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Apply clearing denominators relative to an independent module subset to find $0\ne a$ with $aM$ lying in a finite free module. Multiplication by $a$ is injective on a torsion-free module, giving the required embedding. A submodule need not itself be free over a general integral domain.

An invertible maximal-rank minor in the Jacobian matrix eliminates generators and gives a local surjection from ${\mathcal{O}}_X^n$ to the Kähler differential sheaf. For an integral smooth variety the map is an isomorphism at the generic point. Its kernel is a submodule of a free module over an integral domain, so it is a torsion-free module; vanishing generically forces it to be zero. This gives local free frames for differentials.

A Euclidean domain is a principal ideal domain. The structure theorem for finitely generated modules over a principal ideal domain therefore gives, for every finitely generated module $M$ over $R$,where the $d_i$ are nonzero nonunits and $d_1\mid d_2\mid \cdots \mid d_s$. The factors $d_i$ are unique up to multiplication by units; $r$ is uniquely determined.

The rank of a module is this integer $r$. Equivalently, with $K$ the field of fractions of $R$,The tensor product of modules kills each displayed torsion summand after extending scalars to $K$, while sending $R^r$ to $K^r$, explaining the equality.

If $M$ is a torsion-free module, no summand $R/(d_i)$ can occur: its nonzero element $1+(d_i)$ is killed by the nonzero scalar $d_i$. Hence $s=0$ andThis also covers $M=0$, with $r=0$.

Let $x,y\in \mathbb{Q}$. Choose a positive integer $D$ clearing their denominators and write $x=m/D$, $y=n/D$ with $m,n\in \mathbb{Z}$. If $x$ and $y$ are not both zero, thenis a relation with integer coefficients not both zero. If both vanish, use $1\cdot x+0\cdot y=0$. Thus every pair is linearly dependent over $\mathbb{Z}$.

A submodule $M\subseteq \mathbb{Q}$ is a torsion-free module: if $kz=0$ for a nonzero integer $k$ and a rational number $z$, then $z=0$. If $M$ is also a finitely generated module, part (a), applied to the Euclidean domain $\mathbb{Z}$, makes it a free module. Its rank cannot exceed one, since two different members of a basis of a module would be linearly independent, contradicting the relation just proved. Since $M\ne 0$, its rank cannot be zero. ThereforeThe denominator-and-greatest common divisor construction in part (c) gives an explicit version of this result for a finitely generated subgroup of the rational additive group.

The printed vanishing claim needs the restriction $i\ge 1$. On $C={\mathbb{P}}^1$ with $\mathcal{F}=\mathcal{O}$ and $D=n\{\infty \}$, Question 3 gives $h^0(\mathcal{F}(D))=n+1$ for every $n\ge 0$. Thus no bound on the degree of a divisor can force vanishing in degree zero. We prove finiteness in every degree and the corrected vanishing in every positive degree, uniformly over all divisors on an algebraic curve of large degree.

First take a finite morphism $\pi :C\to {\mathbb{P}}^1$ as in Question 3 and put $E={\pi }_{\ast }\mathcal{F}$. The direct image sheaf $E$ is coherent: on an affine chart its module is finitely generated over a finite algebra and hence over the chart's coordinate ring. The two affine-cover Čech cochain complexes for $\mathcal{F}$ on $C$ and for $E$ on ${\mathbb{P}}^1$ are identical. Therefore

We need an elementary presentation of any coherent sheaf $E$ on the projective line. Choose finitely many generators on each of its two standard affine charts. A generator on the first chart restricts to an element of the second chart's module localized at $t^{-1}$. Multiplication by a sufficiently large power of $t^{-1}$ brings that element into the second chart's module. Since the transition of $\mathcal{O}(n)$ is $t^n$, this makes the original generator extend to a global section of $E(n)$. A generator on the second chart similarly extends after clearing its denominator at $t=0$. Use one common exponent for all the finitely many generators. The extended sections generate $E(n)$ on both charts. Untwisting gives a surjectionwith coherent kernel $K$. The long exact sequence in sheaf cohomology and Question 3 show that $H^1(E)$ is a quotient of the finite-dimensional $H^1(P)$, since $H^2(K)=0$. Apply the same presentation argument to $K$ to obtain finiteness of $H^1(K)$. Then the segment $H^0(P)\to H^0(E)\to H^1(K)$ proves finiteness of $H^0(E)$. Higher groups vanish by Question 3. Hence every $H^i(C,\mathcal{F})$ is finite-dimensional.

Next establish a useful degree estimate without using the duality requested later. A global regular function on $C$ is constant: a nonconstant one would give a finite morphism $C\to {\mathbb{P}}^1$ by Question 1. Such a morphism is surjective, because its closed image contains the generic point, but a regular function misses $\infty$, a contradiction. Thus $H^0(C,{\mathcal{O}}_C)=k$. Write $g=h^1(C,{\mathcal{O}}_C)$, the genus of a smooth projective curve. For any divisor on an algebraic curve $A$ and point $P$, the local discrete valuation ring givesThe last term is a length-one skyscraper sheaf. Additivity of the Euler characteristic of a coherent sheaf and repeated addition or subtraction of points giveThe inequality follows because $h^1$ is nonnegative, and is sufficient for the following argument.

Fix $P\in C$. For $n\ge g+1$ this estimate gives at least two independent sections of ${\mathcal{O}}_C(nP)$, one of them the constant section. A nonconstant section is a rational function on an algebraic variety whose poles are confined to $P$. It must have a pole there, since a globally regular function is constant. Its extension $C\to {\mathbb{P}}^1$ is finite, and its inverse image of ${\mathbb{A}}^1$ is exactly $C\setminus \{P\}$. Consequently $U=C\setminus \{P\}$ is affine.

Let $T$ be the torsion subsheaf of $\mathcal{F}$ and let $V=\mathcal{F}/T$. The support of $T$ is finite; such a coherent sheaf is a finite direct sum of sheaves supported at points and has zero higher sheaf cohomology. The torsion-free sheaf $V$ is locally free: a finitely generated torsion-free module over a discrete valuation ring is free. Tensoring by any invertible sheaf is exact, so

The inclusions $V(nP)\subset V((n+1)P)$ have point-supported quotients. Their long exact sequences in sheaf cohomology therefore give surjections on $H^1$. Moreovera section regular away from $P$ has some finite pole order at $P$. Compute this direct limit on a finite affine cover of $C$. All its intersections with $U$ are affine by Question 2, and localization commutes with filtered direct limits. The Čech cochain complex consequently givesThe maps from $H^1(C,V)$ to these groups are surjective. Choose a finite basis of this starting vector space. Each basis vector becomes zero at some stage in the direct limit; one common stage kills the entire basis. Thus there exists $n_0\ge 0$ with $H^1(C,V(n_{0}P))=0$.

Now let $D$ be any divisor on an algebraic curve with $\mathrm{deg}D\ge n_0+g$. The estimate above gives $h^0({\mathcal{O}}_C(D-n_{0}P))\ge 1$. A nonzero section identifies $D-n_{0}P$ with an effective divisor $E$ up to linear equivalence of divisors. Hence $V(D)\cong V(n_{0}P+E)$. Adding $E$ gives a point-supported quotient of $V(n_{0}P+E)/V(n_{0}P)$, so its first sheaf cohomology is a quotient of $H^1(C,V(n_{0}P))=0$. We have proved uniform high-degree vanishing on a smooth projective curve:This bound concerns all high-degree divisors, not only multiples of a previously chosen divisor.

On $D_{+}(x_3)$, write $r=x_0/x_3$, $q=x_1/x_3$, and $s=x_2/x_3$. The equations include $q=\pi s^2$ and $r=s^3$; after making these substitutions, the remaining equations are identities. ThereforeThe other chart isThere is a homomorphism $B\to R[t]$ defined by $a\mapsto \pi t$, $b\mapsto t^2$, $c\mapsto t^3$. To prove injectivity rather than merely exhibit a parametrization, reduce monomials using the four relations. Powers of $a$ reduce to at most one $a$; any remaining product of $a$ with $b$ or $c$ eliminates $a$; powers of $c$ reduce to at most one $c$. Thus every element is an $R$-linear combination ofTheir images are $1,\pi t,t^{2n},t^{2n+3}$, with pairwise distinct degrees. Since $R$ is a domain and $\pi \ne 0$, these images are $R$-linearly independent. Hence they are a basis, the map is injective, andIn particular $B$ is torsion-free, indeed free, as an $R$-module.

The two charts cover $X$ by the prime-ideal argument in part i. Both coordinate rings are torsion-free over the discrete valuation ring $R$, hence flat: over a principal ideal domain a torsion-free module is a filtered union of finitely generated free submodules, and filtered colimits preserve flatness. Thus the morphism $X\to \mathrm{Spec}R$ is flat on an open cover and therefore flat everywhere. It is proper because $X$ is a closed subscheme of the proper projective space ${\mathbb{P}}_R^3$. Consequently $X$ is both proper and flat over $R$.

An R-module $M$ is a free module if it has a basis $(e_i{)}_{i\in I}$: every $m\in M$ has a unique expression as a finite sum ${\sum }_i{r}_i{e}_i$, with $r_i\in R$. Equivalently, it is a direct sum of copies of $R$. Let $q:M\to M/N$ be the quotient map. When $M/N$ is free, choose a basis $({\overline{f}}_j)$ and lifts $f_j\in M$. The universal property of a free module defines an $R$-linear map $s:M/N\to M$ by $s({\sum }_j{r}_j{\overline{f}}_j)={\sum }_j{r}_j{f}_j$. It satisfies $q\circ s=\mathrm{id}$, so every $m\in M$ decomposes uniquely asThis gives a split short exact sequence and $M\cong N\oplus (M/N)$. If $N$ is also free, the union of its basis and the lifted quotient basis is a basis of $M$: spanning follows from the decomposition, and applying $q$ to a putative relation kills all quotient coefficients before independence in $N$ kills the remaining ones. Thus free submodule and free quotient imply a free module.

For the first claim, take $R=\mathbb{Z}$ andHere $N\cong N^{\prime }\cong \mathbb{Z}$ are free modules and $M/N\cong M^{\prime }/N^{\prime }\cong \mathbb{Z}/2\mathbb{Z}$. But $M$ is a torsion-free module, whereas $M^{\prime }$ contains the nonzero torsion element $(0,\overline{1})$. Therefore claim (1) is false.

For the second claim, the same lifting construction works without assuming that $N$ is free. Since both isomorphic quotient modules are free, it yieldsThus claim (2) is true. Equivalently, this uses the fact that free modules are projective.

Let $T$ consist of all torsion elements of a module in $M$. If $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs(m+n)=0$. Since $R$ is an integral domain, $rs\ne 0$. Also $r(am)=0$ for every scalar $a\in R$. Thus $T$ is the torsion submodule.

If $r(m+T)=0$ in $M/T$ with $r\ne 0$, then $rm\in T$, so some $s\ne 0$ satisfies $srm=0$. The domain property again gives $sr\ne 0$, hence $m\in T$. Therefore $M/T$ is a torsion-free module.

Finally, any submodule $N$ with the stated properties satisfies $N\subseteq T$. Conversely, if $m\in T$, its coset in $M/N$ is torsion and therefore zero, so $m\in N$. Hence $\boxed{N=T}$ is the unique choice.

The map $\phi :M\to N$, $x\mapsto rx$, is a module homomorphism, and it is injective because $r\ne 0$ and $M$ is a torsion-free module. The linear independence in a module of $T$ makes the map $R^{|T|}\to N$ sending the standard basis to $T$ an isomorphism. Thus $N$ is a finite free module, and $M\cong \phi (M)\subseteq N\cong R^{|T|}$. This proves embedding a finitely generated torsion-free module in a finite free module. No claim that the submodule itself is free over an arbitrary integral domain is needed.

On an affine chart $\mathrm{Spec}A\subseteq V$, the module of Kähler differentials ${\Omega }_{A/k}$ is generated by symbols $da$ subject to $k$-linearity and $d(ab)=adb+bda$. It represents $k$-derivations. These modules commute with localization, so their associated quasi-coherent sheaves glue to the Kähler differential sheaf ${\Omega }_{V/k}^1$. It is coherent: if $A=k[x_1,\dots ,x_N]/I$ and $I=(f_1,\dots ,f_s)$, then it has the finite presentation of a moduleFor a closed point $P$, put $R={\mathcal{O}}_{V,P}$ with maximal ideal $\mathfrak{m}$ and residue field $k$. The Zariski tangent space is ${\mathrm{Hom}}_k(\mathfrak{m}/{\mathfrak{m}}^2,k)$, equivalently ${\mathrm{Der}}_k(R,k)$. A derivation kills constants and ${\mathfrak{m}}^2$, so it factors through $a\mapsto a-a(P)$; conversely every linear functional on $\mathfrak{m}/{\mathfrak{m}}^2$ defines such a derivation by the product rule. The universal property of Kähler differentials therefore identifies this space withIn particular, ${\Omega }_{V,P}^1/{\mathfrak{m}}_P{\Omega }_{V,P}^1\cong {\mathfrak{m}}_P/{\mathfrak{m}}_P^2$, the algebraic cotangent space.

A point is a smooth point of a variety when its local ring is a regular local ring; over this algebraically closed field this says ${\dim }_k{T}_{V,P}=n=\dim V$. Tensor the finite presentation of a module above with $k(P)$. The tangent space is the kernel of the Jacobian matrix $J(P)$, hence has dimension $N-\mathrm{rank}J(P)$. This proves the Jacobian criterionFor a smooth irreducible variety $V$, choose at each point an invertible $(N-n)$-minor and shrink the affine chart so that it remains invertible. Its relations eliminate $N-n$ of the generators of ${\Omega }_{A/k}$, yielding a surjection $A^n\to {\Omega }_{A/k}$. At the generic point, its target has dimension $n$: over a perfect ground field, a separating transcendence basis of the function field has differentials forming a basis. Equivalently this follows from the assumed density of the smooth locus. The kernel therefore becomes zero over the fraction field of the integral domain $A$. As a submodule of $A^n$, it is a torsion-free module, so it is already zero. Thus these maps give local isomorphisms with ${\mathcal{O}}_V^n$, proving local freeness of differentials on a smooth variety with rank $n$.

For an affine chart $\mathrm{Spec}A\subseteq V$, write $W\cap \mathrm{Spec}A=\mathrm{Spec}(A/I)$ and $\mathcal{M}=\tilde{M}$. The restriction of a module sheaf to a closed subvariety is $\widetilde{M{\otimes }_{A}A/I}=\widetilde{M/IM}$. The Conormal exact sequence for Kähler differentials isIt follows from the generators and relations: passing to $A/I$ imposes precisely the additional relations $df=0$ for $f\in I$. The first map is well defined because $d(I^2)$ lies in $I{\Omega }_{A/k}$. Glue these exact module sequences, using exactness of localization, to obtainIn the final assertion, interpret locally principal subvariety as a proper local hypersurface. Its ideal on each chart is $(f)$ with $f\ne 0$. Since $V$ is an irreducible variety and reduced, $f$ is a non-zero-divisor, and $A/(f)\to (f)/(f^2)$, $\bar{a}\mapsto af$, is an isomorphism. These local rank-one descriptions make the conormal sheaf ${\mathcal{I}}_W/{\mathcal{I}}_W^2$ invertible.

Because $W$ is not contained in the singular locus of $V$, there is a dense open subset of $W$ where both $V$ and $W$ are smooth varieties. At a closed point there, $\dim W=\dim V-1$ by the Krull principal ideal theorem. The tangent description $T_{W,P}=\ker (df:T_{V,P}\to k)$ then forces $df\ne 0$. Hence the conormal map is injective at the generic point of $W$. Its kernel is a subsheaf of a line bundle on the integral variety $W$, so it is a torsion-free sheaf; a torsion-free sheaf with zero generic fibre is zero. This proves conormal injectivity for a generically smooth Cartier divisor. If zero equations were allowed in the phrase locally principal, $W=V$ would be a counterexample to invertibility; the proper-hypersurface convention is essential.

Let the torsion submodule of the module over a ring $M$ beThis is a submodule. Indeed, if $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs\ne 0$ because the scalar ring is an integral domain, and $rs(m+n)=0$; scalar multiples and additive inverses are handled similarly.

Take the quotient module $M_0=M/T(M)$ and let $q:M\to M_0$ be the quotient R-module homomorphism. It is torsion-free: if $0\ne r\in R$ and $r(m+T(M))=0$, then $rm\in T(M)$, so $srm=0$ for some $s\ne 0$. Since $sr\ne 0$, this puts $m$ in $T(M)$ and therefore $m+T(M)=0$.

Now let $N$ be torsion-free and let $f:M\to N$ be an $R$-module homomorphism. If $m\in T(M)$ and $rm=0$ for $r\ne 0$, then $rf(m)=f(rm)=0$, so torsion-freeness gives $f(m)=0$. Hence $T(M)\subseteq \ker f$, and the universal property of a quotient module gives a unique homomorphismwith $\boxed{f=f_0\circ q}$. Thus $M/T(M)$ is the maximal torsion-free quotient of $M$.

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.

Yes. The integers form a principal ideal domain, and over a principal ideal domain a module is flat exactly when it is torsion-free. Thus the torsion-free modules $A$ and $B$ are flat modules. The functoris a composite of two exact tensor functors, so $A{\otimes }_{\mathbb{Z}}B$ is flat. Applying the converse direction of the same characterization shows that it is torsion-free. This is the torsion-free module over a principal ideal domain is flat criterion.

Let $A$ be a Noetherian ring that is an integral domain. Ifis exact, $N$ is a finitely generated free module, and $P$ is a torsion-free module, then $M$ is a reflexive module. In particular, the dual of every finitely generated $A$-module is reflexive.

On an integral scheme, a quasi-coherent sheaf is torsion-free when its module on every affine chart is a torsion-free module over that chart's integral domain. It embeds into its generic fibre. Therefore a torsion-free sheaf with zero generic fibre is zero; locally free sheaves and their subsheaves are torsion-free.

For a module $M$ over an integral domain $R$, its torsion submodule isThis is a submodule. If $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs(m+n)=0$ and $rs\ne 0$; also $r(am)=a(rm)=0$ for every $a\in R$. The integral domain hypothesis supplies both the nonzero product and commutativity. These are the conventions of ^[ref].

A mixed example is the direct sum $M={\mathbb{Z}}^2\oplus \mathbb{Z}/6\mathbb{Z}$. If a nonzero integer kills $(a,b,\overline{c})$, then $a=b=0$. Conversely, $6$ kills every $(0,0,\overline{c})$. ThusThe torsion submodule isolates the finite-order part while the quotient module keeps the independent integer directions. In general $M/T(M)$ is a torsion-free module: if $r(m+T(M))=0$ with $r\ne 0$, some $s\ne 0$ kills $rm$, so $sr\ne 0$ kills $m$.

The hypothesis matters. In the ring $R=\mathbb{Z}/6\mathbb{Z}$ viewed as a module over itself, the elements $\overline{2}$ and $\overline{3}$ are killed by nonzero scalars $\overline{3}$ and $\overline{2}$, but their sum $\overline{5}$ is a unit and is killed by no nonzero scalar. Thus using arbitrary nonzero scalars in a ring with zero divisors need not give a submodule. A common broader convention uses regular elements of a ring as annihilators; the integral domain convention here avoids that ambiguity.