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With susceptible, infective, and immune populations $X,Y,Z$, equal per-capita birth and death rate $\mu$, mass-action transmission coefficient $\beta$, and recovery rate $\nu$, the equations areThe conserved population is $N=X+Y+Z$, and the threshold population is $N_c=(\mu +\nu )/\beta$.

The basic reproduction number ${\mathcal{R}}_0$ is the expected number of secondary infections caused by one infective in an otherwise susceptible population.

The equationscombine mass-action infection with an exponential density-dependent birth model. Susceptibles reproduce at per-capita rate $b{e}^{-aS}$, while infectives have no reproductive term and die at total per-capita rate $d+\delta$.

An endemic equilibrium is an equilibrium at which infection persists, so at least one infected compartment is positive.

A disease-free equilibrium is an equilibrium of an epidemic model at which every infected compartment is zero.

A spatial SIR-type model combines compartment transitions with diffusion in the compartments whose members move appreciably.

Under mass action, random pairwise encounters between populations of sizes $X$ and $Y$ occur at a rate proportional to $XY$.

The SEIR model divides a population into susceptible, exposed, infectious, and recovered compartments.

If every transition transfers individuals between compartments without births, deaths, immigration, or emigration, the sum of all compartment sizes is constant.

Let $n\ge 2$, let the characteristic of a field $K$ not divide $n$, and suppose $K$ contains all $n$th roots of unity. For $a\in K^{\times }$, the polynomial $X^n-a$ is irreducible exactly when $a$ is not a $q$th power in $K$ for every prime divisor $q$ of $n$. Equivalently, the class of $a$ has order $n$ in $K^{\times }/(K^{\times }{)}^n$. Kummer theory then gives a cyclic extension of degree $n$.^[ref]

A useful sufficient test is a discrete valuation with $\gcd (n,v(a))=1$: if $a$ were a $q$th power for $q\mid n$, its valuation would be divisible by $q$. The denominator criterion for Newton polygon irreducibility gives the same conclusion directly.

Without the roots-of-unity hypothesis there is an additional fourth-power obstruction. For example, $X^4+4=(X^2-2X+2)(X^2+2X+2)$ over $\mathbb{Q}$, even though $-4$ is not a rational square. The precise hypotheses replace an unspecified appeal to valuation data or an unspecified standard exception.

The irreducible polynomials over $\mathbb{R}$ have degree one, or degree two with negative discriminant.

For an ideal $I$ of a commutative ring $R$, reducing coefficients modulo $I$ is a surjection $R[X]\to (R/I)[X]$ with kernel $I[X]$. Hence

A Boolean ring is a unital ring satisfying $a^2=a$ for every element. Expanding $(a+1{)}^2=a+1$ gives $2a=0$ for every $a$. A nonzero Boolean ring therefore has characteristic two. Expanding $(a+b{)}^2=a+b$ then gives $ab+ba=0$, which in characteristic two says $ab=ba$: Boolean rings are automatically commutative.

The zero ring is Boolean too, but has characteristic one under the least-positive-integer convention used here. Products of copies of ${\mathbb{F}}_2$, such as ${\mathbb{F}}_2\times {\mathbb{F}}_2$, give nonzero examples; squaring acts coordinatewise as the identity. Every prime ideal of a Boolean ring is maximal because its integral-domain quotient has only zero and one.

A proper ideal $P$ of a commutative ring $R$ is prime when $ab\in P$ implies $a\in P$ or $b\in P$. Equivalently, the nonzero quotient ring $R/P$ is an integral domain. The properness condition excludes the unit ideal, which would satisfy the elementwise implication vacuously.

For a positive integer $p>1$, $(p)$ is prime in $\mathbb{Z}$ exactly when $p$ is a prime number, while the zero ideal is also prime because $\mathbb{Z}$ is an integral domain. The ideal $(6)$ is not prime: the product of the classes of two and three vanishes in $\mathbb{Z}/6\mathbb{Z}$, although neither class is zero. Prime ideals are the points of the spectrum of a ring.

Module theory studies abelian groups with scalar multiplication by a ring. It extends linear algebra: a module over a field is a vector space, while ring scalars let the same language describe integer arithmetic and linear operators. The new phenomena are visible before any abstract classification: some nonzero scalars cannot be divided out, some modules have torsion elements of a module, and some have no basis of a module.

The distinction is useful because an integer relation such as $6m=0$ carries information which disappears if one assumes all nonzero scalars have multiplicative inverses. A finitely generated module over a principal ideal domain can be separated into a finite free module and a torsion module; the structure theorem for finitely generated modules over a principal ideal domain unifies the classification of finitely generated abelian groups with the study of linear operators through polynomial rings. See Keith Conrad's Modules over a PID, Theorem 4.1 and the operator applications.

A historical anchor is Emmy Noether's 1921 paper ^[ref]. Its first section studies rings and ideals under the condition that every ideal is a finitely generated module over its ring, and connects this to the ascending chain condition; later sections develop primary decomposition. This is an early source for the structural viewpoint underlying Noetherian rings. ^[ref] makes those statements accessible in English.

Every zero and pole of a square in $K(Y)$ has even order. Thus a rational function with a zero or pole of odd order cannot be a square. For characteristic different from two, this proves that $1-Y^2$ is not a square in $K(Y)$.

For a primitive polynomialif a prime $p$ does not divide $a_n$, divides every $a_j$ for $j<n$, and has $p^2\nmid a_0$, then $F$ is irreducible. If $F=GH$, reduction modulo $p$ forces both $\overline{G}$ and $\overline{H}$ to be positive-degree monomials. Their constant terms are consequently divisible by $p$, making $p^2\mid F(0)$, a contradiction. Gauss lemma for polynomials transfers the result between $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$.

For a nonzero polynomial over a unique factorization domain, its content is a greatest common divisor of its coefficients, defined up to a unit. Over $\mathbb{Z}$, choose the positive gcd. For example, $6X^2+10X+14$ has content two and primitive part $3X^2+5X+7$. Gauss lemma for polynomials says that content is multiplicative up to units in this setting.

For a polynomial over a general commutative ring, another common convention uses the ideal generated by its coefficients as its content. This convention does not require gcds. The two descriptions agree through a principal coefficient ideal when that ideal is principal, but a unit gcd need not make the coefficient ideal the whole ring. For example, the coefficients of $xT+y$ over $k[x,y]$ have gcd one and generate the proper ideal $(x,y)$. State the convention before using content to define a primitive polynomial.

Reduction modulo an ideal $I$ of a commutative ring $R$ gives a group homomorphism $R^{\times }\to (R/I{)}^{\times }$. It need not be surjective: the units $\{1,-1\}$ of $\mathbb{Z}$ do not lift the classes two and three in $(\mathbb{Z}/5\mathbb{Z}{)}^{\times }$.

A useful sufficient condition is $I\subseteq J(R)$, the Jacobson radical. If $ab\equiv 1(\mathrm{mod}I)$, then $ab$ is a unit by the unit criterion for the Jacobson radical, and consequently $a$ is a unit. Every invertible residue class then lifts.

For positive integers $n\mid m$, reduction $(\mathbb{Z}/m\mathbb{Z}{)}^{\times }\to (\mathbb{Z}/n\mathbb{Z}{)}^{\times }$ is always surjective. Start with a representative coprime to $n$ and use the Chinese remainder theorem to choose a congruent representative avoiding each additional prime dividing $m$. This explicit integer result should be distinguished from unrestricted reduction in an arbitrary ring.

Modulo p^e, a class is nilpotent exactly when its reduction modulo p is zero.