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A Smith normal form of an $m\times n$ matrix $A$ over a principal ideal domain $R$ is a diagonal matrixwhere $U\in G{L}_m(R)$ and $V\in G{L}_n(R)$. It is guaranteed to exist for every matrix over a principal ideal domain. The entries $d_i$ are determined up to multiplication by units; over $\mathbb{Z}$ they are conventionally chosen positive.

Put $R=k[x,y,z]/(xy-z^3)$ and let $X=\mathrm{Spec}R$. The ideal $P=(x,z)$ is a height-one prime ideal, because $R/P\cong k[y]$ and $\dim R=2$. It therefore defines a prime Weil divisor $D=V(x,z)$.

Localizing at $x$ eliminates $y$, givinga unique factorization domain. The Nagata theorem for divisor class groups says that $\mathrm{Cl}(X)$ is generated by the height-one primes containing $x$. Sincethe only such prime is $P$, and hence $[D]$ generates. At the generic point of $D$, $y$ is a unit in a ring and $x=z^3/y$, so the order of vanishing is three:Thus $3[D]=0$.

This relation has exact order three. Indeed, if a principal divisor were supported on $D$, its defining rational function would be a unit on $D(x)$. The units of $R_x=k[z][x,x^{-1}]$ are precisely $c{x}^m$ with $c\in k^{\times }$ and $m\in \mathbb{Z}$, whose divisors are $3mD$. Consequentlygenerated by $[V(x,z)]$. This is the $n=3$ case of the Divisor class group of an A-type surface singularity.

Cover $Y$ byThese sets cover because a point of $D_Y$ has $(y,w)\ne (0,0)$. On $U_y$, the relation $x=zw/y$ shows that $D_Y$ has local equation $z$; on $U_w$, the relation $z=xy/w$ gives local equation $x$; and on $U_0$ its local equation is $1$. Thus $D_Y$ is a Cartier divisor.

For the line bundle associated to a divisor ${\mathcal{O}}_Y(D_Y)$, choose local framesThe transition functions areon the corresponding overlaps. Every displayed ratio is a unit in a ring in the corresponding ring of regular functions, and the Čech cocycle condition follows immediately.

An element of the p-adic integers is a unit in a ring exactly when its reduction modulo $p$ is nonzero. More explicitly, if $\alpha \not\equiv 0(\mathrm{mod}p)$, its inverses modulo $p^n$ are unique and compatible, so they define $\beta \in {\mathbb{Z}}_p$ with $\alpha \beta =1$. The converse follows by reducing $\alpha \beta =1$ modulo $p$. Part ii therefore proves that $\alpha$ topologically generates the additive group if and only if $\alpha$ is a p-adic unit.

Consider the symmetric Laurent polynomialIt satisfies $f(1)=1$, so the Alexander polynomial realization theorem gives a knot $K_3$ with ${\Delta }_{K_3}(t)\doteq f(t)$. The polynomial is not a unit of $\mathbb{Z}[t^{\pm 1}]$, whereas

A morphism of schemes is proper when it is a finite type morphism, a separated morphism, and a universally closed morphism. The valuative criterion for properness says, under the usual finite-type and Noetherian hypotheses, that $f:X\to Y$ is proper exactly when every commutative squarewith $R$ a discrete valuation ring and $K=\mathrm{Frac}R$ has a unique diagonal lift $\mathrm{Spec}R\to X$.

For ${\mathbb{P}}_k^2\to \mathrm{Spec}k$, a $K$-point is $[x_0:x_1:x_2]$ with not all $x_i$ zero. If $\pi$ is a uniformizer, multiply all coordinates by one power of $\pi$ so that ${\min }_{i}v(x_i)=0$. The resulting coordinates lie in $R$ and at least one is a unit, so they define an $R$-point extending the given $K$-point. If two extensions exist, on a chart where one coordinate is a unit their affine coordinate ratios agree in $K$ and therefore in the integral domain $R$; hence the extensions agree. This verifies existence and uniqueness directly.

Polynomial division separates a polynomial into a multiple of a chosen divisor and a smaller polynomial remainder. Given $f,g\in F[X]$ over a field $F$, with $g\ne 0$, there are unique polynomials $q,r$ satisfyingHere $q$ is the polynomial quotient. The bound concerns polynomial degree, rather than the sizes of the coefficients.^[ref]

For an original example over the rational numbers, takePolynomial long division first cancels $2X^3$ by subtracting $2Xg$:Subtracting one more copy of $g$ cancels $X^2$:ThusThe final polynomial remainder has polynomial degree one, below the divisor's polynomial degree two. Multiplying out the last identity checks every coefficient, including the constant term $-2+8=6$.

The polynomial degree bound makes this decomposition useful: it replaces a potentially long polynomial with a short polynomial remainder whenever multiples of $g$ can be discarded. At any root of a polynomial of $g$, those multiples vanish under polynomial evaluation. Repeated polynomial division also drives the Euclidean algorithm for finding a greatest common divisor of polynomials.

The coefficient setting matters. Over a commutative ring, the same existence and uniqueness hold when the divisor's leading coefficient is a unit in a ring. A monic polynomial therefore works over the integers, even though they do not form a field; monic polynomial division over a ring explains why zero divisors cause no difficulty in that case.^[ref]

A nonzero divisor alone is insufficient over the integers. Dividing $X$ by $2X+1$ with an integer polynomial quotient and a constant polynomial remainder would require $1=2c$ for an integer $c$, which is impossible. Over the rational numbers the division instead givesThe obstruction is the nonunit leading coefficient $2$. This example separates existence from uniqueness: over an integral domain, the polynomial degree argument still makes any admissible pair unique, if it exists.

For a formal group law over any commutative ring $R$, the invariant differential of a formal group law satisfies $[p{]}^{\ast }\omega =p\omega$. Its coefficient is a unit in a ring, so $[p{]}^{\prime }(T)\in pR[[T]]$. If $[p](T)=\sum a_m{T}^m$, this means $m{a}_m\in pR$. When $p\nmid m$, the integer Bézout's identity implies $a_m\in pR$, even if $p$ is a zero divisor. Put those coefficients into $pf(T)$ and the exponents divisible by $p$ into $g(T^p)$.

A principal fractional ideal of an integral domain $A$ with fraction field $K$ is $aA$ for a nonzero $a\in K$. Its generators differ by a unit in a ring in $A$. For a number field, taking $A={\mathcal{O}}_K$ and quotienting nonzero fractional ideals by principal fractional ideals gives the ideal class group; requiring congruences and signs of the generators instead gives a ray class group.

For a nonconstant morphism between nonsingular curves and a local target uniformizer $t$ at the image of $P$, the index is the order of vanishing of its pullback. Changing $t$ multiplies it by a local unit in a ring and does not change that order. This is a valuation definition and remains meaningful for inseparable morphisms.