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A complex-valued function $f:U\to \mathbb{C}$ is holomorphic on an open set $U$ when it has a complex derivative at every point of $U$. Requiring a derivative throughout a neighbourhood is much stronger than complex differentiability at a point. It forces locally convergent Taylor series and derivatives of every order, as follows from the Cauchy integral formula and Cauchy derivative formula.

To see the restriction in real coordinates, write $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$. The real and imaginary increment quotients must agree, giving the Cauchy-Riemann equationsConversely, if $u,v$ have continuous partial derivatives on $U$ and satisfy the Cauchy-Riemann equations there, then $f$ is a holomorphic function. The equations at a lone point, without a differentiability hypothesis for the real coordinate map, are insufficient. Their role is to force the Jacobian matrix to have the special formSuch a matrix represents multiplication by a complex number; it has only two free real entries. ^[ref]

For $f(z)=z^2$, $u=x^2-y^2$ and $v=2xy$, so $u_x=v_y=2x$ and $u_y=-v_x=-2y$. Directly,At $z_0=1+i$, multiplication by $2+2i$ scales lengths by its modulus $2\sqrt{2}$ and rotates by its argument $\pi /4$. A horizontal increment $h=0.01$ gives the exact displacement $0.0201+0.02i$; a vertical increment $h=0.01i$ gives $-0.0201+0.02i$. The first-order predictions $0.02+0.02i$ and $-0.02+0.02i$ are perpendicular, and the errors have modulus $10^{-4}$.

At every nonzero $z_0$, this local rotation and scaling makes $z^2$ a local conformal map. At zero, the first-order map vanishes: $r{e}^{i\theta }$ becomes $r^2{e}^{2i\theta }$, so the argument doubles and the origin is a critical point. Globally $z$ and $-z$ have the same square. Local preservation of angles therefore coexists with global folding, a useful warning when using holomorphic functions as coordinate changes.

For $|z|<1$, the polylogarithm is ${\mathrm{Li}}_s(z)={\sum }_{n\ge 1}{z}^n/n^s$; contour formulas analytically continue it to a slit plane.

For susceptible, infective, and recovered populations $S,I,R$, mass-action infection coefficient $\beta$, recovery rate $\nu$, and immunity-loss rate $f$, the SIRS equations areThey conserve $S+I+R=N$ and have basic reproduction number ${\mathcal{R}}_0=\beta N/\nu$.

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.

Over a field containing the relevant roots of unity, $X^n-a$ is irreducible when the valuation data prevent $a$ from being a proper prime-divisor power, with the standard fourth-power 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 satisfies $r^2=r$ for every element. Expanding $(r+1{)}^2=r+1$ shows that it has characteristic two.

An ideal $\mathfrak{p}$ is prime when $ab\in \mathfrak{p}$ implies $a\in \mathfrak{p}$ or $b\in \mathfrak{p}$, equivalently when the quotient is an integral domain.

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 $\mathbb{Q}[X]$.