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A simple function is a measurable function with finitely many values, written $s={\sum }_{j=1}^m{a}_j{1}_{A_j}$ using disjoint measurable sets. For nonnegative $s$ its Lebesgue integral is $\mu (s)={\sum }_j{a}_j\mu (A_j)$, with $0\cdot \infty =0$. An integrable signed simple function has ${\sum }_j|a_j|\mu (A_j)<\infty$ and the same signed sum defines its integral.

For a nonnegative measurable $f$ define $\mu (f)=\sup \{\mu (s):s\text{ simple},0\le s\le f\}$. A real measurable $f$ is Lebesgue integrable when $\mu (|f|)<\infty$; then $\mu (f)=\mu (f^{+})-\mu (f^{-})$, both terms finite. The monotone convergence theorem states that if $0\le f_n\uparrow f$ pointwise, or almost everywhere, then $\mu (f_n)\uparrow \mu (f)$, allowing infinity.

For nonnegative $f,g$, choose nonnegative simple approximations $s_n\uparrow f$, $t_n\uparrow g$. Their sums increase to $f+g$; simple-function additivity and monotone convergence give $\mu (f+g)=\mu (f)+\mu (g)$. The same approximation proves $\mu (cf)=c\mu (f)$ for $c\ge 0$.

For integrable signed $f,g$, $|f+g|\le |f|+|g|$ makes their sum integrable. Write $u=f^{+}+g^{+}$, $v=f^{-}+g^{-}$. The identity $u+(f+g{)}^{-}=v+(f+g{)}^{+}$ and nonnegative additivity, with finite integrals, give $\mu (f+g)=\mu (f)+\mu (g)$. Also $(cf{)}^{\pm }=c{f}^{\pm }$ when $c\ge 0$, while negative $c$ interchanges the two parts, so $\mu (cf)=c\mu (f)$ for every real $c$. Integrals do not change upon alteration on a null set, hence this descends to Lebesgue space. The indicated map is linear.

A rational map $V⇢{\mathbb{P}}^m$ is a morphism on a nonempty open subset of $V$, with two descriptions identified if they agree on a dense open subset. It can be represented there by rational homogeneous coordinate functions not all zero. A point is regular for the map if it extends as a morphism on an open neighbourhood of that point.

For the given projective Cremona transformation, the three products vanish simultaneously exactly at the three coordinate vertices. Away from them they define a regular map. At $(1:0:0)$ consider $[1:t:st]$ as $t\to 0$, with $s\ne 0$. Its image tends to $[0:s:1]$, which depends on $s$. No regular extension could have these different limiting values. Algebraically these curves lie in its domain for $t\ne 0$ and a regular extension must give the same specialization at their common vertex, so the obstruction also works over any infinite characteristic-zero field. The other vertices follow by cyclic symmetry. On the torus $X_0{X}_1{X}_2\ne 0$, applying the map twice givesIt is therefore a birational map and is its own rational inverse.

Use coordinates $[A:B:C]$ on the target. Substitution in the quintic givesThus on the torus its image is the plane quartic $Q=0$. This quartic is nonsingular even over the algebraic closure. Its partial derivatives are $C^3+3A^{2}B$, $A^3+3B^{2}C$ and $3A{C}^2+B^3$. If one coordinate vanishes at a common zero of these partials, all vanish. If none vanishes, multiplying the three equations gives $A^3{B}^3{C}^3=-27A^3{B}^3{C}^3$, impossible in characteristic zero. Hence there is no projective singular point.

A smooth plane curve is geometrically irreducible: if it had two positive-degree components, Bézout theorem over the algebraic closure would give an intersection point, and the product equation has zero gradient there; a repeated factor also makes the gradient vanish along its component. Therefore this quartic is irreducible. Its torus part is dense and is identified by the projective Cremona transformation with the torus part of $V$. No coordinate line is a component of $V$, since restricting $F$ to each coordinate line leaves a nonzero monomial. Consequently every component meets the torus, and its torus part is already irreducible. $V$ is irreducible and birational to the nonsingular quartic $A{C}^3+A^{3}B+B^{3}C=0$.

Each Möbius strip deformation retracts onto its core circle. Its boundary travels twice around that circle. Thicken the two strips slightly across their common boundary to obtain an open cover whose intersection deformation retracts onto that boundary circle. The Seifert-van Kampen theorem therefore givesDepending on the boundary orientation the relation may first read $a^2=b^{-2}$; replacing $b$ by its inverse gives the displayed presentation.

The map $a,b\mapsto 1$ respects the relation and surjects onto the infinite cyclic group $\mathbb{Z}$, so the fundamental group is infinite. It also surjects onto the infinite dihedral group: map $a$ and $b$ to the reflections $s(x)=-x$ and $t(x)=2-x$ of the real line. Both squares are the identity, but $st(x)=x-2$ and $ts(x)=x+2$ differ. An abelian group cannot have this nonabelian quotient. The fundamental group is infinite and nonabelian.

The Chinese remainder theorem says that for pairwise coprime positive moduli $m_1,\dots ,m_s$, every choice of residues determines exactly one residue class modulo $M={\prod }_j{m}_j$. Equivalently $\mathbb{Z}/M\mathbb{Z}\to {\prod }_j\mathbb{Z}/m_j\mathbb{Z}$ is an isomorphism under reduction.

The first two modular congruences combine to $x=2+15k$. Reducing this expression modulo seven gives $2+k\equiv 6$, so $k=4+7j$. Consequently all integer solutions areThe moduli are pairwise coprime, so the Chinese remainder theorem guarantees that this is the entire solution class.

The permutation representation on ${\mathbb{C}}^X$ sends the basis vector $e_x$ to $e_{gx}$. Its character is ${\pi }_X(g)=|X^g|$, because precisely the fixed basis vectors contribute to the trace. Burnside lemma says that the number of orbits is $|G{|}^{-1}{\sum }_g|X^g|$. To prove it, double-count $(g,x)$ with $gx=x$: the count is ${\sum }_g|X^g|={\sum }_x|G_x|$. Each orbit contributes $|Gx||G_x|=|G|$ by the orbit-stabilizer theorem.

A pair is fixed under the diagonal action exactly when both entries are fixed. Since permutation characters are real,by Burnside lemma on the product.

The general linear group ${\mathrm{GL}}_2(q)$ acts on the $q+1$ lines of ${\mathbb{F}}_q^2$. It is doubly transitive: choosing nonzero representatives of any two distinct lines gives a basis, and a linear map can send that basis to representatives of any other ordered pair. Thus there are two orbits on ordered pairs, the diagonal and its complement. The character inner product of this permutation character with itself is two, while the multiplicity of the trivial character is one by transitivity. The sum of squares of irreducible multiplicities is therefore two; the remaining summand is one irreducible representation, of dimension $(q+1)-1$. Its dimension is $\boxed{q}$.

For $S_n$ acting on two-element subsets, an ordered pair of subsets is classified by intersection size $0,1,2$. All three occur for $n\ge 4$, and a permutation carries any pair of a given type to any other. Hence $⟨{\pi }_Z,{\pi }_Z⟩=3$. The natural point permutation representation is the trivial representation plus the standard representation of the symmetric group, of dimension $n-1$: its irreducibility also follows from double transitivity and inner product two.

Embed this point representation into ${\mathbb{C}}^Z$ by $e_i\mapsto {\sum }_{j\ne i}{e}_{\{i,j\}}$. The map is equivariant, and its kernel consists of vectors with $c_i+c_j=0$ for every distinct $i,j$. Three distinct indices force every $c_i=0$, so it is injective for $n\ge 3$. Thus the trivial and standard representations each occur in ${\mathbb{C}}^Z$. The norm three leaves exactly one further irreducible, with multiplicity one. It is the orthogonal complement of that embedded subspace, equivalently the vectors $a_{ij}=a_{ji}$ satisfying ${\sum }_{j\ne i}{a}_{ij}=0$ for every $i$. Its dimension isThe constituent dimensions are $\boxed{1,n-1,n(n-3)/2}$, each once. For $n=3$, complementation identifies two-element subsets with points, so there are only the trivial and 2-dimensional standard constituents.

A K-isomorphism $L\to L^{\prime }$ is a field isomorphism fixing every element of $K$. The field automorphism group ${\mathrm{Aut}}_K(L)$ consists of such isomorphisms $L\to L$, with composition.

If an isomorphism sends $\alpha$ to $\beta$, it takes every polynomial relation over $K$ for $\alpha$ to the same relation for $\beta$; the monic minimal polynomials coincide. Conversely, if their common minimal polynomial is $m$, evaluation induces field isomorphisms $K[X]/(m)\cong K(\alpha )$ and $K[X]/(m)\cong K(\beta )$. Their composite sends $\alpha$ to $\beta$ and fixes $K$.

An automorphism of $K(\alpha )$ sends $\alpha$ to a root of $m$ inside $K(\alpha )$ and is determined by this image. More precisely the field automorphism group permutes the set of roots of $m$ lying in this field, a set of size $s\le d$. This action is faithful because fixing $\alpha$ fixes the generated field. Thus ${\mathrm{Aut}}_K(K(\alpha ))$ embeds in $S_s$, and then in $S_d$ by fixing $d-s$ extra labels. In particular it is finite. This argument also covers inseparable minimal polynomials.

For an example not isomorphic even as abstract fields, take $K={\mathbb{F}}_p(t)$, let $\alpha$ generate ${\mathbb{F}}_{p^p}$ over ${\mathbb{F}}_p$, and put $\beta =t^{1/p}$. The constant-field minimal polynomial of $\alpha$ remains irreducible over the rational function field: if it factored over ${\mathbb{F}}_p(t)$, each monic factor's coefficients would be algebraic over ${\mathbb{F}}_p$, being symmetric functions of constant algebraic roots. A rational function algebraic over ${\mathbb{F}}_p$ is constant, so those coefficients would lie in ${\mathbb{F}}_p$, contradicting the original irreducibility. The polynomial $X^p-t$ is irreducible by Eisenstein criterion at $t$ in ${\mathbb{F}}_p[t]$. Both degrees are $p$, butThe first field has $p^p$ roots of $X^{p^p}-X$; the second has only $p$. Indeed a rational function algebraic over a finite constant field must be constant: a nonconstant rational function makes the transcendental variable algebraic over its own rational-function field, and therefore cannot itself be algebraic over the constants. The root counts are isomorphism invariants. These extensions are not isomorphic.

If $K={\mathbb{F}}_q$ is finite, every degree-$d$ extension is a field with $q^d$ elements. Such fields are splitting fields of $X^{q^d}-X$ over $K$, so uniqueness of the splitting field gives a K-isomorphism. There is no finite-base-field example of this kind.