# Algebra II

$\texttt{Dr. Kuncham Syam Prasad}$

###### tags: `Notes`
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$\newcommand{\defeq}{\overset{\text{def}}{=}}$ $\newcommand{\qed}{\qquad\square}$ $\DeclareMathOperator{\lcm}{lcm}$ $\DeclareMathOperator{\Ann}{Ann}$
<span style="color:#aa9090">Note: These are based on the lecture notes I wrote while attending Dr. Syam Prasad's Algebra II classes. I have made some minor changes in some place, and the exercise solutions given are my own. Therefore, only I am responsible for any errors or other deficiencies in these notes.</span>

## Lecture 01 (Preliminaries)

**Definition.** A *ring* $(A, +, \cdot)$ is a non-empty set with binary operations $+$ and $\cdot$ satisfying:
1. $(A, +)$ is an Abelian group.
2. $(a \cdot b) \cdot c = a \cdot (b \cdot c),$ $\forall~ a, b, c \in A.$
3. both distributive laws: $\forall~ a, b, c \in A,$
$a \cdot (b + c) = a \cdot b + a \cdot c$
$(a + b) \cdot c = a \cdot c + b \cdot c.$

$A$ is *commutative* if $a \cdot b = b \cdot a,$ $\forall~ a, b \in A.$
If $1 \in A$ (multiplicative identity: $1 \cdot a = a \cdot 1 = a,$ $\forall~ a \in A$), then $A$ is a *ring with unity*.

**Note.**
1. A ring can have at most one unity.
2. $a \cdot 0 = 0,$ $\forall~ a \in A.$
3. If $A$ is a ring with unity, then $1 = 0 \iff A = \{0\}.$
That is, in order for a ring to be non-zero, we must have $1 \ne 0.$

**Examples.**
1. $\mathbb Z$ is a commutative ring with $1.$
2. $2 \mathbb Z$ is a commutative ring without $1.$
3. $\mathbb R$ is a commutative ring with $1.$
4. $\left( M_{n \times n}(A), +, \cdot \right)$ is a non-commutative ring, where $A$ is a ring, $M_{n \times n}(A)$ is the set of all $n \times n$ matrices over $A,$ and $+$ and $\cdot$ are matrix addition and multiplication respectively.
5. Let $F = \{\, f \colon \mathbb R \to \mathbb R \,\},$ and define $(f + g)(x) = f(x) + g(x)$ and $(f \cdot g)(x) = f(x)g(x),$ $\forall~ x \in \mathbb R.$ Then $(F, +, \cdot)$ is a commutive ring with unity ($x \mapsto 1$).
6. Define $F$ and $+$ as in 5. Let $(f \circ g)(x) = f(g(x))$ $\forall~ x \in \mathbb R.$
$(F, +, \circ)$ is *not* a ring, as it is not distributive:
With $f(x) = x^2,$ $g(x) = x + 1,$ and $h(x) = x,$ $f \circ (g + h) \ne f \circ g + f \circ h.$

**Definition.** A subset $S$ of a commutative ring $A$ with unity is a *subring* of $A$ if $S$ is a commutative ring with respect to the restricted binary operators and contains the identity element of $A.$

**Note.** A subset $S$ of $A$ is a subring iff the following hold:
1. $0, 1 \in S.$
2. $a - b \in S,$ $\forall~ a, b \in S.$ [Subgroup condition]
3. $ab \in S,$ $\forall~ a, b \in S.$ [Subsemigroup condition]

**Definition.** Let $A$ be a ring with $1 \ne 0.$ An element $u \in A$ is called a *unit* if it has a multiplicative inverse in $A.$ A ring in which every non-zero element is a unit is a *division ring* or *skew field*. A commutative division ring is a *field*.

---

## Lecture 02 (Ideals in Rings)

**Definition.** Let $A$ and $B$ be two rings. A map $f \colon A \to B$ is a *homomorphism* if
1. $f(x + y) = f(x) + f(y),$ $\forall~ x, y \in A$
2. $f(xy) = f(x)f(y),$ $\forall~ x, y \in A$
3. $f(1_A) = 1_B.$

**Note.** Let $f \colon A \to B$ be a homomorphism. Then
1. $f(a - b) = f(a) - f(b),$ $\forall~ a, b \in A.$
2. $f(-a) = -f(a),$ $\forall~ a \in A.$
3. $f(0_A) = 0_B.$

**Result.** Let $A$ be a ring and $S$ a subring of $A.$ Then the map $f \colon S \to A$ defined by $f(x) = x,$ $\forall~ x \in S,$ is a homomorphism (called *inclusion*).
*Proof.* Let $x, y \in S.$
$f(x + y) = x + y = f(x) + f(y).$
$f(xy) = xy = f(x)f(y).$
$f(1_S) = 1_S = 1_A.$ $\qed$

**Result.** The composition of ring homomorphisms is again a homomorphism. That is, if $f \colon A \to B$ and $g \colon B \to C$ are homomorphisms, then so is $g \circ f \colon A \to C.$
*Proof.* Let $x, y \in A.$
\begin{align*}
(g \circ f)(x + y) & = g(f(x + y)) \\
& = g(f(x) + f(y)) \\
& = g(f(x)) + g(f(y)) \\
& = (g \circ f)(x) + (g \circ f)(y).
\end{align*}

\begin{align*}
(g \circ f)(xy) & = g(f(xy)) \\
& = g(f(x)f(y)) \\
& = g(f(x))g(f(y)) \\
& = (g \circ f)(x) (g \circ f)(y).
\end{align*}

\begin{align*}
(g \circ f)(1) & = g(f(1_A)) \\
& = g(1_B) \\
& = 1_C. \qed
\end{align*}

**Definition.** Let $A$ be a ring. A non-empty subset $I$ of $A$ is said to be an *ideal* if it satisfies:
1. $(I, +)$ is a subgroup of $(A, +).$ That is, if $x, y \in I,$ then $x - y \in I.$
2. For all $x \in I$ and $a \in A,$ $xa \in I.$

We write $I \unlhd A$ to denote that $I$ is an ideal of $A.$

**Result.** Let $A$ be a ring. Then for any $a \in A,$
\begin{equation*}
\langle a \rangle \defeq \{\, ax \mid x \in A\,\}
\end{equation*}

is an ideal of $A$ (called the *principal ideal* generated by $a$ in $A$).
*Proof.* $0 \in A \implies 0 = a \cdot 0 \in \langle a \rangle \implies \langle a \rangle \ne \varnothing.$
Take $ax, ay \in A.$ Then $ax - ay = a(x - y) \in \langle a \rangle.$
If $ax \in \langle a \rangle$ and $t \in A,$ then $(ax) t = a(xt) \in \langle a \rangle.$
Therefore, $\langle a \rangle$ is an ideal of $A.$ $\qed$

**Lemma.** Let $A$ be a ring, and $I$ an ideal of $A.$ We define a relation $\Theta$ on $A$ as
\begin{equation*}
(x, y) \in \Theta \iff x - y \in I, \forall~ x, y \in A.
\end{equation*}
Then $\Theta$ is an equivalence relation on $A.$
*Proof.* $\Theta$ is reflexive, since $0 \in I \implies x - x \in I \implies x \Theta x,$ $\forall~ x \in A.$
$\Theta$ is symmetric. For, suppose $x \Theta y.$
Then $x - y \in I \implies -(x - y) \in I \implies y - x \in I \implies y \Theta x.$
$\Theta$ is transitive. For, if $x \Theta y$ and $y \Theta z,$ then $(x - y), (y - z) \in I \implies (x - y) + (y - z) = x - z \in I \implies x \Theta z.$ $\qed$

**Note.** If $\Theta_x$ denotes the equivalence class containing $x \in A$ with respect to this equivalence relation, then
\begin{align*}
\Theta_x & = \{\, y \in A \mid (x, y) \in \Theta \,\} \\
& = \{\, y \in A \mid (y, x) \in \Theta \,\} \\
& = \{\, y \in A \mid y - x \in I \,\} \\
& = \{\, y \in A \mid y - x = i, \exists~ i \in I \,\} \\
& = \{\, y \in A \mid y = x + i, \exists~ i \in I \,\} \\
& = \{\, x + i \mid i \in I \,\} \\
& \defeq x + I.
\end{align*}
1. $(x, y) \in \Theta \iff \Theta_x = \Theta_y.$ That is,
\begin{equation*}
x - y \in I \iff x + I = y + I.
\end{equation*}
2. $(x, 0) \in \Theta \iff \Theta_x = \Theta_0.$ That is,
\begin{align*}
x - 0 \in I \iff x + I & = 0 + I\ \text{or,}\\
x \in I \iff x + I & = I.
\end{align*}

**Theorem.** Let $A$ be a ring, and $I$ an ideal of $A.$ Write
\begin{equation*}
A/I = \{\, x + I \mid x \in A \,\}.
\end{equation*}

Define $+$ and $\cdot$ on $A / I$ as:
\begin{align*}
(x + I) + (y + I) & = (x + y) + I \\
(x + I)(y + I) & = xy + I.
\end{align*}

Then $A/I$ is a ring under $+$ and $\cdot.$
*Proof.* First, note that the operations are well defined. For, if $x_1 + I = x_2 + I$ and $y_1 + I = y_2 + I,$ then
\begin{align*}
& x_1 - x_2, y_1 - y_2 \in I \\
& \implies (x_1 - x_2) + (y_1 - y_2) \in I \\
& \implies (x_1 + y_1) - (x_2 + y_2) \in I \\
& \implies (x_1 + y_1) + I = (x_2 + y_2) + I.
\end{align*}

Similarly,
\begin{align*}
x_1 y_1 - x_2 y_2 & = x_1(y_1 - y_2) + (x_1 - x_2) y_2 \\
& \in x_1 I + I y_2 \\
& = I + I \implies \\
x_1 y_1 + I & = x_2 y_2 + I.
\end{align*}

Now to see that $A/I$ is a ring, observe that $+$ and $\cdot$ are associative and commutative binary operations on $A/I,$ with $I$ as the additive identity and $1 + I$ as the multiplicative identity, and for $x \in A,$ $(x + I) + ((-x) + I) = (x + (-x)) + I = 0 + I = I,$ so every element has an additive inverse. Further, $\cdot$ distributes over $+,$ since for $x, y, z \in A,$
\begin{align*}
(x + I)((y + I) + (z + I)) & = (x + I)((y + z) + I) \\
& = x(y + z) + I\\
& = (xy + xz) + I\\
& = (xy + I) + (xz + I). \qed
\end{align*}

**Lemma.** Let $A$ be a ring with $I$ an ideal of $A.$ The map $f \colon A \to A/I$ defined by $f(x) = x + I,$ $\forall~ x \in A,$ is a surjective homomorphism (called the *canonical homomorphism* from $A$ to $A/I$).
*Proof.* Let $x, y \in A.$ Then $f(x + y) = (x + y) + I = (x + I) + (y + I) = f(x) + f(y)$ and $f(xy) = xy + I = (x + I)(y + I) = f(x)f(y).$ $f(1) = 1 + I,$ which is the multiplicative identity of $A/I.$ Therefore, $f$ is a homomorphism. To see that $f$ is surjective, observe that for any $x + I \in A/I,$ $f(x) = x + I$ by definition. $\qed$

---

## Lecture 03 (Correspondence Theorem)

**Correspondence Theorem.** Let $A$ be a ring, and $I$ and ideal of $A.$ Then there is an order-preserving, one-to-one correspondence between the set of all ideals of $A$ containing $I$ and the set of all ideals of $A/I.$
*Proof.* Let
\begin{align*}
\mathbb G & = \left\{\, \overline J \mid \overline J \unlhd A/I \,\right\} \\
\mathbb H & = \left\{\, J \mid I \subseteq J \unlhd A \,\right\}.
\end{align*}

Define a map $\psi \colon \mathbb G \to \mathbb H$ by $\psi(\overline J) = \pi^{-1}(\overline J),$ $\forall~ \overline J \in G,$ where $\pi \colon A \to A/I$ is the canonical homomorphism.

Now, we prove that $\psi$ is well-defined -- i.e., we show that $\pi^{-1}(\overline J)$ is an ideal containing $I.$
\begin{align*}
\pi^{-1}(\overline J) = \left\{\, x \in A \mid \pi(x) \in \overline J \,\right\}
\end{align*}

$I = 0 + I \in \overline J \implies \pi(0) \in \overline J \implies 0 \in \pi^{-1}(\overline J).$
Thus, $\pi(\overline J) \ne \varnothing.$

Take $x, y \in \pi^{-1}(\overline J).$ Then $\pi(x), \pi(y) \in \overline J \implies$ $\pi(x - y) = \pi(x) - \pi(y) \in \overline J \implies$ $x - y \in \pi^{-1}(\overline J).$

Take $x \in \pi^{-1}(\overline J)$ and $a \in A.$ Then $\pi(x) \in \overline J,$ $a \in A,$ $\pi(a) \in A/I \implies$ $\pi(xa) = \pi(x)\pi(a) \in \overline J$ (since $\overline J \unlhd A/I$) $\implies xa \in \pi^{-1}(\overline J).$
Thus, $\pi^{-1}(\overline J) \unlhd A.$

Take $x \in I.$ Then $x + I = I \in \overline J \implies$ $x + I \in \overline J \implies$ $\pi(x) \in \overline J \implies$ $x \in \pi^{-1}(\overline J).$ Thus, $I \subseteq \pi^{-1}(\overline J) = \psi(\overline J).$

Therefore, $\psi$ is well defined.

Now, let $\overline J_1, \overline J_2 \in \mathbb G$ be such that $\psi(\overline J_1) = \psi(\overline J_2).$

Take $x + I \in \overline J_1.$ Then $\pi(x) \in \overline J_1 \implies x \in \pi^{-1}(\overline J_1) = \pi(\overline J_2) \implies x + I \in \overline J_2.$ Therefore, $\overline J_1 \subseteq \overline J_2.$ Similarly, $\overline J_2 \subseteq \overline J_1.$
Thus, $\psi$ is $1$-$1.$

Let $J \in \mathbb H.$ Then $J$ is an ideal of $A$ containing $I.$ First, we show that $\pi(J)$ is an ideal of $A/I.$

$\pi(J) \ne \varnothing,$ since $\pi(0) \in \pi(J).$

Let $\pi(x), \pi(y) \in \pi(J).$ Then $\pi(x) - \pi(y) = \pi(x - y) \in \pi(J),$ since $x - y \in J.$

Let $\pi(x) \in J$ and $a + I \in A/I.$ Then $x \in J,$ $a \in A,$ $J \unlhd A \implies$ $xa \in J \implies$ $\pi(xa) \in \pi(J) \implies$ $\pi(x)\pi(a) \in \pi(J) \implies$ $\pi(x)(a + I) \in \pi(J).$
Thus, $\pi(J)$ is an ideal of $A/I.$

Now, we show that $\psi(\pi(J)) = J.$ That is, $\pi^{-1}(\pi(J)) = J.$

Let $x \in \pi^{-1}(\pi(J)).$ Then $\pi(x) \in \pi(J) \implies \pi(x) = \pi(y),$ $\exists~ y \in J$ $\implies \pi(x - y) = 0 I$ $\implies (x - y) + I = 0 + I$ $\implies x - y \in I \subseteq J.$ Then $x - y \in J$ and $y \in J \implies$ $x \in J.$
Thus, $\pi^{-1}(\pi(J)) \subseteq J.$

If $x \in J,$ then $\pi(x) \in \pi(J) \implies x \in \pi^{-1}(\pi(J)).$
Therefore, $\pi^{-1}(\pi(J)) = J.$

Hence, $\psi$ is a bijection.

Suppose that $\overline J_1 \subseteq \overline J_2 \in \mathbb G.$ We will show that $\psi(\overline J_1) \subseteq \psi(\overline J_2).$

Let $x \in \pi^{-1}(\overline J_1).$ Then $\pi(x) \in \overline J_1 \subseteq \overline J_2 \implies \pi(x) \in \overline J_2 \implies x \in \pi^{-1}(\overline J_2).$

Thus, $\pi^{-1}(\overline J_1) \subseteq \pi^{-1}(\overline J_2).$ $\qed$

**Exercise.** Find all ideals in $\mathbb Z / 12\mathbb Z.$
:::spoiler *Solution.*
By the Correspondence Theorem, this is equivalent to computing all ideals of $\mathbb Z$ that contain $12\mathbb Z.$ Since $\mathbb Z = \langle 1 \rangle$ is a principal ideal, all ideals of $\mathbb Z$ are of the form $\langle n \rangle,$ $\exists~ n \in \mathbb Z.$
$I = \langle n \rangle$ contains $12\mathbb Z = \langle 12 \rangle$ iff $n \mid 12.$
Hence, the ideals that contain $12$ are
$\langle 1 \rangle,$ $\langle 2 \rangle,$ $\langle 3 \rangle,$ $\langle 4 \rangle,$ $\langle 6 \rangle,$ and $\langle 12 \rangle.$
Therefore, the ideals of $\mathbb Z / 12\mathbb Z$ are
$\mathbb Z / \langle 12 \rangle,$ $2\mathbb Z / \langle 12 \rangle,$ $3\mathbb Z / \langle 12 \rangle,$ $4\mathbb Z / \langle 12 \rangle,$ $6\mathbb Z / \langle 12 \rangle,$ and $12\mathbb Z / \langle 12 \rangle = 0.$
:::
<p></p>

**Exercise.** Let $A$ be the non-commutative ring
\begin{equation*}
M_{2 \times 2}(\mathbb Z) = \left\{\, \begin{bmatrix} a & b \\ c & d \end{bmatrix} ~\bigg|~ a, b, c, d \in \mathbb Z \,\right\}.
\end{equation*}

Let
\begin{equation*}
I = \left\{\, \begin{bmatrix} 2a & 2b \\ 2c & 2d \end{bmatrix} ~\bigg|~ a, b, c, d \in \mathbb Z \,\right\},
\end{equation*}

which is a (two-sided) ideal of $A.$ Then show that
\begin{equation*}
A/I = \left\{\, \begin{bmatrix} p & q \\ r & s \end{bmatrix} + I ~\bigg|~ p, q, r, s \in \{0, 1\} \,\right\},
\end{equation*}

and check if it is commutative.
:::spoiler *Solution.*
Let $x = [a_{ij}] \in A.$ For each $i, j = 1, 2,$ $a_{ij} = 2q_{ij} + r_{ij},$ $\exists~ r_{ij} \in \{0, 1\},$ by the division algorithm. Then, $x = [r_{ij}] + [2q_{ij}] \implies x + I = [r_{ij}] + I,$ since $[2q_{ij}] \in I.$ Therefore, $A/I$ is as claimed.

$A/I$ is not commutative. Let $x = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix},$ $y = \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}.$ Then, $x + I \ne y + I,$ but $(x + I)(y + I) = xy + I = x + I$ and $(y + I)(x + I) = yx + I = y + I.$
:::
<p></p>

**Exercises.**
1. If $f \colon A \to B$ is a ring homomorphism and $I \unlhd A,$ then $f(I)$ is not necessarily an ideal of $B.$
2. Let $m, n \in \mathbb Z,$ and $J = \langle m \rangle,$ $I = \langle n \rangle.$ Then $J \subseteq I \iff n \mid m.$
3. There exists a surjective homomorphism $f: \mathbb Z \to \mathbb Z_n$ such that $\ker f \cong n \mathbb Z.$ Therefore, $\mathbb Z / n \mathbb Z \cong \mathbb Z_n.$
4. What is $\langle 1 \rangle$?
5. An element $x \in A$ is a unit if and only if $\langle x \rangle = A.$
6. If $I \unlhd A$ contains a unit, then $I = A.$

*Solutions.*
~ :::spoiler 1.
Let $A = \mathbb Z,$ $B = \mathbb Q,$ and $f \colon \mathbb A \to \mathbb B$ the inclusion map. Then $A \unlhd A,$ but $f(A) \not\unlhd B,$ since, for example, $1 \in f(A)$ and $\frac 1 2 \in B,$ but $(\frac 1 2) 1 \notin f(A).$
:::
:::spoiler 2.
Suppose $J \subseteq I.$ Then, since $m \in I = \langle n \rangle,$ by definition of $\langle n \rangle,$ $m = kn,$ $\exists~ k \in \mathbb Z,$ so $n \mid n.$
\
Now suppose $n \mid m,$ so $m = kn,$ $\exists~ k \in \mathbb Z.$ Then, for all $x \in J = \langle m \rangle,$ $\exists~ l \in \mathbb Z,$ $x = lm = lkn,$ so $x \in \langle n \rangle = I.$ Therefore, $J \subseteq I.$
:::
:::spoiler 3.
$\mathbb Z_n$ is the ring
\begin{equation*}
\mathbb Z_n \defeq \{0, 1, \ldots, n - 1\}
\end{equation*}
with addition and multiplication defined modulo $n.$
\
Define $f \colon \mathbb Z \to \mathbb Z_n$ by $f(m) = r,$ $\forall~ m \in \mathbb Z,$ where $r$ is the remainder obtained when $m$ is divided by $n.$ By the division algorithm, $0 \le r < n,$ so this map is well defined. Also, for any $r \in \mathbb Z_n,$ $f(r) = r,$ therefore $f$ is surjective.
\
To see that $f$ is a homomorphism, recall that if $m_i \cong r_i \mod n,$ $i = 1, 2,$ then
\begin{align*}
m_1 + m_2 & \cong (r_1 + r_2) \mod n \\
m_1 m_2 & \cong (r_1 r_2) \mod n
\end{align*}
and also, $1 \cong 1 \mod n,$ so $f(1) = 1,$ which is the multiplicative identity of $\mathbb Z_n.$
\
Now, for any $m \in \mathbb Z,$ $f(m) = 0 \iff m \cong 0 \mod n \iff m = kn,$ $\exists~ k \in \mathbb Z$ $\iff m \in n \mathbb Z.$ Thus, $\ker f = n \mathbb Z.$
:::
:::spoiler 4.
$\langle 1 \rangle = A,$ since $a = 1a \implies a \in \langle 1 \rangle$ for every $a \in A$ (and $\langle 1 \rangle \subseteq A$ by definition).
:::
:::spoiler 5.
Suppose $x \in A$ is a unit. Then $\exists~ y \in A$ such that $xy = 1.$ Therefore, $1 = xy \in \langle x \rangle,$ which implies that $\langle x \rangle = A.$
\
Now, suppose $\langle x \rangle = A.$ Then $1 \in \langle x \rangle \implies 1 = xy,$ $\exists~ y \in A,$ and therefore $x$ is a unit.
:::
:::spoiler 6.
If $I$ contains a unit, then $I$ contains $1.$ Then $\forall~ a \in A,$ $a = a1 \implies a \in I,$ therefore $A \subseteq I \implies I = A.$
:::
<p></p>

:::spoiler *Discussion on solution to 1.*
[Siddharth Bhat](https://github.com/bollu) remarked that he thought of the same solution as well and wondered if there was a more "general" example.

Yes, more generally, we could take $A$ to be any proper subring of $B$ and $f$ to be the inclusion homomorphism. Then $I = A$ is (always) an ideal of $A$ but $f(I)$ cannot be an ideal of $B,$ for it contains $1$ but is not equal to $B.$

Still more generally, we could take $I$ to be any ideal of a proper subring $A$ of $B,$ and take $f \colon A \to B$ to be the inclusion map. In some sense, this is the most general counterexample possible, in that any other counterexample can be factorised into some surjection and an injection of this kind. For, if $f \colon A \to B$ is surjective, then it maps ideals of $A$ to ideals of $B.$ And the image of a homomorphism is necessarily a subring of the codomain. Thus, given any homomorphism $f \colon A \to B$ and an ideal $I \unlhd A$ such that $f(I) \not\unlhd B,$ we get $f = \iota \circ \pi,$ with a surjection $\pi \colon A \to f(A)$ and the inclusion $\iota \colon f(A) \to B.$ Then $f(I)$ must be an ideal of the subring $f(A),$ but by assumption it is not an ideal of $B.$ That is, $f(I)$ is an ideal of $f(A),$ but $\iota(f(I))$ is not an ideal of $B,$ which is exactly the "generalised counterexample" described earlier.
:::

---

## Lecture 04 (Ideals and Homomorphisms)

**Note.** Let $I \subseteq A.$ Then
$I \unlhd A$ $\iff$ $\forall~ x, y \in I,$ $\forall~ a \in A,$ $xy, ax \in I$ $\iff$ $\forall~ x, y \in I,$ $\forall~ a \in A,$ $ax + y \in I.$

**Exercise.** Let $f \colon A \to B$ be a homomorphism. Then
1. If $J \unlhd B,$ then $f^{-1}(J)$ is an ideal of $A$ containing $\ker f.$
2. If $I \unlhd A,$ then $f(I) \unlhd f(A).$

:::spoiler *Solution.*
1. Let $a, x, y \in A.$
If $x, y \in f^{-1}(J),$ then $f(x), f(y) \in J$ $\implies f(ax + y) = f(a)f(x) + f(y) \in J$ $\implies ax + y \in f^{-1}(J).$
2. Let $f(a) \in A$ and $f(x), f(y) \in f(I).$ Then, $a \in A,$ $x, y \in I \unlhd A$ $\implies ax + y \in I \implies f(a)f(x) + f(y) = f(ax + y) \in f(I).$
:::
<p></p>

**Result.** Let $f \colon A \to B$ be a homomorphism. Then
\begin{equation*}
\ker f \defeq \{\, x \in A \mid f(x) = 0_B \,\}
\end{equation*}

is a proper ideal of $A.$
*Proof.* $f(0) = 0 \implies 0 \in \ker f \implies \ker f \ne \varnothing.$

Take $x, y \in \ker f.$ Then
$f(x - y) = f(x) - f(y) = 0 - 0 = 0 \implies x - y \in \ker f.$

Take $a \in A,$ $x \in \ker f.$ Then
$f(ax) = f(a)f(x) = f(a) \cdot 0 = 0 \implies ax \in \ker f.$

Thus, $\ker f \unlhd A.$

$f(1) = 1 \ne 0 \implies 1 \notin \ker f \implies \ker f \subsetneq A.$ $\qed$

**Result.** Let $f \colon A \to B$ be a homomorphism. Then $f(A)$ is a subring of $B.$
*Proof.* $f(A) = \{\, f(a) \mid a \in A \,\}.$
$A \ne \varnothing \implies f(A) \ne \varnothing.$

$f(x), f(y) \in f(A) \implies f(x) - f(y) = f(x - y) \in f(A).$

Thus, $(f(A), +)$ is a subgroup of $(B, +).$

Also, $f(x), f(y) \in f(A) \implies f(x)f(y) = f(xy) \in f(A),$ and $1_B = f(1_A) \implies 1_B \in f(A).$

Therefore, $f(A)$ is a subring of $B.$ $\qed$

**Theorem.** Let $f \colon A \to B$ be a homomorphism. Then
\begin{equation*}
A / \ker f \cong f(A).
\end{equation*}
Thus, if $f$ is surjective, then $A / \ker f \cong B.$
*Proof.* Let $I = \ker f,$ which is a proper ideal of $A.$
Define $g \colon A/I \to f(A)$ be $g(a + I) = f(a),$ $\forall~ a + I \in A/I.$

$g$ is well-defined and injective:
Let $a + I, b + I \in A/I.$ Then
\begin{align*}
a + I = b + I & \iff a - b \in I = \ker f \\
& \iff f(a - b) = 0 \\
& \iff f(a) - f(b) = 0 \\
& \iff f(a) = f(b) \\
& \iff g(a + I) = g(b + I).
\end{align*}

Clearly, $g$ is surjective.

$g$ is homomorphism:
\begin{align*}
g((a + I) + (b + I)) & = g((a + b) + I) \\
& = f(a + b) \\
& = f(a) + f(b) \\
& = g(a + I) + g(b + I)\\
g((a + I)(b + I)) & = g(ab + I) \\
& = f(ab) \\
& = f(a)f(b) \\
& = g(a + I)g(b + I).
\end{align*}

$g(1 + I) = f(1) = 1.$

Thus, $g$ is an isomorphism. $\qed$

**Definition.** Let $A$ be a ring. An element $x \in A$ is a *zero divisor* if $\exists~ y \in A$ such that $y \ne 0$ and $xy = 0.$

**Definition.** A commutative ring with $1$ without zero divisors is an *integral domain*.

**Examples.**
1. $(\mathbb Z, +, \cdot)$
2. Every field is an integral domain (but the converse is not true, as seen from 1.).

**Definition.** Let $A$ be a ring.
1. An element $x \in A$ is *nilpotent* if $x^n = 0,$ for some positive integer $n.$
2. An element $x \in A$ is a *unit* if $\exists~ y \in$ such that $xy = 1.$ Units are also called *invertible elements*.

**Result.** If $x$ is a unit in a ring, then its inverse is unique.
*Proof.* Let $x$ be a unit, and let $y, z$ be such that $xy = xz = 1.$ Then $xy = 1 \implies xyz = 1z \implies xzy = z \implies 1y = z \implies y = z.$ $\qed$

**Result.** Let $A$ be a ring. Then the set of all units of $A$ is a group with respect to multiplication.
*Proof.* Let $S = \{\, x \in A \mid x\ \text{is a unit} \,\}.$

$1 \in S.$

Take $x, y \in S.$ Then $\exists~ u, v \in A,$ such that $xu = yv = 1.$
Now, $xyuv = (xu)(yv) = 1 \cdot 1 = 1 \implies$ $xy$ is a unit $\implies xy \in S.$

Clearly, $\cdot$ is associative.

By definition, $\forall~ x \in S,$ $\exists~ y \in A,$ $xy = 1$ $\implies$ $y$ is a unit $\implies y \in S.$

Thus, $(S, \cdot)$ is a group. $\qed$

**Theorem.** Let $A$ be a non-zero ring. Then the following are equivalent:
1. $A$ is a field.
2. The only ideals in $A$ are $\langle 0 \rangle$ and $A.$
3. Any homomorphism from $A$ into a non-zero ring is injective.

*Proof.* 1. $\implies$ 2. Let $I \unlhd A.$
If $I = \langle 0 \rangle,$ there is nothing to prove.
Suppose $I \ne \langle 0 \rangle.$ Then $\exists~ a \in I,$ $a \ne 0$ $\implies \exists~ b \in A,$ $ab = 1$ $\implies 1 = ab \in I \implies I = A.$

2\. $\implies$ 3. Let $f \colon A \to B$ be a homomorphism into a non-zero ring $B.$
Then $\ker f \unlhd A,$ $\ker f \ne A \implies \ker f = \langle 0 \rangle.$
Thus, $f$ is injective.

3\. $\implies$ 1. Let $x \in A.$ Suppose $x$ is not a unit.
Then $\langle x \rangle$ is an ideal, and $A / \langle x \rangle$ is a quotient ring.
Claim: $A / \langle x \rangle$ is not a zero ring.
For, if $A / \langle x \rangle = \{ \langle x \rangle \},$ then $1 + \langle x \rangle = \langle x \rangle \implies 1 \in \langle x \rangle \implies$ $x$ is a unit, which is a contradiction.
Therefore, $A / \langle x \rangle$ is a non-zero ring.

Define $f \colon A \to A / \langle x \rangle$ as the canonical homomorphism $(a \mapsto a + \langle x \rangle).$ By 3., $f$ is injective $\implies \ker f = \langle 0 \rangle.$ But
\begin{align*}
\ker f & = \{\, a \in A \mid f(a) = \langle x \rangle \,\} \\
& = \{\, a \in A \mid a + \langle x \rangle = \langle x \rangle \,\} \\
& = \{\, a \in A \mid a \in \langle x \rangle \,\} \\
& = \langle x \rangle.
\end{align*}

$\implies \langle x \rangle = \langle 0 \rangle \implies x = 0.$

Thus, the only non-unit of $A$ is $0,$ and therefore, $A$ is a field. $\qed$

---

## Lecture 05 (Prime and Maximal Ideals)

**Result.** Let $I$ and $J$ be ideals of $A.$ Then
\begin{equation*}
I + J \defeq \{\, x + y \mid x \in I, y \in J\,\}
\end{equation*}
is the smallest ideal containing both $I$ and $J.$
*Proof.* $0 \in I, J \implies 0 = 0 + 0 \in I + J \implies I + J \ne \varnothing.$

Take $x_1 + y_1, x_2 + y_2 \in I + J.$ Then
\begin{equation*}
(x_1 + y_1) - (x_2 + y_2) = \underbrace{(x_1 - x_2)}_{\in I} + \underbrace{(y_1 - y_2)}_{\in J} \in I + J.
\end{equation*}

Let $x + y \in I + J,$ $a \in A.$ Then
\begin{equation*}
a(x + y) = \underbrace{ax}_{\in I} + \underbrace{ay}_{\in J} \in I + J.
\end{equation*}

Therefore, $I + J \unlhd A.$

Take $x \in I.$ Then $x + 0 = x \in I + J \implies I \subseteq I + J.$ Similarly, $J \subseteq I + J.$

Let $K$ be any ideal of $A$ such that $I \subseteq K$ and $J \subseteq K.$ Then $I + J \subseteq K + K \subseteq K.$

Therefore, $I + J$ is the smallest ideal of $A$ containing both $I$ and $J.$ $\qed$

**Definition.** Let $A$ be a ring. An ideal $P$ of $A$ is *prime* if it satisfies:
1. $P \ne A.$ ($P$ is proper)
2. If $x, y \in A,$ such that $xy \in P,$ then $x \in P$ or $y \in P.$

**Definition.** Let $A$ be a ring. An ideal $M$ of $A$ is *maximal* if it satisfies:
1. $M \ne A.$ ($M$ is proper)
2. If $H$ is any ideal of $A$ such that $M \subseteq H \subseteq A,$ then $H = M$ or $H = A.$

**Theorem.** Let $A$ be a ring.
1. An ideal $P$ of $A$ is prime $\iff$ $A/P$ is an integral domain.
2. An ideal $M$ of $A$ is maximal $\iff$ $A/M$ is a field.

*Proof.*
1. Suppose $P$ is a prime ideal of $A.$
Let $x + P, y + P \in A/P$ such that $(x + P)(y + P) = P.$ Then $xy + P = P \implies xy \in P \implies$ $x \in P$ or $y \in P,$ since $P$ is prime.
Thus, $x + P = P$ or $y + P = P$ in $A/P.$
Therefore, $A/P$ has no non-zero zero divisors.
Hence, $A/P$ is an integral domain.
\
Conversely, suppose that $A/P$ is an integral domain.
Clearly, $P \ne A.$
Let $x, y \in P$ such that $xy \in P.$ Then $xy + P = P \implies (x + P)(y + P) = P \implies x + P = P$ or $y + P = P,$ since $A/P$ is an integral domain. Therefore, $x \in P$ or $y \in P.$
Thus, $P$ is a prime ideal of $A.$
2. Suppose $M$ is a maximal ideal of $A.$
Since $M \subsetneq A,$ $A/M$ is a non-zero commutative ring with identity $1 + M.$
Let $a + M \ne M$ in $A/M.$
Then $a \notin M.$
Now consider $\langle a \rangle + M,$ which is the smallest ideal of $A$ containing both $\langle a \rangle$ and $M.$
Thus, $M \subseteq \langle a \rangle + M$ and $a \in \langle a \rangle + M,$ but $a \notin M,$ which implies that $M \subsetneq \langle a \rangle + M.$
Since $M$ is maximal, $\langle a \rangle + M = A.$
Then, $1 \in A = \langle a \rangle + M \implies 1 = ab + m,$ $\exists~ b \in A,$ $\exists~ m \in M$ $\implies 1 - ab = m \in M \implies 1 + M = ab + M = (a + M)(b + M).$
Therefore, $a + M$ is a unit in $A/M.$
Hence, $A/M$ is a field.
\
Conversely, suppose that $A/M$ is a field.
To prove that $M$ is maximal, let $H \unlhd A$ such that $M \subseteq H \subseteq A,$ and suppose $M \ne H.$
Let $h \in H$ and $h \notin M.$
Then $h + M$ is a non-zero element in $A/M$
$\implies \exists~ t + M \in A/M$ such that $(t + M)(h + M) = 1 + M \implies$ $th + M = 1 + M \implies$ $1 - th \in M \subseteq H \implies$ $1 - th \in H \implies $1 \in H,$ since $th \in H.$
Therefore, $H = A.$
Thus, $M$ is maximal. $\qed$

**Note.** $P$ is maximal $\iff$ $A/P$ is a field $\implies$ $A/P$ is an integral domain $\iff$ $P$ is prime.
Thus, every maximal ideal is a prime ideal, but not conversely.

**Exercise.** Let $A$ be a ring. Then $\langle 0 \rangle$ is prime $\iff$ $A$ is an integral domain.
:::spoiler *Solution.*
Suppose $\langle 0 \rangle$ is prime. Let $a, b \in A,$ $ab = 0 \in \langle 0 \rangle.$ Then $a \in \langle 0 \rangle$ or $b \in \langle 0 \rangle$ $\implies a = 0$ or $b = 0$ $\implies$ $A$ is an integral domain.

Suppose $A$ is an integral domain. If $ab \in \langle 0 \rangle,$ then $ab = 0 \implies$ $a = 0$ or $b = 0$ $\implies a \in \langle 0 \rangle$ or $b \in \langle 0 \rangle$ $\implies$ $\langle 0 \rangle$ is prime.
:::
<p></p>

**Note.** $\mathbb Z$ is an integral domain $\implies$ $\langle 0 \rangle$ is a prime ideal. But $\langle 0 \rangle$ is not maximal in $\mathbb Z,$ since $\mathbb Z / \langle 0 \rangle \cong \mathbb Z$ is not a field. $\langle 0 \rangle \subsetneq \langle 2 \rangle \subsetneq \mathbb Z,$ and $\langle 2 \rangle$ is a maximal ideal in $\mathbb Z.$

**Exercise.** Prove the following:
1. Let $A = \{\, f \colon \mathbb R \to \mathbb R \mid f\ \text{is continuous} \,\}.$ Then $I = \{\, f \in A \mid f(0) = 0 \,\}$ is a maximal ideal of $A.$
2. Let $A = \mathbb Z[x],$ the ring of polynomials in $x$ with integer coefficients. Then $I = \{\, f \in A \mid f(0) = 0 \,\}$ is a prime ideal of $A,$ but is not maximal.
3. Let $A = \mathbb Z \times \mathbb Z.$ Then $I = \{\, (a, 0) \mid a \in \mathbb Z \,\}$ is a prime ideal of $A,$ but is not maximal.

*Solutions.*
~ :::spoiler 1.
Let $f + I \in A/I,$ $f + I \ne I.$ Then $f \notin I \implies f(0) \ne 0.$
Now, observe that $f + I = g + I,$ where $g$ is defined by $g(x) = f(0),$ $\forall~ x \in \mathbb R,$ since $g(0) - f(0) = 0 \implies g - f \in I.$
But $g(x) \ne 0,$ $\forall~ x \in \mathbb R,$ and therefore we can define $h \colon \mathbb R \to \mathbb R$ by $h(x) = \frac{1}{g(x)},$ $\forall~ x \in \mathbb R,$ which is continuous, and $g(x)h(x) = 1 \implies (f + I)(h + I) = (g + I)(h + I) = 1 + I.$ Thus, $f$ is a unit in $A/I.$ Therefore, $A/I$ is a field, which implies that $I$ is maximal in $A.$
**Alternative solution:** Define $\varphi \colon A \to \mathbb R$ be $\varphi(f) = f(0),$ $\forall~ f \in A.$ Then $\varphi$ is a surjecive homomorphism and $\ker \varphi = \{\, f \in A \mid f(0) = 0 \,\} = I.$ Thus, $A/I = A / \ker \varphi \cong \mathbb R,$ which is a field. Therefore, $I$ is maximal in $A.$
:::
:::spoiler 2.
Let $p(x), q(x) \in A,$ such that $p(x)q(x) \in I.$ Then $p(0)q(0) = 0 \implies p(0) = 0$ or $q(0) = 0$ $\implies$ $p(x) \in I$ or $q(x) \in I.$ Therefore, $I$ is a prime ideal.
To see that it is not a maximal ideal, consider $p(x) = 2 + x.$ $p(0) = 2 \ne 0 \implies p(x) \notin I \implies p(x) + I \ne I$ in $A/I.$ Suppose that $p(x) + I$ is a unit. Then there exists $q(x) \in A$ such that $(p(x) + I)(q(x) + I) = 1 + I \implies$ $p(x)q(x) + I = 1 + I \implies$ $p(x)q(x) - 1 \in I$ \implies$ $p(0)q(0) - 1 = 0$ $\implies 2q(0) = 1.$ But since $q(0)$ is an integer, this is not possible. Thus, $p(x) + I$ is non-zero non-unit in $A/I,$ and therefore $A/I$ is not a field. Hence, $I$ is not maximal in $A.$
**Alternative solution:** Define $\varphi \colon A \to \mathbb Z$ by $\varphi(f) = f(0),$ $\forall~ f \in A.$ Then $f$ is a surjective homomorphism and $\ker f = \{\, f \in A \mid f(0) = 0 \,\} = I.$ Hence, $A / I = A / \ker f \cong \mathbb Z,$ which is an integral domain but not a field. Therefore, $I$ is a prime ideal but is not maximal.
:::
:::spoiler 3.
Define $f \colon A \to \mathbb Z,$ by $f(a, b) = b,$ $\forall~ (a, b) \in A.$ Then $f$ is a surjective homomorphism, and
\begin{align*}
\ker f & = \{\, (a,b) \in A \mid b = 0 \,\} \\
& = \{\, (a, 0) \mid a \in A \,\} \\
& = I.
\end{align*}

Then $A / I = A / \ker f \cong \mathbb Z,$ which is an integral domain, but not a field. Therefore, $I$ is a prime ideal but is not maximal.
:::

---

## Lecture 06 (Principal Ideal Domains and Maximal Ideals)

**Exercise.** Find all maximal ideals in
1. $\mathbb Z_8$
2. $\mathbb Z / 10 \mathbb Z$
3. $\mathbb Z_{12}$
4. $\mathbb Z / n \mathbb Z,$ $n > 1.$

*Solutions.*
~ :::spoiler 1.
$2 \mathbb Z_8 = \{0, 2, 4, 8\}$ is the unique maximal ideal.
:::
:::spoiler 2.
$2 \mathbb Z / 10 \mathbb Z = \left\{ \bar 0, \bar 2, \bar 4, \bar 6, \bar 8 \right\}$ and $5 \mathbb Z / 10 \mathbb Z = \left\{ \bar 0, \bar 5 \right\}.$
:::
::: spoiler 3.
$2 \mathbb Z_{12} = \{0, 4, 6, 8, 10\}$ and $3 \mathbb Z_{12} = \{0, 3, 6, 9\}.$
:::
:::spoiler 4.
The maximal ideals of $\mathbb Z / n \mathbb Z$ are exactly all the ideals of the form $p (\mathbb Z / n \mathbb Z),$ where $p$ is a prime factor of $n.$
Every ideal of $\mathbb Z / n \mathbb Z$ is of the form $m(\mathbb Z / n \mathbb Z)$ for some $m \mid n,$ and the corresponding quotient is isomorphic to $\mathbb Z / m \mathbb Z.$ But $\mathbb Z / m \mathbb Z$ is a field iff $m$ is prime.
:::

**Exercise.** Let $A = \mathbb Z \times \mathbb Z,$ and $I = \{\, (3x, y) \mid x, y \in \mathbb Z \,\}.$ Prove that $I$ is a maximal ideal of $A.$
:::spoiler *Solution.*
Define $f \colon A \to \mathbb Z_3$ by $f(a, b) = r,$ $\forall~ (a, b) \in A,$ where $r \in \mathbb Z_3$ is such that $r \equiv a \mod 3.$ This is a well-defined surjective homomorphism, and
\begin{equation*}
\ker f = \{\, (a, b) \in A \mid a \equiv 0 \mod 3\,\} = \{\, (3x, b) \mid x, b \in \mathbb Z \,\} = I.
\end{equation*}

Hence, $A/I = A / \ker f \cong \mathbb Z_3,$ which is a field. Therefore, $I$ is a maximal ideal in $A.$
:::
<p></p>

**Definition.** An integral domain in which every ideal is principal is a *principal ideal domain* (PID).

**Lemma.** In a PID, every non-zero prime ideal is maximal.
*Proof.* Let $A$ be a PID, and let $I$ be a non-zero prime ideal of $A.$
Then $I = \langle x \rangle,$ $\exists~ 0 \ne x \in A.$

To show that $I$ is maximal, let $J$ be an ideal of $A,$ such that $I \subsetneq J \subseteq A.$
Then $J = \langle y \rangle,$ $\exists~ y \in A,$ and $\langle x \rangle \subsetneq \langle y \rangle.$
Now, $x \in \langle x \rangle \subsetneq \langle y \rangle \implies x = yz,$ $\exists~ z \in A.$

Then $yz = x \in \langle x \rangle \implies$ $y \in x$ or $z \in x,$ since $I$ is prime.
Since $\langle x \rangle \subsetneq \langle y \rangle,$ we have $z \in \langle x \rangle$ $\implies$
$z = xt,$ $\exists~ t \in A.$
Now, $x = yz = yxt = xyt \implies$
$x - xyt = x(1 - yt) = 0 \implies$
$x = 0$ or $1 - yt = 0.$
But $x \ne 0,$ since $\langle x \rangle \ne 0$ $\implies$ $1 - yt = 0$ $\implies$
$y$ is a unit.
Thus, $J = \langle y \rangle = A.$ $\qed$

**Lemma.** Every non-zero ring contains at least one maximal ideal.
*Proof.* Suppose $A$ is a non-zero ring.
Let $\mathcal S = \{\, I \unlhd A \mid I \ne A \,\}.$
Then $\langle 0 \rangle \in \mathcal S \implies \mathcal S \ne \varnothing.$
Define $\le$ on $\mathcal S$ as $I_1 \le I_2 \iff I_1 \subseteq I_2,$ $\forall~ I_1, I_2 \in \mathcal S.$
Then $\le$ is a partial order on $\mathcal S.$

Let $\mathcal C = \{ I_\alpha \}_{\alpha \in \Delta}$ be a chain of ideals in $\mathcal S$ (i.e., $I_\alpha \subseteq I_\beta$ or $I_\beta \subseteq I_\alpha,$ $\forall~ \alpha, \beta \in \Delta$).
Define $I = \bigcup\limits_{\alpha \in \Delta} I_\alpha.$ Then $I$ is an upper bound of $\mathcal C$ in $\mathcal S.$
:::spoiler *Justification.*
Take $x, y \in I$ and $a \in A.$
Then $x \in I_\alpha,$ $y \in I_\beta,$ $\exists~ \alpha, \beta \in \Delta.$ Without loss of generality, $I_\alpha \subseteq I_\beta.$
Then $x, y \in I_\beta \implies x - y \in I_\beta \subseteq I,$ and $ax \in I_\beta \subseteq I,$ since $I_\beta \unlhd A.$
:::
<p></p>

Claim: $I \subsetneq A.$
For, suppose otherwise that $I = A.$
Then $1 \in A = I \implies 1 \in I_\alpha,$ $\exists~ \alpha \in \Delta$ $\implies I_\alpha = A,$ which is a contradiction.

Hence, $I \in \mathcal S.$
Therefore, every chain in $\mathcal S$ has an upper bound in $\mathcal S.$
Therefore, by Zorn's lemma, $\mathcal S$ has a maximal element, say $M.$
Then $M$ is a maximal ideal of $A.$ $\qed$

**Corollary.** Every proper ideal is contained in some maximal ideal.
*Proof.* Let $I \unlhd A,$ $I \ne A.$
Then $A/I$ is a non-zero ring, and hence has some maximal ideal $M/I.$
By the Correspondence Theorem, $M$ is a maximal ideal of $A$ containing $I.$ $\qed$

---

## Lecture 07 (Local Rings and Nilradicals)

**Corollary.** Every non-unit is contained in a maximal ideal.
*Proof.* Let $x$ be a non-unit of $A.$ Then $\langle x \rangle$ is a proper ideal of $A,$ and hence is contained in some maximal ideal $M.$ Therefore, $x \in \langle x \rangle \subseteq M.$ $\qed$

**Definition.** A ring $A$ with a unique maximal ideal is a *local ring*.

**Example.** Any field is a local ring, since $\langle 0 \rangle$ is its unique maximal ideal.

**Definition.** A ring with only a finite number of maximal ideals is *semi-local*.

**Theorem.** Let $A$ be a ring and $M$ a proper ideal of $A$ such that every element of $A - M$ is a unit. Then $A$ is a local ring with maximal ideal $M.$
*Proof.* Suppose every element of $A - M$ is a unit.
Let $J$ be a proper ideal of $A,$ and let $x \in J.$ If $x \notin M,$ then $x \in A - M,$ and hence it is a unit $\implies J = A,$ which is a contradiction. Therefore, $x \in M.$
Hence, $J \subseteq M.$

Now, let $H$ be a maximal ideal of $A.$ Then $H$ is proper, and hence is contained in $M.$ Since $H$ is maximal, and $M$ is proper, $H = M.$
Therefore, $M$ is the unique maximal ideal of $A.$
Hence, $A$ is local. $\qed$

**Theorem.** Let $A$ be a ring, and $M$ a maximal ideal of $A$ such that every element in $1 + M \defeq \{\, 1 + x \mid x \in M \,\}$ is a unit. Then $A$ is a local ring with unique maximal ideal $M.$
*Proof.* It is enough to show that every element of $A - M$ is a unit.
Let $x \in A - M.$ Then the ideal $\langle x \rangle + M = A,$ for $M \subsetneq \langle x \rangle + M$ and $M$ is maximal.
Now, $1 \in A = \langle x \rangle + M \implies 1 = xy + M,$ $\exists~ y \in A,$ $\exists~ m \in M$ $\implies xy = 1 - m \in 1 + M \implies$ $xy$ is a unit $\implies xyz = 1,$ $\exists~ z \in A \implies$ $x$ is a unit.
Therefore, every element of $A - M$ is a unit.
Hence, $A$ is local, and $M$ is its unique maximal ideal. $\qed$

**Examples of nilpotents.**
1. $0$ is nilpotent in any ring.
2. Let $A = \mathbb Z_8 = \{\bar 0, \bar 1, \bar 2, \bar 3, \bar 4, \bar 5, \bar 6, \bar 7\}.$ Then the nilpotent elements of $A$ are $\bar 0,$ $\bar 2,$ $\bar 4,$ and $\bar 6.$
3. Let $A = \mathbb Z_6.$Then the only nilpotent of $A$ is $\bar 0.$

**Remark.** A nilpotent element is always a zero divisor, but the converse is not true.
*Proof.* Let $x$ be a nilpotent element. If $x = 0,$ it is a zero divisor. If $x \ne 0,$ let $m$ be the least positive integer such that $x^m = 0.$ Then $m \ge 2 \implies x^m = x \cdot x^{m-1} = 0,$ and $x^{m-1} \ne 0.$
Hence, $x$ is a zero divisor.

To see that the converse is not true, note that $\bar 2$ is a zero divisor in $\mathbb Z_6$ (since $\bar 2 \cdot \bar 3 = \bar 0$), but it is not nilpotent. $\qed$

**Theorem.** Let $A$ be a ring, and $N$ the set of all nilpotent elements in $A.$ Then $N$ is an ideal of $A,$ and $A/N$ contains no non-zero nilpotents.
*Proof.* $N = \{\, x \in A \mid x^n = 0, \exists~ n \in \mathbb N \,\}.$
$0 \in N \implies N \ne \varnothing.$

Take $x, y \in N.$ Then $x^m = 0,$ $x^n = 0,$ $\exists~ m, n \in \mathbb N.$ Now,
\begin{equation*}
(x - y)^{m + n} = \sum_{k=0}^{m+n} (-1)^{m+n-k} \binom{m+n}{k} x^k y^{m+n-k}.
\end{equation*}

For every $k \le m + n,$ either $k \ge m$ or $m+n-k \ge n$ $\implies$ either $x^k = 0$ or $y^{m+n-k} = 0.$
Therefore, $(x - y)^{m+n} = 0 \implies x - y \in N.$

Take $a \in A,$ $x \in N.$ Then $x^n = 0,$ $\exists~ n \in \mathbb N$
$\implies (ax)^n = a^n x^n = 0 \implies ax \in N.$

Hence, $N$ is an ideal.

Let $x + N \in A/N,$ such that $x + N$ is nilpotent.
Then $(x + N)^k = N,$ $\exists~ k > 0$
$\implies x^k \in N \implies (x^k)^t = 0,$ $\exists~ t > 0$ $\implies x^{kt} = 0 \implies$
$x$ is nilpotent $\implies x \in N \implies$
$x + N = N.$

Hence, $A/N$ has no non-zero nilpotent elements. $\qed$

**Definition.** The ideal $N$ consisting of all nilpotents of $A$ is the *nilradical* of $A.$

**Theorem.** The nilradical of $A$ is the intersection of all prime ideals of $A.$
*Proof.* Let $N$ be the nilradical of $A.$ We show that $N = \bigcap\limits_{P\ \text{prime}} P.$
Let $x \in N.$ Then $x^n = 0,$ $\exists~ n > 0.$ Thus, for every prime ideal $P,$ $x^n = 0 \in P \implies x \in P,$ by induction on $n.$ Therefore, $x \in \bigcap\limits_{P\ \text{prime}} P.$
Hence, $N \subseteq \bigcap\limits_{P\ \text{prime}} P.$

Now, suppose $x \notin N.$ Then $x^k \ne 0,$ $\forall~ k > 0.$ Define
\begin{equation*}
\mathcal S = \{\, J \mid J \unlhd A,\, x^k \notin J, \forall~ k > 0 \,\}.
\end{equation*}

Clearly, $\langle 0 \rangle \in \mathcal S \implies \mathcal S \ne \varnothing.$

$\mathcal S$ is a poset under set inclusion, and satisfies the hypothesis of Zorn's Lemma. Therefore, by Zorn's Lemma, $\mathcal S$ has a maximal element, say $P.$ $P$ is a proper ideal of $A,$ since $x^k \notin P,$ $\forall~ k > 0.$

In particular, $x \notin P.$

Claim: $P$ is a prime ideal.
Let $a, b \in A,$ $a \notin P,$ $b \notin P.$
Consider the ideals $\langle a \rangle + P$ and $\langle b \rangle + P.$
$P \subsetneq \langle a \rangle + P,$ $P \subsetneq \langle b \rangle + P.$
Since $P$ is maximal in $\mathcal S,$ $\langle a \rangle + P \notin \mathcal S$ and $\langle b \rangle + P \notin \mathcal S$
$\implies x^{k_1} \in \langle a \rangle + P,$ $x^{k_2} \in \langle b \rangle _ P,$ $\exists~ k_1, k_2 > 0$
$\implies x^{k_1} = at_1 + p,$ $x^{k_2} = bt_2 + q,$ $\exists~ t_1, t_2 \in A,$ $\exists~ p, q \in P.$
Then
$x^{k_1+k_2} = x^{k_1} x^{k_2} = abt_1t_2 + \underbrace{at_1q}_{\in P} + \underbrace{bt_2p}_{\in P} + \underbrace{pq}_{\in P}.$
If $ab \in P,$ then $x^{k_1 + k_2} \in P,$ which is a contradiction, since $P \in \mathcal S.$
Therefore, $ab \notin P.$
Hence, $P$ is prime.

Now, $x \notin P \implies x \notin \bigcap\limits_{P\ \text{prime}} P.$

Therefore, $N = \bigcap\limits_{P\ \text{prime}} P.$ $\qed$

---

## Lecture 08 (The Jacobson Radical and Lattice of Ideals)

**Definition.** Let $A$ be a ring. Then the *Jacobson radical* of $A$ is the intersection of all maximal ideals in $A.$

**Theorem.** Let $A$ be a ring, and $M$ the Jacobson radical of $A.$ Then $x \in M \iff 1 - xy$ is a unit for all $y \in A.$
*Proof.* By definition,
\begin{equation*}
M = \bigcap_{M_i\ \text{maximal}} M_i.
\end{equation*}

Let $x \in M,$ and suppose that $1 - xy$ is not a unit, $\exists~ y \in A.$
Then there exists a maximal ideal $M_0$ of $A$ such that $1 - xy \in M_0.$

Clearly, $M \subseteq M_0 \implies x \in M_0 \implies$ $xy \in M_0 \implies 1 - xy + xy = 1 \in M_0 \implies$ $M_0 = A,$ which is a contradiction.

Therefore, $1 - xy$ is a unit for all $y \in A.$

Conversely, suppose that $1 - xy$ is a unit for all $y \in A.$

Claim: $x \in M.$
If $x \notin M = \bigcup\limits_{M_i\ \text{maximal}} M_i \implies x \notin M_0,$ for some maximal ideal $M_0.$

Then, $M_0 \subseteq \langle x \rangle + M_0 \implies \langle x \rangle + M_0 = A,$ since $M_0$ is maximal.

Now, $1 \in A = \langle x \rangle + M_0 \implies 1 = xy + m$ $\exists~ y \in A,$ $m \in M_0.$
$\implies 1 - xy = m \in M_0,$ and $1 - xy$ is a unit $\implies M_0 = A,$ a contradiction.

Therefore, $x \in M.$ $\qed$

**Corollary.** If $x$ is in the Jacobson radical, then $1 - x$ is a unit.

**Theorem.** Let $A$ be a ring and let $\{ I_\alpha \}_{\alpha \in \Delta}$ be a family of ideals in $A.$ Then
\begin{equation*}
\sum_{\alpha \in \Delta} I_\alpha \defeq \left\{\, \sum x_\alpha \mid x_\alpha \in I_\alpha\ \text{and almost all}\ x_\alpha = 0 \,\right\}
\end{equation*}

is the smallest ideal containing each $I_\alpha.$
*Proof.* $0 \in \sum I_\alpha \implies \sum I_\alpha \ne \varnothing.$

Take $\sum x_\alpha, \sum y_\alpha \in \sum I_\alpha.$ Then
$\sum x_\alpha - \sum y_\alpha = \sum (x_\alpha - y_\alpha) \in I_\alpha.$

Take $a \in A,$ $\sum x_\alpha \in \sum I_\alpha.$
Then $\forall~ \alpha \in \Delta,$ $x_\alpha \in I_\alpha \unlhd A \implies ax_\alpha \in I_\alpha \implies a \sum x_\alpha = \sum a x_\alpha \in I_\alpha.$

Thus, $I_\alpha \unlhd A.$

Now, let $J \unlhd A$ such that $I_\alpha \subseteq J,$ $\forall~ \alpha \in \Delta.$
Take $\sum x_\alpha \in \sum I_\alpha.$ Then $\forall~ \alpha \in \Delta,$ $x_\alpha \in I_\alpha \subseteq J \unlhd A \implies \sum x_\alpha \in J.$

Thus, $\sum I_\alpha \subseteq J.$

Therefore, $\sum I_\alpha$ is the smallest ideal containing each $I_\alpha.$ $\qed$

**Lemma.** Let $A$ be a ring, and $\{I_\alpha\}_{\alpha \in \Delta}$ a family of ideals in $A.$ Then $\bigcap\limits_{\alpha \in \Delta} I_\alpha$ is the largest ideal of $A$ contained in all $I_\alpha.$
*Proof.* Let $I = \bigcap\limits_{\alpha \in \Delta} I_\alpha.$

$\forall~ \alpha \in \Delta,$ $0 \in I_\alpha$ $\implies 0 \in I \implies I \ne \varnothing.$

Take $x, y \in \bigcap I_\alpha.$ Then $\forall~ \alpha \in \Delta,$ $x, y \in I_\alpha \unlhd A \implies xy \in I_\alpha \implies xy \in I.$

Take $a \in A$ and $x \in I.$ Then $\forall~ \alpha \in \Delta,$ $x \in I_\alpha \implies ax \in I_\alpha \implies ax \in I.$

Thus, $I \unlhd A.$

If $J$ is any ideal contained in all each $I_\alpha,$ then $J \subseteq I_\alpha,$ $\forall~ \alpha \in \Delta$ $\implies J \subseteq \bigcap\limits_{\alpha \in \Delta} I_\alpha = I.$

Therefore, $I$ is the largest ideal of $A$ contained in all $I_\alpha.$ $\qed$

**Theorem.** The set of all ideals of a ring $A$ is a complete lattice with respect to set inclusion.
*Proof.* Let $\mathcal S = \{\, I \mid I \unlhd A \,\}.$
$\langle 0 \rangle, A \in \mathcal S \implies \mathcal S \ne \varnothing.$
Define $\le$ on $\mathcal
S$ as $I_1 \le I_2 \iff I_1 \subseteq I_2.$
Clearly, $\le$ is a partial order relation on $\mathcal S.$

Let $\{I_\alpha\}_{\alpha \in \Delta}$ be a non-empty subset of $\mathcal S.$
Then $\sum\limits_{\alpha \in \Delta} I_\alpha$ and $\bigcap\limits_{\alpha \in \Delta} I_\alpha$ are the least upper bund and the greatest lower bound, respectively, of $\{I_\alpha\}_{\alpha \in \Delta}.$

Therefore, $(\mathcal S, \le)$ is a complete lattice. $\qed$

---

## Lecture 09 (Products of Ideals and Direct Products of Rings)

**Definition.** Let $A$ be a ring, and $I$ and $J$ ideals of $A.$ Then the *product* of $I$ and $J$ is defined as
\begin{equation*}
IJ = \left\{\, \sum_{i=1}^n x_i y_i ~\bigg|~ x_1, \ldots, x_n \in I,\, y_1, \ldots, y_n \in J, n \in \mathbb N\,\right\}.
\end{equation*}

**Definition.** Let $A$ be a ring, and $I$ and $J$ ideals of $A.$ Then $I$ and $J$ are *coprime* if $I + J = A.$

**Note.** $IJ \subseteq I \cap J.$
If $x-i \in I,$ $x_iy_i \in I,$ and if $y_i \in J,$ $x_iy_i \in J.$ Therefore, $x_i y_i \in I \cap J \implies \sum x_i y_i \in I \cap J.$

**Note.** Let $I$ and $J$ be ideals of $A.$ Then $I$ and $J$ are coprime iff $\exists~ x \in I,$ $\exists~ y \in J,$ such that $x + y = 1.$
($\implies$) $I + J = A \implies 1 \in I + J \implies \exists~ x \in I,$ $y \in J,$ $1 = x + y.$
($\impliedby$) $\exists~ x \in I,$ $y \in J,$ $x + y = 1 \implies \forall~ a \in A,$ $a = a \cdot 1 = ax + ay \in I + J.$

**Note.** If $I$ and $J$ are coprime, then $IJ = I \cap J.$
*Proof.* Clearly, $IJ \subseteq I \cap J.$
take $x \in I \cap J.$ Then $x \in I,$ $x \in J.$
$A = I + J \implies \exists~ a \in I,$ $b \in J,$ $1 = a + b$
$\implies x = xa + xb \in IJ.$
Hence, $I \cap J \subseteq IJ.$ $\qed$

**Theorem.** Let $A_1, \ldots, A_n$ be rings. Then
\begin{equation*}
A = \prod_{i=1}^n A_i \defeq \left\{, (x_1, \ldots, x_n) \mid x_i \in A_i,\, i = 1, \ldots, n\,\right\}
\end{equation*}

is a ring called the *direct product* of $A_1, \ldots, A_n,$ under the operations:
1. $(x_1, \ldots, x_n) + (y_1, \ldots, y_n) = (x_1 + y_1, \ldots, x_n + y_n)$
2. $(x_1, \ldots, x_n)(y_1, \ldots, y_n) = (x_1 y_1, \ldots, x_n y_n).$

In particular, if each $A_i$ is commutative with identity, then $A$ is also commutative with identity.
*Proof.* $(A, +)$ is an Abelian group, being the direct product of the groups $(A_1, +), \ldots, (A_n, +).$
$(A, \cdot)$ is clearly a semigroup, and $\cdot$ distributes over $+.$
Commutativity of all $A_i$ implies commutativity of componentwise multiplication.
If each $A_i$ has $1,$ then the identity of $A$ is $(1, \ldots, 1).$ $\qed$

**Theorem.** Let $A_1, \ldots, A_n$ be rings, and $\prod\limits_{i=1}^n A_i$ their direct product. For $1 \le i \le n,$ define
\begin{equation*}
p_i \colon \prod_{i=1}^n A_i \to A_i
\end{equation*}

by $p_i(x_1, \ldots, x_n) = x_i,$ $\forall~ (x_1, \ldots, x_n) \in \prod\limits_{i=1}^n A_i,$ is an epimorphism.
*Proof.* Let $(x_1, \ldots, x_n), (y_1, \ldots, y_n) \in \prod\limits_{i=1}^n A_i.$
Note that both $+$ and $\cdot$ are defined componentwise, and let $*$ denote either in the following:
\begin{align*}
p_i(x_1, \ldots, x_n)*(y_1, \ldots, y_n) & = p_i(x_1 * y_1, \ldots, x_n * y_n) \\
& = x_i * y_i \\
& = p(x_1, \ldots, x_n) * p(y_1, \ldots, y_n).
\end{align*}

$p_i(1, \ldots, 1) = 1 \in A_i.$

Hence, $p_i$ is a homomorphism.
Take $x_i \in A_i.$ Then $p_i(0_1, \ldots, 0_{i-1}, x_i, 0_{i+1}, \ldots, 0_n) = x_i.$
Therefore, $p_i$ is an epimorphism. $\qed$

**Lemma.** Let $I_1, \ldots, I_n$ be ideals of a ring $A.$ Define $f \colon A \to \prod\limits_{i=1}^n A/I_i$ by $f(x) = (x + I_1, \ldots, x + I_n),$ $\forall~ x \in A.$ Then $f$ is a homomorphism.
*Proof.* Take $x, y \in A.$ Let $*$ denote either $+$ or $\cdot$ in the following.
\begin{align*}
f(x*y) & = ((x*y) + I_1, \ldots, (x*y) + I_n) \\
& = ((x + I_1)*(y + I_1), \ldots, (x + I_n)*(y + I_n)) \\
& = (x + I_1, \ldots, x + I_n) * (y + I_1, \ldots, y + I_n) \\
& = f(x) * f(y).
\end{align*}

$f(1) = (1 + I_1, \ldots, 1 + I_n),$ which is the identity in $\prod\limits_{i=1}^n A/I_i,$ since $1 + I_i$ is the identity of $A/I_i.$ $\qed$

**Lemma.** Let $I_1, \ldots, I_n$ be ideals of a ring $A.$ If $I_i$ and $I_j$ are coprime or all $i \ne j,$ then
\begin{equation*}
\prod_{i=1}^n I_i = \bigcap_{i=1}^n I_i.
\end{equation*}

*Proof.* The proof is by induction of $n.$ The case $n = 1$ is trivial, and the case $n = 2$ is already established.

Inductive hypothesis: $\prod\limits_{i=1}^{n-1} I_i = \bigcap\limits_{i=1}^{n-1} I_i.$

We show that $\prod\limits_{i=1}^n I_i = \bigcap\limits_{i=1}^n I_i.$

\begin{align*}
1 \in A = I_1 + I_n & \implies 1 = x_1 + y_1, \qquad x_1 \in I_1,\, y_1 \in I_n.\\
1 \in A = I_2 + I_n & \implies 1 = x_2 + y_2, \qquad x_2 \in I_2,\, y_2 \in I_n.\\
& \quad \vdots \\
1 \in A = I_{n-1} + I_n & \implies 1 = x_{n-1} + y_{n-1}, \qquad x_{n-1} \in I_{n-1},\, y_{n-1} \in I_n.\\
\end{align*}

Then $x_i = 1 - y_i,$ $i = 1, \ldots, n - 1$ $\implies \prod\limits_{i=1}^{n-1} x_i = \prod\limits_{i=1}^{n-1} (1 - y_i).$

Now, noting that $1 - \prod\limits_{i=1}^{n-1} (1 - y_i) \in I_n,$
\begin{align*}
\prod_{i=1}^{n-1} x_i + 1 - \prod_{i=1}^{n-1} (1 - y_i) & \in \prod_{i=1}^{n-1} I_i + I_n \implies \\
\prod_{i=1}^{n-1} x_i + 1 - \prod_{i=1}^{n-1} x_i & \in \prod_{i=1}^{n-1} I_i + I_n \implies \\
1 & \in \prod_{i=1}^{n-1} I_i + I_n \implies \\
\prod_{i=1}^{n-1} I_i + I_n & = A.
\end{align*}

Therefore,
\begin{align*}
\prod_{i=1}^n I_i & = \left( \prod_{i=1}^{n-1} I_i \right) I_n \\
& = \left( \prod_{i=1}^{n-1} \right) \cap I_n \\
& = \left( \bigcap_{i=1}^{n-1} \right) \cap I_n \\
& = \bigcap_{i=1}^n I_i. \qed
\end{align*}

## Lecture 10 (Coprime Ideals)

**Theorem.** Let $I_1, \ldots, I_n$ be ideals of a ring $A,$ and $f \colon A \to \prod\limits_{i=1}^n A/I_i$ the canonical homomorphism defined by
\begin{equation*}
f(x) = (x + I_1, \ldots, x + I_n), \quad \forall~ x \in A.
\end{equation*}

Then
1. $f$ is surjective iff $I_1, \ldots, I_n$ are pairwise coprime.
2. $f$ is injective iff $\bigcap\limits_{i=1}^n I_i = \{0\}.$

*Proof.* Given $f \colon A \to \prod\limits_{i=1}^n A/I_i$ by $f(x) = (x + I_1, \ldots, x + I_n),$ $\forall~ x \in A.$
1. Suppose $f$ is surjective.
We show that the $I_i$ are pairwise coprime.
$(0 + I_1, \ldots, 1 + I_i, \ldots, 0 + I_n) \in \prod\limits_{i=1}^n A/I_i,$ and $f$ is surjective $\implies$
$\exists~ x \in A$ such that $f(x) = (0 + I_1, \ldots, 1 + I_i, \ldots, 0 + I_n) \implies$
$(x + I_1, \ldots, x + I_i, \ldots, x + I_n) = (0 + I_1, \ldots, 1 + I_i, \ldots, 0 + I_n) \implies$
$x \in I_1,$ $\ldots,$ $x - 1 \in I_i,$ $\ldots,$ $x \in I_n \implies$
$x \in I_j,$ $\forall~ j \ne i,$ and $1 - x \in I_i \implies$
$1 = x + (1 - x) \in I_j + I_i,$ $\forall~ j \ne i \implies$
$I_j + I_i = A,$ $\forall~ j \ne i.$
Therefore, the $I_i$s are coprime.
\
Conversely, suppose that the $I_i$s are pairwise coprime.
We show that $f$ is surjective.
Let $(x_1 + I_1, \ldots, x_n + I_n) \in \prod\limits_{i=1}^n A/I_i.$
Note that we can write
\begin{equation*}
(x_1 + I_1, \ldots, x_n + I_n) = \sum_{i=1}^n (x_i + I_1, \ldots, x_i + I_n)(I_1, \ldots, 1 + I_i, \ldots, I_n).
\end{equation*} First, we show that $(1 + I_1, I_2 \ldots, I_n)$ has a preimage.
By hypothesis,
\begin{align*}
I_1 + I_2 = A & \implies 1 \in I_1 + I_2 \implies 1 = z_2 + y_2, \quad \exists~ z_2 \in I_1,\, y_2 \in I_2 \\
I_1 + I_3 = A & \implies 1 \in I_1 + I_3 \implies 1 = z_3 + y_3, \quad \exists~ z_3 \in I_1,\, y_3 \in I_3 \\
& \quad \vdots \\
I_1 + I_n = A & \implies 1 \in I_1 + I_n \implies 1 = z_n + y_n, \quad \exists~ z_n \in I_1,\, y_n \in I_n.
\end{align*}
Now, $q - z_i = y_i,$ $i = 2, \ldots, n \implies \prod\limits_{i=2}^n y_i = \prod\limits_{i=2}^n (1 - z_i).$
Let $v_1 = \prod\limits_{i=2}^n y_i.$ Clearly, $v_1 \in I_j,$ $\forall~ j \ne 1 \implies$
$1 - v_1 = 1 - \prod\limits_{i=2}^n y_i = 1 - \prod\limits_{i=2}^n (1 - z_i) \in I_1 \implies$
$1 - v_1 \in I_1,$ $v_1 \in I_2,$ $\ldots,$ $v_1 \in I_n \implies$
$1 + I_1 = v_1 + I_1,$ $I_2 = v_1 + I_2,$ $\ldots,$ $I_n = v_1 + I_n \implies$
$(1 + I_1, I_2, \ldots, 1 + I_n) = (v_1 + I_1, v_1 + I_2, \ldots, v_1 + I_n) = f(v_1).$
Therefore, $v_1$ is the preimage if $(1 + I_1, I_2, \ldots, I_n).$
Similarly, $(I_1, 1 + I_2, \ldots, I_n)$ has a preimage $v_2.$
$\vdots$
$(I_1, \ldots, I_{n-1}, 1 + I_n)$ has a preimage $v_n.$
Take $v = x_1 v_1 + \ldots x_n v_n \in A.$ Then
\begin{align*}
f(v) & = f(x_1)f(y_1) + \cdots + f(x_n)f(y_n) \\
& = \sum\limits_{i=1}^n (x_i + I_1, \ldots, x_i + I_i, \ldots, x_i + I_n) (I_1, \ldots, 1 + I_i, \ldots, I_n) \\
& = (x_1 + I_1, \ldots, x_n + I_n).
\end{align*} Thus $f$ is surjective.
2. $\ker f = \{\, x \in A \mid f(x) = (I_1, \ldots, I_n) \,\}$
\begin{align*}
& = \{\, a \in A \mid (x + I_1, \ldots, x + I_n) = (I_1, \ldots, I_n) \,\}\\
& = \{\, x \in A \mid x \in I_1, \ldots, x \in I_n \,\} \\
& = \bigcap_{i=1}^n I_i.
\end{align*} Thus, $f$ is injective $\iff$ $\ker f = \{0\}$ $\iff$ $\bigcap\limits_{i=1}^n I_i = \{0\}.$ $\qed$

**Definition.** Let $S$ be a non-empty subset of $A.$ Then the *ideal generated by* $S$ is
\begin{equation*}
\langle S \rangle \defeq \left\{\, \sum_{i=1}^n a_i s_i ~\bigg| a_1, \ldots, a_n \in A,\, s_1, \ldots, s_n \in S \,\right\}.
\end{equation*}

**Note.**
1. $I \cap J = J \cap I.$
2. $(I \cap J) \cap K = I \cap (J \cap K).$
3. $I \cap A = I.$
4. $I + J = J + I.$
5. $(I + J) + K = I + (J + K)$
6. $I + \langle 0 \rangle = I.$
7. $IJ = JI.$
8. $IA = I.$

**Proposition.** Let $I,$ $J,$ and $K$ be ideals of a ring $A.$ Then
1. $I(J + K) = IJ + IK.$
2. $I \cap (J + K) = I \cap J + I \cap K$ if $J \subseteq I$ or $K \subseteq I.$
3. $(I + J)(I \cap J) \subseteq IJ.$

*Proof.*
1. Let $a \in A.$ Then $a \in I(J + K) \iff a = \sum\limits_{i = 1}^n x_i(y_i + z_i),$ $\exists~ x_i \in I,$ $y_i \in J,$ $z_i \in K$ $\iff a = \sum\limits_{i=1}^n x_i y_i + \sum\limits_{i=1}^n x_i z_i \iff a \in IJ + IK.$ Thus, $I(J + K) = IJ + IK.$ Note that in reverse direction, the summations may originally have different numbers of terms, but can be written with a common number of terms $n$ by a suitable addition of zeroes.
2. Suppose $J \subseteq I.$
Let $a \in I \cap (J + K).$ Then $a \in I,$ and $a \in J + K \implies a = \sum\limits_{i=1}^n (x_i + y_i),$ $\exists~ x_i \in J,$ $y_i \in K.$ Then $a = \sum\limits_i x_i + \sum\limits_i y_i \in I$ and $\sum\limits_i x_i \in J \subseteq I \implies \sum\limits_i y_i \in I$ $\implies$ $\sum\limits_i y_i \in I \cap K.$ Therefore, $a \in I \cap J + I \cap K.$
On the other hand, if $a \in I \cap J + I \cap K,$ then $a = x + y,$ where $x \in I,\, J$ and $y \in I,\,K$ $\implies a = x + y \in I,\, J + K$ $\implies a \in I \cap (J + K).$ Therefore, $I \cap (J + K) = I \cap J + I \cap K.$ The proof is similar when $K \subseteq I.$
3. Let $a \in (I + J)(I \cap J).$ Then $a = \sum\limits_{i=1}^n (x_i + y_i) z_i,$ $\exists~ x_i \in I,$ $y_i \in J,$ $z_i \in I \cap J.$
Then $a = \sum_i x_i y_i + \sum_i x_i z_i.$ But $\sum_i x_i y_i \in I(I \cap J) \subseteq IJ,$ and $\sum_i x_i y_i \in J(I \cap J) \subseteq JI = IJ$ $\implies a \in IJ.$ Therefore, $(I + J)(I \cap J) \subseteq IJ.$ $\qed$

## Lecture 11 (Ideal Quotients and Annihilator of an Ideal)

**Note.** If $I = \langle a_1, \ldots, a_m \rangle$ and $J = \langle b_1, \ldots, b_n \rangle,$ then
\begin{align*}
I + J & = \langle a_1, \ldots, a_m, b_1, \ldots, b_n \rangle \\
IJ & = \langle a_1 b_1, \ldots, a_1 b_n, a_2 b_1, \ldots, a_2 b_n, \ldots, a_m b_1, \ldots, a_m b_n \rangle.
\end{align*}

**Exercise.** Let $I = \langle m \rangle$ and $J = \langle n \rangle$ be two non-zero ideals of $\mathbb Z.$ Then
1. $I + J = \langle \gcd(m, n) \rangle.$
2. $I \cap J = \langle \lcm(m, n) \rangle.$
3. $IJ = \langle mn \rangle.$

*Solution.*
~ :::spoiler 1.
Since $\mathbb Z$ is a PID and $I + J \unlhd \mathbb Z,$ $I + J = \langle d \rangle,$ $\exists~ d \in \mathbb N.$ Clearly,
$m \in \langle m \rangle = I \subseteq I + J = \langle d \rangle,$
$n \in \langle n \rangle = I \subseteq I + J = \langle d \rangle$
$\implies d \mid m$ and $d \mid n.$
Let $c \in \mathbb N,$ such that $c \mid m$ and $c \mid n.$ Then
$\langle m \rangle \subseteq \langle c \rangle$ and $\langle n \rangle \subseteq \langle c \rangle.$
Since $I + J = \langle m \rangle + \langle n \rangle$ is the smallest ideal containing both $I$ and $J,$ we have $\langle d \rangle \subseteq \langle c \rangle.$
Thus, $c \mid d.$
Therefore, $d = \gcd(m, n).$
:::
:::spoiler 2.
$I \cap J = \langle d \rangle,$ $\exists~ d \in \mathbb N.$
$\langle d \rangle = I \cap J \subseteq I = \langle m \rangle \implies m \mid d.$
$\langle d \rangle = I \cap J \subseteq J = \langle n \rangle \implies n \mid d.$
Suppose $m \mid c$ and $n \mid c,$ $\exists~ c \in \mathbb N.$ Then
$\langle c \rangle \subseteq \langle m \rangle$ and $\langle c \rangle \subseteq \langle n \rangle.$
Since $I \cap J$ is the largest ideal contained in both $I$ and $J,$ this implies that $\langle c \rangle \subseteq I \cap J = \langle d \rangle.$
Thus, $d \mid c.$
Therefore, $d = \lcm(m, n).$
:::
:::spoiler 3.
Clearly, $mn \in IJ \implies \langle mn \rangle \subseteq IJ.$
Now, any element of $IJ$ is of the form $\sum\limits_{i=1}^k x_i y_i,$ where $x_i \in I,$ $y_i \in J,$ $i = 1, \ldots, k,$ $\exists~ k \in \mathbb N.$ Then
$I = \langle m \rangle \implies m \mid x_i,$ $i = 1, \ldots, k,$ and
$J = \langle n \rangle \implies n \mid y_i,$ $i = 1, \ldots, k$
$\implies mn \mid \sum\limits_{i=1}^k x_iy_i \implies \sum\limits_{i=1}^k x_i y_i \in \langle mn \rangle.$
Thus, $IJ \subseteq \langle mn \rangle.$
Therefore, $IJ = \langle mn \rangle.$
:::

**Example.** In the ring of integers,
1. $\langle 2 \rangle + \langle 3 \rangle = \langle \gcd(2, 3) \rangle = \langle 1 \rangle = \mathbb Z.$
2. $\langle 3 \rangle + \langle 6 \rangle = \langle \gcd(5, 6) \rangle = \langle 3 \rangle = 3 \mathbb Z.$
3. $\langle 9 \rangle + \langle 30 \rangle = \langle \gcd(9, 30) \rangle = \langle 3 \rangle = 3 \mathbb Z.$
4. $\langle 3 \rangle \langle 4 \rangle = \langle 12 \rangle = 12 \mathbb Z.$
5. $\langle 6 \rangle \langle 8 \rangle = \langle 48 \rangle = 48 \mathbb Z.$
6. $\langle 6 \rangle \cap \langle 8 \rangle = \langle \lcm(6, 8)\rangle = \langle 24 \rangle = 24 \mathbb Z.$

**Definition.** if $I$ and $J$ are ideals of a ring $A,$ then their *ideal quotient* is
\begin{equation*}
(I : J) = \{\, x \in A \mid xJ \subseteq I \,\}
\end{equation*}

In particular, the *annihilator* of $J$ is
\begin{equation*}
\Ann J = (\langle 0 \rangle : J) = \{\, x \in A \mid xt = 0,\, \forall~ t \in J \,\}
\end{equation*}

**Remark.** Let $I J \unlhd A.$ Then
1. $(I : \langle 0 \rangle) = A.$
2. $(I : A) = I.$
3. $(A : J) = A.$
4. $\Ann(1) = \langle 0 \rangle,$ i.e., $(\langle 0 \rangle : \langle 1 \rangle) = \langle 0 \rangle.$

**Exercise.** Let $m, n \in \mathbb N,$ and let $I = \langle m \rangle,$ and $J = \langle n \rangle$ be two non-zero ideals of $\mathbb Z.$ Then $(I : J) = \left\langle \dfrac m {\gcd(m, n)} \right\rangle.$
:::spoiler *Solution.*
Let $K = \left\langle \dfrac m {\gcd(m, n)} \right\rangle.$
If $x \in K,$ then $x = \frac{am}{\gcd(m, n)},$ $\exists~ a \in \mathbb Z.$ Now, for any $y \in J,$ $y = bn,$ $\exists~ b \in \mathbb Z.$ Therefore, $xy = \left(\frac{abn}{\gcd(m,n)}\right)m \mid m \implies xy \in I.$ Thus, $xJ \subseteq I,$ and hence $x \in (I : J).$ Therefore, $K \subseteq (I : J).$
Now, let $x$ be any element of $(I : J).$ Then $n \in J,\, xJ \subseteq I \implies xn \in I \implies xn \mid m \implies x \mid \frac m {\gcd(m, n)} \implies x \in K.$ Hence, $(I : J) \subseteq K.$
:::
<p></p>

**Example.** In $\mathbb Z$:
1. $(\langle 4 \rangle : \langle 12 \rangle) = \left\langle \dfrac 4 {\gcd(4, 12)} \right\rangle = \left\langle \dfrac 4 4 \right \rangle = \langle 1 \rangle = \mathbb Z.$
2. $(\langle 18 \rangle : \langle 4 \rangle) = \left\langle \dfrac{18}{\gcd(18, 4)} \right\rangle = \left\langle \dfrac {18} 2 \right\rangle = \langle 9 \rangle = 9 \mathbb Z.$

**Exercise.** Let $x$ be a nilpotent element of $A.$ Show that $1 + x$ is a unit of $A.$ Deduce that the sum of a nilpotent element and a unit is a unit.
:::spoiler *Solution.*
Since $x$ is nilpotent, $x^n = 0,$ $\exists~ n > 0.$ Then
$(1 - x)(1 + x + \cdots + x^{n-1}) = 1 - x^n = 1.$
Thus, $1 - x$ is a unit.
If $x$ is a unit, then so is $-x.$ Therefore, $1 + x$ is a unit.
Let $a$ be a unit of $A.$ Then
$a + x = a + aa^{-1}x = a(1 + a^{-1}x).$
Since the set of all nilpotents is an ideal, $a^{-1}x$ is also nilpotent $\implies 1 + a^{-1}x$ is a unit $\implies a(1 + a^{-1}x) = a + x$ is a unit.
**Alternative solution:** Since the nilradical of $A$ is contained in the Jacobson radical, if $x$ is nilpotent, then $x$ is in the Jacobson radical, which implies that $1 - xy$ is a unit for all $y \in A,$ so in particular, for $y = -1,$ $1 + x$ is a unit. Similarly, if $a$ is a unit, then $1 + a^{-1}x$ is a unit, and therefore $a(1 + a^{-1}x) = a + x$ is a unit.
:::
<p></p>

**Exercise.** Let $A$ be a ring in which every element $x$ satisfies $x^n = x,$ $\exists~ n > 1.$ Show that every prime ideal is maximal.
:::spoiler *Solution.*
Let $I$ be any non-zero prime ideal of $A,$ and consider $x + I \in A/I,$ $x + I \ne I.$ Then $\exists~ n > 1,$ $x^n = x \implies (x + I)^n = x^n + I = x + I.$
Since $A/I$ is an integral domain, $(x + I)^{n - 1} = 1 + I,$ and therefore, $x + I$ is a unit of $A/I.$
Therefore, $A/I$ is a field, which proves that $I$ is maximal.
:::

## Lecture 12 (Radicals of Ideals)

**Theorem.** Let $A$ be a ring, and $I,$ $J,$ and $K$ ideals of $A.$ Then
1. $I \subseteq (I : J)$
2. $(I : J) J \subseteq I$
3. $((I: J) : K) = (I : JK) = ((I : K) : J)$
4. $\left( \bigcap\limits_{i=1}^n I_i : J \right) = \bigcap\limits_{i=1}^n (I_i : J)$
5. $(I : \sum\limits_{i=1}^n J_i) = \bigcap\limits_{i=1}^n (I : J_i).$

*Proof.*
1. If $x \in I,$ then $xt \in I,$ $\forall~ t \in A,$ hence in particular $\forall~ t \in J.$
$\implies xJ \subseteq I \implies x \in (I : J).$ Therefore, $I \subseteq (I : J).$
2. Let $\sum x_i y_i \in (I : J)K,$ where $x_i \in (I : J)$ and $y_i \in J.$ Then $x_i y_i \in I,$ $\forall~ i$ $\implies$ $\sum x_i y_i \in I.$ Hence, $(I: J)J \subseteq I.$
3. Take $x \in ((I : J) : K).$ Then $xK \subseteq (I : J) \iff xKJ \subseteq I \iff x \in (I : JK).$ Thus, $((I : J) : K) = (I : JK).$ Similarly, $(I : JK) = ((I : K) : J).$
4. Take $x \in \left( \bigcap\limits_{i=1}^n I_i : J \right).$ Then $xJ \subseteq \bigcap\limits_{i=1}^n I_i \subseteq I_i,$ $\forall~ i$ $\iff$ $x \in (I_i : J),$ $\forall~ i$ $\iff$ $x \in \bigcap\limits_{i=1}^n (I_i : J).$
5. Let $x \in \left( I : \sum\limits_{i=1}^n J_i \right).$ Then $x \sum\limits_{i=1}^n J_i \subseteq I \iff xJ_i \subseteq I,$ $\forall~ i$ $\iff x \in (I : J_i),$ $\forall~ i$ $\iff$ $x \in \bigcap\limits_{i=1}^n (I : J_i).$ $\qed$

**Definition.** Let $A$ be a ring, and $I$ an ideal of $A.$ Then the *radical* of $I,$ denoted by $r(I),$ is defined as
\begin{equation*}
r(I) = \{\, x \in A \mid x^n \in I,\, \exists n > 0 \,\}.
\end{equation*}

**Result.** For any ideal $I$ of a ring $A,$ $r(I)$ is an ideal of $A.$
*Proof.* Clearly, $0 \in r(I),$ since $0 \in I.$
Since $I \unlhd A,$ $A/I$ is a quotient ring.
Let $f \colon A \to A/I$ be the canonical homomorphism $f(x) = x + I,$ $\forall x \in A.$
Suppose $N$ is the nilradical of $A/I,$ which is an ideal of $A/I.$ Then $f^{-1}(N)$ is an ideal of $A.$ Now,
\begin{align*}
f^{-1}(N) & = \{\, x \in A \mid f(x) \in N \,\} \\
& = \{\, x \in A \mid x + I \in N \,\} \\
& = \{\, x \in A \mid (x + I)^n = I,\, \exists n > 0 \,\} \\
& = \{\, x \in A \mid x^n + I = I,\, \exists n > 0 \,\}\\
& = \{\, x \in A \mid x^n \in I \,\} \\
& = r(I).
\end{align*}

Therefore, $r(I)$ is an ideal of $A.$

**Theorem.** Let $A$ be a ring, and $D$ the set of all zero divisors in $A.$ Then
1. $D = \bigcup\limits_{x \ne 0} \Ann(x).$
2. $D = r(D).$
3. $D = \bigcup\limits_{x \ne 0} r(\Ann(x)).$

*Proof.*
1. Let $x \in D.$ Then $x$ is a zero divisor $\implies$
$\exists y \in A,$ $y \ne 0,$ such that $xy = 0$ $\implies$ $x \in \Ann(y),$ $y \ne 0$ $\implies$
$x \in \bigcup\limits_{y \ne 0} \Ann(y).$
\
Let $t \in \bigcup\limits_{x \ne 0} \Ann(x).$ Then $t \in \Ann(x),$ $\exists x \in A,$ $x \ne 0$ $\implies$
$tx = 0,$ $\exists x \ne 0$ $\implies$
$t$ is a zero divisor $\implies$ $t \in D.$
2. Clearly, $D \subseteq r(D).$
Now, let $x \in r(D).$ By definition, $x^n \in D,$ $\exists n > 0$ $\implies$ $x^n$ is a zero divisor $\implies$ $\exists y \in A,$ $y \ne 0,$ x^n y = 0$ $\implies$ $x \cdot x^{n-1} y = 0.$
If $x^{n-1} y \ne 0,$ then $x \in D.$
Otherwise, if $x^{n-1} y = 0,$ then $x \cdot x^{n-2} y = 0.$
If $x^{n-2} y \ne 0,$ then $x \in D.$
Otherwise, continue similarly until $x \cdot x^k y = 0$ and $x^k y \ne 0.$ Then $x \in D.$
Therefore, $r(D) \subseteq D.$
Thus, $r(D) = D.$
3. If $\{\, E_\alpha \,\}_{\alpha \in \Delta}$ is a family of subsets of $A,$ then
\begin{equation*}
r \left( \bigcup_{\alpha \in \Delta} E_\alpha \right) = \bigcup_{\alpha \in \Delta} r\left( E_\alpha \right).
\end{equation*}

Take $x \in r \left( \bigcup\limits_{\alpha \in \Delta} E_\alpha \right).$ Then $x^k \in \bigcup\limits_\alpha E_\alpha,$ $\exists k > 0$ $\iff$
$x^k \in E_\alpha,$ $\exists \alpha \in \Delta,$ $\exists k > 0$ $\iff$
$x \in r(E_\alpha),$ $\exists \alpha \in \Delta$ $\iff$
$x \in \bigcup\limits_{\alpha \in \Delta} E_\alpha.$
Now, $D = r(D) = r \left( \bigcup\limits_{x \ne 0} \Ann(x) \right) = \bigcup\limits_{x \ne 0} r\left( \Ann(x) \right).$ $\qed$