2Rings and Arithmetic

Proof. As for groups (Theorems 1.3 and 1.5), noting that all maps in sight also respect products: $\bar f$ is well defined, bijective onto the image, and multiplicative; the correspondence $J \mapsto J/I$, $\bar J \mapsto \pi^{-1}(\bar J)$ preserves ideals in both directions because $\pi$ is a surjective ring morphism. ∎

Proof. Write $\bar a$ for classes in $A/I$. “$I$ prime” translates verbatim to “$\bar a\bar b = 0 \Rightarrow \bar a = 0$ or $\bar b = 0$”, and $A/I \neq 0$ to $I \neq A$: that is the definition of a domain. For maximality, use the correspondence theorem: no ideal strictly between $I$ and $A$ $\iff$ $A/I$ has no ideal other than $0$ and itself $\iff$ $A/I$ is a field — for the last step: in a field the only ideals are $0$ and everything (an ideal containing $x \ne 0$ contains $x^{-1}x = 1$); conversely if every nonzero $x$ generates the unit ideal, then $xy = 1$ for some $y$. Fields are domains, so maximal ideals are prime. ∎

Proof. Order by inclusion the set $\mathcal E$ of proper ideals containing $I$; it is nonempty ($I \in \mathcal E$). A chain $(J_\lambda)$ in $\mathcal E$ has upper bound $J = \bigcup J_\lambda$: an ideal (any $a, b \in J$ lie in a common $J_\lambda$ by totality), proper ($1 \notin J_\lambda$ for all $\lambda$), containing $I$. Zorn’s lemma yields a maximal element of $\mathcal E$, which is a maximal ideal containing $I$ (an ideal strictly above it and proper would lie in $\mathcal E$). ∎

Proof. The map $f(a) = (a + I_k)_k$ is a ring morphism with kernel $\bigcap I_k$; by Theorem 2.2 it suffices to prove surjectivity. Fix $k$; for each $l \neq k$ write $1 = u_l + v_l$ with $u_l \in I_k$, $v_l \in I_l$ (comaximality). Then

so $f(e_k) = (0, \dots, 1, \dots, 0)$; given a target $(a_k + I_k)_k$, the element $\sum_k a_k e_k$ maps to it.

Products vs intersection: $I_1\cdots I_n \subseteq \bigcap I_k$ always. Conversely, by induction it suffices to treat $n = 2$ (one checks $I_1$ and $I_2\cdots I_n$ are comaximal: multiplying $1 = u_l + v_l$ over $l \geq 2$ gives $1 \in I_1 + I_2\cdots I_n$). For $n = 2$: write $1 = u + v$, $u \in I_1$, $v \in I_2$; for $x \in I_1 \cap I_2$, $x = xu + xv \in I_2I_1 + I_1I_2 = I_1I_2$. ∎

Proof. Let $p$ be prime and $p = ab$. Then $p \mid ab$, say $p \mid a$: $a = pc$, so $p = pcb$, and cancelling $p$ (domain!) gives $cb = 1$: $b \in A^\times$.

In $\Z[\iu\sqrt5]$, use the norm $N(x + \iu y\sqrt 5) = x^2 + 5y^2$, which is multiplicative (it is $\abs z^2$). If $2 = ab$ with $a, b$ nonunits, then $4 = N(a)N(b)$ with $N(a), N(b) \neq 1$ (norm-$1$ elements are $\pm1$, the units), so $N(a) = 2$: impossible, $x^2 + 5y^2 = 2$ has no integer solution. So $2$ is irreducible. But $2 \mid 6 = (1 + \iu\sqrt5)(1 - \iu\sqrt5)$ while $2$ divides neither factor ($\frac12 \pm \frac{\iu\sqrt5}2 \notin \Z[\iu\sqrt5]$): not prime. ∎

Proof. Let $I \neq (0)$ be an ideal and $b \in I \setminus\{0\}$ with $\nu(b)$ minimal. For $a \in I$, divide: $a = bq + r$; then $r = a - bq \in I$, and $\nu(r) < \nu(b)$ would contradict minimality, so $r = 0$ and $a \in (b)$: $I = (b)$. ∎

Proof. $I = \bigcup_n I_n$ is an ideal (the union is increasing), so $I = (a)$; the element $a$ lies in some $I_N$, and then $I = (a) \subseteq I_N \subseteq I_n \subseteq I$ for $n \geq N$. ∎

Proof. $(a) + (b)$ is an ideal, hence $(d)$; $a, b \in (d)$ gives $d \mid a, b$; and $d = au + bv \in (a) + (b)$. A common divisor $c$ of $a, b$ divides $au + bv = d$.

Euclid: let $p$ be irreducible, $p \mid ab$, $p \nmid a$. A gcd $d$ of $p$ and $a$ divides $p$, so $d$ is a unit or an associate of $p$ (irreducibility); associate is excluded by $p \nmid a$. So $1 = pu + av$, whence $b = pub + abv$, and $p$ divides both terms: $p \mid b$. ∎

Proof. Existence. Suppose some nonzero nonunit $a$ has no factorization into irreducibles. Then $a$ is not irreducible: $a = a_1b_1$ with both factors nonunits; at least one of them, say $a_1$, again has no factorization (a product of two factorizable elements is factorizable). Iterating, we get $a = a_0, a_1, a_2, \dots$, each a proper divisor of the last with no factorization, so $(a_0) \subsetneq (a_1) \subsetneq (a_2) \subsetneq \cdots$ — the inclusions are strict because $a_n = a_{n+1}c$ with $c$ a nonunit means $(a_n) = (a_{n+1})$ would force $c \in A^\times$ (cancel in a domain). This contradicts Lemma 2.16.

Uniqueness. Let $p_1 \cdots p_r = q_1 \cdots q_s$ with all factors irreducible, $r \leq s$, by induction on $r$. The prime (Lemma 2.17) $p_1$ divides the right-hand side, so divides some $q_j$; renumber $j = 1$. As $q_1$ is irreducible and $p_1$ is not a unit, $q_1 = u p_1$ with $u \in A^\times$: $p_1, q_1$ are associates. Cancel $p_1$: $p_2 \cdots p_r = (u q_2) q_3\cdots q_s$ and conclude by induction ($r = 1$ forces $s = 1$: a unit times irreducibles cannot be $1$). ∎

Proof. Let $P, Q$ be primitive and suppose some irreducible (= prime, UFD) $p$ divides all coefficients of $PQ$. Reduce modulo $p$: in $(A/(p))[X]$, $\bar P \bar Q = 0$. But $A/(p)$ is a domain ($(p)$ prime), so $(A/(p))[X]$ is a domain (leading coefficients multiply), forcing $\bar P = 0$ or $\bar Q = 0$: $p$ divides all coefficients of $P$ or all of $Q$, contradicting primitivity. For the consequence, write $P = c(P)P_1$, $Q = c(Q)Q_1$: $PQ = c(P)c(Q) P_1Q_1$ with $P_1Q_1$ primitive. ∎

Proof. (1) ($\Leftarrow$) If $P = QR$ in $A[X]$ with $Q, R$ nonunits, then neither factor is constant (a constant factor of a primitive polynomial is a unit), so the factorization is proper in $K[X]$. ($\Rightarrow$) Suppose $P = QR$ with $Q, R \in K[X]$ of degrees $\geq 1$. Write $Q = \lambda Q_1$, $R = \mu R_1$ with $Q_1, R_1 \in A[X]$ primitive: $P = \lambda\mu\, Q_1R_1$, and $Q_1R_1$ is primitive by Gauss. Taking contents, $\lambda\mu \in A^\times$ (both sides have unit content; formally, $\lambda\mu = c(P) \in A^\times$ up to units, and in particular $\lambda \mu \in A$): $P = (\lambda\mu Q_1) R_1$ is a proper factorization in $A[X]$.

(2) Existence: given $P \neq 0$ nonunit, factor $P = c(P)P_1$, factor $c(P)$ into irreducibles of $A$, and factor $P_1$ in the UFD $K[X]$ as $\prod Q_i$ with $Q_i \in K[X]$ irreducible; writing $Q_i = \lambda_i R_i$ with $R_i \in A[X]$ primitive (hence irreducible over $K$, hence in $A[X]$ by (1)), the product $\prod \lambda_i$ is a unit of $A$ as before, and $P_1 = u\prod R_i$. Uniqueness: compare a factorization’s constant part and polynomial part; the constants multiply to $c(P)$ (Gauss), unique by factoriality of $A$; the polynomial parts give two factorizations in $K[X]$ of the same polynomial, so they match up to constants of $K^\times$ (factoriality of $K[X]$, Theorem 2.18), and matching primitive polynomials associated in $K[X]$ are associated in $A[X]$: if $R = \lambda R'$ with $R, R'$ primitive and $\lambda \in K^\times$, then taking contents forces $\lambda \in A^\times$. ∎

Proof. By Theorem 2.24(1), a proper factorization over $K$ yields $P = QR$ with $Q, R \in A[X]$, $\deg Q, \deg R \geq 1$ (constants are excluded: they would be units or spoil primitivity).

(1) Reduce mod $p$: $\bar P = \bar Q\bar R$ in $(A/(p))[X]$. Since $p \nmid a_n$ and $\deg$ can only drop under reduction, $\deg \bar Q = \deg Q \geq 1$ and $\deg\bar R = \deg R \geq 1$ (their leading coefficients multiply to $\bar a_n \neq 0$, so neither drops): $\bar P$ factors properly — contradiction.

(2) Reduce mod $p$: $\bar Q \bar R = \bar P = \bar a_n X^n$ (all lower coefficients die). In the domain $(A/(p))[X]$, the factorizations of $cX^n$ ($c \ne 0$) are into constants and pure powers $c'X^k$: indeed if $\bar Q\bar R = \bar a_nX^n$, and say $\bar Q$ had a nonzero coefficient in degree $< \deg\bar Q$, take lowest nonzero terms: $\operatorname{val}(\bar Q\bar R) = \operatorname{val}\bar Q + \operatorname{val}\bar R$ (domain), which must equal $n = \deg\bar Q + \deg\bar R$, forcing $\operatorname{val} = \deg$ for both: both are monomials. As above, degrees do not drop, so $Q$ and $R$ have their constant terms $Q(0), R(0)$ divisible by $p$ — both, since both reductions are monomials of degree $\geq 1$. Then $p^2 \mid Q(0)R(0) = a_0$: contradiction. ∎

Proof. (FG $\Rightarrow$ ACC): for a chain $I_1 \subseteq I_2 \subseteq \cdots$, the union $I$ is an ideal, generated by $x_1, \dots, x_r$; all $x_i$ lie in some $I_N$, so $I = I_N = I_n$ for $n \geq N$. (ACC $\Rightarrow$ maximal elements): if a nonempty family $\mathcal F$ had no maximal element, pick $I_1 \in \mathcal F$, then inductively $I_{n+1} \supsetneq I_n$ in $\mathcal F$ (possible since $I_n$ is not maximal): an infinite strictly increasing chain. (This uses the axiom of dependent choices, a weak form of choice we do not fuss over.) (Maximal elements $\Rightarrow$ FG): given an ideal $I$, the family of finitely generated ideals contained in $I$ is nonempty ($(0)$); a maximal element $J = (x_1, \dots, x_r)$ must equal $I$: otherwise, adding $x \in I \setminus J$ to the generators produces a strictly bigger member of the family. ∎

Proof. Let $I$ be an ideal of $A[X]$, and suppose $I$ is not finitely generated. Build a sequence: $f_1 \in I \setminus \{0\}$ of minimal degree, and inductively $f_{k+1} \in I \setminus (f_1, \dots, f_k)$ of minimal degree (the set is nonempty by assumption). Degrees $d_k = \deg f_k$ are nondecreasing (by minimality of each choice: $f_{k+1}$ was available at step $k+1$... precisely, $f_{k+1} \notin (f_1,\dots,f_k) \supseteq (f_1, \dots, f_{k-1})$, so $f_{k+1}$ competed at step $k$ and lost or tied: $d_{k+1} \geq d_k$). Let $a_k \in A$ be the leading coefficient of $f_k$. The chain of ideals $(a_1) \subseteq (a_1, a_2) \subseteq \cdots$ stabilizes: $a_{n+1} \in (a_1, \dots, a_n)$ for some $n$, say $a_{n+1} = \sum_{k\leq n} u_k a_k$. Consider

Then $g \in I \setminus (f_1, \dots, f_n)$ (the sum lies in the ideal, $f_{n+1}$ does not), yet the coefficient of degree $d_{n+1}$ cancels: $\deg g < d_{n+1}$, contradicting the minimality of $d_{n+1} = \deg f_{n+1}$.

Iterating, $A[X_1, \dots, X_n] = (A[X_1, \dots, X_{n-1}])[X_n]$ is Noetherian; a quotient $A/I$ is Noetherian because its ideals $J/I$ lift to ideals of $A$ (correspondence), where finitely many generators project onto generators. ∎