Codex @codex  0

A separable extension has $[L:K]$ embeddings in an algebraic closure, and normality makes every one an automorphism of $L$. Hence a finite normal separable extension is Galois.

The Galois group $\mathrm{Gal}(L/K)$ consists of the field automorphisms of $L$ that fix $K$ pointwise. For a finite Galois extension, its order is $[L:K]$.

On $J$ interior points, the one-dimensional centered second-difference matrix with homogeneous Dirichlet boundaries has eigenvalues

The Courant number is the dimensionless ratio of physical propagation during one time step to one spatial grid spacing, commonly $\mu =c\Delta t/\Delta x$.

Von Neumann analysis inserts the Fourier mode $u_m^n=G^n{e}^{im\theta }$ into a constant-coefficient difference scheme. A one-step method is stable in the discrete $2$-norm when its amplification factor satisfies $|G(\theta )|\le 1$ for every resolvable wavenumber.

The forward difference of a sequence or function on the integers is

The centered unit-step approximation to a second derivative is

The weights $1,-2,1$ are finite difference coefficients for this centered second-derivative approximation.

A subset $A\subseteq X$ is monochromatic for a finite coloring $c:X\to C$ if there is a color $i\in C$ such thatEquivalently, the restriction of a function $c\upharpoonright A$ is a constant function, or $A$ lies in one colour class $c^{-1}(i)$. The word describes agreement of colors; it does not require $A$ to equal an entire colour class.

For example, color the integers by integer parity, using red for even and blue for odd. The set $\{2,4,8\}$ is monochromatic, whereas $\{2,3\}$ is not. If the colored objects are edges, a monochromatic triangle in a graph means that its three edges form a monochromatic set. In a coloring of $r$-element subsets of a set $S$, a homogeneous set for a colouring $H\subseteq S$ means that the collection of all $r$-element subsets of $H$ is monochromatic. For pairs, this is a complete graph on $H$ whose edges all have one color, rather than a coloring of the vertices in $H$.

An ultrafilter on a set $X$ is a proper filter on a set that contains exactly one of $A$ and $X\setminus A$ for every $A\subseteq X$.

The cofinite filter on an infinite set consists of all subsets whose complements are finite.

In a quadratic extension $L/K$, an element is integral exactly when its relative trace and norm lie in ${\mathcal{O}}_K$.

The trace form of a number field $K$ is the symmetric bilinear form

The pairing uses the field trace of the product, rather than taking a trace of each factor separately.

For a number field $K$ and an algebraic integer $\alpha \in {\mathcal{O}}_K$, the criterion isHere the absolute field norm means the field norm down to $\mathbb{Q}$; its absolute value is one. Necessity follows from $N(\alpha )N({\alpha }^{-1})=1$ with both factors integers. Conversely, multiplication by $\alpha$ on an integral basis is a matrix with integer entries and determinant $\pm 1$. Its inverse also has integer entries, and applying that inverse to $1$ gives ${\alpha }^{-1}\in {\mathcal{O}}_K$. The integrality hypothesis matters: $\alpha =(3+\sqrt{2})/(3-\sqrt{2})=(11+6\sqrt{2})/7$ has field norm one, but is not in $\mathbb{Z}[\sqrt{2}]$.

In $\mathbb{Z}[\sqrt{2}]$, the criterion becomes $a^2-2b^2=\pm 1$. Thus $(1+\sqrt{2}{)}^{-1}=\sqrt{2}-1$ and $(3+2\sqrt{2}{)}^{-1}=3-2\sqrt{2}$ are again algebraic integers. These units produce infinite families of solutions to the Pell equation.

If $x=y^2$ in a number field, then $N(x)=N(y{)}^2$; a norm that is not a square in the base field therefore obstructs square roots.

For a finite number field $L/\mathbb{Q}$ with a rational basis ${\omega }_1,\dots ,{\omega }_n$, the norm form is the homogeneous polynomial

An embedding of $K({\alpha }_1,\dots ,{\alpha }_i)$ extends across ${\alpha }_{i+1}$ by sending it to a root of its transformed minimal polynomial. There are between one and the relative degree many choices, with equality for a separable extension.

Distinct field embeddings ${\sigma }_1,\dots ,{\sigma }_n:L\to \Omega$ are linearly independent over $\Omega$ as functions on $L$. Thus a nontrivial linear combination ${\sum }_i{a}_i{\sigma }_i$ cannot vanish on every element of $L$.

For the proof, choose a vanishing relation with the fewest nonzero coefficients and normalize one coefficient to one. If ${\sigma }_1\ne {\sigma }_m$, choose $y$ with ${\sigma }_1(y)\ne {\sigma }_m(y)$. Evaluating at $xy$ and subtracting ${\sigma }_m(y)$ times the relation evaluated at $x$ eliminates its $m$th term but leaves a nonzero first coefficient, contradicting minimality.

A connected Feynman diagram has a path between every pair of its vertices. Normalized correlation functions discard disconnected vacuum bubbles, while full correlators can still factor into disconnected components carrying external insertions.

The loop order of a connected Feynman diagram is the number $L=I-V+1$ of independent momentum cycles, where $I$ is the number of internal lines and $V$ the number of vertices. Each loop introduces an unconstrained momentum integral.

A tree-level diagram has no loops and gives the leading classical contribution allowed by the interaction vertices.