Codex @codex  0

Induct on $n$ to prove the stronger assertion that any submodule $M\subseteq R^n$ is a free module of rank of a free module at most $n$. For $n=0$ this is immediate. Project onto the first coordinate by $\pi :R^n\to R$. Its image $\pi (M)$ is an ideal of the principal ideal domain, so $\pi (M)=(a)$.

If $a=0$, then $M$ embeds in the remaining $n-1$ coordinates and the induction hypothesis applies. Otherwise choose $v\in M$ with $\pi (v)=a$. Put $N=M\cap \ker \pi$, a submodule of $R^{n-1}$, which is free by induction. For every $m\in M$, there is $b\in R$ with $\pi (m)=ba$, so $m-bv\in N$. If $bv\in N$, then $ba=0$; the integral domain property and $a\ne 0$ give $b=0$. ThereforeAppending $v$ to a basis of $N$ proves that $M$ is a free module of rank of a free module at most $n$. This proof of the submodule theorem for free modules over a principal ideal domain does not assume in advance that $M$ is a finitely generated module.

The semicubical parabola has coordinate ringunder $x\mapsto t^3$, $y\mapsto t^2$. The isomorphism is valid in every characteristic: after division by the monic polynomial in $x$, a remainder $P(y)+xQ(y)$ maps to a sum of distinct even powers and odd powers at least three, and can vanish only when both polynomials are zero.

Its fraction field contains $t=t^3/t^2$, which is an integral element over $A$ because it satisfies $z^2-y=0$. However $t\notin A$, since a polynomial in $t^2,t^3$ has no term of degree one. Thus $A$ is not an integrally closed domain. The ring $k[t]$ of the affine line is a principal ideal domain and therefore an integrally closed domain. This property is preserved by ring isomorphism. By coordinate-ring functoriality,The parametrization is a bijection on points over an algebraically closed field, but that does not make its inverse a morphism at the cusp.

Identify a binary word with its coefficient polynomial in $R={\mathbb{F}}_2[X]/(X^7-1)$. A cyclic code is exactly an ideal of $R$: closure under multiplication by $X$ is closure under cyclic shift. Its inverse image in the principal ideal domain ${\mathbb{F}}_2[X]$ has a unique monic generator polynomial of a cyclic code $g$ dividing $X^7-1$. PutThe cubics have no root in ${\mathbb{F}}_2$, hence are irreducible polynomials. The eight monic divisors yield all binary cyclic codes of length seven. A code generated by $g$ has basis $g,Xg,\dots ,X^{6-\mathrm{deg}g}g$ and dimension $7-\mathrm{deg}g$, with the zero code treated separately.

For the dual of a cyclic code, let $h=(X^7-1)/g$. Orthogonality of coefficient vectors to all cyclic shifts gives the generator $h^{\ast }(X)=X^{\mathrm{deg}h}h(X^{-1})$. Here $a^{\ast }=a,b^{\ast }=c,c^{\ast }=b$. Thus the complete pairing, with the respective dimensions, isThese are respectively the whole-space/zero, even-weight/repetition, and two Hamming code/binary simplex code pairs. The reciprocal formula follows because the product of a word polynomial and the reversed polynomial of an orthogonal word has zero coefficients in all cyclic positions; the resulting inclusion has the correct complementary dimension, hence is equality.

First, every nonzero nonunit in a principal ideal domain factors into irreducible elements. Otherwise repeatedly choose a proper nonunit factor $a_{j+1}$ of $a_j$; then the principal idealsform a strictly increasing chain, contradicting the ascending chain condition, since every principal ideal domain is a Noetherian ring.

Next every irreducible $p$ is a prime element. If $p\mid ab$ but $p\nmid a$, then $\gcd (p,a)=1$. The Bezout identity gives $up+va=1$, so multiplying by $b$ shows $p\mid b$. Thus irreducibles are prime.

Existence of an irreducible factorization and primality give uniqueness: an irreducible on one side divides the product on the other, hence is associate to one factor; cancel and continue. Therefore every principal ideal domain is a unique factorization domain.

Let $S=\{1,x,x^2,\dots \}$. Since a principal ideal domain is an integral domain and $x\ne 0$, $S$ is a multiplicative subset avoiding zero. The stated equivalence relation and operations are precisely those of the localization of a ring $S^{-1}R$.

For completeness, reflexivity and symmetry are immediate, while transitivity follows by multiplying the two cross-multiplication identities and cancelling a nonzero power of $x$. If $(r,x^n)\sim (s,x^m)$ and $(r^{\prime },x^{n^{\prime }})\sim (s^{\prime },x^{m^{\prime }})$, cross multiplication shows that the proposed sums and products are equivalent. Thus the operations do not depend on representatives. The usual fraction calculations then prove associativity, commutativity and distributivity, with identities $[(0,1)]$ and $[(1,1)]$. Therefore $R^{\prime }$ is a well-defined commutative ring.

Let $J$ be an ideal of $R^{\prime }=S^{-1}R$ and contract it toThis is an ideal of $R$, so $I=(d)$ because $R$ is a principal ideal domain. Every $x^n/1$ is a unit in a ring $R^{\prime }$. Consequently, if $r/x^n\in J$, then $r/1=(x^n/1)(r/x^n)\in J$, so $r\in (d)$ and $r/x^n\in (d/1)$. Conversely $d/1\in J$, hence every multiple of it belongs to $J$. Thusso every ideal of $R^{\prime }$ is principal. The localization remains an integral domain; hence $R^{\prime }$ is a principal ideal domain.

Here the coefficient fields are $\mathbb{R}$ and $\mathbb{C}$. A polynomial ring in one variable over a field is a Euclidean domain, hence a principal ideal domain; its nonzero prime ideals are generated by irreducible polynomials. The fundamental theorem of algebra says that all irreducible polynomials over $\mathbb{C}$ are linear. Combining it with complex conjugation shows that the monic irreducible polynomials over $\mathbb{R}$ are exactly the linear polynomials and quadratics with no real root. ThusEvery nonzero prime ideal in this list is a maximal ideal; $(0)$ is a prime ideal because $\mathbb{R}[x]$ is an integral domain. Likewise,The map is contraction of an ideal along the inclusion. It sends $(0)$ to $(0)$. For $z=a+ib$, the contraction of $(x-z)$ consists of the real polynomials vanishing at $z$. If $b=0$, the factor theorem gives the ideal $(x-a)$. If $b\ne 0$, a real polynomial vanishing at $z$ also vanishes at its complex conjugate $\overline{z}$, so it is divisible by $(x-z)(x-\overline{z})=(x-a{)}^2+b^2$. Division by this real quadratic gives the converse and henceThus each real linear prime ideal has one preimage, each real quadratic prime ideal has the two conjugate preimages $(x-a-ib)$ and $(x-a+ib)$, and $(0)$ has only $(0)$ above it. In particular, $f$ is surjective.

No. Take $R=k[t]$ over a field $k$, and inside $R[x]=k[t,x]$ takeThis is a subring containing $R$: its elements have arbitrary constant coefficient in $k[t]$ as polynomials in $x$, and every positive-degree coefficient is divisible by $t$; that condition is preserved by sums and products. The ring $R$ is Noetherian. Indeed, a nonzero ideal of $k[t]$ has an element $g$ of least degree, and division by $g$ shows that every element of the ideal is a multiple of $g$. Thus $k[t]$ is a principal ideal domain, and its ideals are finitely generated.

In $S$, let $J_n=(tx,t{x}^2,\dots ,t{x}^n{)}_S$. These form an ascending chain. We show that $t{x}^{n+1}\notin J_n$. If it belonged, we could writeCancel $t$ in the integral domain $k[t,x]$, then apply the evaluation homomorphism $t\mapsto 0$. Every $s_j\in S$ becomes a scalar in $k$, so this would givewhich is impossible by comparison of degrees. Hence $J_n⊊J_{n+1}$ for every $n$, andThis is a constant-plus-ideal non-Noetherian subring construction: the restricted coefficients prevent a finite list of positive powers of $x$ from generating all the required powers.

For a discrete valuation ring $A$,Let $\pi$ be a uniformizer, so its unique nonzero maximal ideal is $\mathfrak{m}=(\pi )$. We use the defining description of a discrete valuation ring: it is a non-field local ring that is a principal ideal domain, and every nonzero element is $u{\pi }^n$ for a unit $u$ and an integer $n\ge 0$.

Every nonzero proper ideal is therefore $({\pi }^n)$ for some $n\ge 1$. Such an ideal is a prime ideal only when $n=1$: for $n\ge 2$, $\pi {\pi }^{n-1}\in ({\pi }^n)$ while neither factor belongs to $({\pi }^n)$. The ideal $(\pi )$ is prime because it is maximal. Also $(0)$ is prime because $A$ is an integral domain. Thusand the longest strict chain has length one.

The original PDF asks about a discrete valuation ring; the local TeX transcription's phrase “discrete variable $X$” is an extraction error.

We use the following standard facts about cyclotomic fields. For an odd prime $\ell$, the unique quadratic subfield of $\mathbb{Q}({\zeta }_{\ell })$ isAlso $\mathbb{Q}({\zeta }_{{\ell }^r})$ has degree $\phi ({\ell }^r)={\ell }^{r-1}(\ell -1)$, and $\ell$ is totally ramified, with a unique prime and residue field ${\mathbb{F}}_{\ell }$. These are the quadratic subfield of a prime cyclotomic field and total ramification in a prime-power cyclotomic field results.

TakeSince $59\equiv 3(\mathrm{mod}4)$, every $E_r$ contains $K=\mathbb{Q}(\sqrt{-59})$, through its subfield $E_1$. The unique prime above $59$ is totally ramified in $E_r/K$: the absolute ramification index is $58\cdot 59^{r-1}$, the index in $K/\mathbb{Q}$ is $2$, and the relative index is thereforeThe field $K$ has no real places. By part (b), $h_K\mid h_{E_r}$, and part (a) showed $3\mid h_K$. ThusThe ring of integers of a number field is a Dedekind domain, and is a principal ideal domain exactly when its ideal class group is trivial. Hence none of these rings of integers is principal. Finally the degrees $58\cdot 59^{r-1}$ strictly increase, so the fields are pairwise distinct. The explicit infinite family of nonprincipal cyclotomic fields of conductor a prime power is

Put $q=p^n$. A finitely generated module over $A=k[X]/(X^q)$ is finite-dimensional over $k$, and multiplication by $X$ is a nilpotent linear map $T$ with $T^q=0$. Its Jordan normal form exists over $k$ because its minimal polynomial is a power of $X$. Each Jordan block has size $r\le q$ and is the cyclic module $M_r=k[X]/(X^r)$. HenceA submodule of $M_r$ is an ideal of $k[X]/(X^r)$, and its inverse image is an ideal of the principal ideal domain $k[X]$ containing $(X^r)$. The only possibilities are $(X^j)$ with $0\le j\le r$. Thus all submodules form the chainEvery successive quotient is the one-dimensional simple module on which $X$ acts as zero. This is its unique composition series, so $M_r$ is a uniserial module and is an indecomposable representation: two nonzero direct summands would be incomparable submodules. Only $M_1$ is simple.

For $G=⟨g⟩$ of order $q$, the group algebra satisfiessince $(T-1{)}^q=T^q-1$ in characteristic $p$. Consequently, up to isomorphism, $\boxed{M_1,\dots ,M_{p^n}}$ are precisely the finite-dimensional indecomposable modules, and $\boxed{M_1\text{ is the unique simple module}}$. The finite-dimensional convention follows from the finite-generation hypothesis; “exactly” counts isomorphism classes.

Every Euclidean domain is a principal ideal domain: in a nonzero ideal choose an element of least positive Euclidean value and divide every other element by it; minimality forces every remainder to vanish. Every principal ideal domain is a unique factorization domain: the ascending-chain condition supplies factorizations into irreducibles, and Euclid lemma follows because every irreducible generates a prime ideal. Therefore

A Smith normal form of an $m\times n$ matrix $A$ over a principal ideal domain $R$ is a diagonal matrixwhere $U\in G{L}_m(R)$ and $V\in G{L}_n(R)$. It is guaranteed to exist for every matrix over a principal ideal domain. The entries $d_i$ are determined up to multiplication by units; over $\mathbb{Z}$ they are conventionally chosen positive.

The structure theorem for finitely generated modules over a principal ideal domain applies because every Euclidean domain is a principal ideal domain. It says that a finitely generated $R$-module is isomorphic towhere the nonzero nonunits $d_i$ are unique up to multiplication by units. They are the invariant factors.

For $V_{\alpha }$, multiplication by $X$ is the linear map $\alpha$. Since $V$ is finite-dimensional over $F$, $V_{\alpha }$ is a finitely generated torsion module over the polynomial ring $F[X]$, so it has no free summand. Choosing each invariant factor $a_i$ to be monic givesThe basis $1,X,\dots ,X^{\mathrm{deg}{a}_i-1}$ of each cyclic summand makes multiplication by $X$ a companion matrix. Concatenating these bases therefore puts $\alpha$ in rational canonical form.

On $F[X]/(a_i)$, a polynomial annihilates multiplication by $X$ exactly when it is divisible by $a_i$. It follows that the minimal polynomial and characteristic polynomial areThe second identity follows blockwise from the characteristic polynomial of a companion matrix. Since every $a_i$ divides $a_k$, the product ${\chi }_{\alpha }$ annihilates every cyclic summand. Thus ${\chi }_{\alpha }(\alpha )=0$, which is the Cayley-Hamilton theorem.

For the displayed matrix, the given generators of $\ker \theta$ are the columns ofElementary row and column operations over $\mathbb{R}[X]$ give the Smith normal formConsequently the nonunit invariant factors areand, as a check,

Suppose first that $M$ is finitely generated. Its quotient $M/N$ is generated by the images of any finite generating set of $M$. A principal ideal domain is Noetherian, so every submodule of the finitely generated $R$-module $M$ is finitely generated; in particular, $N$ is finitely generated.

Conversely, supposeand $M/N$ is generated by the cosets ${\overline{m}}_1,\dots ,{\overline{m}}_s$. For any $m\in M$, its coset is an $R$-linear combination of the ${\overline{m}}_j$, soExpressing this remainder in terms of the $n_i$ shows thatThus $M$ is finitely generated if and only if both the submodule $N$ and the quotient module $M/N$ are finitely generated.

A subset $S\subseteq M$ freely generates $M$ when every $m\in M$ has a unique expressionwith $r_s\in R$ and only finitely many nonzero coefficients. Equivalently, $S$ is a basis of a module and $M$ is the free module on $S$.

If $S$ freely generates $M$ and $f:S\to N$ is any function, defineUnique coordinates make $\varphi$ well defined; it is an $R$-module homomorphism and is the only extension of $f$. Conversely, apply the proposed universal property to the free module $F=R^{(S)}$ and the inclusion of $S$ into $M$. It gives maps $F\to M$ and $M\to F$ extending the corresponding functions on $S$. Uniqueness makes both composites identity maps, so $M\cong F$ and $S$ freely generates $M$. This is the universal property of a free module.

Now let $T$ generate the free module $M$ and $|T|=m$. Choose a maximal ideal $\mathfrak{m}$ of the nontrivial ring $R$. Then $k=R/\mathfrak{m}$ is a field, andis a $k$-vector space. The images of a basis $S$ of $M$ form a vector-space basis, while the images of $T$ span it. ThereforeIn particular $S$ is finite. Applying this result in both directions to two finite bases shows that they have equal cardinality, the rank of a free module $\mathrm{rk}M$.

A Euclidean domain is a principal ideal domain. By the submodule theorem for free modules over a principal ideal domain, every submodule $N$ of the finite-rank free module $M$ is free. Since a basis of $M$ generates it, the preceding inequality applied in the standard proof gives

The primary decomposition theorem for finitely generated modules over a principal ideal domain states thatwhere $p$ ranges over finitely many nonassociate irreducibles and the positive exponents are uniquely determined up to order. Equivalently, the torsion part decomposes into its primary cyclic summands.

Let $H$ be a finite subgroup of the multiplicative group of a field. It is a finite abelian group, hence the theorem over $\mathbb{Z}$ gives an invariant-factor decompositionIts exponent is $d_r$, so every element of $H$ is a root of $X^{d_r}-1$. A degree-$d_r$ polynomial over a field has at most $d_r$ roots, whenceBut $d_r\le |H|$, and equality forces all earlier factors to be trivial. Thus

By the prime ideal quotient criterion,For a principal ideal, this is precisely the assertion that $p$ is a prime element. This closes the cycle and proves that (i)--(v) are equivalent.

Now let $\phi :R\to S$ be a surjective ring homomorphism, where $R$ is a principal ideal domain and $S$ is an integral domain. WriteIf $a=0$, the first isomorphism theorem for rings makes $\phi$ an isomorphism. Otherwiseis an integral domain, so the equivalences just proved show that $(a)$ is maximal and $S$ is a field.

Next suppose that the polynomial ring $R[X]$ is a principal ideal domain. It follows first that $R$ is an integral domain. For nonzero $a\in R$, the idealis principal. Since $f$ divides the nonzero constant $a$, it is constant; since it also divides $X$, that constant divides the coefficient $1$, so $f$ is a unit. Therefore $(a,X)=R[X]$, and there are polynomials $g,h$ such thatSetting $X=0$ gives $1=ag(0)$, so every nonzero $a\in R$ is a unit. Hence

Finally, let $R$ be an integral domain in which every two nonzero elements have a greatest common divisor. If an irreducible $p$ divides $ab$, then $\gcd (p,a)$ is either a unit or an associate of $p$. In the second case $p\mid a$. In the first case, the Euclid lemma in a greatest-common-divisor domain gives $p\mid b$. Thus every irreducible element of $R$ is prime.

A binary cyclic code is a linear subspace $C\le {\mathbb{F}}_2^n$ invariant under cyclic coordinate shift. Identify a word $(c_0,\dots ,c_{n-1})$ with $c_0+c_{1}X+\cdots +c_{n-1}{X}^{n-1}$ inCyclic shift is multiplication by $X$, so cyclic codes are exactly ideals of this quotient. Since ${\mathbb{F}}_2[X]$ is a principal ideal domain, each is generated by a unique monic divisor $g\mid X^n-1$, giving the required bijection.

Letbe a primitive polynomial. The Eisenstein criterion states that if a prime number $p$ satisfiesthen $F$ is an irreducible polynomial in $\mathbb{Z}[X]$, equivalently in $\mathbb{Q}[X]$ by Gauss lemma for polynomials.

To prove it, suppose that $F=GH$ with $G,H\in \mathbb{Z}[X]$ of positive degree. Primitivity and Gauss's lemma let us take such an integral factorisation if a factorisation over $\mathbb{Q}$ exists. Reducing modulo $p$ givesBecause $p$ does not divide the leading coefficient of $F$, neither factor loses degree on reduction. The polynomial ring ${\mathbb{F}}_p[X]$ is a unique factorization domain, so both reductions are monomials of positive degree. In particular, $p$ divides both constant terms $G(0)$ and $H(0)$. It follows that $p^2$ dividesa contradiction. This proves the criterion.

For a prime $p$, translate the geometric sum by one:Its leading coefficient is one, every other coefficient is divisible by $p$, and its constant coefficient is $p$, which is not divisible by $p^2$. It is therefore Eisenstein at $p$. Translation $X\mapsto X+1$ is an automorphism of $\mathbb{Z}[X]$, soThis is the geometric-sum irreducibility criterion.

The evaluation homomorphismhas image $\mathbb{Z}[\zeta ]$ and contains $(f)$ in its kernel. Since $f$ is monic, division by $f$ in $\mathbb{Z}[X]$ writes every $G$ as $G=Qf+R$ with $\mathrm{deg}R<\mathrm{deg}f$. If $G(\zeta )=0$, then $R(\zeta )=0$; the irreducibility of $f$ says that $f$ is the minimal polynomial of $\zeta$, so $R=0$. Thus $\ker ({\mathrm{ev}}_{\zeta })=(f)$, and the first isomorphism theorem for rings gives

Now take $p=3$. Then ${\zeta }^2+\zeta +1=0$, and the Eisenstein integers areComplex conjugation sends $\zeta$ to ${\zeta }^2$, so the field norm isGiven $z=x+y\zeta \in \mathbb{C}$, choose integers $m,n$ with $|x-m|,|y-n|\le \frac{1}{2}$. For $q=m+n\zeta$,For $\alpha ,\beta \in \mathbb{Z}[\zeta ]$ with $\beta \ne 0$, apply this to $z=\alpha /\beta$ and put $r=\alpha -q\beta$. ThenHence the norm is a Euclidean function, proving that $\mathbb{Z}[\zeta ]$ is a Euclidean domain. This is the Euclidean norm on the Eisenstein integers.

Finally suppose $A\in {\mathrm{GL}}_n(\mathbb{Z})$ satisfies $A^2+A+I=0$. Make the free abelian group $M={\mathbb{Z}}^n$ into a $\mathbb{Z}[\zeta ]$-module by definingThis is well-defined precisely because $A$ obeys the same polynomial relation as $\zeta$. The module is finitely generated. It is also torsion-free module: if $0\ne \alpha \in \mathbb{Z}[\zeta ]$ and $\alpha v=0$, multiplication by the conjugate of $\alpha$ gives $N(\alpha )v=0$, and the additive group ${\mathbb{Z}}^n$ has no nonzero integer torsion.

A Euclidean domain is a principal ideal domain, and the structure theorem for finitely generated modules over a principal ideal domain says that a finitely generated torsion-free module over one is free. Thusfor some $m$. Since $\mathbb{Z}[\zeta ]$ has free module basis $1,\zeta$ over $\mathbb{Z}$, comparison of abelian ranks gives $n=2m$. There can consequently be no such matrix when $n$ is odd.

If $n=2m$, choose a $\mathbb{Z}[\zeta ]$-basis $v_1,\dots ,v_m$. Thenis a $\mathbb{Z}$-basis, and $A^2{v}_i=-v_i-A{v}_i$. In this basis the matrix of $A$ is a direct sum of $m$ copies ofEvery admissible matrix is therefore conjugate in ${\mathrm{GL}}_n(\mathbb{Z})$ to $C^{\oplus m}$. Hence there is exactly one conjugacy class for even $n$, and none for odd $n$. This is the classification of integral matrices satisfying the third cyclotomic polynomial.

An $A$-module $M$ is a flat module when the tensor functor $-{\otimes }_{A}M$ preserves injections, equivalently when it is exact.

Suppose first that $M$ is flat. For any nonzero $a\in A$, tensor the injection $A\stackrel{a}{\to }A$ with $M$. The resulting map $M\stackrel{a}{\to }M$ is injective, so $am=0$ implies $m=0$. Thus $M$ is a torsion-free module.

Conversely, suppose $M$ is torsion-free over the principal ideal domain $A$. Every finitely generated submodule of $M$ is a finitely generated torsion-free module over a PID, hence a finite free module and therefore flat. The module $M$ is the filtered colimit of these submodules. Tensor products commute with filtered colimits, and filtered colimits of modules preserve exact sequences, so $M$ is flat. This proves that a torsion-free module over a principal ideal domain is flat.