Chapter 6 • Theory & Derivations

Unit 6: Ring Theory: Subrings, Ideals, Quotient Rings & Ring Homomorphisms

Comprehensive axiomatization of rings, commutative rings, integral domains, division rings and fields, characteristic of a ring, subrings, two-sided ideals, principal ideals, quotient rings R/I, ring homomorphisms, and the decisive distinction between prime ideals and maximal ideals.

§6.1 Axioms of Rings, Ring Classes & Basic Properties

1. Axiomatic Definition of a Ring

A ring is an algebraic system endowed with two binary operations: addition $(+)$ and multiplication $(\cdot)$.

Definition 6.1 (Ring): A set $R$ equipped with two binary operations $+: R \times R \to R$ and $\cdot: R \times R \to R$ is called a ring $(R, +, \cdot)$ if it satisfies:

$(R, +)$ is an Abelian Group:

Associativity: $(a + b) + c = a + (b + c)$ for all $a, b, c \in R$.

Additive Identity: There exists an element $0 \in R$ such that $a + 0 = 0 + a = a$ for all $a \in R$.

Additive Inverses: For every $a \in R$, there exists $-a \in R$ such that $a + (-a) = (-a) + a = 0$.

Commutativity of Addition: $a + b = b + a$ for all $a, b \in R$.

$(R, \cdot)$ is a Semigroup:

Associativity: $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ for all $a, b, c \in R$.

Distributive Laws (Addition over Multiplication):

Left Distributivity: $a \cdot (b + c) = a \cdot b + a \cdot c$ for all $a, b, c \in R$.

Right Distributivity: $(a + b) \cdot c = a \cdot c + b \cdot c$ for all $a, b, c \in R$.

2. Ring Classifications

Ring with Unity (Identity): There exists an element $1 \in R$ ($1 \ne 0$) such that $a \cdot 1 = 1 \cdot a = a$ for all $a \in R$.
Commutative Ring: Multiplication is commutative: $a \cdot b = b \cdot a$ for all $a, b \in R$.
Division Ring (Skew Field): A ring with unity $1 \ne 0$ in which every non-zero element $a \in R \setminus \{0\}$ has a multiplicative inverse $a^{-1} \in R$ such that $a \cdot a^{-1} = a^{-1} \cdot a = 1$. (e.g., the Quaternions $\mathbb{H}$).
Field: A commutative division ring. (e.g., $\mathbb{Q}, \mathbb{R}, \mathbb{C}, \mathbb{F}_p$).

3. Fundamental Arithmetic Properties

Proposition 6.1 (Elementary Ring Arithmetic): In any ring $(R, +, \cdot)$, for all $a, b, c \in R$:

$a \cdot 0 = 0 \cdot a = 0$.

$a \cdot (-b) = (-a) \cdot b = -(a \cdot b)$.

$(-a) \cdot (-b) = a \cdot b$.

$a \cdot (b - c) = a \cdot b - a \cdot c$.

If $R$ has unity $1$, then $(-1) \cdot a = -a$ and $(-1) \cdot (-1) = 1$.

Complete Proof of 1 and 2:

Proof of 1 ($a \cdot 0 = 0$):

We have $0 + 0 = 0$. Multiplying on the left by $a$:

a \cdot (0 + 0) = a \cdot 0

By left distributivity:

a \cdot 0 + a \cdot 0 = a \cdot 0 = a \cdot 0 + 0

By additive cancellation in the abelian group $(R, +)$:

a \cdot 0 = 0 \quad \blacksquare

Proof of 2 ($a \cdot (-b) = -(a \cdot b)$):

Consider $a \cdot b + a \cdot (-b)$:

a \cdot b + a \cdot (-b) = a \cdot (b + (-b)) = a \cdot 0 = 0

By uniqueness of additive inverses in $(R, +)$, $a \cdot (-b)$ must be the additive inverse of $a \cdot b$:

a \cdot (-b) = -(a \cdot b) \quad \blacksquare

§6.2 Subrings, The Subring Test & Ring Characteristic

1. Subrings and the Subring Criterion

Definition 6.2 (Subring): A subset $S \subseteq R$ of a ring $(R, +, \cdot)$ is a subring if $S$ is itself a ring under the operations of addition and multiplication restricted from $R$.

Theorem 6.1 (The Subring Test): A non-empty subset $S \subseteq R$ is a subring of $R$ if and only if for all $a, b \in S$:

$a - b \in S$ (closed under subtraction / additive group test).

$a \cdot b \in S$ (closed under multiplication).

If $R$ has unity $1_R$, a subring with unity often requires $1_R \in S$.

Proof:

By the one-step subgroup test (Theorem 2.1), condition 1 guarantees that $(S, +)$ is a subgroup of $(R, +)$. Since $(R, +)$ is abelian, $(S, +)$ is automatically abelian. Condition 2 ensures that multiplication is a closed binary operation on $S$. Associativity of multiplication and both distributive laws hold on $R$, and hence automatically hold on the subset $S$. Therefore, $S$ satisfies all ring axioms. $\blacksquare$

2. The Characteristic of a Ring

Definition 6.3 (Characteristic): The characteristic of a ring $R$, denoted $\text{char}(R)$, is the smallest positive integer $n \in \mathbb{Z}^+$ such that:

n \cdot a = \underbrace{a + a + \dots + a}_{n \text{ times}} = 0, \quad \forall a \in R

If no such positive integer exists, we define $\text{char}(R) = 0$.

Theorem 6.2 (Characteristic in Rings with Unity): Let $R$ be a ring with unity $1_R$.

$\text{char}(R) = n > 0 \iff n$ is the additive order of $1_R$ in $(R, +)$.

If $R$ has no zero divisors (in particular, if $R$ is an integral domain or field), then $\text{char}(R)$ is either $0$ or a prime number $p$.

Complete Proof of Part 2:

Suppose $R$ has no zero divisors and $\text{char}(R) = n > 0$. Suppose for contradiction that $n$ is composite: $n = a \cdot b$ where $1 < a, b < n$. Then:

n \cdot 1_R = 0 \implies (a \cdot b) \cdot 1_R = 0

Using the distributive law in rings with unity:

(a \cdot 1_R) \cdot (b \cdot 1_R) = (a \cdot b) \cdot 1_R = n \cdot 1_R = 0

Since $R$ has no zero divisors, the product of two non-zero elements cannot be zero. Therefore, either $a \cdot 1_R = 0$ or $b \cdot 1_R = 0$. However, $1 < a, b < n$, and $n$ was defined as the smallest positive integer such that $n \cdot 1_R = 0$! This is a direct contradiction. Thus $n$ cannot be composite; $n$ must be a prime number $p$. $\blacksquare$

§6.3 Ideals: Left, Right, Two-Sided & Principal Ideals

1. Why Subrings Fail to Form Quotients

In group theory, we needed normal subgroups to ensure that coset multiplication $(aN)(bN) = (ab)N$ was well-defined. In ring theory, if $S$ is merely a subring, coset multiplication on $R/S$:

(r + S)(s + S) \stackrel{?}{=} rs + S

fails! Let $s_1 \in S$. Then $(r + s_1)(s) = rs + s_1 s$. For $(rs + s_1 s) \in rs + S$, we require $s_1 s \in S$ for all $s \in R$. A subring is only closed under multiplication by elements of itself ($S \cdot S \subseteq S$), but to absorb coset representatives, it must "absorb" multiplication from the entire ring $R$! This motivates the concept of an ideal.

2. Definition of Ideals

Definition 6.4 (Ideals): Let $R$ be a ring. A non-empty subset $I \subseteq R$ is called:

A left ideal if $(I, +) \le (R, +)$ and for all $r \in R, x \in I$, $r x \in I$ ($R \cdot I \subseteq I$).

A right ideal if $(I, +) \le (R, +)$ and for all $r \in R, x \in I$, $x r \in I$ ($I \cdot R \subseteq I$).

A two-sided ideal (or simply an ideal, denoted $I \trianglelefteq R$) if it is both a left ideal and a right ideal:

r x \in I \quad \text{and} \quad x r \in I, \quad \forall r \in R, \; x \in I

In a commutative ring, every left ideal is a right ideal, so all ideals are two-sided.

3. Principal Ideals

Definition 6.5 (Principal Ideal): Let $R$ be a commutative ring with unity. The principal ideal generated by an element $a \in R$, denoted $\langle a \rangle$ or $(a)$, is:

\langle a \rangle = R a = \{ r a : r \in R \}

More generally, the ideal generated by a set $X = \{a_1, a_2, \dots, a_k\}$ is:

\langle a_1, \dots, a_k \rangle = \{ r_1 a_1 + \dots + r_k a_k : r_i \in R \}

Theorem 6.3 (Ideals in $\mathbb{Z}$): Every ideal in the ring of integers $\mathbb{Z}$ is principal. That is, for every ideal $I \trianglelefteq \mathbb{Z}$, there exists an integer $n \ge 0$ such that:

I = n\mathbb{Z} = \langle n \rangle

Complete Proof:

If $I = \{0\}$, then $I = \langle 0 \rangle = 0\mathbb{Z}$. Suppose $I \ne \{0\}$. Since $I$ is an additive subgroup, if $x \in I$ then $-x \in I$. Thus $I$ contains positive integers. By the Well-Ordering Principle of $\mathbb{Z}^+$, the set $I \cap \mathbb{Z}^+$ has a smallest element; call it $n$. We claim $I = n\mathbb{Z}$:

Since $n \in I$ and $I$ is an ideal, $k \cdot n \in I$ for all $k \in \mathbb{Z}$. Thus $n\mathbb{Z} \subseteq I$.
Conversely, let $m \in I$ be any element.

By the Division Algorithm in $\mathbb{Z}$:

m = q \cdot n + r, \quad \text{where } 0 \le r < n

Then $r = m - qn$. Since $m \in I$ and $qn \in I$, and $I$ is closed under subtraction:

r = m - qn \in I

If $r > 0$, then $r \in I \cap \mathbb{Z}^+$ with $r < n$, contradicting the minimality of $n$! Therefore, $r = 0$, which implies $m = qn \in n\mathbb{Z}$. Thus $I \subseteq n\mathbb{Z}$. Combining both inclusions gives $I = n\mathbb{Z} = \langle n \rangle$. $\blacksquare$

§6.4 Quotient Rings & The First Isomorphism Theorem for Rings

1. Construction of the Quotient Ring $R/I$

Let $I$ be a two-sided ideal of a ring $R$. The set of additive cosets:

R/I = \{ r + I : r \in R \}

forms a ring under the natural coset operations:

Addition: $(a + I) + (b + I) = (a + b) + I$
Multiplication: $(a + I) \cdot (b + I) = (ab) + I$

Theorem 6.4 (Well-Definedness and Ring Axioms of $R/I$): If $I \trianglelefteq R$, then $R/I$ is a well-defined ring under coset addition and multiplication. Its zero element is $0 + I = I$. If $R$ has unity $1_R$, then $1_R + I$ is the unity of $R/I$. If $R$ is commutative, then $R/I$ is commutative.

Proof of Well-Definedness of Multiplication:

Suppose $a_1 + I = a_2 + I$ and $b_1 + I = b_2 + I$. Then $a_2 = a_1 + i_1$ and $b_2 = b_1 + i_2$ for some $i_1, i_2 \in I$. Compute the product $a_2 b_2$:

$$a_2 b_2 = (a_1 + i_1)(b_1 + i_2) = a_1 b_1 + a_1 i_2 + i_1 b_1 + i_1 i_2$$

Because $I$ is a two-sided ideal:

$a_1 \in R$ and $i_2 \in I \implies a_1 i_2 \in I$.
$b_1 \in R$ and $i_1 \in I \implies i_1 b_1 \in I$.
$i_1 \in I$ and $i_2 \in I \implies i_1 i_2 \in I$.

Thus $i_3 = a_1 i_2 + i_1 b_1 + i_1 i_2 \in I$ (by additive closure of $I$). Therefore:

a_2 b_2 = a_1 b_1 + i_3 \in a_1 b_1 + I \implies a_2 b_2 + I = a_1 b_1 + I

Thus coset multiplication is strictly well-defined! $\blacksquare$

2. Ring Homomorphisms & The First Isomorphism Theorem

Definition 6.6 (Ring Homomorphism): A map $\phi: R \to S$ between rings is a ring homomorphism if for all $a, b \in R$:

$\phi(a + b) = \phi(a) + \phi(b)$

$\phi(a \cdot b) = \phi(a) \cdot \phi(b)$

If $R$ and $S$ have unity, we require $\phi(1_R) = 1_S$.

The kernel is $\ker(\phi) = \{ r \in R : \phi(r) = 0_S \}$. Just as in group theory, $\ker(\phi)$ is a two-sided ideal of $R$, and $\text{im}(\phi)$ is a subring of $S$.

Theorem 6.5 (First Isomorphism Theorem for Rings): Let $\phi: R \to S$ be a ring homomorphism. Then:

\frac{R}{\ker(\phi)} \cong \text{im}(\phi)

via the isomorphism $\Phi(r + \ker(\phi)) = \phi(r)$.

Proof:

Since $\phi$ is an additive group homomorphism, by Theorem 5.2, $\Phi$ is an isomorphism of abelian groups $(R/\ker(\phi), +) \to (\text{im}(\phi), +)$. It remains only to verify multiplicativity:

\Phi((a + \ker(\phi)) \cdot (b + \ker(\phi))) = \Phi(ab + \ker(\phi)) = \phi(ab) = \phi(a)\phi(b) = \Phi(a+\ker(\phi))\Phi(b+\ker(\phi))

Thus $\Phi$ is a ring isomorphism. $\blacksquare$

§6.5 Prime Ideals vs. Maximal Ideals & Field Quotient Criteria

1. Definitions of Prime and Maximal Ideals

Let $R$ be a commutative ring with unity $1_R$.

Definition 6.7 (Prime Ideal): A proper ideal $P \subsetneq R$ is called a prime ideal if whenever $a, b \in R$ with $a \cdot b \in P$, then either $a \in P$ or $b \in P$.

Definition 6.8 (Maximal Ideal): A proper ideal $M \subsetneq R$ is called a maximal ideal if there are no ideals strictly between $M$ and $R$. That is, if $I \trianglelefteq R$ such that $M \subseteq I \subseteq R$, then either $I = M$ or $I = R$.

2. The Grand Quotient Characterization Theorems

The structural power of prime and maximal ideals lies in how they classify quotient rings:

Theorem 6.6 ($R/P$ is an Integral Domain $\iff P$ is Prime): Let $R$ be a commutative ring with unity $1 \ne 0$, and let $P \subsetneq R$ be an ideal. Then $R/P$ is an integral domain if and only if $P$ is a prime ideal.

Proof:

$(\implies)$ Suppose $R/P$ is an integral domain. Let $a, b \in R$ such that $ab \in P$.

Then $(a + P)(b + P) = ab + P = 0 + P$. Since $R/P$ has no zero divisors, either $a + P = 0 + P$ or $b + P = 0 + P$. This means $a \in P$ or $b \in P$. Hence $P$ is prime.

$(\impliedby)$ Suppose $P$ is prime. Since $P \ne R$, $1 + P \ne 0 + P$ in $R/P$, so $R/P$ is a commutative ring with non-zero unity.

Suppose $(a + P)(b + P) = 0 + P$. Then $ab + P = 0 + P \implies ab \in P$. Because $P$ is prime, $a \in P$ or $b \in P$. Therefore, $a + P = 0 + P$ or $b + P = 0 + P$. Thus $R/P$ has no zero divisors, making $R/P$ an integral domain. $\blacksquare$

Theorem 6.7 ($R/M$ is a Field $\iff M$ is Maximal): Let $R$ be a commutative ring with unity $1 \ne 0$, and let $M \subsetneq R$ be an ideal. Then $R/M$ is a field if and only if $M$ is a maximal ideal.

Complete Proof:

$(\implies)$ Suppose $R/M$ is a field. Let $I \trianglelefteq R$ such that $M \subsetneq I \subseteq R$.

Since $M \subsetneq I$, there exists an element $x \in I \setminus M$. Then $x + M \ne 0 + M$ in $R/M$. Because $R/M$ is a field, every non-zero element has a multiplicative inverse. Thus there exists $y + M \in R/M$ such that:

(x + M)(y + M) = 1 + M \implies xy + M = 1 + M \implies 1 - xy \in M

Since $M \subseteq I$, $1 - xy \in I$. Also $x \in I \implies xy \in I$ (by ideal property). Therefore:

1 = (1 - xy) + xy \in I

If an ideal contains the unity $1$, then for any $r \in R$, $r = r \cdot 1 \in I$, so $I = R$. Thus no ideal exists strictly between $M$ and $R$. $M$ is maximal.

$(\impliedby)$ Suppose $M$ is maximal. Let $a + M \in R/M$ with $a + M \ne 0 + M$, so $a \notin M$.

Consider the ideal generated by $M$ and $a$:

I = M + \langle a \rangle = \{ m + r a : m \in M, r \in R \}

Clearly $M \subseteq I$, and since $a = 0 + 1 \cdot a \in I$ with $a \notin M$, $M \subsetneq I$. By maximality of $M$, it must be that $I = R$. In particular, the unity $1 \in I$. Thus there exist $m \in M$ and $r \in R$ such that:

m + r a = 1 \implies r a - 1 = -m \in M \implies ra + M = 1 + M

$$(r + M)(a + M) = 1 + M$$

Thus $r + M$ is the multiplicative inverse of $a + M$ in $R/M$. Since every non-zero element of $R/M$ is invertible, $R/M$ is a field. $\blacksquare$

Corollary 6.7.1 (Maximal Ideals are Prime): In any commutative ring with unity, every maximal ideal is a prime ideal:

M \text{ maximal} \implies M \text{ prime}

Proof:

If $M$ is maximal, then $R/M$ is a field. Since every field is an integral domain, $R/M$ is an integral domain. By Theorem 6.6, $M$ is a prime ideal. $\blacksquare$

Tiered Solved Examination Problems & Rigorous Derivations

Step-by-step rigorous derivations with unskipped proofs, categorized into Foundational Concepts, Advanced Structural Analysis, and Honors / Proof Challenge tiers.

Foundational Example 6.1: Ideals, Quotients and Matrix Rings in Direct Products

In the direct product ring $R = \mathbb{Z} \times \mathbb{Z}$, consider the subset $I = \{(a, 0) : a \in \mathbb{Z}\}$. Prove that $I$ is an ideal of $R$. Determine whether $I$ is a prime ideal or a maximal ideal by analyzing the quotient ring $R/I$. 2. In the ring of $2 \times 2$ real matrices $M_2(\mathbb{R})$, prove that the only two-sided ideals are $\{0\}$ and $M_2(\mathbb{R})$ itself (i.e., $M_2(\mathbb{R})$ is a simple ring).

Part 1: The Ideal $I = \mathbb{Z} \times \{0\}$ in $\mathbb{Z} \times \mathbb{Z}$

Define the projection map:

\pi: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}, \quad \pi(a, b) = b

1. Ring Homomorphism:

$\pi((a, b) + (c, d)) = \pi(a+c, b+d) = b + d = \pi(a, b) + \pi(c, d)$
$\pi((a, b) \cdot (c, d)) = \pi(ac, bd) = bd = \pi(a, b) \cdot \pi(c, d)$
$\pi(1, 1) = 1$

Thus $\pi$ is a surjective ring homomorphism.

2. Kernel:

\ker(\pi) = \{ (a, b) \in \mathbb{Z} \times \mathbb{Z} : \pi(a, b) = 0 \} = \{ (a, b) : b = 0 \} = I

Since $I = \ker(\pi)$, $I$ is an ideal of $R$.

3. Application of First Isomorphism Theorem:

By Theorem 6.5:

\frac{\mathbb{Z} \times \mathbb{Z}}{I} \cong \text{im}(\pi) = \mathbb{Z}

4. Prime vs. Maximal Analysis:

The quotient ring is isomorphic to $\mathbb{Z}$.
$\mathbb{Z}$ is an integral domain (it has no zero divisors). Therefore, by Theorem 6.6, $I$ is a prime ideal.
However, $\mathbb{Z}$ is not a field (e.g., $2 \in \mathbb{Z}$ has no multiplicative inverse in $\mathbb{Z}$).
Therefore, by Theorem 6.7, $I$ is not a maximal ideal!

(For example, $I \subsetneq \mathbb{Z} \times 2\mathbb{Z} \subsetneq \mathbb{Z} \times \mathbb{Z}$).

Part 2: Simplicity of the Matrix Ring $M_2(\mathbb{R})$

Let $J \trianglelefteq M_2(\mathbb{R})$ be a non-zero two-sided ideal. We must prove $J = M_2(\mathbb{R})$. Since $J \ne \{0\}$, there exists a non-zero matrix $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \in J$. At least one entry $a_{ij} \ne 0$. Let $E_{kl}$ be the matrix with $1$ in row $k$, column $l$, and $0$ elsewhere. Recall the fundamental matrix unit multiplication rule:

E_{pi} E_{kl} = \delta_{ik} E_{pl}

Now compute $E_{1i} A E_{j1}$:

E_{1i} A E_{j1} = E_{1i} \left( \sum_{r, s} a_{rs} E_{rs} \right) E_{j1} = a_{ij} E_{11}

Since $J$ is a two-sided ideal and $A \in J$, for all $P, Q \in M_2(\mathbb{R})$, $P A Q \in J$. Choosing $P = E_{1i}$ and $Q = E_{j1}$:

a_{ij} E_{11} \in J

Since $a_{ij} \ne 0$ is a real scalar, we can multiply by the scalar $\frac{1}{a_{ij}} I_2 \in M_2(\mathbb{R})$:

E_{11} = \frac{1}{a_{ij}} (a_{ij} E_{11}) \in J

Now we can generate all matrix units:

$E_{22} = E_{21} E_{11} E_{12} \in J$

Summing them:

I_2 = E_{11} + E_{22} \in J

Since the identity matrix $I_2 \in J$, for any matrix $M \in M_2(\mathbb{R})$:

M = M \cdot I_2 \in J \implies J = M_2(\mathbb{R})

Therefore, the only two-sided ideals of $M_2(\mathbb{R})$ are $\{0\}$ and $M_2(\mathbb{R})$. $\blacksquare$

Advanced Example 6.2: Ideals in $\mathbb{R}[x]$ and $\mathbb{Z}[x]$: Constructing $\mathbb{C}$ and Distinguishing Prime from Maximal

In the polynomial ring $\mathbb{R}[x]$, let $I = \langle x^2 + 1 \rangle$. Prove that $I$ is a maximal ideal and deduce that $\mathbb{R}[x] / \langle x^2 + 1 \rangle \cong \mathbb{C}$. 2. In the polynomial ring $\mathbb{Z}[x]$, let $J = \langle x^2 + 1 \rangle$. Prove that $J$ is a prime ideal, but NOT a maximal ideal, by explicitly exhibiting an ideal strictly between $J$ and $\mathbb{Z}[x]$.

Part 1: $\mathbb{R}[x] / \langle x^2 + 1 \rangle \cong \mathbb{C}$

Define the evaluation homomorphism at the imaginary unit $i \in \mathbb{C}$:

\phi: \mathbb{R}[x] \to \mathbb{C}, \quad \phi(p(x)) = p(i)

1. Homomorphism:

Evaluation preserves polynomial addition and multiplication:

\phi(p + q) = p(i) + q(i) = \phi(p) + \phi(q)

\phi(p \cdot q) = p(i) \cdot q(i) = \phi(p) \cdot \phi(q)

\phi(1) = 1

2. Surjectivity:

Any complex number has the form $a + bi$ with $a, b \in \mathbb{R}$. Consider the linear polynomial $p(x) = bx + a \in \mathbb{R}[x]$. Then $\phi(p(x)) = bi + a = a + bi$. Thus $\phi$ is surjective: $\text{im}(\phi) = \mathbb{C}$.

3. Kernel:

\ker(\phi) = \{ p(x) \in \mathbb{R}[x] : p(i) = 0 \}

Clearly $x^2 + 1 \in \ker(\phi)$ since $i^2 + 1 = -1 + 1 = 0$. Thus $\langle x^2 + 1 \rangle \subseteq \ker(\phi)$. Conversely, let $p(x) \in \ker(\phi)$. By the division algorithm in $\mathbb{R}[x]$:

p(x) = q(x)(x^2 + 1) + (rx + s), \quad \text{where } r, s \in \mathbb{R}

Evaluating at $x = i$:

0 = p(i) = q(i)(i^2 + 1) + (ri + s) = 0 + (s + ri) \implies s + ri = 0

Since $r, s \in \mathbb{R}$, this requires $s = 0$ and $r = 0$. Therefore, the remainder is identically zero, so $p(x) = q(x)(x^2 + 1) \in \langle x^2 + 1 \rangle$. Thus $\ker(\phi) = \langle x^2 + 1 \rangle$.

4. Application of First Isomorphism Theorem:

By Theorem 6.5:

\frac{\mathbb{R}[x]}{\langle x^2 + 1 \rangle} \cong \text{im}(\phi) = \mathbb{C}

Since $\mathbb{C}$ is a field, Theorem 6.7 implies that $\langle x^2 + 1 \rangle$ is a maximal ideal of $\mathbb{R}[x]$. $\blacksquare$

Part 2: The Ideal $J = \langle x^2 + 1 \rangle$ in $\mathbb{Z}[x]$

Consider the same evaluation map restricted to integer polynomials:

\psi: \mathbb{Z}[x] \to \mathbb{Z}[i] = \{ a + bi : a, b \in \mathbb{Z} \}, \quad \psi(p(x)) = p(i)

The image is the ring of Gaussian integers $\mathbb{Z}[i]$. By the identical division algorithm argument, $\ker(\psi) = \langle x^2 + 1 \rangle = J$. By the First Isomorphism Theorem:

\frac{\mathbb{Z}[x]}{\langle x^2 + 1 \rangle} \cong \mathbb{Z}[i]

Analysis:

1. Is $J$ a prime ideal?

$\mathbb{Z}[i]$ is a subring of the field $\mathbb{C}$. Since $\mathbb{C}$ has no zero divisors, $\mathbb{Z}[i]$ has no zero divisors, making $\mathbb{Z}[i]$ an integral domain. By Theorem 6.6, $J = \langle x^2 + 1 \rangle$ is a prime ideal in $\mathbb{Z}[x]$.

2. Is $J$ a maximal ideal?

The quotient $\mathbb{Z}[i]$ is not a field! For example, $2 \in \mathbb{Z}[i]$ is not invertible, since $\frac{1}{2} \notin \mathbb{Z}[i]$. By Theorem 6.7, $J$ is not a maximal ideal.

3. Explicit Intermediate Ideal:

Consider the ideal generated by $x^2 + 1$ and $2$:

K = \langle 2, x^2 + 1 \rangle = \{ 2 f(x) + (x^2+1) g(x) : f, g \in \mathbb{Z}[x] \}

Clearly $J \subseteq K$.
$2 \in K$, but $2 \notin J$ (since any element in $J$ has degree $\ge 2$ or is 0). Thus $J \subsetneq K$.
Is $K = \mathbb{Z}[x]$? If $1 \in K$, then $2 f(x) + (x^2+1)g(x) = 1$. Evaluating at $x=i$ gives $2f(i) = 1 \implies f(i) = 1/2$, impossible in $\mathbb{Z}[i]$!

Thus $J \subsetneq K \subsetneq \mathbb{Z}[x]$. This confirms explicitly that $J$ is not maximal. $\blacksquare$

Honors / Proof Challenge Example 6.3: The Nilradical $\text{Nil}(R)$ and Krull's Theorem on Prime Ideals

Let $R$ be a commutative ring with unity $1 \ne 0$. An element $x \in R$ is called nilpotent if $x^n = 0$ for some positive integer $n \ge 1$. 1. Prove that the set of all nilpotent elements, $\text{Nil}(R) = \{ x \in R : \exists n \ge 1, \; x^n = 0 \}$, forms an ideal of $R$ (called the nilradical of $R$). 2. Prove that if $P$ is any prime ideal of $R$, then $\text{Nil}(R) \subseteq P$. 3. Prove that the nilradical of $R$ is equal to the intersection of all prime ideals of $R$: $\text{Nil}(R) = \bigcap_{P \text{ prime}} P$.

Part 1: Proof that $\text{Nil}(R)$ is an Ideal

1. Non-emptiness: $0^1 = 0 \implies 0 \in \text{Nil}(R)$.

2. Additive closure: Let $x, y \in \text{Nil}(R)$. Then there exist $n, m \ge 1$ such that $x^n = 0$ and $y^m = 0$.

Since $R$ is commutative, we expand $(x - y)^{n + m}$ using the Binomial Theorem:

(x - y)^{n + m} = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k (-y)^{n+m-k}

For each term in the sum:

If $k \ge n$, then $x^k = x^{k-n} x^n = x^{k-n} \cdot 0 = 0$.
If $k < n$, then $(n + m - k) > (n + m - n) = m$, so $(-y)^{n+m-k} = \pm y^{n+m-k-m} y^m = 0$.

Thus every single term in the sum is $0$! Therefore, $(x - y)^{n+m} = 0$, which proves $x - y \in \text{Nil}(R)$.

3. Ideal absorption: Let $r \in R$ and $x \in \text{Nil}(R)$ with $x^n = 0$.

Since $R$ is commutative:

(rx)^n = r^n x^n = r^n \cdot 0 = 0

Thus $rx \in \text{Nil}(R)$. Therefore, $\text{Nil}(R)$ is an ideal of $R$. $\blacksquare$

Part 2: $\text{Nil}(R) \subseteq P$ for every Prime Ideal $P$

Let $P$ be a prime ideal of $R$, and let $x \in \text{Nil}(R)$. Then $x^n = 0$ for some $n \ge 1$. Since $0 \in P$, we have $x^n \in P$. We prove $x \in P$ by induction on $n$:

If $n = 1$, $x \in P$ immediately.
Assume that for any $k < n$, $x^k \in P \implies x \in P$.
Write $x^n = x \cdot x^{n-1} \in P$.
Since $P$ is a prime ideal, $a b \in P \implies a \in P$ or $b \in P$.
Here $x \in P$ or $x^{n-1} \in P$.
In either case, by induction hypothesis, $x \in P$.

Therefore, every nilpotent element belongs to every prime ideal:

\text{Nil}(R) \subseteq \bigcap_{P \text{ prime}} P \quad \blacksquare

Part 3: Krull's Theorem: $\text{Nil}(R) = \bigcap_{P \text{ prime}} P$

To prove $\bigcap_{P \text{ prime}} P \subseteq \text{Nil}(R)$, we prove the contrapositive: Suppose $f \in R$ and $f \notin \text{Nil}(R)$. We must construct a prime ideal $P$ such that $f \notin P$.

Consider the multiplicative set generated by powers of $f$:

S = \{ f^k : k \ge 0 \} = \{ 1, f, f^2, f^3, \dots \}

Since $f \notin \text{Nil}(R)$, $0 \notin S$. Let $\Sigma$ be the collection of all ideals of $R$ that are disjoint from $S$:

\Sigma = \{ I \trianglelefteq R : I \cap S = \emptyset \}

$\Sigma \ne \emptyset$: The zero ideal $\{0\} \in \Sigma$, because $0 \notin S$.
Partially ordered set: $\Sigma$ is partially ordered by set inclusion $\subseteq$.
Zorn's Lemma hypothesis: Let $\{I_\alpha\}_{\alpha \in A}$ be a totally ordered chain of ideals in $\Sigma$.

Let $J = \bigcup_{\alpha \in A} I_\alpha$. $J$ is an ideal of $R$, and if $J \cap S \ne \emptyset$, then some $f^k \in I_\alpha$, contradicting $I_\alpha \in \Sigma$. Thus $J \in \Sigma$, serving as an upper bound for the chain. By Zorn's Lemma, $\Sigma$ contains a maximal element; call it $P$. By definition of $\Sigma$, $P \cap S = \emptyset$, so in particular $f \notin P$.

We claim $P$ is a prime ideal of $R$:

Suppose for contradiction that $P$ is not prime. Then there exist $a, b \in R$ such that $ab \in P$, but $a \notin P$ and $b \notin P$. Consider the strictly larger ideals:

I_a = P + \langle a \rangle \quad \text{and} \quad I_b = P + \langle b \rangle

Since $a, b \notin P$, we have $P \subsetneq I_a$ and $P \subsetneq I_b$. By the maximality of $P$ in $\Sigma$, neither $I_a$ nor $I_b$ can belong to $\Sigma$! Therefore, both must intersect $S$:

\exists s_1 = f^u \in I_a \quad \text{and} \quad \exists s_2 = f^v \in I_b

Thus we can write:

f^u = p_1 + r_1 a \quad \text{and} \quad f^v = p_2 + r_2 b \quad (p_1, p_2 \in P, \; r_1, r_2 \in R)

Now multiply these two elements:

f^{u+v} = (p_1 + r_1 a)(p_2 + r_2 b) = p_1 p_2 + p_1 r_2 b + p_2 r_1 a + r_1 r_2 (ab)

Examine each term:

$p_1 p_2 \in P$ (since $p_1 \in P$).
$p_1 r_2 b \in P$ (since $p_1 \in P$).
$p_2 r_1 a \in P$ (since $p_2 \in P$).
$r_1 r_2 (ab) \in P$ (since by hypothesis $ab \in P$!).

Thus the entire sum lies in $P$:

f^{u+v} \in P

However, $f^{u+v} \in S$, which means $P \cap S \ne \emptyset$, a direct contradiction!

Therefore, our assumption was false: $P$ must be a prime ideal. Since $f \notin P$, we have found a prime ideal $P$ not containing $f$. Consequently, $f \notin \bigcap_{P \text{ prime}} P$. This completes the proof:

\text{Nil}(R) = \bigcap_{P \text{ prime}} P \quad \blacksquare