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An affine scheme is a scheme isomorphic to $\mathrm{Spec}A$ for some commutative ring $A$. Homomorphisms $A\to B$ correspond contravariantly to morphisms $\mathrm{Spec}B\to \mathrm{Spec}A$.

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 homomorphism of commutative rings $R\to A$ and two $A$-modules, the tensor product of modules over $A$ is the quotient of the tensor product of modules over $R$ imposing all relations $am\otimes n=m\otimes an$. The quotient map is $A$-linear when $A$ acts through the first factor on the source. This expresses the extra balancing imposed by change of rings.

A comonoid in the monoidal category of modules over a commutative ring, with tensor ${\otimes }_k$ and unit $k$.

For a monic degree-$n$ polynomial $g$ over a commutative ring $R$, let $A=R[X]/(g)$ and let $\lambda$ take the coefficient of $X^{n-1}$ in the reduced representative. The pairing $(a,b)\mapsto \lambda (ab)$ is unimodular on $1,X,\dots ,X^{n-1}$: reversing one basis order gives a triangular matrix with diagonal ones. When $A$ embeds in a separable field extension, Lagrange interpolation identifies $\lambda (a)$ with $\mathrm{Tr}(a/g^{\prime }(x))$. This yields the derivative formula for the inverse different.

For an associative algebra $A$ over a field $k$, a derivation is a $k$-linear map $D:A\to A$ satisfying $D(ab)=D(a)b+aD(b)$. It is the case $M=A$ of a derivation into a bimodule; the order of factors matters when $A$ is noncommutative. Substituting $a=b=1$ gives $D(1)=0$. Such maps form a Lie algebra under the commutator $[D,E]=D\circ E-E\circ D$.

More generally, for a commutative ring $R$, a derivation from a commutative $R$-algebra $B$ into a $B$-module $M$ is an $R$-linear map $D:B\to M$ with $D(r)=0$ for $r\in R$ and $D(b{b}^{\prime })=bD(b^{\prime })+b^{\prime }D(b)$. The second term has this order because the algebra is commutative. Derivations of smooth functions are vector fields provide a geometric example.

A division ring is a nonzero ring in which every nonzero element has a multiplicative inverse. Its multiplication need not be commutative.

For a multiplicative subset $S$ of a commutative ring $A$, localization preserves short exact sequences of $A$-modules. Surjectivity follows by lifting a numerator. If a fraction maps to zero, some element of $S$ annihilates its image, so multiplying its numerator by that element puts it into the original kernel. For an injective original map, the same criterion shows its localized kernel is zero. Thus $S^{-1}A$ is a flat module over $A$.

For a finite family generating the unit ideal in a commutative ring, this augmented Čech cochain complex of any module is exact. One proof clears the finitely many denominators and cocycle relations by powers $f_i^N$, writes $1={\sum }_i{a}_i{f}_i^N$, and contracts an alternating cocycle by the weighted insertions ${\sum }_i{a}_i{f}_i^N{c}_{iI}$. The argument uses vanishing criterion in a module localization and does not require the module to be finitely generated.

An element $a$ of a commutative ring $R$ is zero if and only if its image in every localization at a prime ideal is zero. If $a\ne 0$, its proper annihilator lies in a maximal ideal $\mathfrak{m}$. No element outside $\mathfrak{m}$ can annihilate $a$, so $a/1\ne 0$ in $R_{\mathfrak{m}}$. This also applies to elements of an $R$-module using localization of a module.

A local ring is a commutative ring with exactly one maximal ideal.

The smooth sections of a vector bundle form a module over the commutative ring $C^{\infty }(M)$ by pointwise multiplication. A vector bundle morphism over the identity induces a $C^{\infty }(M)$-linear map between these modules. Bundle morphisms from maps of smooth sections gives the converse, without assuming the bundle has a global frame.

The nilradical of a commutative ring is its ideal of nilpotent elements, equivalently the radical of an ideal applied to its zero ideal.

For every commutative ring $R$,Every nilpotent element lies in every prime ideal. If $a$ is not nilpotent, an ideal maximal among those disjoint from $\{1,a,a^2,\dots \}$ is a prime ideal avoiding $a$, by the Zorn lemma and the maximality argument.