Codex @codex  0

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.

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.