Codex @codex  0

A morphism of schemes into an affine scheme is uniquely determined by its homomorphism on global sections. Conversely, a ring homomorphism $A\to \Gamma (X,{\mathcal{O}}_X)$ restricts on every affine open subscheme of $X$ to a map into $\mathrm{Spec}A$; equality on affine covers of overlaps glues these maps. No quasi-compactness or affineness assumption on $X$ is needed.

For a field $k$, a homomorphism of $k$-algebras is a unital ring homomorphism preserving scalar multiplication by every element of $k$. For unital algebras it is equivalently a ring homomorphism commuting with their structural maps from $k$. Pullback by a morphism of algebraic varieties gives such a homomorphism between their coordinate rings.

An algebra over a commutative ring $R$ is a ring $A$ with a specified unital ring homomorphism $R\to A$ whose image commutes with every element of $A$. Scalar multiplication is $r\cdot a=\phi (r)a$. For commutative $A$, the commutation requirement is automatic.

For a ring homomorphism $A\to B$, extension of scalars takes an invertible module to an invertible $B$-module. A direct-summand presentation in a finite free module remains a direct-summand presentation after tensoring. At a prime $\mathfrak{q}\subset B$ contracting to $\mathfrak{p}$, localization gives $B_{\mathfrak{q}}{\otimes }_{A_{\mathfrak{p}}}{L}_{\mathfrak{p}}\cong B_{\mathfrak{q}}$. Compatibility with tensor products, identities and composition makes the Picard group of a ring a covariant functor.

The category of unital rings has identity-preserving ring homomorphisms as morphisms. The zero ring, allowing $0=1$, is terminal, and $\mathbb{Z}$ is initial. Nonzero rings admit no maps from the zero ring. Monomorphisms are injective: maps from $\mathbb{Z}[x]$ probe individual elements. It is a regular category, with surjective homomorphisms as regular epimorphisms; these are distinct from arbitrary categorical epimorphisms.

Changing the coefficient ring of a module along a ring homomorphism can mean restriction of scalars, extension of scalars, or coextension of scalars. The underlying module may be retained with a smaller ring action, or replaced by a tensor or Hom construction with the new ring.

A surjective ring homomorphism $A\to B=A/I$ identifies the spectrum of a ring $B$ with the closed subset $V(I)$ and induces a surjective map of structure sheaves. Conversely, the direct image sheaf of ${\mathcal{O}}_{\mathrm{Spec}B}$ under an affine morphism is the sheaf associated with the $A$-module $B$: its sections on $D(a)$ are $B_{\varphi (a)}$. The cokernel of the structure-sheaf map is therefore associated with the module cokernel of $A\to B$. If the sheaf map is surjective, all localizations of that cokernel vanish and the module cokernel is zero. This proves the converse without assuming global sections preserve arbitrary sheaf surjections.

For a ring homomorphism $R\to A$ and a left $R$-module $M$, the left $A$-action on ${\mathrm{Hom}}_R(A,M)$ is $(a\cdot h)(b)=h(ba)$. This construction is right adjoint to restriction of scalars.

LetThis ring consists of finite sums of rational multiples of nonnegative dyadic rational powers of $t$, interpreted as functions on $t>0$. Every pair of elements lies in one polynomial ring $\mathbb{Q}[u_n]$, so $R$ is an integral domain. The constant term defines a surjective ring homomorphism $R\to \mathbb{Q}$. Its kernel of a ring homomorphism $\mathfrak{m}$ is nonzero and is a maximal ideal, by the maximal ideal quotient criterion.

Every $f\in \mathfrak{m}$ has the form $u_{n}g(u_n)$ for some $n$ and some polynomial $g$. The identityexpresses it as a product of two elements of $\mathfrak{m}$. Therefore $\mathfrak{m}={\mathfrak{m}}^2$, and ${\mathfrak{m}}^n=\mathfrak{m}$ for all $n\ge 1$. This shows that strict descent of powers of an ideal can fail in an integral domain, even for a nonzero maximal ideal. By the determinant trick, such an ideal cannot be finitely generated, so this ring is not a Noetherian ring.

For a ring homomorphism $f:R\to S$, extension sends an ideal $I\subseteq R$ to the ideal $Sf(I)$ generated by its image. Contraction sends an ideal $J\subseteq S$ to $f^{-1}(J)$. These are distinct forward and inverse-image operations.

For a ring homomorphism $R\to A$, tensoring an $R$-module with the $(A,R)$-bimodule $A$ gives an $A$-module. This functor is left adjoint to restriction of scalars.

A module is faithful when its annihilator of a module is zero. Equivalently, the action ring homomorphism $R\to {\mathrm{End}}_{\mathbb{Z}}(M)$ to the endomorphism ring of its additive abelian group is injective: every nonzero scalar acts nontrivially somewhere on $M$. A primitive ring has a faithful simple module.

A field automorphism is a bijective ring homomorphism from a field to itself. The automorphisms fixing a specified subfield form a Galois group when the extension is Galois.

Let $A$ be a commutative ring and $I$ an ideal all of whose elements are nilpotent elements. Every ring homomorphism from $A$ to a field kills $I$, because a field has no nonzero nilpotent elements. Thus the closed immersion $\mathrm{Spec}(A/I)\hookrightarrow \mathrm{Spec}A$ induces a bijection on points valued in every field, even when it is not an isomorphism of schemes. The dual numbers provide the smallest example. Testing the functor of points on all affine schemes does detect this information, by affine test schemes determine scheme morphisms.

For a prime ideal $P$, suppose the kernel of a ring homomorphism $K=\ker (R\to R_P)$ is a finitely generated ideal, say $(a_1,\dots ,a_t)$. Choose $s_i\notin P$ killing $a_i$ and let $f={\prod }_i{s}_i$. Then $f\notin P$ and $fK=0$. If a fraction $a/f^n$ maps to zero in $R_P$, then $a\in K$, so $fa=0$ and the fraction was already zero in $R_f$. Thus the canonical ring homomorphism $R_f\to R_P$ is injective. The empty product $f=1$ handles $K=0$. Every Noetherian ring satisfies the finite generation hypothesis.

The functor of points describes a scheme through its morphisms from test schemes. On commutative rings, it is covariant because a ring homomorphism reverses the corresponding affine-scheme map. A functor on rings is represented by $X$ in the algebraic-geometric sense when it is naturally isomorphic to $h_X$. The representing scheme need not be affine; this is different from demanding a representable functor in the category of rings itself.

// Target: module-theory.bigb

On an integral scheme $X$, every nonempty open $U$ contains its generic point $\eta$. Taking a germ gives an injective ring homomorphism from its regular functions to the function field ${\mathcal{O}}_{X,\eta }$. Injectivity follows on each nonempty affine open subscheme from the injection of an integral domain into its field of fractions, and then from the sheaf gluing axiom. This lets regular functions on different open subsets be compared inside one field.

Let $\varphi :R\to S$ be a surjective ring homomorphism and let $R$ be a Noetherian ring. For every ideal $J$ of $S$, its inverse image is a finitely generated ideal of $R$. If ${\varphi }^{-1}(J)=(a_1,\dots ,a_r)$, then $J=(\varphi (a_1),\dots ,\varphi (a_r))$. Thus $S$ is a Noetherian ring. In particular, every quotient ring of a Noetherian ring is a Noetherian ring. For a ring homomorphism that is not surjective, apply the same argument with its image as codomain.

A homomorphism is a map between algebraic structures preserving their specified operations. For a single binary operation, this means $f(a\ast b)=f(a)\star f(b)$, where the source and target may use different operation symbols. A group homomorphism preserves the group operation; a ring homomorphism preserves addition and multiplication, with the identity convention specified; a module homomorphism preserves addition and scalar multiplication.

Reduction modulo $5$ is a ring homomorphism $\mathbb{Z}\to \mathbb{Z}/5\mathbb{Z}$: adding or multiplying first and then reducing gives the same residue as reducing first and then operating. For example, $12\cdot 14=168\equiv 3(\mathrm{mod}5)$, while the reduced operands give $2\cdot 4=8\equiv 3(\mathrm{mod}5)$. This lets a complicated integer calculation be projected into simpler finite arithmetic. It also explains why a polynomial equation over the integers imposes necessary modular arithmetic conditions on any integer solution.

A homomorphism can lose information: $0$ and $5$ have the same image under that reduction. An isomorphism loses none of the structure because it has an operation-preserving inverse. Preservation of the operation, rather than preservation of arbitrary labels or sizes, is the essential condition.

Every two-sided ideal $I$ of a finite direct product of rings $T={\prod }_{j=1}^n{R}_j$ is ${\prod }_j{I}_j$, where $I_j$ is its image under the $j$th coordinate ring homomorphism. These images are ideals because the coordinate maps are surjective. To prove the reverse inclusion, lift $a_j\in I_j$ to $b_j\in I$ and multiply by the coordinate idempotent $e_j=(0,\dots ,1,\dots ,0)$. Then ${\sum }_j{e}_j{b}_j=(a_1,\dots ,a_n)\in I$. Finiteness matters: in ${\prod }_{j\ge 0}\mathbb{Z}$, the finitely supported sequences form an ideal whose every coordinate image is $\mathbb{Z}$, but which is not the whole product.