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The splitting field of $f\in K[t]$ is the smallest extension $L/K$ in which $f$ is a product of linear factors; equivalently, $L$ is generated over $K$ by all its roots of a polynomial. To prove existence and uniqueness of splitting fields, adjoin a root of an irreducible polynomial dividing $f$ and repeat until $f$ splits. If $L$ and $L^{\prime }$ are two resulting fields, an embedding between the fields generated so far extends by sending each newly adjoined root to a root of its transformed minimal polynomial. Repetition gives a $K$-embedding $L\to L^{\prime }$, whose image contains every root of $f$ and is therefore all of $L^{\prime }$. Thus the splitting field exists and is unique up to $K$-isomorphism.

Now write $K={\mathbb{F}}_q$, and let the distinct irreducible factors of $f$ have degrees $d_1,\dots ,d_r$. The irreducible factors of a finite-field Frobenius polynomial show that a root of the degree-$d_i$ factor lies in ${\mathbb{F}}_{q^n}$ exactly when $d_i\mid n$. Hence the splitting field over a finite field isEvery finite field is a perfect field, so $L/K$ is separable; it is normal because it is a splitting field. It is consequently a Galois extension. More explicitly, the Galois group of a finite field extension is the cyclic group of order $m$ generated by the finite-field Frobenius automorphism $x\mapsto x^q$.

Finally, let $N_2(n)$ denote the number of monic irreducible polynomials over a finite field of degree $n$ over ${\mathbb{F}}_2$. Sinceand $\mu (1)=1$, $\mu (3)=-1$, $\mu (9)=0$, we obtain

We first prove the required case of Menger's theorem directly. Split each vertex $v$ into $v_{min}$ and $v_{mout}$ joined by a directed edge of capacity one. Replace every graph edge $uv$ by the two directed edges $u_{mout}{v}_{min}$ and $v_{mout}{u}_{min}$ of capacity $k$, add a source joined to every $a_{\mathrm{in}}$ for $a\in A$, and join every $b_{\mathrm{out}}$ for $b\in B$ to a sink, using unit capacities at the ends.

Starting with zero integral flow, repeatedly augment by one along a source-to-sink path in the residual directed graph. If fewer than $k$ augmentations have occurred and no augmenting path remains, let $R$ be the residual-reachable vertices. The capacity from $R$ to its complement equals the current flow: every forward edge in this cut is saturated and every reverse edge carries zero flow, by residual reachability. Since the cut has capacity below $k$, it uses no capacity-$k$ graph edge. The original vertices whose unit-capacity split or endpoint edges cross the cut form an $AB$-separator of size below $k$, contrary to the hypothesis. Hence at least $k$ augmentations occur. The resulting integral flow decomposes into $k$ source-to-sink paths, and the unit split capacities make their corresponding $AB$ paths vertex-disjoint. This proves

The same augmenting-path proof, with the source vertex given capacity $k$, yields the fan lemma: from a vertex $x$ to any set of at least $k$ vertices in a k-connected graph, there is a $k$-fan with distinct endpoints.

Now let $C$ be a longest cycle in a finite $k$-connected graph $G$. If $C$ is not Hamiltonian, choose $x\notin C$. A $k$-fan from $x$ to $C$ has $k$ distinct attachment vertices. If $|C|<2k$, two consecutive attachment vertices on $C$ are joined by an edge of $C$, because the $k$ cyclic gaps cannot all have length at least two. The two corresponding fan paths form, through $x$, an alternative path of length at least two between those adjacent vertices. Replacing their edge on $C$ by this path creates a longer cycle, a contradiction. ThereforeThis is the Dirac circumference theorem; in particular, a $k$-connected graph contains a cycle of length at least $k$, and when $|G|\ge 2k$ it must contain one of length at least $2k$.

For $k=3$, every 3-connected graph with at least six vertices consequently has a cycle of length at least six. This cannot be improved: the complete bipartite graph $K_{3,m}$ is 3-connected for $m\ge 3$, but every cycle alternates between its two parts and therefore has length at most six. Taking $m$ arbitrarily large defeats every proposed bound above six. Hence

The Knaster–Tarski fixed-point theorem states that if $L$ is a complete lattice and $f:L\to L$ is order-preserving, then the fixed points of $f$ form a complete lattice; in particular, $f$ has least and greatest fixed points.

LetFor every $x\in P$, monotonicity gives $f(a)\le f(x)\le x$, so $f(a)\le a$. Applying $f$ once more shows $f(f(a))\le f(a)$, hence $f(a)\in P$. Therefore $a\le f(a)$ by the definition of $a$, and $f(a)=a$. Every fixed point belongs to $P$, so $a$ is the least fixed point. Dually, $\bigvee \{x:x\le f(x)\}$ is the greatest fixed point. For any family $S$ of fixed points, apply the same argument inside the upper interval above $\bigvee S$ to obtain the least fixed point above every member of $S$; this is their join in the fixed-point set. The dual construction supplies meets, proving the full statement.

To deduce the Cantor-Schröder-Bernstein theorem, let $i:A\to B$ and $j:B\to A$ be injections. On the complete lattice $\mathcal{P}(A)$ defineThis map is order-preserving, so it has a fixed point $X$. The relationshows thatis a bijection: its two pieces map bijectively onto the disjoint sets $i(X)$ and $B\setminus i(X)$. Thus injections both ways imply a bijection.

Let $P$ be the poset of countable subsets of $\mathbb{R}$. The family of all singletons $\{\{r\}:r\in \mathbb{R}\}$ has no upper bound in $P$, since any upper bound would contain every real number and would be uncountable. Therefore

Finally, use the well-ordering theorem to fix a well-order $\prec$ of $\mathbb{R}$. Every countable $A$ omits some real number; let $m(A)$ be its $\prec$-least omitted element and defineIf $A\subseteq B$, then either $m(A)\in B$, or every point preceding $m(A)$ lies in $A\subseteq B$ and $m(A)=m(B)$. In either case $f(A)\subseteq f(B)$, so $f$ is order-preserving. Yet $m(A)\notin A$, and henceThis does not contradict Knaster–Tarski because $P$ is not a complete lattice.

A principal body frame is an orthonormal basis ${\mathbf{e}}_a(t)$ fixed in the rotating body and aligned with its principal axes. If $R(t)$ is the rotation matrix whose columns are these vectors, differentiating $R^{T}R=I$ shows that $\dot{R}{R}^T$ is skew-symmetric. Every skew-symmetric linear map in three dimensions is cross product with a unique vector $\mathbf{\omega }$, so

For any vector $\mathbf{A}=A_a{\mathbf{e}}_a$,Applying this to the conserved angular momentumgives the Euler equations for a torque-free rigid body:

Multiplying these equations respectively by ${\omega }_i$ and summing proves conservation of kinetic energy; multiplying by $I_i{\omega }_i$ and summing proves conservation of squared angular momentum. Thus the torque-free rigid-body invariants areSolving these two linear equations for the first two squared components givesUsing $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2$ yields the quartic equation for one component of torque-free angular velocity:

In the RSA cryptosystem, choose distinct primes $p,q$, set $N=pq$, and choose $e$ with $\gcd (e,\varphi (N))=1$. The public key is $(N,e)$; the private exponent satisfiesA message $m$ is encrypted and decrypted by

Suppose $\varphi (N)\mid 2^{a}b$, where $b$ is odd. For a unit $x$ modulo $N$, both $O_p(x^b)$ and $O_q(x^b)$ are powers of two dividing $2^a$. If these orders differ, assume without loss of generality thatTaking $t=r<a$ givesand hencea nontrivial factor.

To count successful $x$, write $p-1=2^{r}u$ and $q-1=2^{s}v$ with $u,v$ odd. Raising to the odd power $b$, which is divisible by the odd parts, maps a uniformly chosen element of each cyclic multiplicative group uniformly onto its Sylow two-subgroup. In a cyclic group of order $2^r$, the probability of order $1$ is $2^{-r}$ and that of order $2^j$ is $2^{j-1-r}$ for $1\le j\le r$. If $r\le s$, the probability that the two resulting orders agree isThus at least half of the $\varphi (N)$ units satisfy $O_p(x^b)\ne O_q(x^b)$.

Knowing $d$ reveals the multiple $ed-1$ of $\varphi (N)$. Writing $ed-1=2^{a}b$ with $b$ odd and trying random units therefore yields a factor with probability at least one half per trial; nonunits already reveal a factor by taking their greatest common divisor with $N$. Conversely, factoring $N$ gives $\varphi (N)$ and hence $d=e^{-1}(\mathrm{mod}\varphi (N))$. This proves that recovering the private key is computationally equivalent, up to randomized polynomial-time work, to factoring the RSA modulus, as summarized by RSA private exponent reveals the factorization.

For $N=77$ and $e=43$, $\varphi (N)=60$ andTherefore

Finally, $e$ is odd because it is coprime to the even integer $\varphi (N)$. If $m^e\equiv m(\mathrm{mod}N)$, thenThus fixed points occur in complementary pairs $m,N-m$, which is the Complementary RSA fixed points property.

Let $\mathcal{H}=a^{\prime }/a=\dot{a}$, where a prime denotes differentiation with respect to conformal time $d\tau =dt/a$. Multiplying the Friedmann equation by $a^2$ givesThe critical density is the value obtained when $k=0$, so

Since ${\mathcal{H}}^{\prime }=a\ddot{a}$, the pressure-free Raychaudhuri equation givesorDifferentiating $\Omega =1+k{c}^2/{\mathcal{H}}^2$ and substituting this equation yields

For an expanding universe $\mathcal{H}>0$. The Einstein-de Sitter universe $\Omega =1$ is an unstable constant solution: curves starting with $0<\Omega <1$ decrease towards $0$, while curves starting with $\Omega >1$ rise away from $1$ and diverge as the closed model reaches maximum expansion. This is the flatness problem.

Use the first-order actionVarying $q$ with fixed endpoints and varying $p$ freely givesThe principle of stationary action therefore yields Hamilton's equations

For $H=p^2+q^{-2}$ these become $\dot{q}=2p$ and $\dot{p}=2q^{-3}$. Along their Hamiltonian flow, $H=E$ is conserved andThus $F=pq-ctH$ is an integral of motion exactly when $c=2$.

Write its constant value as $F_0$. The surfaces $pq-2tE=F_0$ in extended $(q,p,t)$-space giveUsing $E{q}^2=p^2{q}^2+1$,The sign of $q$ is fixed. In phase space, each energy orbit is one of the two branchesEach branch approaches $p=-\sqrt{E}$ and $p=+\sqrt{E}$ as $|q|\to \infty$ and turns at $(p,q)=(0,\pm E^{-1/2})$.