Ideals and Quotient Rings

9.1 Ideals

A subset $I \subseteq R$ is an ideal if:

1. $(I, +)$ is a subgroup of $(R, +)$.
2. For all $r \in R$ and $a \in I$: $ra \in I$ and $ar \in I$.

If only $ra \in I$ for all $r \in R$ and $a \in I$Then $I$ is a left ideal. Similarly for right Ideals. A two-sided ideal (or ideal) satisfies both.

Proposition 9.1. Every ideal is a subring. The converse is false.

Example. $n\mathbb{Z} = \{nk : k \in \mathbb{Z}\}$ is an ideal of $\mathbb{Z}$.

Example. If $\phi : R \to S$ is a ring homomorphism, then $\ker(\phi)$ is an ideal of $R$.

9.2 Quotient Rings

If $I$ is an ideal of $R$The quotient ring $R/I$ has elements $\{a + I : a \in R\}$ (cosets) With operations $(a + I) + (b + I) = (a + b) + I$ and $(a + I)(b + I) = ab + I$.

Theorem 9.2 (Ring Isomorphism Theorem). If $\phi : R \to S$ is a surjective ring homomorphism, Then $R / \ker(\phi) \cong S$.

The …/1-number-and-algebra/3_proof-and-logic follows the same pattern as the first isomorphism theorem for groups.

9.3 Prime and Maximal Ideals

An ideal $I \neq R$ is prime if $ab \in I$ implies $a \in I$ or $b \in I$.

An ideal $I \neq R$ is maximal if there is no ideal $J$ with $I \subsetneq J \subsetneq R$.

Theorem 9.3. In a commutative ring $R$ with unity:

1. $I$ is a prime ideal if and only if $R/I$ is an integral domain.
2. $I$ is a maximal ideal if and only if $R/I$ is a field.

Corollary 9.4. Every maximal ideal is prime.

Proof. A field is an integral domain. $\blacksquare$

Example. In $\mathbb{Z}$The ideal $(p)$ is maximal (hence prime) if and only if $p$ is prime. The ideal $(6)$ is neither prime nor maximal. The ideal $(0)$ is prime ($\mathbb{Z}$ is an integral domain) But not maximal ($\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field).

Example. In $\mathbb{R}[x]$The ideal $(x^2 + 1)$ is maximal since $\mathbb{R}[x]/(x^2 + 1) \cong \mathbb{C}$ is a field.

Problem. Show that $(2)$ is a maximal ideal of $\mathbb{Z}$ but $(4)$ is not.

9.4 The Chinese Remainder Theorem

Theorem 9.5 (Chinese Remainder Theorem for Rings). Let $R$ be a commutative ring with unity and Let $I, J$ be ideals with $I + J = R$. Then

$R/(I \cap J) \cong R/I \times R/J$

Proof. Define $\phi : R \to R/I \times R/J$ by $\phi(r) = (r + I, r + J)$. This is a ring homomorphism. It is surjective: since $I + J = R$There exist $a \in I$ and $b \in J$ with $a + b = 1$. For any $(r_1 + I, r_2 + J)$Take $r = r_1b + r_2a$. Then $r \equiv r_1b \equiv r_1(1-a) \equiv r_1 \pmod{I}$ And $r \equiv r_2a \equiv r_2(1-b) \equiv r_2 \pmod{J}$.

The kernel is $\ker(\phi) = \{r : r \in I\ \mathrm{and\ r} \in J\} = I \cap J$. By the ring isomorphism theorem, $R/(I \cap J) \cong R/I \times R/J$. $\blacksquare$

Corollary 9.6. If $m, n \in \mathbb{Z}$ are coprime, then $\mathbb{Z}/(mn) \cong \mathbb{Z}/(m) \times \mathbb{Z}/(n)$.

Proof. Apply Theorem 9.5 with $I = (m)$, $J = (n)$. Since $\gcd(m, n) = 1$We have $(m) + (n) = (1) = \mathbb{Z}$. Also $(m) \cap (n) = (\mathrm{lcm}(m, n)) = (mn)$. $\blacksquare$

Problem. Find all solutions to $x \equiv 2 \pmod{3}$, $x \equiv 3 \pmod{5}$, $x \equiv 1 \pmod{7}$.