September 20, 2026

Abstract Algebra and Ring Theory

abstract algebragroup theoryring theorymath

Syllabus:
Abstract Algebra: Coset Decomposition, Lagrange's Theorem, Homomorphism and Isomorphism, Elementary Idea of Normal Subgroups, Quotient Groups, Centre of a Group, Fundamental Theorem of Group Homomorphism, Automorphism, Inner Automorphism, Automorphism Groups and Their Computations, Conjugacy Relation and Normalizer.

Ring Theory: Definition and Examples of Rings, Properties of Rings, Subrings, Characteristic of a Ring, Ideals and Quotient Ring, Ideal Generated by a Subset of a Ring, Factor Rings, Operations on Ideals, Prime and Maximal Ideals, Ring Homomorphism, Properties of Ring Homomorphism, Field of Quotients of an Integral Domain, Polynomial Rings.

In this notes, we'll cover -
Part 1. Abstract Algebra

Coset Decomposition
Lagrange's Theorem
Homomorphism and Isomorphism
Normal Subgroups
Quotient Groups
Centre of a Group
Fundamental Theorem of Group Homomorphism
Second and Third Isomorphism Theorems
Automorphism and Inner Automorphism
Automorphism Groups and Computations
Conjugacy Relation and Normalizer

Part 2. Ring Theory

Rings — Definition, Examples, Properties
Subrings
Characteristic of a Ring
Ideals, Ideal Generated by a Set, Operations on Ideals
Quotient (Factor) Rings
Prime and Maximal Ideals
Ring Homomorphisms and Their Properties
Field of Quotients of an Integral Domain
Polynomial Rings

Part 1. Abstract Algebra

Coset Decomposition

Definition 1. Left Coset:
Let $G$ be a group and $H$ be a subgroup of $G$ . Let $a \in G$ . Then the set
$aH = \{ah : h \in H\}$
is called the left coset of $H$ in $G$ determined by (or containing) the element $a$ .

Definition 2. Right Coset:
Let $G$ be a group and $H$ be a subgroup of $G$ . Let $a \in G$ . Then the set
$Ha = \{ha : h \in H\}$
is called the right coset of $H$ in $G$ determined by the element $a$ .

Notes:

Since $e \in H$ , we have $a = ae \in aH$ and $a = ea \in Ha$ . Thus every element belongs to the coset it determines.
Taking $a = e$ : $eH = H = He$ .
In general, left cosets and right cosets need not coincide, i.e., $aH \neq Ha$ is possible.

E.g., Let $G = S_3$ and $H = \{e, (1\,2)\}$ . Then
$(1\,3)H = \{(1\,3)\cdot e,\ (1\,3)(1\,2)\} = \{(1\,3),\ (1\,3\,2)\},$
$H(1\,3) = \{e\cdot(1\,3),\ (1\,2)(1\,3)\} = \{(1\,3),\ (1\,2\,3)\}.$
Hence $(1\,3)H \neq H(1\,3)$ , so $H$ is not normal in $S_3$ .

Theorem 1 (Properties of Cosets)

Let $G$ be a group and $H$ a subgroup of $G$ . Let $a, b \in G$ . Then:
(i) $aH = H$ if and only if $a \in H$ .
(ii) $aH = bH$ if and only if $a^{-1}b \in H$ if and only if $b \in aH$ .
(iii) Either $aH = bH$ or $aH \cap bH = \emptyset$ .
(iv) There is a one-one correspondence between $H$ and $aH$ ; in particular, if $H$ is finite, $|aH| = |H|$ . (Similarly for right cosets.)

Proof:

Let $G$ be a group and $H \leq G$ . Let $a, b \in G$ .

(i) First suppose $aH = H$ .
Since $a = ae \in aH$ and $aH = H$ , we get $a \in H$ .

Conversely, suppose $a \in H$ .
We show $aH = H$ .

Claim 1: $aH \subseteq H$ .
Let $x \in aH$ . Then $x = ah$ for some $h \in H$ . Since $a \in H$ and $h \in H$ and $H$ is a subgroup, $ah \in H$ . Thus $x \in H$ . So $aH \subseteq H$ .

Claim 2: $H \subseteq aH$ .
Let $h \in H$ . Since $a \in H$ and $H$ is a subgroup, $a^{-1} \in H$ , so $a^{-1}h \in H$ .
Therefore $h = a(a^{-1}h) \in aH$ . So $H \subseteq aH$ .

Hence $aH = H$ .

(ii) We prove the equivalences in a chain.

$aH = bH$
$\iff$ $a^{-1}(aH) = a^{-1}(bH)$
$\iff$ $H = (a^{-1}b)H$
$\iff$ $a^{-1}b \in H$ (by part (i)).

Next: $a^{-1}b \in H$
$\iff$ $b = a(a^{-1}b) \in aH$ (since $a^{-1}b \in H$ ).

Thus $aH = bH \iff a^{-1}b \in H \iff b \in aH$ .

(iii) Suppose $aH \cap bH \neq \emptyset$ .
We have to show that $aH = bH$ .

Let $c \in aH \cap bH$ . Then $c \in aH$ and $c \in bH$ .
So there exist $h_1, h_2 \in H$ such that
$c = ah_1 = bh_2.$
Hence $a = bh_2 h_1^{-1}$ .
Since $h_2, h_1^{-1} \in H$ and $H$ is a subgroup, $h_2 h_1^{-1} \in H$ .
Therefore $a = b(h_2 h_1^{-1}) \in bH$ .

By part (ii), $a \in bH \implies aH = bH$ .

Thus either $aH \cap bH = \emptyset$ , or $aH = bH$ .

(iv) Define a mapping $f: H \to aH$ by
$f(h) = ah \quad \text{for all } h \in H.$

We show that $f$ is bijective.

$f$ is one-one:
Let $f(h_1) = f(h_2)$ . Then $ah_1 = ah_2$ .
Multiplying both sides on the left by $a^{-1}$ (left cancellation), we get $h_1 = h_2$ .
Hence $f$ is one-one.

$f$ is onto:
Let $y \in aH$ be arbitrary. Then $y = ah$ for some $h \in H$ .
So $y = f(h)$ . Hence $f$ is onto.

Therefore $f$ is a bijection between $H$ and $aH$ .
In particular, if $H$ is finite, $|aH| = |H|$ .

(The proof for right cosets is similar: define $f: H \to Ha$ by $f(h) = ha$ .)

Theorem 2 (Coset Decomposition)

Let $G$ be a group and $H$ a subgroup of $G$ . Then the set of all distinct left cosets of $H$ in $G$ forms a partition of $G$ . That is,
$G = \bigcup aH,$
where the union runs over a complete set of distinct left coset representatives, and any two distinct left cosets are disjoint. A similar result holds for right cosets.

Proof:

Let $G$ be a group and $H \leq G$ .

Step 1. Every element of $G$ belongs to some left coset.
Let $a \in G$ be arbitrary. Then $a = ae \in aH$ .
Hence $a$ belongs to the left coset $aH$ .
Therefore
$G \subseteq \bigcup_{a \in G} aH.$
Clearly $\bigcup_{a \in G} aH \subseteq G$ . So $G = \bigcup_{a \in G} aH$ .

Step 2. Distinct left cosets are disjoint.
This follows immediately from Theorem 1(iii): if two left cosets have a common element, they are equal; otherwise their intersection is empty.

Therefore the distinct left cosets of $H$ in $G$ form a partition of $G$ .

(The argument for right cosets is identical.)

Definition 3. Index of a Subgroup:
Let $G$ be a group and $H$ a subgroup of $G$ . The number of distinct left cosets of $H$ in $G$ is called the index of $H$ in $G$ and is denoted by $[G:H]$ .
(It can be shown that the number of distinct right cosets equals the number of distinct left cosets, so the index is well-defined.)

Lagrange's Theorem

Theorem 3 (Lagrange's Theorem)

Let $G$ be a finite group and $H$ a subgroup of $G$ . Then the order of $H$ divides the order of $G$ . In fact,
$|G| = [G:H] \cdot |H|.$

Proof:

Let $G$ be a finite group and $H$ a subgroup of $G$ .
Let $|G| = n$ and $|H| = m$ . We have to prove that $m$ divides $n$ .

Since $G$ is finite, the number of distinct left cosets of $H$ in $G$ is finite.
Let the distinct left cosets of $H$ in $G$ be
$a_1 H,\ a_2 H,\ \ldots,\ a_k H,$
where $k = [G:H]$ .

By the coset decomposition theorem,
$G = a_1 H \cup a_2 H \cup \cdots \cup a_k H,$
and this union is disjoint (no two distinct cosets have a common element).

By Theorem 1(iv), each left coset has exactly $m$ elements:
$|a_i H| = |H| = m \quad \text{for } i = 1,2,\ldots,k.$

Counting the number of elements in $G$ :
$|G| = |a_1 H| + |a_2 H| + \cdots + |a_k H| = k \cdot m.$
That is,
$n = k \cdot m = [G:H] \cdot |H|.$

Hence $m$ divides $n$ , i.e., $|H|$ divides $|G|$ .

Corollaries:

Order of an element divides order of the group.
Let $a \in G$ and let $o(a) = r$ . Then $H = \langle a \rangle = \{e, a, a^2, \ldots, a^{r-1}\}$ is a subgroup of order $r$ . By Lagrange, $r$ divides $|G|$ .
Groups of prime order are cyclic.
Let $|G| = p$ , where $p$ is prime. Let $a \in G$ , $a \neq e$ . Then $o(a)$ divides $p$ , so $o(a) = p$ (since $o(a) \neq 1$ ). Thus $\langle a \rangle = G$ , so $G$ is cyclic.
$a^{|G|} = e$ for all $a \in G$ .
Since $o(a)$ divides $|G|$ , we have $a^{|G|} = (a^{o(a)})^{|G|/o(a)} = e$ .

Remark (Converse is false):
The converse of Lagrange's theorem is not true in general.
E.g., The alternating group $A_4$ has order $12$ , but $A_4$ has no subgroup of order $6$ .

Homomorphism and Isomorphism

Definition 4. Group Homomorphism:
Let $G$ and $G'$ be two groups. A mapping $\phi: G \to G'$ is called a homomorphism (or group homomorphism) if
$\phi(ab) = \phi(a)\,\phi(b) \quad \text{for all } a, b \in G.$
That is, $\phi$ preserves the group operation.

Definition 5. Kernel and Image:
Let $\phi: G \to G'$ be a homomorphism, and let $e'$ denote the identity of $G'$ . Then:

The kernel of $\phi$ is $\ker\phi = \{a \in G : \phi(a) = e'\}$ .
The image of $\phi$ is $\operatorname{Im}\phi = \{\phi(a) : a \in G\}$ .

Definition 6. Isomorphism:
A homomorphism $\phi: G \to G'$ which is both one-one and onto is called an isomorphism.
If there exists an isomorphism from $G$ onto $G'$ , we say $G$ is isomorphic to $G'$ and write $G \cong G'$ .

E.g.,

Define $\phi: (\mathbb{Z}, +) \to (\mathbb{Z}, +)$ by $\phi(n) = 2n$ . Then $\phi(m+n) = 2(m+n) = 2m+2n = \phi(m)+\phi(n)$ , so $\phi$ is a homomorphism. It is one-one but not onto.
Define $\phi: (\mathbb{R}, +) \to (\mathbb{R}^+, \cdot)$ by $\phi(x) = e^x$ . Then $\phi$ is an isomorphism.
The sign homomorphism $\operatorname{sgn}: S_n \to \{1,-1\}$ has kernel $A_n$ .

Theorem 4 (Properties of a Homomorphism)

Let $\phi: G \to G'$ be a group homomorphism. Let $e$ and $e'$ be the identities of $G$ and $G'$ respectively. Then:
(i) $\phi(e) = e'$ .
(ii) $\phi(a^{-1}) = \big(\phi(a)\big)^{-1}$ for all $a \in G$ .
(iii) $\phi(a^n) = \big(\phi(a)\big)^n$ for all $a \in G$ and all integers $n$ .
(iv) $\ker\phi$ is a subgroup of $G$ , and $\operatorname{Im}\phi$ is a subgroup of $G'$ .
(v) $\phi$ is one-one if and only if $\ker\phi = \{e\}$ .

Proof:

Let $\phi: G \to G'$ be a homomorphism.

(i) We have
$\phi(e) = \phi(ee) = \phi(e)\,\phi(e).$
Multiplying both sides on the left by $\big(\phi(e)\big)^{-1}$ in $G'$ , we get
$e' = \phi(e).$
Hence $\phi(e) = e'$ .

(ii) Consider
$\phi(a)\,\phi(a^{-1}) = \phi(aa^{-1}) = \phi(e) = e'.$
Similarly, $\phi(a^{-1})\,\phi(a) = e'$ .
Therefore $\phi(a^{-1})$ is the inverse of $\phi(a)$ in $G'$ , i.e.,
$\phi(a^{-1}) = \big(\phi(a)\big)^{-1}.$

(iii) For $n > 0$ , the result follows by induction on $n$ .
For $n = 0$ : $\phi(a^0) = \phi(e) = e' = \big(\phi(a)\big)^0$ .
For $n < 0$ , write $n = -m$ with $m > 0$ . Then
$\phi(a^{-m}) = \phi\big((a^{-1})^m\big) = \big(\phi(a^{-1})\big)^m = \big(\phi(a)^{-1}\big)^m = \big(\phi(a)\big)^{-m}.$

(iv) First we show $\ker\phi \leq G$ .
Clearly $e \in \ker\phi$ , since $\phi(e) = e'$ . So $\ker\phi$ is nonempty.

Let $a, b \in \ker\phi$ . Then $\phi(a) = e'$ and $\phi(b) = e'$ .
Now
$\phi(ab^{-1}) = \phi(a)\,\phi(b^{-1}) = \phi(a)\,\big(\phi(b)\big)^{-1} = e'\,(e')^{-1} = e'.$
Hence $ab^{-1} \in \ker\phi$ .
By the subgroup test, $\ker\phi \leq G$ .

Next, $\operatorname{Im}\phi \leq G'$ .
Let $\phi(a), \phi(b) \in \operatorname{Im}\phi$ . Then
$\phi(a)\,\big(\phi(b)\big)^{-1} = \phi(a)\,\phi(b^{-1}) = \phi(ab^{-1}) \in \operatorname{Im}\phi.$
Also $\phi(e) = e' \in \operatorname{Im}\phi$ . Hence $\operatorname{Im}\phi \leq G'$ .

(v) Suppose $\phi$ is one-one.
Let $a \in \ker\phi$ . Then $\phi(a) = e' = \phi(e)$ .
Since $\phi$ is one-one, $a = e$ . Thus $\ker\phi = \{e\}$ .

Conversely, suppose $\ker\phi = \{e\}$ .
Let $\phi(a) = \phi(b)$ . Then
$\phi(a^{-1}b) = \phi(a)^{-1}\phi(b) = \phi(a)^{-1}\phi(a) = e'.$
So $a^{-1}b \in \ker\phi = \{e\}$ , which gives $a^{-1}b = e$ , i.e., $a = b$ .
Hence $\phi$ is one-one.

Normal Subgroups

Definition 7. Normal Subgroup:
Let $G$ be a group and $N$ a subgroup of $G$ . Then $N$ is said to be a normal subgroup of $G$ (written $N \triangleleft G$ or $N \unlhd G$ ) if
$aN = Na \quad \text{for all } a \in G.$

Equivalent conditions (any one may be used as definition):

$aNa^{-1} = N$ for all $a \in G$ .
$aNa^{-1} \subseteq N$ for all $a \in G$ .
$ana^{-1} \in N$ for all $a \in G$ and all $n \in N$ .

E.g.,

For any group $G$ , the trivial subgroups $\{e\}$ and $G$ are always normal in $G$ .
Every subgroup of an abelian group is normal (since $ah = ha$ for all $a \in G$ , $h \in H$ ).
$A_n$ is normal in $S_n$ , because $[S_n : A_n] = 2$ .
In $S_3$ , $H = \{e,(1\,2)\}$ is a subgroup but not normal (as shown in the coset example).

Theorem 5

Let $G$ be a group and $H$ a subgroup of $G$ such that $[G:H] = 2$ . Then $H$ is normal in $G$ .

Proof:

Let $G$ be a group and $H \leq G$ with $[G:H] = 2$ .
We have to prove that $H \triangleleft G$ , i.e., $aH = Ha$ for all $a \in G$ .

Since the index is $2$ , there are exactly two distinct left cosets of $H$ in $G$ , namely
$H \quad \text{and} \quad G \setminus H.$
(Indeed, one left coset is $H = eH$ ; the other must be the set of all remaining elements.)

Similarly, the two distinct right cosets are
$H \quad \text{and} \quad G \setminus H.$

Now let $a \in G$ be arbitrary.

Case 1: $a \in H$ .
Then by Theorem 1(i), $aH = H$ and $Ha = H$ . So $aH = Ha$ .

Case 2: $a \notin H$ .
Then $aH \neq H$ , so $aH = G \setminus H$ .
Similarly $Ha = G \setminus H$ .
Hence $aH = Ha$ .

In both cases, $aH = Ha$ . Therefore $H \triangleleft G$ .

Theorem 6

Let $\phi: G \to G'$ be a group homomorphism. Then $\ker\phi$ is a normal subgroup of $G$ .

Proof:

Let $\phi: G \to G'$ be a homomorphism.
We already know from Theorem 4 that $\ker\phi \leq G$ .
We have to prove that $\ker\phi \triangleleft G$ .

Let $a \in G$ and let $k \in \ker\phi$ be arbitrary.
We must show that $aka^{-1} \in \ker\phi$ .

Now
$\phi(aka^{-1}) = \phi(a)\,\phi(k)\,\phi(a^{-1}) = \phi(a)\,\phi(k)\,\big(\phi(a)\big)^{-1}.$
Since $k \in \ker\phi$ , we have $\phi(k) = e'$ . Therefore
$\phi(aka^{-1}) = \phi(a)\, e'\, \big(\phi(a)\big)^{-1} = e'.$
Hence $aka^{-1} \in \ker\phi$ .

Thus $a\,(\ker\phi)\,a^{-1} \subseteq \ker\phi$ for all $a \in G$ .
Therefore $\ker\phi \triangleleft G$ .

Quotient Groups

Definition 8. Quotient Group (Factor Group):
Let $G$ be a group and $N$ a normal subgroup of $G$ . Let $G/N$ denote the set of all (left) cosets of $N$ in $G$ :
$G/N = \{aN : a \in G\}.$
Define multiplication of cosets by
$(aN)(bN) = (ab)N.$
Then $G/N$ (G mod N) is a group under this operation, called the quotient group (or factor group) of $G$ by $N$ .

The identity element of $G/N$ is $eN = N$ .
The inverse of $aN$ is $a^{-1}N$ .

Theorem 7 (Product of Cosets is Well-Defined)

Let $N \triangleleft G$ . Then the product $(aN)(bN) = (ab)N$ is well-defined, i.e., it does not depend on the choice of representatives.

Proof:

Let $N \triangleleft G$ . Suppose $aN = a'N$ and $bN = b'N$ .
We have to show that $(ab)N = (a'b')N$ .

Since $aN = a'N$ , we have $a^{-1}a' \in N$ (Theorem 1).
Since $bN = b'N$ , we have $b^{-1}b' \in N$ .

Consider
$(ab)^{-1}(a'b') = b^{-1}a^{-1}a'b' = b^{-1}(a^{-1}a')b \cdot (b^{-1}b').$

Since $N$ is normal and $a^{-1}a' \in N$ , we have
$b^{-1}(a^{-1}a')b \in N.$
Also $b^{-1}b' \in N$ .
Since $N$ is a subgroup, their product
$b^{-1}(a^{-1}a')b \cdot (b^{-1}b') \in N.$
Hence $(ab)^{-1}(a'b') \in N$ , which means $(ab)N = (a'b')N$ .

Therefore the product of cosets is well-defined.

E.g., Take $G = (\mathbb{Z}, +)$ and $N = n\mathbb{Z}$ . Then $N \triangleleft G$ (since $\mathbb{Z}$ is abelian), and
$\mathbb{Z}/n\mathbb{Z} = \{0+n\mathbb{Z},\ 1+n\mathbb{Z},\ \ldots,\ (n-1)+n\mathbb{Z}\} \cong \mathbb{Z}_n.$

Centre of a Group

Definition 9. Centre of a Group:
Let $G$ be a group. The centre of $G$ , denoted $Z(G)$ , is the set of all elements of $G$ that commute with every element of $G$ :
$Z(G) = \{z \in G : zg = gz \text{ for all } g \in G\}.$

Theorem 8

Let $G$ be a group. Then $Z(G)$ is an abelian normal subgroup of $G$ .

Proof:

Let $G$ be a group. We prove the result in three steps.

Step 1. $Z(G)$ is a subgroup of $G$ .

Clearly $e \in Z(G)$ , since $eg = ge = g$ for all $g \in G$ . So $Z(G) \neq \emptyset$ .

Let $z_1, z_2 \in Z(G)$ . We show that $z_1 z_2^{-1} \in Z(G)$ .
Let $g \in G$ be arbitrary. Since $z_2 \in Z(G)$ , we have $z_2 g = g z_2$ , so $g z_2^{-1} = z_2^{-1} g$ (multiply by $z_2^{-1}$ on both sides appropriately).
Now

(z_1 z_2^{-1})g = z_1(z_2^{-1}g) = z_1(gz_2^{-1}) = (z_1 g)z_2^{-1} = (g z_1)z_2^{-1} = g(z_1 z_2^{-1})

(using $z_1 \in Z(G)$ in the second-last step).
Hence $z_1 z_2^{-1} \in Z(G)$ . By the subgroup test, $Z(G) \leq G$ .

Step 2. $Z(G)$ is abelian.

Let $z_1, z_2 \in Z(G)$ . Since $z_1$ commutes with every element of $G$ , in particular $z_1 z_2 = z_2 z_1$ .
Hence $Z(G)$ is abelian.

Step 3. $Z(G)$ is normal in $G$ .

Let $a \in G$ and $z \in Z(G)$ . Then
$aza^{-1} = a(a^{-1}z) = (aa^{-1})z = ez = z \in Z(G),$
where we used $za^{-1} = a^{-1}z$ because $z \in Z(G)$ .
(Alternatively: $aza^{-1} = aa^{-1}z = z$ .)

Thus $a Z(G) a^{-1} \subseteq Z(G)$ for all $a \in G$ .
Therefore $Z(G) \triangleleft G$ .

E.g.,

If $G$ is abelian, then $Z(G) = G$ .
$Z(S_3) = \{e\}$ .
$Z\big(GL_n(\mathbb{R})\big) = \{\lambda I : \lambda \in \mathbb{R},\ \lambda \neq 0\}$ (scalar matrices).

Fundamental Theorem of Group Homomorphism

Theorem 9 (Fundamental Theorem / First Isomorphism Theorem)

Let $\phi: G \to G'$ be a group homomorphism. Then
$\frac{G}{\ker\phi} \ \cong\ \operatorname{Im}\phi.$
In particular, if $\phi$ is onto, then $G/\ker\phi \cong G'$ .

Proof:

Let $\phi: G \to G'$ be a group homomorphism.
Let $K = \ker\phi$ . We know that $K \triangleleft G$ (Theorem 6), so the quotient group $G/K$ exists.
Also $\operatorname{Im}\phi \leq G'$ (Theorem 4).

Define a mapping $\psi: G/K \to \operatorname{Im}\phi$ by
$\psi(aK) = \phi(a) \quad \text{for all } aK \in G/K.$

We show that $\psi$ is an isomorphism.

Step 1. $\psi$ is well-defined.
Suppose $aK = bK$ . Then $a^{-1}b \in K = \ker\phi$ , so $\phi(a^{-1}b) = e'$ .
Thus $\phi(a)^{-1}\phi(b) = e'$ , i.e., $\phi(a) = \phi(b)$ .
Hence $\psi(aK) = \psi(bK)$ . So $\psi$ is well-defined.

Step 2. $\psi$ is a homomorphism.

\psi\big((aK)(bK)\big) = \psi(abK) = \phi(ab) = \phi(a)\,\phi(b) = \psi(aK)\,\psi(bK).

Step 3. $\psi$ is one-one.
Suppose $\psi(aK) = \psi(bK)$ . Then $\phi(a) = \phi(b)$ , so $\phi(a^{-1}b) = e'$ .
Thus $a^{-1}b \in \ker\phi = K$ , which means $aK = bK$ .
Hence $\psi$ is one-one.
(Equivalently: $\ker\psi = \{K\}$ , the identity of $G/K$ .)

Step 4. $\psi$ is onto.
Let $y \in \operatorname{Im}\phi$ be arbitrary. Then $y = \phi(a)$ for some $a \in G$ .
So $y = \psi(aK)$ . Hence $\psi$ is onto.

Therefore $\psi$ is an isomorphism, and $G/\ker\phi \cong \operatorname{Im}\phi$ .

If $\phi$ is onto, then $\operatorname{Im}\phi = G'$ , so $G/\ker\phi \cong G'$ .

Second and Third Isomorphism Theorems

The Fundamental Theorem of Group Homomorphism (Theorem 9) is also known as the First Isomorphism Theorem, and we label it I for reference. Its two classical companions are labelled II and III below. (These three theorems are given Roman numerals so that the running theorem numbering of these notes is not disturbed.)

Isomorphism Theorem II (Second Isomorphism Theorem — Diamond Theorem)

Let $G$ be a group, let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$ . Then:
(i) $HN = \{hn : h \in H,\ n \in N\}$ is a subgroup of $G$ .
(ii) $N$ is normal in $HN$ , and $H \cap N$ is normal in $H$ .
(iii) $\dfrac{H}{H \cap N} \cong \dfrac{HN}{N}$ .

Proof:

Let $G$ be a group with $H \leq G$ and $N \triangleleft G$ .

(i) First, $HN \neq \emptyset$ , since $e = e \cdot e \in HN$ .

Let $h_1 n_1,\ h_2 n_2 \in HN$ , where $h_1, h_2 \in H$ and $n_1, n_2 \in N$ . Then
$(h_1 n_1)(h_2 n_2) = h_1 (n_1 h_2) n_2 = h_1 (h_2 h_2^{-1} n_1 h_2) n_2 = (h_1 h_2)(h_2^{-1} n_1 h_2\, n_2).$
Since $N \triangleleft G$ and $h_2 \in G$ , we have $h_2^{-1} n_1 h_2 \in N$ , and therefore $h_2^{-1} n_1 h_2\, n_2 \in N$ . Also $h_1 h_2 \in H$ . Hence the product lies in $HN$ .

Next, for any $hn \in HN$ ,
$(hn)^{-1} = n^{-1} h^{-1} = h^{-1}\,(h n^{-1} h^{-1}),$
and $h n^{-1} h^{-1} \in N$ because $N \triangleleft G$ . Hence $(hn)^{-1} \in HN$ .

By the subgroup test, $HN \leq G$ .

(ii) First, $N \triangleleft HN$ . Since $N \triangleleft G$ , we have $xNx^{-1} = N$ for every $x \in G$ ; in particular for every $x \in HN \subseteq G$ . Hence $N$ is normal in the subgroup $HN$ .

Next, $H \cap N \triangleleft H$ . Since $H \leq G$ and $N \leq G$ , their intersection $H \cap N$ is a subgroup of $G$ contained in $H$ , i.e., $H \cap N \leq H$ .

Let $h \in H$ and let $x \in H \cap N$ . Since $x \in N$ and $N \triangleleft G$ , we get $hxh^{-1} \in N$ . Since $x \in H$ and $H$ is a subgroup, we get $hxh^{-1} \in H$ . Therefore $hxh^{-1} \in H \cap N$ . Hence $H \cap N \triangleleft H$ .

(iii) Note that every coset in $HN/N$ has the form $(hn)N = hN$ with $h \in H$ , so
$\frac{HN}{N} = \{hN : h \in H\}.$

Define a mapping $\phi: H \to HN/N$ by
$\phi(h) = hN \quad \text{for all } h \in H.$

$\phi$ is a homomorphism:
For $h_1, h_2 \in H$ ,
$\phi(h_1 h_2) = (h_1 h_2)N = (h_1 N)(h_2 N) = \phi(h_1)\,\phi(h_2),$
where the second equality uses that $N \triangleleft G$ (coset multiplication is well-defined, Theorem 7).

$\phi$ is onto:
Every element of $HN/N$ is of the form $hN = \phi(h)$ with $h \in H$ , by the description above.

Kernel of $\phi$ :
For $h \in H$ ,
$h \in \ker\phi \iff hN = N \iff h \in N \iff h \in H \cap N.$
Hence $\ker\phi = H \cap N$ .

By the First Isomorphism Theorem (Theorem 9),
$\frac{H}{\ker\phi} \cong \operatorname{Im}\phi, \qquad \text{i.e.,} \qquad \frac{H}{H \cap N} \cong \frac{HN}{N}.$

E.g., Let $G = S_3$ , $H = \{e, (1\,2)\}$ and $N = A_3 = \{e, (1\,2\,3), (1\,3\,2)\}$ . Then $N \triangleleft G$ (since $[S_n : A_n] = 2$ ), so the theorem applies. Here $HN = S_3$ and $H \cap N = \{e\}$ , and indeed
$\frac{H}{H \cap N} = H \cong \mathbb{Z}_2 \cong \frac{S_3}{A_3} = \frac{HN}{N}.$

Isomorphism Theorem III (Third Isomorphism Theorem)

Let $G$ be a group and let $N$ and $H$ be normal subgroups of $G$ with $N \subseteq H$ . Then $H/N$ is a normal subgroup of $G/N$ , and
$\frac{G/N}{H/N} \ \cong\ \frac{G}{H}.$

Proof:

Let $G$ be a group with $N \triangleleft G$ , $H \triangleleft G$ and $N \subseteq H$ .

Step 1. $H/N$ is a normal subgroup of $G/N$ .

First, $H/N = \{hN : h \in H\}$ is a subset of $G/N$ , and $N = eN \in H/N$ , so $H/N \neq \emptyset$ .

Let $h_1 N,\ h_2 N \in H/N$ . Then
$(h_1 N)(h_2 N)^{-1} = (h_1 h_2^{-1})N \in H/N,$
because $h_1 h_2^{-1} \in H$ . Hence $H/N \leq G/N$ .

Now let $aN \in G/N$ and let $hN \in H/N$ . Then
$(aN)(hN)(aN)^{-1} = (a h a^{-1})N.$
Since $H \triangleleft G$ and $h \in H$ , we have $a h a^{-1} \in H$ , so $(a h a^{-1})N \in H/N$ .
Hence $H/N \triangleleft G/N$ , and the quotient group $(G/N)/(H/N)$ exists.

Step 2. Define a mapping $\psi: G/N \to G/H$ by
$\psi(aN) = aH \quad \text{for all } aN \in G/N.$

$\psi$ is well-defined:
Suppose $aN = bN$ . Then $a^{-1}b \in N$ . Since $N \subseteq H$ , we get $a^{-1}b \in H$ , so $aH = bH$ , i.e., $\psi(aN) = \psi(bN)$ .

$\psi$ is a homomorphism:
$\psi\big((aN)(bN)\big) = \psi(abN) = (ab)H = (aH)(bH) = \psi(aN)\,\psi(bN).$

$\psi$ is onto:
Every element of $G/H$ is of the form $aH = \psi(aN)$ for some $a \in G$ .

Kernel of $\psi$ :
For $aN \in G/N$ ,
$aN \in \ker\psi \iff aH = H \iff a \in H \iff aN \in H/N.$
Hence $\ker\psi = H/N$ .

By the First Isomorphism Theorem (Theorem 9),
$\frac{G/N}{\ker\psi} \cong \operatorname{Im}\psi, \qquad \text{i.e.,} \qquad \frac{G/N}{H/N} \cong \frac{G}{H}.$

E.g., Let $G = (\mathbb{Z}, +)$ , $N = 12\mathbb{Z}$ and $H = 4\mathbb{Z}$ . Since $\mathbb{Z}$ is abelian, both $N$ and $H$ are normal in $G$ , and $N \subseteq H$ . Then $H/N = 4\mathbb{Z}/12\mathbb{Z} \cong \mathbb{Z}_3$ , and
$\frac{\mathbb{Z}/12\mathbb{Z}}{4\mathbb{Z}/12\mathbb{Z}} \cong \frac{\mathbb{Z}}{4\mathbb{Z}} \cong \mathbb{Z}_4.$

Automorphism and Inner Automorphism

Definition 10. Automorphism:
Let $G$ be a group. An isomorphism of $G$ onto itself is called an automorphism of $G$ .
The set of all automorphisms of $G$ is denoted by $\operatorname{Aut}(G)$ .

Theorem 10

Let $G$ be a group. Then $\operatorname{Aut}(G)$ is a group under composition of mappings.

Proof:

Let $G$ be a group, and let $\operatorname{Aut}(G)$ be the set of all automorphisms of $G$ .
We verify the group axioms under composition $\circ$ .

Closure: Let $\phi, \psi \in \operatorname{Aut}(G)$ . Then $\phi \circ \psi: G \to G$ is the composition of two bijections, hence a bijection. Also
$(\phi \circ \psi)(ab) = \phi\big(\psi(ab)\big) = \phi\big(\psi(a)\psi(b)\big) = \phi(\psi(a))\,\phi(\psi(b)) = (\phi \circ \psi)(a)\,(\phi \circ \psi)(b).$
So $\phi \circ \psi$ is a homomorphism, and hence an automorphism. Thus $\operatorname{Aut}(G)$ is closed.

Associativity: Composition of functions is always associative.

Identity: The identity map $I: G \to G$ given by $I(x) = x$ is clearly an automorphism, and $I \circ \phi = \phi \circ I = \phi$ .

Inverse: Let $\phi \in \operatorname{Aut}(G)$ . Since $\phi$ is a bijection, $\phi^{-1}: G \to G$ exists.
We show $\phi^{-1}$ is a homomorphism. Let $a, b \in G$ . Put $a = \phi(x)$ , $b = \phi(y)$ for unique $x, y \in G$ . Then
$\phi^{-1}(ab) = \phi^{-1}\big(\phi(x)\,\phi(y)\big) = \phi^{-1}\big(\phi(xy)\big) = xy = \phi^{-1}(a)\,\phi^{-1}(b).$
Also $\phi^{-1}$ is bijective. Hence $\phi^{-1} \in \operatorname{Aut}(G)$ .

Therefore $\operatorname{Aut}(G)$ is a group under composition.

Definition 11. Inner Automorphism:
Let $G$ be a group and let $g \in G$ . Define a mapping $T_g: G \to G$ by
$T_g(x) = gxg^{-1} \quad \text{for all } x \in G.$
Then $T_g$ is called the inner automorphism of $G$ determined by $g$ .
The set of all inner automorphisms of $G$ is denoted by $I(G)$ or $\operatorname{Inn}(G)$ .

Theorem 11

Let $G$ be a group and $g \in G$ . Then $T_g$ is an automorphism of $G$ .

Proof:

Let $G$ be a group and $g \in G$ . Define $T_g: G \to G$ by $T_g(x) = gxg^{-1}$ .

$T_g$ is a homomorphism:

T_g(xy) = g(xy)g^{-1} = gx(g^{-1}g)yg^{-1} = (gxg^{-1})(gyg^{-1}) = T_g(x)\,T_g(y).

$T_g$ is bijective:
The mapping $T_{g^{-1}}$ satisfies
$T_g \circ T_{g^{-1}} = T_{gg^{-1}} = T_e = I = T_{g^{-1}} \circ T_g.$
Hence $T_g$ is invertible (with inverse $T_{g^{-1}}$ ), so $T_g$ is bijective.

Therefore $T_g$ is an automorphism of $G$ .

Theorem 12

Let $G$ be a group. Then $I(G)$ is a normal subgroup of $\operatorname{Aut}(G)$ , and
$I(G) \cong \frac{G}{Z(G)}.$

Proof:

Let $G$ be a group.

Step 1. Define $\theta: G \to I(G)$ by $\theta(g) = T_g$ .

$\theta$ is a homomorphism:
For any $x \in G$ ,
$T_{gh}(x) = (gh)x(gh)^{-1} = g(hxh^{-1})g^{-1} = T_g\big(T_h(x)\big) = (T_g \circ T_h)(x).$
So $T_{gh} = T_g \circ T_h$ , i.e., $\theta(gh) = \theta(g)\,\theta(h)$ .

$\theta$ is onto: By definition, every element of $I(G)$ is $T_g$ for some $g \in G$ .

Kernel of $\theta$ :

g \in \ker\theta \iff T_g = I \iff gxg^{-1} = x \text{ for all } x \in G \iff gx = xg \text{ for all } x \in G \iff g \in Z(G).

Hence $\ker\theta = Z(G)$ .

Step 2. By the Fundamental Theorem of Group Homomorphism,
$\frac{G}{\ker\theta} \cong \operatorname{Im}\theta \implies \frac{G}{Z(G)} \cong I(G).$

Step 3. $I(G) \triangleleft \operatorname{Aut}(G)$ .

Let $\phi \in \operatorname{Aut}(G)$ and $T_g \in I(G)$ . We show that $\phi \circ T_g \circ \phi^{-1} \in I(G)$ .
For any $x \in G$ :

(\phi \circ T_g \circ \phi^{-1})(x) = \phi\Big(T_g\big(\phi^{-1}(x)\big)\Big) = \phi\Big(g\,\phi^{-1}(x)\,g^{-1}\Big) = \phi(g)\, x\, \phi(g)^{-1} = T_{\phi(g)}(x).

Hence $\phi \circ T_g \circ \phi^{-1} = T_{\phi(g)} \in I(G)$ .
Therefore $I(G) \triangleleft \operatorname{Aut}(G)$ .

Automorphism Groups — Computations

Theorem 13

$\operatorname{Aut}(\mathbb{Z}_n) \cong U(n)$ , where
$U(n) = \{k : 1 \leq k \leq n,\ \gcd(k,n) = 1\}$
is the multiplicative group of integers modulo $n$ that are coprime to $n$ . In particular,
$|\operatorname{Aut}(\mathbb{Z}_n)| = \varphi(n),$
where $\varphi$ is Euler's totient function.

Proof:

Let $G = \mathbb{Z}_n = \{\bar{0}, \bar{1}, \ldots, \overline{n-1}\}$ under addition modulo $n$ .
Then $G = \langle \bar{1} \rangle$ , and $G$ is cyclic of order $n$ .

Let $\phi \in \operatorname{Aut}(\mathbb{Z}_n)$ . Since $\phi$ is an isomorphism, it must send a generator to a generator.
The generators of $\mathbb{Z}_n$ are precisely those $\bar{k}$ for which $\gcd(k,n) = 1$ .
Thus $\phi(\bar{1}) = \bar{k}$ for some $k$ with $\gcd(k,n) = 1$ , and then
$\phi(\bar{m}) = \phi(m\cdot\bar{1}) = m\,\phi(\bar{1}) = m\bar{k} = \overline{mk}.$

Conversely, for each $k$ with $\gcd(k,n) = 1$ , the map $\phi_k: \mathbb{Z}_n \to \mathbb{Z}_n$ defined by $\phi_k(\bar{m}) = \overline{mk}$ is an automorphism.

Define $\Psi: \operatorname{Aut}(\mathbb{Z}_n) \to U(n)$ by
$\Psi(\phi) = k, \quad \text{where } \phi(\bar{1}) = \bar{k}.$

$\Psi$ is a homomorphism:
If $\phi(\bar{1}) = \bar{k}$ and $\psi(\bar{1}) = \bar{\ell}$ , then
$(\phi \circ \psi)(\bar{1}) = \phi(\bar{\ell}) = \overline{k\ell},$
so $\Psi(\phi \circ \psi) = k\ell \bmod n = \Psi(\phi)\,\Psi(\psi)$ in $U(n)$ .

$\Psi$ is bijective: From the correspondence above between automorphisms and such integers $k$ .

Hence $\operatorname{Aut}(\mathbb{Z}_n) \cong U(n)$ , and $|\operatorname{Aut}(\mathbb{Z}_n)| = |U(n)| = \varphi(n)$ .

E.g.,

$\operatorname{Aut}(\mathbb{Z}_5) \cong U(5) = \{1,2,3,4\} \cong \mathbb{Z}_4$ .
$\operatorname{Aut}(\mathbb{Z}_6) \cong U(6) = \{1,5\} \cong \mathbb{Z}_2$ .
$|\operatorname{Aut}(\mathbb{Z}_8)| = \varphi(8) = 4$ , and $\operatorname{Aut}(\mathbb{Z}_8) \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ .

Theorem 14

Let $G = \langle a \rangle$ be an infinite cyclic group. Then $\operatorname{Aut}(G) \cong \mathbb{Z}_2$ .

Proof:

Let $G = \langle a \rangle = \{\ldots, a^{-2}, a^{-1}, e, a, a^2, \ldots\}$ be infinite cyclic.
The generators of $G$ are only $a$ and $a^{-1}$ .

Any automorphism $\phi$ of $G$ is completely determined by $\phi(a)$ , and $\phi(a)$ must be a generator.
So there are exactly two possibilities:

$\phi_1(a) = a$ , so $\phi_1 = I$ (the identity automorphism);
$\phi_2(a) = a^{-1}$ , so $\phi_2(a^n) = a^{-n}$ .

Now $\phi_2 \circ \phi_2 = I$ , so $\{\,I,\ \phi_2\,\}$ is a group of order $2$ , isomorphic to $\mathbb{Z}_2$ .
Hence $\operatorname{Aut}(G) \cong \mathbb{Z}_2$ .

Conjugacy Relation and Normalizer

Definition 12. Conjugacy Relation:
Let $G$ be a group. Two elements $a, b \in G$ are said to be conjugate in $G$ if there exists an element $g \in G$ such that
$b = gag^{-1}.$
The relation “ $a$ is conjugate to $b$ ” is an equivalence relation on $G$ .
The equivalence class of $a$ is called the conjugacy class of $a$ :
$\operatorname{Cl}(a) = \{gag^{-1} : g \in G\}.$

Definition 13. Normalizer:
Let $G$ be a group and $a \in G$ . The normalizer of $a$ in $G$ is
$N(a) = \{g \in G : ga = ag\} = \{g \in G : gag^{-1} = a\}.$

More generally, if $H$ is a subgroup of $G$ , the normalizer of $H$ in $G$ is
$N(H) = \{g \in G : gHg^{-1} = H\}.$
Note that $H \triangleleft G$ if and only if $N(H) = G$ .

Theorem 15

Conjugacy is an equivalence relation on $G$ .

Proof:

Let $G$ be a group. Define $a \sim b$ iff $b = gag^{-1}$ for some $g \in G$ .

Reflexive: $a = eae^{-1}$ , so $a \sim a$ .

Symmetric: Suppose $a \sim b$ . Then $b = gag^{-1}$ for some $g$ .
Multiplying, $a = g^{-1}bg = g^{-1}b(g^{-1})^{-1}$ . So $b \sim a$ (using $g^{-1}$ ).

Transitive: Suppose $a \sim b$ and $b \sim c$ . Then $b = gag^{-1}$ and $c = hbh^{-1}$ for some $g,h \in G$ .
Thus $c = h(gag^{-1})h^{-1} = (hg)a(hg)^{-1}$ . So $a \sim c$ .

Hence conjugacy is an equivalence relation.

Theorem 16

Let $G$ be a group and $a \in G$ . Then $N(a)$ is a subgroup of $G$ .

Proof:

Let $G$ be a group and $a \in G$ .
Clearly $e \in N(a)$ , since $ea = ae$ . So $N(a) \neq \emptyset$ .

Let $g, h \in N(a)$ . Then $ga = ag$ and $ha = ah$ .
We show $gh^{-1} \in N(a)$ .

Since $ha = ah$ , we get $h^{-1}a = ah^{-1}$ (multiply both sides by $h^{-1}$ on left and right).
Now

(gh^{-1})a(gh^{-1})^{-1} = g(h^{-1}ah)g^{-1} = g a g^{-1} = (ga)g^{-1} = (ag)g^{-1} = a,

(using $h^{-1}ah = a$ and $g \in N(a)$ ).
Hence $gh^{-1} \in N(a)$ .
By the subgroup test, $N(a) \leq G$ .

Theorem 17

Let $G$ be a finite group and $a \in G$ . Then
$|\operatorname{Cl}(a)| = [G : N(a)] = \frac{|G|}{|N(a)|}.$

Proof:

Let $G$ be a finite group and $a \in G$ .
Let $N(a)$ be the normalizer of $a$ . We define a mapping
$f: \frac{G}{N(a)} \to \operatorname{Cl}(a)$
by
$f\big(g\,N(a)\big) = gag^{-1}.$

Well-defined:
Suppose $g\,N(a) = h\,N(a)$ . Then $h^{-1}g \in N(a)$ , so $(h^{-1}g)a = a(h^{-1}g)$ .
Multiplying, $gag^{-1} = hah^{-1}$ .
Thus $f(g\,N(a)) = f(h\,N(a))$ .

One-one:
Suppose $f(g\,N(a)) = f(h\,N(a))$ . Then $gag^{-1} = hah^{-1}$ , so $h^{-1}ga = ah^{-1}g$ , i.e., $h^{-1}g \in N(a)$ .
Hence $g\,N(a) = h\,N(a)$ .

Onto:
Every element of $\operatorname{Cl}(a)$ is of the form $gag^{-1} = f(g\,N(a))$ for some $g \in G$ .

Therefore $f$ is a bijection, and
$|\operatorname{Cl}(a)| = \left|\frac{G}{N(a)}\right| = [G:N(a)] = \frac{|G|}{|N(a)|}.$

Notes:

$a \in Z(G)$ if and only if $\operatorname{Cl}(a) = \{a\}$ if and only if $N(a) = G$ .
The conjugacy classes partition $G$ , and the class equation is
$|G| = |Z(G)| + \sum |\operatorname{Cl}(a_i)|,$
where the sum is over representatives of conjugacy classes with more than one element.

Part 2. Ring Theory

Rings — Definition, Examples, Properties

Definition 14. Ring:
A nonempty set $R$ together with two binary operations, addition ( $+$ ) and multiplication ( $\cdot$ ), is called a ring if the following axioms hold:

$(R, +)$ is an abelian group, i.e.,
- $a + b \in R$ for all $a,b \in R$ (closure);
- $(a+b)+c = a+(b+c)$ (associativity);
- there exists $0 \in R$ such that $a + 0 = a$ for all $a$ (zero element);
- for each $a \in R$ there exists $-a \in R$ such that $a + (-a) = 0$ (additive inverse);
- $a + b = b + a$ (commutativity of addition).
Multiplication is associative: $(ab)c = a(bc)$ for all $a,b,c \in R$ .
Distributive laws: For all $a,b,c \in R$ ,
$a(b+c) = ab + ac, \qquad (b+c)a = ba + ca.$

Definition 15. Special types of rings:

Commutative ring: $ab = ba$ for all $a,b \in R$ .
Ring with unity (or ring with identity): There exists $1 \in R$ , $1 \neq 0$ , such that $1\cdot a = a\cdot 1 = a$ for all $a \in R$ .
Zero ring: The ring consisting of only the zero element. (We usually exclude this case and assume $1 \neq 0$ .)

E.g.,

$\mathbb{Z}$ , $\mathbb{Q}$ , $\mathbb{R}$ , $\mathbb{C}$ are commutative rings with unity.
$\mathbb{Z}_n$ (integers modulo $n$ ) is a commutative ring with unity.
$M_n(\mathbb{R})$ , the set of all $n \times n$ real matrices, is a non-commutative ring with unity ( $I_n$ ).
$2\mathbb{Z} = \{0, \pm 2, \pm 4, \ldots\}$ is a commutative ring without unity.
$C[0,1]$ , the set of continuous real-valued functions on $[0,1]$ , with pointwise addition and multiplication, is a commutative ring with unity.

Theorem 18 (Basic Properties of a Ring)

Let $R$ be a ring. Then for all $a, b \in R$ :
(i) $a \cdot 0 = 0 \cdot a = 0$ .
(ii) $a(-b) = (-a)b = -(ab)$ .
(iii) $(-a)(-b) = ab$ .
(iv) $a(b-c) = ab - ac$ and $(b-c)a = ba - ca$ .

Proof:

Let $R$ be a ring and let $a, b, c \in R$ .

(i) We have
$a \cdot 0 = a \cdot (0 + 0) = a\cdot 0 + a\cdot 0$
(by the distributive law).
Adding $-(a\cdot 0)$ to both sides,
$0 = a\cdot 0.$
Similarly, $0 \cdot a = 0$ .

(ii) Consider
$ab + a(-b) = a\big(b + (-b)\big) = a\cdot 0 = 0.$
Therefore $a(-b)$ is the additive inverse of $ab$ , i.e., $a(-b) = -(ab)$ .

Similarly,
$ab + (-a)b = \big(a + (-a)\big)b = 0\cdot b = 0,$
so $(-a)b = -(ab)$ .

(iii) Using (ii) twice:
$(-a)(-b) = -\big(a(-b)\big) = -\big(-(ab)\big) = ab.$

(iv)
$a(b-c) = a\big(b + (-c)\big) = ab + a(-c) = ab + \big(-(ac)\big) = ab - ac.$
Similarly, $(b-c)a = ba - ca$ .

Subrings

Definition 16. Subring:
Let $R$ be a ring and $S$ a nonempty subset of $R$ . Then $S$ is called a subring of $R$ if $S$ itself is a ring under the same operations of addition and multiplication as defined on $R$ .

Theorem 19 (Subring Criterion / Subring Test)

Let $R$ be a ring and $S$ a nonempty subset of $R$ . Then $S$ is a subring of $R$ if and only if for all $a, b \in S$ :
$a - b \in S \quad \text{and} \quad ab \in S.$

Proof:

Necessity: Suppose $S$ is a subring of $R$ .
Then $(S, +)$ is a subgroup of $(R, +)$ , so $a, b \in S$ implies $a - b \in S$ .
Also multiplication is closed in $S$ , so $ab \in S$ .

Sufficiency: Suppose $S \neq \emptyset$ and $a - b \in S$ , $ab \in S$ for all $a,b \in S$ .
We verify that $S$ is a ring.

Since $S \neq \emptyset$ , there exists some $a \in S$ . Then $a - a = 0 \in S$ .
Also $0 - a = -a \in S$ .
For $a, b \in S$ : $a + b = a - (-b) \in S$ (since $-b \in S$ ).
So $(S, +)$ is closed, contains $0$ and inverses. Associativity and commutativity of addition are inherited from $R$ .
Thus $(S, +)$ is an abelian group.

Associativity of multiplication and the distributive laws hold in $S$ because they hold in $R$ .
Multiplication is closed in $S$ by hypothesis.

Hence $S$ is a subring of $R$ .

E.g.,

$n\mathbb{Z}$ is a subring of $\mathbb{Z}$ for any integer $n$ .
$\mathbb{Z}$ is a subring of $\mathbb{Q}$ ; $\mathbb{Q}$ is a subring of $\mathbb{R}$ .
Let $S$ be the set of all matrices $\begin\{pmatrix\} a & b \\ 0 & c \end\{pmatrix\}$ with $a,b,c \in \mathbb{Z}$ . Then $S$ is a subring of $M_2(\mathbb{Z})$ .
(Verify: difference and product of two such matrices again have the same form.)

Theorem 20

The intersection of any collection of subrings of a ring $R$ is again a subring of $R$ .

Proof:

Let $\{S_i : i \in I\}$ be a family of subrings of a ring $R$ , and let
$S = \bigcap_{i \in I} S_i.$

Clearly $0 \in S_i$ for each $i$ , so $0 \in S$ and $S \neq \emptyset$ .

Let $a, b \in S$ . Then $a, b \in S_i$ for every $i \in I$ .
Since each $S_i$ is a subring, $a - b \in S_i$ and $ab \in S_i$ for every $i$ .
Hence $a - b \in S$ and $ab \in S$ .

By the subring test, $S$ is a subring of $R$ .

Remark: The union of two subrings need not be a subring.
E.g., $2\mathbb{Z} \cup 3\mathbb{Z}$ is not a subring of $\mathbb{Z}$ , because $2 \in 2\mathbb{Z}$ , $3 \in 3\mathbb{Z}$ , but $2 + 3 = 5 \notin 2\mathbb{Z} \cup 3\mathbb{Z}$ .

Characteristic of a Ring

Definition 17. Characteristic of a Ring:
Let $R$ be a ring. The characteristic of $R$ , denoted $\operatorname{char}(R)$ , is defined as follows:

If there exists a positive integer $n$ such that $n\cdot a = 0$ for all $a \in R$ (where $n\cdot a = a + a + \cdots + a$ , $n$ times), then the smallest such positive integer $n$ is called the characteristic of $R$ .
If no such positive integer exists, we say $\operatorname{char}(R) = 0$ .

If $R$ has unity $1$ , then $\operatorname{char}(R)$ is the additive order of $1$ (or $0$ if this order is infinite). Equivalently, $\operatorname{char}(R)$ is the smallest positive integer $n$ such that $n\cdot 1 = 0$ , if it exists.

E.g., $\operatorname{char}(\mathbb{Z}) = 0$ , $\operatorname{char}(\mathbb{Q}) = 0$ , $\operatorname{char}(\mathbb{Z}_n) = n$ , $\operatorname{char}(\mathbb{Z}_6) = 6$ .

Definition 18. Integral Domain:
A ring $D$ is called an integral domain if:

$D$ is commutative;
$D$ has unity $1 \neq 0$ ;
$D$ has no zero divisors, i.e., whenever $a, b \in D$ and $ab = 0$ , then either $a = 0$ or $b = 0$ .

Definition 19. Field:
A ring $F$ is called a field if:

$F$ is commutative;
$F$ has unity $1 \neq 0$ ;
Every nonzero element of $F$ has a multiplicative inverse, i.e., for each $a \in F$ , $a \neq 0$ , there exists $a^{-1} \in F$ such that $a \cdot a^{-1} = 1$ .

E.g., $\mathbb{Z}$ is an integral domain but not a field. $\mathbb{Q}$ , $\mathbb{R}$ , $\mathbb{C}$ are fields. $\mathbb{Z}_n$ is a field if and only if $n$ is prime. $\mathbb{Z}_4$ is not an integral domain (since $\bar{2}\cdot\bar{2} = \bar{0}$ ).

Theorem 21

Every field is an integral domain.

Proof:

Let $F$ be a field.
By definition, $F$ is a commutative ring with unity $1 \neq 0$ .
We only need to show that $F$ has no zero divisors.

Suppose $a, b \in F$ with $ab = 0$ and $a \neq 0$ .
Since $F$ is a field and $a \neq 0$ , there exists $a^{-1} \in F$ such that $a^{-1}a = 1$ .
Multiplying $ab = 0$ on the left by $a^{-1}$ :
$a^{-1}(ab) = a^{-1}\cdot 0 \implies (a^{-1}a)b = 0 \implies 1\cdot b = 0 \implies b = 0.$

Similarly, if $b \neq 0$ and $ab = 0$ , then $a = 0$ .
Hence $F$ has no zero divisors, so $F$ is an integral domain.

Note: The converse is not true. $\mathbb{Z}$ is an integral domain but not a field (e.g., $2$ has no multiplicative inverse in $\mathbb{Z}$ ).

Theorem 22

Every finite integral domain is a field.

Proof:

Let $D$ be a finite integral domain.
Let $D = \{a_1, a_2, \ldots, a_n\}$ , where the $a_i$ are distinct.
By definition, $D$ is a commutative ring with unity and has no zero divisors.
To prove that $D$ is a field, it is enough to show that every nonzero element of $D$ has a multiplicative inverse.

Let $a \in D$ with $a \neq 0$ .
Consider the products
$a a_1,\ a a_2,\ \ldots,\ a a_n.$

Claim: These $n$ products are all distinct.
Suppose $a a_i = a a_j$ . Then $a(a_i - a_j) = 0$ .
Since $D$ has no zero divisors and $a \neq 0$ , we get $a_i - a_j = 0$ , i.e., $a_i = a_j$ .
Hence the products are distinct.

Since $D$ has exactly $n$ elements, and we have $n$ distinct products $aa_i$ , all lying in $D$ , we conclude
$\{aa_1, aa_2, \ldots, aa_n\} = D.$

In particular, the unity $1$ belongs to this set. So there exists some $a_k \in D$ such that
$a \cdot a_k = 1.$
Thus $a_k = a^{-1}$ .

Therefore every nonzero element of $D$ is a unit, and $D$ is a field.

Theorem 23

The characteristic of an integral domain is either $0$ or a prime number.

Proof:

Let $D$ be an integral domain, and let $\operatorname{char}(D) = n$ .

If $n = 0$ , there is nothing to prove.

Suppose $n > 0$ . We claim that $n$ is prime.
Suppose, on the contrary, that $n$ is composite. Then we can write
$n = rs,$
where $r$ and $s$ are integers satisfying $1 < r < n$ and $1 < s < n$ .

Now
$(r \cdot 1)(s \cdot 1) = (rs) \cdot 1 = n \cdot 1 = 0,$
where $1$ is the unity of $D$ .

Since $D$ is an integral domain, it has no zero divisors. Therefore
$r \cdot 1 = 0 \quad \text{or} \quad s \cdot 1 = 0.$

If $r \cdot 1 = 0$ , then $r \cdot a = (r \cdot 1)a = 0$ for all $a \in D$ , which means $\operatorname{char}(D) \leq r < n$ , contradicting the minimality of $n$ .
Similarly, $s \cdot 1 = 0$ leads to a contradiction.

Hence $n$ cannot be composite. Since $n > 1$ (characteristic at least $1$ if positive; actually $n \geq 2$ because $1 \neq 0$ ), $n$ must be a prime number.

Ideals

Definition 20. Ideal:
Let $R$ be a ring and $I$ a nonempty subset of $R$ . Then $I$ is called an ideal (or two-sided ideal) of $R$ if:

$a, b \in I$ implies $a - b \in I$ (i.e., $(I, +)$ is a subgroup of $(R, +)$ );
For all $r \in R$ and all $a \in I$ , both $ra \in I$ and $ar \in I$ .

If condition 2 is replaced by “ $ra \in I$ only”, then $I$ is a left ideal. If “ $ar \in I$ only”, then $I$ is a right ideal.
In a commutative ring, every left (or right) ideal is automatically two-sided.

E.g.,

For any ring $R$ , $\{0\}$ and $R$ itself are ideals, called the trivial (or improper) ideals. Any other ideal is called a proper (or nontrivial) ideal.
For any integer $n$ , $n\mathbb{Z}$ is an ideal of $\mathbb{Z}$ .
In $\mathbb{Z}$ , every ideal is of the form $n\mathbb{Z}$ for some $n \geq 0$ .

Theorem 24

Let $F$ be a field. Then the only ideals of $F$ are $\{0\}$ and $F$ itself.

Proof:

Let $F$ be a field and let $I$ be an ideal of $F$ .

If $I = \{0\}$ , we are done.

Suppose $I \neq \{0\}$ . Then there exists an element $a \in I$ with $a \neq 0$ .
Since $F$ is a field, $a$ has a multiplicative inverse $a^{-1} \in F$ .
Because $I$ is an ideal and $a \in I$ , we have
$1 = a^{-1} \cdot a \in I.$

Now let $r \in F$ be arbitrary. Since $1 \in I$ and $I$ is an ideal,
$r = r \cdot 1 \in I.$
Hence $F \subseteq I$ . But $I \subseteq F$ , so $I = F$ .

Therefore the only ideals of $F$ are $\{0\}$ and $F$ .

Theorem 25

A commutative ring $R$ with unity is a field if and only if $R$ has no proper ideals.

Proof:

Let $R$ be a commutative ring with unity.

( $\Rightarrow$ ) Suppose $R$ is a field.
By Theorem 24, the only ideals of $R$ are $\{0\}$ and $R$ .
Hence $R$ has no proper ideals.

( $\Leftarrow$ ) Suppose $R$ has no proper ideals.
Let $a \in R$ with $a \neq 0$ .
Consider the principal ideal
$(a) = \{ra : r \in R\}.$
This is a nonzero ideal of $R$ (since $a = 1\cdot a \in (a)$ ).
Since $R$ has no proper ideals, we must have $(a) = R$ .

In particular, $1 \in (a)$ , so there exists $r \in R$ such that $ra = 1$ .
Thus $r = a^{-1}$ .

Therefore every nonzero element of $R$ has a multiplicative inverse, and since $R$ is commutative with unity, $R$ is a field.

Theorem 26

Let $R$ be a ring with unity and $I$ an ideal of $R$ . If $1 \in I$ , then $I = R$ .

Proof:

Let $R$ be a ring with unity and $I$ an ideal of $R$ such that $1 \in I$ .
Let $r \in R$ be arbitrary.
Since $I$ is an ideal and $1 \in I$ , we have $r = r \cdot 1 \in I$ .
Hence $R \subseteq I$ . But $I \subseteq R$ , so $I = R$ .

Ideal Generated by a Subset

Definition 21. Ideal Generated by a Set:
Let $R$ be a ring and $S$ a subset of $R$ . The ideal generated by $S$ , denoted by $(S)$ or $\langle S \rangle$ , is the smallest ideal of $R$ containing $S$ .
Equivalently, $(S)$ is the intersection of all ideals of $R$ that contain $S$ .

If $S = \{a\}$ , we write $(a)$ and call it the principal ideal generated by $a$ .
If $S = \{a_1, \ldots, a_n\}$ , we write $(a_1, \ldots, a_n)$ .

Theorem 27

Let $R$ be a commutative ring with unity, and let $a \in R$ . Then
$(a) = \{ra : r \in R\} = Ra.$
More generally, if $a_1, \ldots, a_n \in R$ , then
$(a_1, \ldots, a_n) = \{r_1 a_1 + r_2 a_2 + \cdots + r_n a_n : r_i \in R\}.$

Proof:

Let $R$ be a commutative ring with unity and $a \in R$ .
Let $I = \{ra : r \in R\}$ .

Step 1. $I$ is an ideal of $R$ containing $a$ .

$0 = 0\cdot a \in I$ .
If $r_1 a,\, r_2 a \in I$ , then $r_1 a - r_2 a = (r_1 - r_2)a \in I$ .
If $s \in R$ and $ra \in I$ , then $s(ra) = (sr)a \in I$ .
$a = 1\cdot a \in I$ .

Step 2. $I$ is the smallest such ideal.
Let $J$ be any ideal of $R$ containing $a$ .
For any $r \in R$ , since $a \in J$ and $J$ is an ideal, $ra \in J$ .
Hence $I \subseteq J$ .

Therefore $(a) = I = Ra$ .

The proof for $(a_1,\ldots,a_n)$ is similar: the set of all $R$ -linear combinations of $a_1,\ldots,a_n$ is an ideal containing each $a_i$ , and it is contained in every ideal that contains all the $a_i$ .

Definition 22. Principal Ideal Domain (PID):
An integral domain $R$ in which every ideal is a principal ideal (i.e., of the form $(a)$ for some $a \in R$ ) is called a principal ideal domain.

E.g., $\mathbb{Z}$ is a PID. Every ideal of $\mathbb{Z}$ is $n\mathbb{Z} = (n)$ for some $n \geq 0$ .
$\mathbb{Z}[x]$ is not a PID (proved later: $(2,x)$ is not principal).

Operations on Ideals

Definition 23. Let $I$ and $J$ be ideals of a ring $R$ .

Sum: $I + J = \{a + b : a \in I,\ b \in J\}$ . This is an ideal of $R$ , and it is the smallest ideal containing both $I$ and $J$ .
Product: $IJ = \left\{ \sum_{k=1}^{n} a_k b_k : a_k \in I,\ b_k \in J,\ n \geq 1 \right\}$ . This is an ideal of $R$ .
Intersection: $I \cap J$ is an ideal of $R$ .

In general: $IJ \subseteq I \cap J \subseteq I \subseteq I + J$ and $I \cap J \subseteq J \subseteq I + J$ .

Quotient (Factor) Rings

Definition 24. Quotient Ring (Factor Ring):
Let $R$ be a ring and $I$ an ideal of $R$ . Let $R/I$ denote the set of all cosets of $I$ in $R$ :
$R/I = \{a + I : a \in R\},$
where $a + I = \{a + x : x \in I\}$ .

Define addition and multiplication on $R/I$ by
$(a + I) + (b + I) = (a + b) + I,$
$(a + I)(b + I) = (ab) + I.$

Then $R/I$ is a ring, called the quotient ring (or factor ring, or ring of residue classes) of $R$ modulo $I$ .

The zero element is $0 + I = I$ .
The additive inverse of $a + I$ is $(-a) + I$ .
If $R$ has unity $1$ , then $R/I$ has unity $1 + I$ .

(Well-definedness of multiplication uses that $I$ is an ideal: if $a - a' \in I$ and $b - b' \in I$ , then $ab - a'b' = a(b-b') + (a-a')b' \in I$ .)

E.g., $\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}_n$ .

Prime and Maximal Ideals

Throughout this section, unless stated otherwise, $R$ denotes a commutative ring with unity.

Definition 25. Prime Ideal:
A proper ideal $P$ of $R$ is called a prime ideal if whenever $a, b \in R$ and $ab \in P$ , then either $a \in P$ or $b \in P$ .

Definition 26. Maximal Ideal:
A proper ideal $M$ of $R$ is called a maximal ideal if there is no proper ideal of $R$ that strictly contains $M$ .
In other words: if $I$ is an ideal of $R$ such that $M \subseteq I \subseteq R$ , then either $I = M$ or $I = R$ .

E.g.,

In $\mathbb{Z}$ : the ideal $(p) = p\mathbb{Z}$ is prime if and only if $p$ is a prime number; and in that case it is also maximal. The ideal $(0)$ is prime in $\mathbb{Z}$ but not maximal.
In $\mathbb{Z}[x]$ : $(x)$ is a prime ideal (because $\mathbb{Z}[x]/(x) \cong \mathbb{Z}$ , an integral domain), but $(x)$ is not maximal (because $\mathbb{Z}$ is not a field).
$(2, x)$ is a maximal ideal of $\mathbb{Z}[x]$ , since $\mathbb{Z}[x]/(2,x) \cong \mathbb{Z}_2$ , a field.

Theorem 28

Let $R$ be a commutative ring with unity and $P$ an ideal of $R$ . Then $P$ is a prime ideal if and only if the quotient ring $R/P$ is an integral domain.

Proof:

Let $R$ be a commutative ring with unity and $P$ an ideal of $R$ .

( $\Rightarrow$ ) Suppose $P$ is a prime ideal.
Then $P \neq R$ , so $1 + P \neq P$ (i.e., the unity of $R/P$ is nonzero).
Also $R/P$ is commutative (since $R$ is).

We show that $R/P$ has no zero divisors.
Suppose $(a + P)(b + P) = P$ (the zero of $R/P$ ).
Then $ab + P = P$ , so $ab \in P$ .
Since $P$ is prime, $a \in P$ or $b \in P$ .
Hence $a + P = P$ or $b + P = P$ .
Therefore $R/P$ has no zero divisors, and $R/P$ is an integral domain.

( $\Leftarrow$ ) Suppose $R/P$ is an integral domain.
Then in particular $R/P \neq \{P\}$ , so $P \neq R$ ( $P$ is proper).

Let $a, b \in R$ with $ab \in P$ .
Then $(a + P)(b + P) = ab + P = P$ .
Since $R/P$ has no zero divisors, $a + P = P$ or $b + P = P$ .
Hence $a \in P$ or $b \in P$ .

Therefore $P$ is a prime ideal.

Theorem 29

Let $R$ be a commutative ring with unity and $M$ an ideal of $R$ . Then $M$ is a maximal ideal if and only if the quotient ring $R/M$ is a field.

Proof:

Let $R$ be a commutative ring with unity and $M$ an ideal of $R$ .

( $\Rightarrow$ ) Suppose $M$ is maximal.
Then $M \neq R$ , so $1 + M \neq M$ . Also $R/M$ is commutative with unity.
We show that every nonzero element of $R/M$ has a multiplicative inverse.

Let $a + M \in R/M$ with $a + M \neq M$ , i.e., $a \notin M$ .
Consider the set
$I = \{m + ra : m \in M,\ r \in R\}.$
(This is the ideal generated by $M$ and $a$ , often written $(M, a)$ or $M + (a)$ .)

One verifies that $I$ is an ideal of $R$ , and $M \subsetneq I$ (since $a = 0 + 1\cdot a \in I$ but $a \notin M$ ).
Since $M$ is maximal, $I = R$ .

In particular, $1 \in I$ , so there exist $m \in M$ and $r \in R$ such that
$1 = m + ra.$
Passing to cosets:
$1 + M = (m + M) + (r + M)(a + M) = (r + M)(a + M),$
since $m + M = M$ .
Hence $r + M$ is the multiplicative inverse of $a + M$ in $R/M$ .

Therefore $R/M$ is a field.

( $\Leftarrow$ ) Suppose $R/M$ is a field.
Then $R/M \neq \{M\}$ , so $M \neq R$ ( $M$ is proper).

Let $I$ be an ideal of $R$ such that $M \subseteq I \subseteq R$ .
Suppose $I \neq M$ . We show that $I = R$ .

Take $a \in I$ with $a \notin M$ . Then $a + M \neq M$ in $R/M$ .
Since $R/M$ is a field, $a + M$ has an inverse: there exists $b \in R$ such that
$(a + M)(b + M) = 1 + M,$
i.e., $ab - 1 \in M$ .
Since $M \subseteq I$ , we have $ab - 1 \in I$ .
Also $a \in I$ and $I$ is an ideal, so $ab \in I$ .
Therefore
$1 = ab - (ab - 1) \in I.$
By Theorem 26, $I = R$ .

Hence $M$ is maximal.

Theorem 30

Every maximal ideal of a commutative ring with unity is a prime ideal.

Proof:

Let $R$ be a commutative ring with unity and let $M$ be a maximal ideal of $R$ .
By Theorem 29, $R/M$ is a field.
By Theorem 21, every field is an integral domain.
So $R/M$ is an integral domain.
By Theorem 28, $M$ is a prime ideal.

Converse is false:
A prime ideal need not be maximal.
E.g., $(0)$ is a prime ideal of $\mathbb{Z}$ (since $\mathbb{Z}/(0) \cong \mathbb{Z}$ is an integral domain), but $(0)$ is not maximal (e.g., $(0) \subsetneq (2) \subsetneq \mathbb{Z}$ ).
Another example: $(x)$ is prime but not maximal in $\mathbb{Z}[x]$ .

Ring Homomorphisms

Definition 27. Ring Homomorphism:
Let $R$ and $R'$ be rings. A mapping $\phi: R \to R'$ is called a ring homomorphism if for all $a, b \in R$ :
$\phi(a + b) = \phi(a) + \phi(b),$
$\phi(ab) = \phi(a)\,\phi(b).$

(Many authors also require $\phi(1_R) = 1_{R'}$ when both rings have unity; we state this separately when needed.)

Definition 28. Kernel and Image:
$\ker\phi = \{a \in R : \phi(a) = 0'\}, \qquad \operatorname{Im}\phi = \{\phi(a) : a \in R\},$
where $0'$ is the zero element of $R'$ .

E.g.,

$\phi: \mathbb{Z} \to \mathbb{Z}_n$ given by $\phi(k) = \bar{k}$ is a surjective ring homomorphism with $\ker\phi = n\mathbb{Z}$ .
Evaluation at $0$ : $\phi: \mathbb{Z}[x] \to \mathbb{Z}$ , $\phi(f) = f(0)$ , is a surjective homomorphism with $\ker\phi = (x)$ .
The natural map $\pi: R \to R/I$ , $\pi(a) = a + I$ , is a surjective homomorphism with kernel $I$ .

Theorem 31 (Properties of a Ring Homomorphism)

Let $\phi: R \to R'$ be a ring homomorphism. Then:
(i) $\phi(0) = 0'$ .
(ii) $\phi(-a) = -\phi(a)$ for all $a \in R$ .
(iii) $\ker\phi$ is an ideal of $R$ .
(iv) $\operatorname{Im}\phi$ is a subring of $R'$ .
(v) $\phi$ is one-one if and only if $\ker\phi = \{0\}$ .
(vi) If $\phi$ is onto and $R$ has unity $1$ , then $\phi(1)$ is the unity of $R'$ .

Proof:

Let $\phi: R \to R'$ be a ring homomorphism.

(i) $\phi(0) = \phi(0 + 0) = \phi(0) + \phi(0)$ .
Adding $-\phi(0)$ to both sides: $0' = \phi(0)$ .

(ii) $\phi(a) + \phi(-a) = \phi\big(a + (-a)\big) = \phi(0) = 0'$ .
Hence $\phi(-a) = -\phi(a)$ .

(iii) First, $\ker\phi$ is an additive subgroup of $R$ (same argument as for group homomorphisms, using (i) and (ii)).

Now let $a \in \ker\phi$ and $r \in R$ . Then
$\phi(ra) = \phi(r)\,\phi(a) = \phi(r)\cdot 0' = 0',$
so $ra \in \ker\phi$ . Similarly $ar \in \ker\phi$ .
Hence $\ker\phi$ is an ideal of $R$ .

(iv) Let $\phi(a), \phi(b) \in \operatorname{Im}\phi$ . Then
$\phi(a) - \phi(b) = \phi(a - b) \in \operatorname{Im}\phi, \qquad \phi(a)\,\phi(b) = \phi(ab) \in \operatorname{Im}\phi.$
By the subring test, $\operatorname{Im}\phi$ is a subring of $R'$ .

(v) Suppose $\phi$ is one-one. If $a \in \ker\phi$ , then $\phi(a) = 0' = \phi(0)$ , so $a = 0$ . Thus $\ker\phi = \{0\}$ .

Conversely, suppose $\ker\phi = \{0\}$ . If $\phi(a) = \phi(b)$ , then $\phi(a - b) = 0'$ , so $a - b \in \ker\phi = \{0\}$ , hence $a = b$ . So $\phi$ is one-one.

(vi) Let $r' \in R'$ be arbitrary. Since $\phi$ is onto, $r' = \phi(a)$ for some $a \in R$ . Then
$\phi(1)\,r' = \phi(1)\,\phi(a) = \phi(1\cdot a) = \phi(a) = r'.$
Similarly $r'\,\phi(1) = r'$ . Hence $\phi(1)$ is the unity of $R'$ .

Theorem 32 (First Isomorphism Theorem for Rings)

Let $\phi: R \to R'$ be a ring homomorphism. Then
$\frac{R}{\ker\phi} \ \cong\ \operatorname{Im}\phi.$

Proof:

Let $\phi: R \to R'$ be a ring homomorphism, and let $K = \ker\phi$ .
We know $K$ is an ideal of $R$ , so the quotient ring $R/K$ exists.
Also $\operatorname{Im}\phi$ is a subring of $R'$ .

Define $\psi: R/K \to \operatorname{Im}\phi$ by
$\psi(a + K) = \phi(a).$

Well-defined: If $a + K = b + K$ , then $a - b \in K$ , so $\phi(a - b) = 0'$ , i.e., $\phi(a) = \phi(b)$ .

Homomorphism:
$\psi\big((a+K)+(b+K)\big) = \psi((a+b)+K) = \phi(a+b) = \phi(a)+\phi(b) = \psi(a+K)+\psi(b+K),$
$\psi\big((a+K)(b+K)\big) = \psi(ab+K) = \phi(ab) = \phi(a)\phi(b) = \psi(a+K)\psi(b+K).$

One-one: $\psi(a+K) = 0' \iff \phi(a) = 0' \iff a \in K \iff a + K = K$ .
So $\ker\psi = \{K\}$ , hence $\psi$ is one-one.

Onto: If $y \in \operatorname{Im}\phi$ , then $y = \phi(a) = \psi(a+K)$ for some $a \in R$ .

Therefore $\psi$ is a ring isomorphism, and $R/\ker\phi \cong \operatorname{Im}\phi$ .

Field of Quotients of an Integral Domain

Theorem 33 (Existence of the Field of Quotients)

Let $D$ be an integral domain. Then there exists a field $F$ containing a subring isomorphic to $D$ such that every element of $F$ can be expressed as $ab^{-1}$ for some $a \in D$ and $b \in D$ , $b \neq 0$ . This field $F$ is called the field of quotients (or field of fractions) of $D$ , and it is unique up to isomorphism.

Proof (Construction):

Let $D$ be an integral domain. Let
$S = D \times (D \setminus \{0\}) = \{(a,b) : a \in D,\ b \in D,\ b \neq 0\}.$

Define a relation $\sim$ on $S$ by
$(a,b) \sim (c,d) \iff ad = bc.$

Step 1. $\sim$ is an equivalence relation.

Reflexive: $ab = ba$ (since $D$ is commutative), so $(a,b) \sim (a,b)$ .
Symmetric: If $ad = bc$ , then $cb = da$ , so $(c,d) \sim (a,b)$ .
Transitive: Suppose $(a,b) \sim (c,d)$ and $(c,d) \sim (e,f)$ , so $ad = bc$ and $cf = de$ .
Then $adf = bcf = bde$ . Since $d \neq 0$ and $D$ is an integral domain, we may “cancel” $d$ in the sense: $adf = bde \implies d(af - be) = 0 \implies af = be$ .
Thus $(a,b) \sim (e,f)$ .

Denote the equivalence class of $(a,b)$ by $\dfrac{a}{b}$ .
Let $F = \left\{ \dfrac{a}{b} : a \in D,\ b \neq 0 \right\}$ .

Step 2. Define operations on $F$ :
$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}, \qquad \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}.$

These are well-defined (verification: if $\frac{a}{b} = \frac{a'}{b'}$ and $\frac{c}{d} = \frac{c'}{d'}$ , then the sums and products of the two representations give equivalent pairs — use $ab' = a'b$ and $cd' = c'd$ ).

Step 3. $F$ is a field.

Addition and multiplication are associative and commutative (inherited from $D$ , with routine checks).
Distributivity holds.
Zero element: $\dfrac{0}{1}$ . Additive inverse of $\dfrac{a}{b}$ is $\dfrac{-a}{b}$ .
Unity: $\dfrac{1}{1}$ .
If $\dfrac{a}{b} \neq \dfrac{0}{1}$ , then $a \neq 0$ , so $\dfrac{b}{a}$ exists in $F$ and $\dfrac{a}{b} \cdot \dfrac{b}{a} = \dfrac{1}{1}$ .

Step 4. Embedding of $D$ into $F$ .
Define $\iota: D \to F$ by $\iota(a) = \dfrac{a}{1}$ .
Then $\iota$ is an injective ring homomorphism, so $D$ may be identified with its image $\{\frac{a}{1} : a \in D\} \subseteq F$ .
Under this identification, $\dfrac{a}{b} = \dfrac{a}{1} \cdot \left(\dfrac{b}{1}\right)^{-1} = ab^{-1}$ .

E.g., The field of quotients of $\mathbb{Z}$ is $\mathbb{Q}$ . The field of quotients of $F[x]$ (where $F$ is a field) is the field $F(x)$ of rational functions.

Polynomial Rings

Definition 29. Polynomial Ring:
Let $R$ be a ring. A polynomial in $x$ with coefficients in $R$ is a formal expression
$f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n,$
where $n$ is a non-negative integer and $a_i \in R$ .
The set of all such polynomials is denoted by $R[x]$ .

Two polynomials are equal if and only if corresponding coefficients are equal (after padding with zeros if needed).

Addition and multiplication:
If $f(x) = \sum a_i x^i$ and $g(x) = \sum b_j x^j$ , then
$f + g = \sum (a_i + b_i)x^i, \qquad fg = \sum_k c_k x^k \quad \text{where } c_k = \sum_{i+j=k} a_i b_j.$

Definition 30. Degree:
If $f(x) = a_0 + a_1 x + \cdots + a_n x^n$ with $a_n \neq 0$ , then the degree of $f$ is $\deg(f) = n$ , and $a_n$ is the leading coefficient.
The zero polynomial has degree defined as $-\infty$ by convention (or left undefined).
A polynomial is monic if its leading coefficient is $1$ .

Theorem 34

Let $R$ be a ring. Then $R[x]$ is a ring under the above operations. If $R$ is commutative, so is $R[x]$ . If $R$ has unity $1$ , then $R[x]$ has unity (the constant polynomial $1$ ).

Theorem 35

Let $D$ be an integral domain. Then $D[x]$ is also an integral domain. Moreover, for nonzero $f, g \in D[x]$ :
$\deg(fg) = \deg(f) + \deg(g).$

Proof:

Let $D$ be an integral domain. Then $D[x]$ is a commutative ring with unity (Theorem 34).

Let $f, g \in D[x]$ be nonzero. Write
$f = a_m x^m + \cdots + a_0, \quad a_m \neq 0,$
$g = b_n x^n + \cdots + b_0, \quad b_n \neq 0.$

The leading term of $fg$ is $a_m b_n x^{m+n}$ .
Since $D$ is an integral domain and $a_m \neq 0$ , $b_n \neq 0$ , we have $a_m b_n \neq 0$ .
Therefore $fg \neq 0$ , and $\deg(fg) = m + n = \deg(f) + \deg(g)$ .

Hence $D[x]$ has no zero divisors, so $D[x]$ is an integral domain.

Theorem 36 (Division Algorithm)

Let $F$ be a field and let $f(x), g(x) \in F[x]$ with $g(x) \neq 0$ . Then there exist unique polynomials $q(x), r(x) \in F[x]$ such that
$f(x) = q(x)\, g(x) + r(x),$
where either $r(x) = 0$ or $\deg\big(r(x)\big) < \deg\big(g(x)\big)$ .

Proof:

Let $F$ be a field and $f, g \in F[x]$ with $g \neq 0$ .
Let $\deg(g) = m$ and let the leading coefficient of $g$ be $b_m \neq 0$ .

Existence: We proceed by induction on $n = \deg(f)$ (and handle $f = 0$ separately).

If $f = 0$ or $\deg(f) < \deg(g)$ , take $q = 0$ and $r = f$ . Then the conditions hold.

Now suppose $\deg(f) = n \geq m = \deg(g)$ , and assume the result holds for all polynomials of degree less than $n$ .
Let $f = a_n x^n + \cdots$ with $a_n \neq 0$ .
Define
$f_1(x) = f(x) - a_n b_m^{-1} x^{n-m}\, g(x).$
(Here $b_m^{-1}$ exists because $F$ is a field and $b_m \neq 0$ .)

The leading terms cancel, so either $f_1 = 0$ or $\deg(f_1) < n$ .
By the induction hypothesis, there exist $q_1, r \in F[x]$ such that
$f_1 = q_1 g + r,$
with $r = 0$ or $\deg(r) < \deg(g)$ .

Therefore
$f = f_1 + a_n b_m^{-1} x^{n-m}\, g = \big(q_1 + a_n b_m^{-1} x^{n-m}\big)g + r.$
Take $q = q_1 + a_n b_m^{-1} x^{n-m}$ . This proves existence.

Uniqueness: Suppose
$f = qg + r = q'g + r',$
with $r = 0$ or $\deg(r) < \deg(g)$ , and $r' = 0$ or $\deg(r') < \deg(g)$ .

Then $(q - q')g = r' - r$ .
If $q \neq q'$ , then the left side has degree $\deg(q - q') + \deg(g) \geq \deg(g)$ , while the right side has degree $< \deg(g)$ (or is zero) — a contradiction.
Hence $q = q'$ , and then $r = r'$ .

Theorem 37 (Remainder Theorem / Factor Theorem)

Let $F$ be a field, $f(x) \in F[x]$ , and $a \in F$ . Then:
(i) When $f(x)$ is divided by $x - a$ , the remainder is the constant $f(a)$ .
(ii) $(x - a)$ divides $f(x)$ if and only if $f(a) = 0$ .

Proof:

Let $F$ be a field, $f \in F[x]$ , and $a \in F$ .
By the Division Algorithm, dividing $f$ by $g(x) = x - a$ (degree $1$ ), we get
$f(x) = q(x)\,(x - a) + r,$
where $r \in F$ is a constant (since $\deg(r) < 1$ or $r = 0$ ).

Put $x = a$ :
$f(a) = q(a)\,(a - a) + r = r.$
Hence the remainder is $f(a)$ . This proves (i).

For (ii): $x - a$ divides $f(x)$ if and only if the remainder is $0$ , if and only if $f(a) = 0$ .

Theorem 38

If $F$ is a field, then $F[x]$ is a principal ideal domain (PID).

Proof:

Let $F$ be a field and let $I$ be an ideal of $F[x]$ .

If $I = \{0\}$ , then $I = (0)$ , a principal ideal.

Suppose $I \neq \{0\}$ . Among all nonzero elements of $I$ , choose $g \in I$ of minimal degree.
We claim that $I = (g)$ .

Clearly $(g) \subseteq I$ , because $I$ is an ideal and $g \in I$ .

Conversely, let $f \in I$ be arbitrary.
By the Division Algorithm, $f = qg + r$ with $r = 0$ or $\deg(r) < \deg(g)$ .
Then $r = f - qg \in I$ (since $f, g \in I$ and $I$ is an ideal).

If $r \neq 0$ , then $r \in I$ is nonzero with $\deg(r) < \deg(g)$ , contradicting the minimality of $\deg(g)$ .
Hence $r = 0$ , so $f = qg \in (g)$ .

Therefore $I \subseteq (g)$ , and $I = (g)$ .

Hence every ideal of $F[x]$ is principal, so $F[x]$ is a PID.

Theorem 39

$\mathbb{Z}[x]$ is not a principal ideal domain.

Proof:

Consider the ideal $I = (2, x) = \{2f(x) + x\, g(x) : f,g \in \mathbb{Z}[x]\}$ of $\mathbb{Z}[x]$ .
(Explicitly, $I$ consists of all polynomials in $\mathbb{Z}[x]$ whose constant term is even.)

Suppose, for contradiction, that $I$ is principal, say $I = (d(x))$ for some $d(x) \in \mathbb{Z}[x]$ .

Since $2 \in I$ , we have $2 = d(x)\, q(x)$ for some $q(x) \in \mathbb{Z}[x]$ .
Taking degrees: $\deg(d) + \deg(q) = \deg(2) = 0$ , so $\deg(d) = 0$ .
Thus $d(x)$ is a constant polynomial, say $d(x) = d \in \mathbb{Z}$ .

Since $d$ divides $2$ in $\mathbb{Z}[x]$ , we have $d = \pm 1$ or $d = \pm 2$ .

If $d = \pm 1$ , then $(d) = \mathbb{Z}[x]$ . But $I \neq \mathbb{Z}[x]$ , because the constant polynomial $1$ does not belong to $I$ (constant term of elements of $I$ must be even). Contradiction.

If $d = \pm 2$ , then $(d) = (2) = \{2f(x) : f \in \mathbb{Z}[x]\}$ .
But $x \in I$ , and $x \notin (2)$ (the coefficients of $x$ in $(2)$ are all even; the coefficient of $x$ in $x$ is $1$ , which is odd). Contradiction.

Therefore $I = (2,x)$ is not principal, and $\mathbb{Z}[x]$ is not a PID.

Question Bank

1-Mark Questions

Definitions:

Define left coset and right coset of a subgroup.
Define index of a subgroup.
Define a normal subgroup.
Define the centre $Z(G)$ of a group $G$ .
Define a group homomorphism and its kernel.
Define an automorphism and an inner automorphism.
Define conjugacy class and normalizer of an element.
Define a ring / commutative ring / ring with unity.
Define a subring.
Define characteristic of a ring.
Define an integral domain and a field.
Define an ideal of a ring.
Define a principal ideal and a PID.
Define a prime ideal and a maximal ideal.
Define a ring homomorphism and its kernel.
Define the field of quotients of an integral domain.
Define degree and leading coefficient of a polynomial.
What is a monic polynomial?

Short facts:

State Lagrange's theorem.
If $|G| = p$ (prime), what can you say about $G$ ?
If $[G:H] = 2$ , what can you say about $H$ ?
Write $I(G)$ in terms of $G$ and $Z(G)$ .
$|\operatorname{Aut}(\mathbb{Z}_n)| =$ ?
When is $\mathbb{Z}_n$ a field?
$\operatorname{char}(\mathbb{Z}) =$ ? $\operatorname{char}(\mathbb{Z}_n) =$ ?
State the relation between prime and maximal ideals.
When is $R/I$ an integral domain? A field?
State the remainder theorem for $F[x]$ .

True / False / Counterexample:

Converse of Lagrange's theorem is true.
Every subgroup of a non-abelian group is normal.
Every finite integral domain is a field.
Every integral domain is a field.
Every prime ideal is maximal.
$\mathbb{Z}[x]$ is a PID.
Union of two subrings is a subring.
Characteristic of an integral domain can be $4$ .

5-Mark Questions

Group theory:

Prove that any two left cosets of $H$ in $G$ are either equal or disjoint.
Prove Lagrange's theorem.
Prove: if $[G:H] = 2$ , then $H \triangleleft G$ .
Prove that $\ker\phi$ is a normal subgroup of $G$ .
Prove that $Z(G)$ is an abelian normal subgroup of $G$ .
State and prove: $\phi$ is injective iff $\ker\phi = \{e\}$ .
Prove that conjugacy is an equivalence relation on $G$ .
Prove that $N(a)$ is a subgroup of $G$ .
Prove $|\operatorname{Cl}(a)| = [G:N(a)]$ .
Show that $T_g(x) = gxg^{-1}$ is an automorphism of $G$ .
Prove $\operatorname{Aut}(\mathbb{Z}_n) \cong U(n)$ .
If $G = \langle a \rangle$ is infinite cyclic, prove $\operatorname{Aut}(G) \cong \mathbb{Z}_2$ .
State and prove the Second Isomorphism Theorem (Diamond Theorem): if $N \triangleleft G$ and $H \leq G$ , then $H/(H \cap N) \cong HN/N$ .
State and prove the Third Isomorphism Theorem: if $N \triangleleft G$ and $H \triangleleft G$ with $N \subseteq H$ , then $(G/N)/(H/N) \cong G/H$ .

Ring theory:

Prove: in a ring $R$ , $a\cdot 0 = 0$ and $(-a)(-b) = ab$ .
State and prove the subring test.
Prove that the intersection of two subrings is a subring.
Prove that every field is an integral domain.
Prove that a finite integral domain is a field.
Prove that characteristic of an integral domain is $0$ or prime.
Prove that a field has no proper ideals.
Prove: commutative ring with unity having no proper ideals is a field.
If $I$ is an ideal and $1 \in I$ , prove $I = R$ .
Prove that $R/P$ is an integral domain iff $P$ is prime.
Prove that $R/M$ is a field iff $M$ is maximal.
Prove that every maximal ideal is prime. Give a counterexample for the converse.
If $\phi: R \to R'$ is a ring homomorphism, prove $\phi(0) = 0'$ and $\phi(-a) = -\phi(a)$ .
Prove that $\ker\phi$ is an ideal of $R$ .
Prove: if $D$ is an integral domain, then $D[x]$ is an integral domain.
State and prove the remainder theorem.

Numerical / Examples:

Find all left cosets of $H = \{0,3\}$ in $\mathbb{Z}_6$ .
In $S_3$ , find left and right cosets of $H = \{e,(1\,2)\}$ and show $H$ is not normal.
Find $Z(S_3)$ and $Z(\mathbb{Z}_8)$ .
Compute $\operatorname{Aut}(\mathbb{Z}_8)$ and its order.
Show that the set of all matrices $\begin\{pmatrix\} a & b \\ 0 & c \end\{pmatrix\}$ with $a,b,c \in \mathbb{Z}$ is a subring of $M_2(\mathbb{Z})$ .
Find $\operatorname{char}(\mathbb{Z}_6)$ and $\operatorname{char}(\mathbb{Z}_2 \times \mathbb{Z}_2)$ .
Is $(x)$ maximal in $\mathbb{Z}[x]$ ? In $\mathbb{Q}[x]$ ? Justify.
Divide $f(x) = x^3 + 2x + 4$ by $g(x) = 3x + 2$ in $\mathbb{Z}_5[x]$ ; find $q,r$ .

7 or 8-Mark Questions

Group theory:

State and prove Lagrange's theorem. Hence prove that if $|G|$ is prime, $G$ is cyclic.
State and prove the Fundamental Theorem of Group Homomorphism.
Prove that $I(G) \cong G/Z(G)$ and that $I(G) \triangleleft \operatorname{Aut}(G)$ .
Define cosets. Prove that distinct left cosets partition $G$ , and deduce Lagrange's theorem.
Define normal subgroup and quotient group. Prove that the product of cosets is well-defined iff $N$ is normal.
Prove that $\operatorname{Aut}(G)$ is a group under composition. Compute $\operatorname{Aut}(\mathbb{Z}_n)$ .
Define conjugacy class and normalizer. Prove $|\operatorname{Cl}(a)| = |G|/|N(a)|$ . Hence show $a \in Z(G)$ iff $\operatorname{Cl}(a) = \{a\}$ .
State and prove the Second and Third Isomorphism Theorems. Illustrate each with an example.

Ring theory:

Define ring, integral domain, and field. Prove: (i) every field is an integral domain; (ii) every finite integral domain is a field.
Define characteristic. Prove that characteristic of an integral domain is $0$ or a prime.
Define ideal and quotient ring. Prove that $R/P$ is an integral domain iff $P$ is prime, and $R/M$ is a field iff $M$ is maximal. Deduce that every maximal ideal is prime.
State and prove the First Isomorphism Theorem for rings.
Construct the field of quotients of an integral domain. Verify that addition and multiplication are well-defined and that it is a field.
Prove the Division Algorithm in $F[x]$ . Hence prove that $F[x]$ is a PID.
Prove that $\mathbb{Z}[x]$ is not a PID. Show that $(2,x)$ is not principal.
Define ring homomorphism. Prove: (i) $\ker\phi$ is an ideal; (ii) $\phi$ injective iff $\ker\phi = \{0\}$ ; (iii) $R/\ker\phi \cong \operatorname{Im}\phi$ .

Combined / Application:

Prove the Fundamental Theorem of Group Homomorphism. Using it, show that $S_n / A_n \cong \mathbb{Z}_2$ .
Prove every maximal ideal is prime. Show by example that a prime ideal need not be maximal.
Prove finite integral domain $\implies$ field. Is an infinite integral domain necessarily a field? Justify.
State Division Algorithm. Use it to find gcd of $x^4 + x^3 + 2x^2 + x + 1$ and $x^2 + 1$ in $\mathbb{Q}[x]$ (via Euclidean algorithm).
Prove $I(G) \cong G/Z(G)$ . Hence find $|I(S_3)|$ and $|I(\mathbb{Z}_6)|$ .
Show that a commutative ring with unity is a field iff it has no proper ideals. Use this to prove Theorem 29 (maximal $\iff$ quotient is field) in one direction.