ as a prime ideal of $A$, we denote it by $\mathfrak{p}_x$.

The set $\{x\}$ is closed in $\operatorname{Spec}A$ $\Leftrightarrow$ $\mathfrak{p}$ is maximal.
$\overline{\{x\}}=V{\mathfrak{p}_x}$.
$y\in\overline{\{x\}}\Leftrightarrow\mathfrak{p}_x\subseteq\mathfrak{p}_y$
$X$ is a $T_0$-space.

Remark

Remark. The Zariski topology is generally not Hausdorff; its closed sets are typically large. The points corresponding to maximal ideals are called closed points.

Proposition 19 (Irreducible) A topology space $X$ is said irreducible if $X\neq\emptyset$ and if every pair of non-empty open sets in $X$ intersect, or equivalently if every non-emtpy open set is dense in $X$.

$\operatorname{Spec}A$ is irreducible if and only if the nilradical of $A$ is a prime ideal.
If $Y$ is an irreducible subspace of $X$, then the closure $\overline{Y}$ of $Y$ in $X$ is irreducible.
Every irreducible subspace of $X$ is contained in a maximal irreducible subspace.
The maximal irreducible subspaces of $X$ are closed and cover $X$. They are called the irreducible components of $X$.
The irreducible components of $\operatorname{Spec}A$ are the closed sets $V(\mathfrak{p})$, where $\mathfrak{p}$ is a minimal prime ideal of $A$.

Remark

Remark. Let $A\neq0$ is ring. Then $A$ has the minimal prime ideal with respect to inclusion. (You can consider Zorn’s lemma to prove this remark)

Definition 34 (Induced Map on Spectra) The map $f: A \to B$ induces a map on spectra: \[ f^*: \operatorname{Spec}B \to \operatorname{Spec}A,\quad \mathfrak{q} \mapsto f^{-1}(\mathfrak{q}) \] where $\operatorname{Spec}A$ denotes the set of all prime ideals of $A$.

1.9 Affine Algebraic Varieties

Let $k$ be a field. An affine algebraic variety over $k$ is a subset $V \subseteq k^n$ defined as the common zeros of a set of polynomials: \[ V = V(S) = \{\, x \in k^n \mid f(x) = 0\ \forall f \in S \,\} \] for some subset $S \subseteq k[x_1, \ldots, x_n]$.

The set of all polynomials vanishing on $V$ is an ideal: \[ I(V) = \{\, f \in k[x_1, \ldots, x_n] \mid f(x) = 0\ \forall x \in V \,\} \]

There is a correspondence between affine varieties and radical ideals of $k[x_1, \ldots, x_n]$ (Hilbert’s Nullstellensatz).

The coordinate ring of $V$ is defined as \[ k[V] = k[x_1, \ldots, x_n]/I(V) \] which encodes the algebraic structure of $V$.

2 Modules

2.1 Modules and Module Hom

Definition 35 (Module) Let $A$ be a ring. An $A$-module $M$ is an abelian group $(M, +)$ together with an action $A \times M \to M$, $(a, m) \mapsto a m$, such that for all $a, b \in A$ and $m, n \in M$:

$a(m + n) = a m + a n$
$(a + b)m = a m + b m$
$(a b)m = a(b m)$
$1 m = m$ (if $A$ has $1$)

Definition 36 (Submodule) A submodule $N$ of an $A$-module $M$ is a subgroup $N \leq M$ such that $a n \in N$ for all $a \in A$, $n \in N$.

Definition 37 (Module Homomorphism) Let $M, N$ be $A$-modules. A map $f: M \to N$ is an $A$-module homomorphism if for all $m, m' \in M$ and $a \in A$:

$f(m + m') = f(m) + f(m')$
$f(a m) = a f(m)$

The set of all $A$-module homomorphisms from $M$ to $N$ is denoted $\operatorname{Hom}_A(M, N)$.

Moreover, the set $\operatorname{Hom}_A(M, N)$ forms an abelian group under pointwise addition: \[ (f + g)(m) = f(m) + g(m) \] for all $f, g \in \operatorname{Hom}_A(M, N)$ and $m \in M$.

If $A$ is commutative, then $\operatorname{Hom}_A(M, N)$ is itself an $A$-module, with scalar multiplication defined by \[ (a f)(m) = a \cdot f(m) \] for $a \in A$, $f \in \operatorname{Hom}_A(M, N)$, and $m \in M$.

2.2 Submodules and Quotient Modules

Definition 38 (Quotient Module) If $N \leq M$ is a submodule, the quotient module $M/N$ is the abelian group of cosets $m + N$ with $A$-action $a(m + N) = a m + N$.

Theorem 3 (Correspondence Theorem for Submodules) Let $M$ be an $A$-module and $N \leq M$ a submodule. There is a bijective correspondence between the set of submodules of $M$ containing $N$ and the set of submodules of the quotient module $M/N$.

Definition 39 (Kernel, Image and Cokernel) Let $f: M \to N$ be an $A$-module homomorphism.

The kernel is $\ker f = \{ m \in M \mid f(m) = 0 \}$, a submodule of $M$.
The image is $\operatorname{Im} f = \{ f(m) \mid m \in M \}$, a submodule of $N$.
The cokernel is $\operatorname{Coker} f=N/\operatorname{Im}f$.

Proposition 20 (First Isomorphism Theorem) Let $f: M \to N$ be an $A$-module homomorphism. Then \[ M/\ker f \cong \operatorname{Im} f \] as $A$-modules.

Proof

Proof. Define $\varphi: M/\ker f \to \operatorname{Im} f$ by $\varphi(m + \ker f) = f(m)$. This map is well-defined, $A$-linear, and bijective.

2.3 Operation of Submodule

Let $M$ be an $A$-module, and let $\{N_i\}_{i \in I}$ be a family of submodules of $M$.

Definition 40 (Sum of Submodules) The sum of submodules $\{N_i\}$ is defined as: \[ \sum_{i \in I} N_i = \left\{ n_1 + \cdots + n_k \mid n_j \in N_{i_j},\; i_j \in I,\; k \geq 1 \right\} \] This is the smallest submodule of $M$ containing all the $N_i$.

Definition 41 (Intersection of Submodules) The intersection of submodules $\{N_i\}$ is: \[ \bigcap_{i \in I} N_i = \{ m \in M \mid m \in N_i\ \text{for all}\ i \in I \} \] This is the largest submodule contained in all the $N_i$.

Proposition 21 (Lattice Structure) The set of submodules of $M$ forms a lattice under sum and intersection:

$N_1 + N_2$ is the least upper bound (join) of $N_1$ and $N_2$.
$N_1 \cap N_2$ is the greatest lower bound (meet).

Proposition 22 (Second Isomorphism Theorem) Let $M$ be an $A$-module, and let $N, P$ be submodules of $M$. Then \[ (N + P)/P \cong N/(N \cap P) \] as $A$-modules.

Proof

Proof. Define the map $\varphi: N \to (N + P)/P$ by $\varphi(n) = n + P$. This is an $A$-module homomorphism with kernel $N \cap P$, and it is surjective. By the First Isomorphism Theorem, $N/(N \cap P) \cong (N + P)/P$.

Proposition 23 (Third Isomorphism Theorem) Let $M$ be an $A$-module, and let $N \subseteq P \subseteq M$ be submodules. Then \[ (M/N)/(P/N) \cong M/P \] as $A$-modules.

Proof

Proof. Consider the natural map $\varphi: M/N \to M/P$ given by $m + N \mapsto m + P$. This is a well-defined $A$-module homomorphism with kernel $P/N$. By the First Isomorphism Theorem, $(M/N)/(P/N) \cong M/P$.

Definition 42 (Submodule Generated by a Subset) Given a subset $S \subseteq M$, the submodule generated by $S$ is: \[ \langle S \rangle = \left\{ \sum_{j=1}^n a_j s_j \mid a_j \in A,\ s_j \in S,\ n \geq 1 \right\} \] This is the smallest submodule of $M$ containing $S$.

Definition 43 (Product of Ideal and Submodule) Let $A$ be a ring, $M$ an $A$-module, $\mathfrak{a}\subseteq A$ an ideal, and $N \leq M$ a submodule. The product $\mathfrak{a}N$ is defined as the submodule of $M$ generated by all products $a n$ with $a \in \mathfrak{a}$, $n \in N$: \[ \mathfrak{a}N = \left\{ \sum_{i=1}^k a_i n_i \mid a_i \in \mathfrak{a},\ n_i \in N,\ k \geq 1 \right\} \] This is the smallest submodule of $M$ containing all elements $a n$ with $a \in \mathfrak{a}$, $n \in N$.

Definition 44 (Quotient of Submodules) $N, P \leq M$, then $(N:P):=\{\ a\in A \mid aP\subseteq N\ \}$ is an ideal of $A$.

Definition 45 (Annihilator of a Module) Let $M$ be an $A$-module. The annihilator of $M$ is \[ \operatorname{Ann}_A(M) :=(0:M) = \{ a \in A \mid a m = 0 \text{ for all } m \in M \} \] which is an ideal of $A$.

Proposition 24 If $\mathfrak{a}\subseteq \operatorname{Ann}(M)$, then $M$ is also $A/\mathfrak{a}$-module. The multiplication defined by $\bar{a}m=am$, It’s easy to check well-defined.

Definition 46 If $\operatorname{Ann}(M)=0$, then $A$-module $M$ is faithful.

If $\operatorname{Ann}(M)=\mathfrak{a}$, then $M$ is faithful as a $A/\mathfrak{a}$-module.

2.4 Direct Sum and Direct Product

Definition 47 (Direct Sum and Direct Product of Modules) Let $\{M_i\}_{i \in I}$ be a family of $A$-modules.

The direct product $\prod_{i \in I} M_i$ is the set of all tuples $(m_i)_{i \in I}$ with $m_i \in M_i$, with addition and scalar multiplication defined componentwise.
The direct sum $\bigoplus_{i \in I} M_i$ is the subset of the direct product consisting of tuples $(m_i)_{i \in I}$ such that $m_i = 0$ for all but finitely many $i$.

Both $\prod_{i \in I} M_i$ and $\bigoplus_{i \in I} M_i$ are $A$-modules.

2.5 Finitely Generated Module

Definition 48 (Finitely Generated Module) An $A$-module $M$ is finitely generated if there exist elements $m_1, \ldots, m_n \in M$ such that every $m \in M$ can be written as \[ m = a_1 m_1 + \cdots + a_n m_n \] for some $a_1, \ldots, a_n \in A$. In other words, $M = \langle m_1, \ldots, m_n \rangle$.

Definition 49 (Free Module) Let $A$ be a ring and $S$ a set. The free $A$-module on $S$, denoted $F = \bigoplus_{s \in S} A$, is the set of all functions $f: S \to A$ such that $f(s) = 0$ for all but finitely many $s \in S$. Equivalently, elements of $F$ are finite formal sums \[ \sum_{i=1}^n a_i e_{s_i} \] where $a_i \in A$, $s_i \in S$, and $e_{s}$ is the function with $e_{s}(t) = \delta_{s,t}$.

$F$ is an $A$-module with addition and scalar multiplication defined componentwise.

If $S$ is finite with $n$ elements, then $F \cong A^n$ as $A$-modules.

A module $M$ is free if it is isomorphic to a free module on some set $S$; that is, $M \cong \bigoplus_{s \in S} A$ for some $S$.

Proposition 25 An $A$-module $M$ is finitely generated if and only if there exists an integer $n \geq 0$ and a submodule $N \leq A^n$ such that $M \cong A^n / N$.

Proof (Proof Sketch:)

Proof (Proof Sketch:). If $M$ is finitely generated by $m_1, \ldots, m_n$, define a surjective $A$-module homomorphism $\varphi: A^n \to M$ by $\varphi(a_1, \ldots, a_n) = a_1 m_1 + \cdots + a_n m_n$. Then $M \cong A^n / \ker \varphi$. Conversely, any quotient of $A^n$ is finitely generated.

Proposition 26 A quotient of a finitely generated module is finitely generated.

Proof

Proof. Hint: Let $M$ be generated by $m_1, \ldots, m_n$ and $N \leq M$. Then $M/N$ is generated by the images of $m_1, \ldots, m_n$ in $M/N$.

Theorem 4 (Hamilton-Cayley Theorem) Let $M$ be a finitely generated $A$-module. Let $\mathfrak{a}\vartriangleleft A$, and let $\phi: M \to M$ be an $A$-module endomorphism such that $\phi(M) \subseteq \mathfrak{a}M$. Then there exist $a_1, \ldots, a_n \in \mathfrak{a}$ (for some $n$) such that \[ \phi^n + a_1 \phi^{n-1} + \cdots + a_n = 0 \] as endomorphisms of $M$.

Proof

Proof. Let $M$ be generated by $m_1, \ldots, m_n$. Since $\phi(M) \subseteq \mathfrak{a}M$, for each $i$, \[ \phi(m_i) = \sum_{j=1}^n a_{ij} m_j \] with $a_{ij} \in \mathfrak{a}$. Let $A = (a_{ij})$ be the $n \times n$ matrix over $\mathfrak{a}$ representing $\phi$ in this basis.

Consider the $A$-module homomorphism $\Phi: M^n \to M^n$ given by $\Phi = \phi \cdot I - A$, where $I$ is the identity. By the Cayley-Hamilton theorem for modules, the characteristic polynomial $f(x) = x^n + a_1 x^{n-1} + \cdots + a_n$ with $a_i \in \mathfrak{a}$ annihilates $\phi$: \[ f(\phi) = \phi^n + a_1 \phi^{n-1} + \cdots + a_n = 0 \] as endomorphisms of $M$.

Corollary 3 Let $M$ be a finitely generated $A$-module and $\mathfrak{a}\vartriangleleft A$ such that $\mathfrak{a}M = M$. Then there exists $x \in A$ with $x \equiv 1 \pmod{\mathfrak{a}}$ such that $x M = 0$.

Proof

Proof. Take $\phi=\operatorname{id}$. There exists $1+a_1+a_2+\cdots+a_n=0$ since Theorem 4, let $x=1+a_1+a_2+\cdots+a_n$.

Theorem 5 (Nakayama’s lemma) Let $M$ be a finitely generated $A$-module and $\mathfrak{a}\vartriangleleft A$, if $\mathfrak{a}\subseteq \mathfrak{R}$, then $\mathfrak{a}M = M$ implies $M = 0$.

Proof

Proof. By Corollary 3, if $\mathfrak{a}M = M$ and $\mathfrak{a}\subseteq \mathfrak{R}$, then there exists $x \in A$ with $x \equiv 1 \pmod{\mathfrak{a}}$ such that $x M = 0$. That is, $x = 1 + a$ for some $a \in \mathfrak{a}$, and $x M = 0$.

But $1 + a$ is a unit in $A$ (since $a \in \mathfrak{R}$ and Proposition 8). Therefore, $x$ is invertible, so $x M = 0$ implies $M = 0$.

Corollary 4 Let $M$ be a finitely generated $A$-module, $N$ is a submodule of $M$, $\mathfrak{a}\vartriangleleft A$, if $\mathfrak{a}\subseteq \mathfrak{R}$, then $M=\mathfrak{a}M+N$ implies $N=M$.

Proof

Proof. Consider the quotient module $M/N$. Since $M = \mathfrak{a}M + N$, we have \[ M/N = (\mathfrak{a}M + N)/N \cong \mathfrak{a}M / (\mathfrak{a}M \cap N) \subseteq \mathfrak{a}(M/N) \] so $M/N = \mathfrak{a}(M/N)$. By Theorem 5, since $\mathfrak{a}\subseteq \mathfrak{R}$ and $M/N$ is finitely generated, it follows that $M/N = 0$, i.e., $M = N$.

References

[1]

M. F. Atiyah and I. G. Macdonald, Introduction to commutative algebra. CRC Press, 2018.

[2]

聂灵沼, 代数学引论. 高等教育出版社, 2021.