Take $\boxed{K=F[x]/(f)}$ and let $\alpha$ be the residue class of $x$. Division with remainder represents every class by a polynomial of degree less than $n$, so every element is a polynomial in $\alpha$ and $f(\alpha )=0$. A nonzero class represented by $u$ has $\gcd (u,f)=1$ because $f$ is irreducible. Bézout's identity gives $au+bf=1$, making the class of $a$ its inverse. Thus $K$ is a field, containing $F$ through its constant classes.

A splitting field of a nonconstant polynomial is an extension in which it factors into linear factors and which is generated over the original field by its roots. To construct one, choose an irreducible factor and use the preceding quotient construction to adjoin a root. Divide off a linear factor in that extension and repeat on the remaining polynomial. Each stage lowers the remaining degree, so finitely many stages suffice. The field generated by all roots in the resulting field is the required splitting field. This induction works with repeated roots and does not assume separability. The two roman clauses in the source specify properties of this one construction; they are not separate questions.

The permutation inverse is ${\rho }_{g^{-1}}$: applying the two maps successively sends $\omega$ to $(\omega g)g^{-1}=\omega$. With the paper's right-action convention, permutation products also act on the right, andThus $\boxed{\rho (gh)=\rho (g)\rho (h)}$. If function composition is written on the left instead, the same right action is represented by $g\mapsto {\rho }_{g^{-1}}$; keeping these conventions consistent avoids an artificial reversal. The kernel consists precisely of elements fixing every point, so

The rotational tetrahedral group acts transitively on its four vertices. A rotation fixing a vertex must preserve its opposite face and rotate its three vertices cyclically; its stabilizer has order three. The orbit-stabilizer theorem gives $|G|=4\cdot 3=12$. The rotations are the identity, two nontrivial rotations about each of the four vertex-to-opposite-face axes, and a half-turn about each of the three opposite-edge-midpoint axes. Therefore the counts by element order areA rotation fixing all vertices fixes three linearly independent vertex vectors and is the identity. The vertex permutation action is therefore faithful. The listed rotations induce the identity, 3-cycles and double transpositions, all even permutations. Its injective image has order 12, the full order of $A_4$. Hence $\boxed{G\cong A_4}$.

Fubini's theorem states that an absolutely integrable measurable function on a product of two $\sigma$-finite measure spaces has integrable sections almost everywhere, the corresponding iterated integrals exist, and both orders equal its product integral. In particular, for $f\in L^1({\mathbb{R}}^2)$,To establish integrability of the nonnegative function here, first use Tonelli's theorem: for a nonnegative measurable function the product integral equals either iterated integral with infinity allowed. This avoids assuming the conclusion needed for Fubini's theorem.

If $\alpha \ge -1$, ${\int }_0^{\infty }(1+x+y{)}^{\alpha }dx=\infty$ for every fixed $y$. If $\alpha <-1$, it isThe outer integral is finite exactly when $\alpha +1<-1$, that is, $\alpha <-2$. In that caseFor $-2\le \alpha <-1$ the outer integral diverges, including logarithmic divergence at $\alpha =-2$. Nonnegativity means finiteness of this integral is exactly absolute integrability; after proving it, Fubini's theorem allows either order.

For a nonnegative measurable function, the Lebesgue integral is defined through nonnegative simple functions:The value may be infinite. A real measurable $g$ is Lebesgue integrable when $\int |g|dx<\infty$; then $\int g=\int g^{+}-\int g^{-}$, where $g^{+}=\max (g,0)$ and $g^{-}=\max (-g,0)$.

If $|g|\le h$, every simple function below $|g|$ is also below $h$. Taking the defining suprema proves monotonicity of the Lebesgue integral:Hence $g$ is integrable, and both its positive and negative parts have finite integrals.

The oscillatory function $k$ is measurable because it is continuous off the singleton $\{0\}$ and has a specified value there. The elementary bounds $|\sin u|\le \min (1,|u|)$ give the integrable majorantIndeed, $\int H=2+2{\int }_1^{\infty }{x}^{-3}dx=3$. The comparison just proved gives $\int |k|\le 3$, so $k$ is Lebesgue integrable over the whole real line, despite its oscillations near zero. Its odd symmetry also gives $\int k=0$, now legitimately because absolute integrability has been established.

Euler's totient function isEuler's theorem states that $\gcd (a,n)=1$ implies $\boxed{a^{\phi (n)}\equiv 1(\mathrm{mod}n)}$. For $n>1$, let $r_1,\dots ,r_{\phi (n)}$ be the invertible residue classes modulo $n$. Multiplication by $a$ permutes these classes, since it has an inverse modulo $n$. Multiplying the resulting congruences givesEvery factor is a unit, so the product can be cancelled, proving the theorem. The case $n=1$ is immediate. For distinct prime numbers $p,q$, removing multiples of either prime from $1,\dots ,pq$ gives

For the RSA cryptosystem, choose distinct large primes $p,q$, put $N=pq$, and choose $e$ coprime to $\phi (N)$. Compute $d$ with $ed\equiv 1(\mathrm{mod}\phi (N))$. The public key is $(N,e)$; the private exponent $d$ is retained secretly. Represent a message by $0\le m<N$, encrypt by $c\equiv m^e(\mathrm{mod}N)$, and decrypt by $m^{\prime }\equiv c^d(\mathrm{mod}N)$.

To prove correct decryption even for messages not coprime to $N$, write $ed=1+k(p-1)(q-1)$. Modulo $p$, if $p\mid m$ then $m^{ed}\equiv m\equiv 0$; otherwise Fermat's little theorem gives $m^{ed}\equiv m$. The same reasoning holds modulo $q$. The Chinese remainder theorem therefore gives $m^{ed}\equiv m(\mathrm{mod}N)$, and the chosen representative recovers $m$. Thus decryption works for every message residue, not merely the units.