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The spectrum $\mathrm{Spec}A$ of a commutative ring $A$ is the set of its prime ideals, with closed sets $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$ and a structure sheaf whose sections locally look like fractions.

The Tor functor is the homology of a tensor product of modules with a free resolution: ${\mathrm{Tor}}_i^R(M,N)=H_i(F_{\bullet }{\otimes }_{R}N)$ for a free resolution $F_{\bullet }$ of $M$. It measures the failure of tensoring to preserve exact sequences. For a commutative ring, one may resolve either argument; the double complex of the two resolutions proves that both computations agree.

Given a dual pair of free lattices $N,M$ and a rational fan, glue $\mathrm{Spec}A[{\sigma }^{\vee }\cap M]$ over common-face charts. The monoid rings are semigroup algebras. A lattice map respecting cones gives a toric morphism. This construction works over a commutative ring $A$, including $\mathbb{Z}$, and also for an infinite fan in toric geometry.

// Target: toric-geometry.bigb

Here $S$ consists of the non-zero-divisors of a commutative ring $A$. The localization map $A\to Q(A)$ is injective: a fraction $a/1$ is zero only if some $s\in S$ annihilates $a$. For a reduced ring with finitely many minimal prime ideals, its zero divisors form the union of these primes. Indeed, if $ab=0$ and $a$ avoids every minimal prime, reduction to each domain quotient forces $b$ into their intersection, which is zero. Conversely, for $a$ in a minimal prime ${\mathfrak{p}}_i$, choose $b$ in every other minimal prime but outside ${\mathfrak{p}}_i$, using their incomparability and multiplying suitable elements. Then $b\ne 0$ and $ab=0$.

For a commutative ring $R$ and $n\ge 1$, this quotient ring imposes $Y^n=0$. Each class has a unique representative $a_0+a_{1}Y+\cdots +a_{n-1}{Y}^{n-1}$ by monic polynomial division over a ring. Multiplication is ordinary polynomial multiplication followed by discarding all powers of $Y$ of exponent at least $n$.

Consider the real matrixThe Cayley-Hamilton theorem gives $A^2=3A-I$. ConsequentlyThusAll four entries come from just two scalar coefficients. If $A^n=a_{n}A+b_{n}I$, multiplying once by $A$ and replacing $A^2$ gives the linear recurrence relationEquivalently $a_{n+2}=3a_{n+1}-a_n$, with $a_0=0,a_1=1$. The appearance of Fibonacci numbers is explained by $A=Q^2$, where $Q=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ advances the Fibonacci recurrence. No eigenvectors need to be computed.

The same identity computes an inverse. Rearrange it as $A(3I-A)=I$, givingThis works because the constant term of this characteristic polynomial is nonzero and the coefficients lie in a field. Generally, when $c_0\ne 0$ over a field, rearranging the Cayley-Hamilton theorem expresses $A^{-1}$ as a polynomial evaluated at a matrix. Over a general commutative ring, a nonzero constant need not be a unit; invertibility requires that extra condition.

For a positive integer modulus $n$, a residue class $[a{]}_n$ is a unit when it has a modular multiplicative inverse: some integer $u$ satisfies $au\equiv 1(\mathrm{mod}n)$. Modulo twelve, the units are exactlyIn particular, five is its own modular inverse, since $5\cdot 5=25\equiv 1(\mathrm{mod}12)$. Multiplying all twelve residue classes by five gives a permutation; multiplying the outputs by five recovers the inputs. By contrast, multiplying by six gives only zero and six, with six inputs for each output. It cannot be reversed.

In general, the exact criterion isFor the forward direction, $au-kn=1$ implies every common divisor of $a,n$ divides one. Conversely, if their greatest common divisor is one, Bézout's identity supplies $ua+vn=1$, so $u$ is a modular inverse. The extended Euclidean algorithm computes these coefficients. The units form the unit group $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$ under multiplication. ^[ref]

Multiplicative cancellation while keeping the same integer modulus is valid for a unit modulo n: from $ca\equiv cb(\mathrm{mod}n)$, multiply by a modular inverse of $c$ to get $a\equiv b(\mathrm{mod}n)$. A nonunit cannot be cancelled as a general rule. For example, $6\cdot 0\equiv 6\cdot 2(\mathrm{mod}12)$, yet $0\not\equiv 2(\mathrm{mod}12)$. Thus division by six cannot undo this multiplication; solving a linear congruence instead keeps track of all possible classes.

This distinction marks the boundary between commutative rings and fields. Modulo a prime number $p>1$, every nonzero residue class is a unit, so $\mathbb{Z}/p\mathbb{Z}$ is a finite field. Modulo twelve it is a commutative ring with zero divisors, and is not a field. For $n=1$, the sole class satisfies $[0{]}_1=[1{]}_1$ and is a unit; this one-element zero ring is not a field. ^[ref]

A ring is von Neumann regular if every element $a$ has an element $b$ with $a=aba$. In a commutative ring, this is $a=a^{2}b$. An arbitrary direct product of rings whose factors are fields has this property by taking inverses in each nonzero coordinate. Every localization at a prime ideal of a commutative von Neumann regular ring is a field.

On the opposite of the category of finitely presented unital commutative rings, a basic covering family is $A\to A[f_i^{-1}]$, where finitely many $f_i$ generate the unit ideal. Include the empty cover of the zero ring. Localization commutes with base change, so these families define a coverage on a category. Their sheaf topos classifies the coherent theory of local rings. This coverage is subcanonical: representable functors obey the usual localization gluing equations.