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Uniqueness in the defining characterization gives $\wp (-z)=\wp (z)$: the reflected function has the same poles, periodicity and prescribed principal part. Hence ${\wp }^{\prime }$ is odd and periodic. At each nonzero half-period $\omega \in \{1/2,\tau /2,(1+\tau )/2\}$, $-\omega \equiv \omega (\mathrm{mod}\Lambda )$, so ${\wp }^{\prime }(\omega )=-{\wp }^{\prime }(\omega )=0$.

An elliptic function has equally many zeros and poles, counted with multiplicities, in a fundamental parallelogram. This standard property applies to ${\wp }^{\prime }$, whose sole pole modulo the lattice has order three. The three distinct nonzero half-periods already supply three zeros, so these are all its zeros and each is simple. ThusEvenness and the prescribed expansion give $\wp (z)=z^{-2}+a_2{z}^2+a_4{z}^4+O(z^6)$. There is no constant term because the defining difference tends to zero, and only even powers occur. ConsequentlyAdding $20a_2\wp$ cancels the last pole. The resulting elliptic function is holomorphic everywhere and therefore constant, by boundedness on a fundamental parallelogram and Liouville theorem. Its value at zero is $-28a_4$. ThusAt the three half-periods $Q(\wp (\omega ))=0$. Their values are distinct: if two such points had the same value $e$, then $\wp -e$ would have a zero of multiplicity at least two at each, because its derivative vanishes there. That gives at least four zeros, contrary to its sole double pole and the zero-pole counting property. Hence the cubic has three distinct roots and leading coefficient four, provingAll properties specific to $\wp$ used here were derived from its defining characterization; only the general elliptic zero-pole count and holomorphic constancy were used as standard facts.

Two norms are Lipschitz equivalent norms if constants $c,C>0$ satisfy $c\Vert u{\Vert }_1\le \Vert u{\Vert }_2\le C\Vert u{\Vert }_1$ for every $u$. On the vector space of finitely supported sequences, the norms $\Vert x{\Vert }_1=\sum |x_i|$ and $\Vert x{\Vert }_{\infty }=\max |x_i|$ are not equivalent: a vector with $N$ entries equal to one has the two norms $N$ and one.

For dimension $d<\infty$, choose a basis and write $u=\sum x_i{e}_i$. Any norm satisfies $\Vert u\Vert \le \sum |x_i|\Vert e_i\Vert \le C|x{|}_2$. This also implies its continuity in the Euclidean coordinates, by $|\Vert u\Vert -\Vert v\Vert |\le \Vert u-v\Vert \le C|x-y{|}_2$. Its positive continuous values on the compact Euclidean unit sphere have a positive minimum $c$. Homogeneity gives $c|x{|}_2\le \Vert u\Vert$, proving equivalence with the Euclidean norm and hence between any two norms.

If a finite-dimensional subspace $U$ of a normed space contains a sequence converging in the ambient norm, its coefficient vectors are Cauchy by this equivalence. The coefficients converge in ${\mathbb{R}}^d$ or ${\mathbb{C}}^d$, and the reconstructed vector lies in $U$ and is the ambient limit. Thus finite-dimensional subspaces are closed.

A finite-dimensional normed space has nonempty open balls with compact closures by equivalence and Heine-Borel. Conversely let an open set $O$ have compact closure. Choose $w\in O$ and $r>0$ such that $B(w,r)\subset O$. Its closed ball of radius $r/2$ is a closed subset of the compact closure of $O$ and is therefore compact. Translation and scaling make the closed unit ball compact.

If the space were infinite-dimensional, construct unit vectors $x_j$ with $\mathrm{dist}(x_j,\mathrm{span}(x_1,\dots ,x_{j-1}))>1/2$. Here is the needed Riesz lemma: for a proper closed subspace $M$, choose $z\notin M$, put $d=\mathrm{dist}(z,M)>0$, and choose $m\in M$ with $\Vert z-m\Vert <2d$; then $(z-m)/\Vert z-m\Vert$ has distance greater than $1/2$ from $M$. The finite spans are closed by the earlier argument, so this constructs the sequence. Its members are separated by more than $1/2$, and have no convergent subsequence, contradicting compactness. Therefore the space is finite-dimensional exactly when it has a nonempty open set with compact closure.

Use the usual cell structure on the torus: one vertex, two one-cells $a,b$, and one two-cell attached by the word $ab{a}^{-1}{b}^{-1}$. Take the meridian being collapsed as $a$. The quotient has one vertex, the surviving circle $b$, and a two-cell whose attaching word reduces to $b{b}^{-1}$ and is null-homotopic. Attaching a two-cell to a circle by a null-homotopic loop gives a space homotopy equivalent to $S^1\vee S^2$.

Alternatively its cellular chain groups are $C_2=C_1=C_0=\mathbb{Z}$, and both boundary maps vanish: the two-cell traverses $b$ once in each direction, and the one-cell's endpoints coincide. Thus the integral homology groups areThe pinched torus is connected, so reduced $H_0$ is zero.

There is an error in the printed request: the cubic is not irreducible at $n=2$, and three elements cannot be a field basis in that case. The integral-basis assertion is valid under the necessary hypothesis $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$.

Under that hypothesis, $\alpha$ is an algebraic integer and $\mathbb{Z}[\alpha ]\subseteq {\mathcal{O}}_k$ has finite index $m$. Change of basis in the trace pairing gives the discriminant-index identityThe field discriminant is an integer. If the left side is square-free, $m^2$ cannot divide it unless $m=1$. Thus $1,\alpha ,{\alpha }^2$ is an integral basis when the cubic is irreducible and its discriminant is square-free.

For any root, ${\alpha }^3=n\alpha +1$ givesSince $\alpha +\beta +\gamma =0$, $\alpha \beta +\beta \gamma +\gamma \alpha =-n$ and $\alpha \beta \gamma =1$,On the other hand, the product of $r{f}^{\prime }(r)$ is $-\Delta$: the product of the roots is one, and each of the three unordered root pairs contributes a negative squared difference. Therefore, without needing irreducibility for this polynomial calculation,The three-root product is the field norm only when the cubic is the minimal polynomial.

At $n=1$, the polynomial has no rational root (the only candidates are $\pm 1$), so it is irreducible, and its discriminant is $-23$. Hence $1,\alpha ,{\alpha }^2$ is an integral basis.

At $n=2$,If $\alpha =-1$, then $k=\mathbb{Q}$ with integral basis $\{1\}$. If $\alpha =(1\pm \sqrt{5})/2$, then $k=\mathbb{Q}(\sqrt{5})$ and $\{1,\alpha \}$ is an integral basis, since this quadratic order has square-free discriminant five; also ${\alpha }^2=\alpha +1$. In neither case is the three-element list a basis. Its discriminant five is instead that of the degree-three algebra $\mathbb{Q}[x]/(x^3-2x-1)$; a basis there does not repair the printed assertion about the field $k$.

For every odd prime $p$ and $n\ge 1$, the group of units modulo an integer $(\mathbb{Z}/p^n\mathbb{Z}{)}^{\times }$ is a cyclic group, of order $\phi (p^n)=p^{n-1}(p-1)$; a generator is a primitive root.

Modulo seven, $3,3^2,\dots ,3^6$ are $3,2,6,4,5,1$, so three has order six. Moreover $3^6-1=728=7\cdot 104$ is divisible by seven exactly once. We prove the lifting needed here. If $u=1+7^{r}v$ with $r\ge 1$ and $7\nmid v$, the binomial expansion gives $u^7=1+7^{r+1}v+O(7^{r+2})$, so $v_7(u^7-1)=r+1$. Induction therefore givesAny exponent giving one modulo $7^n$ is divisible by six. The order also divides $6\cdot 7^{n-1}$ by Lagrange theorem, while the displayed valuation rules out each proper divisor of this form. Consequently three is a primitive root modulo every $7^n$.

Since $2^3=1(\mathrm{mod}7)$ and two is not one, its order modulo seven is three. A generator modulo $7^n$ would reduce to a generator modulo seven, because reduction is a surjective group homomorphism. Thus two is never a primitive root modulo $7^n$. Finally every unit modulo eight is odd, and $(2k+1{)}^2=1+4k(k+1)\equiv 1(\mathrm{mod}8)$. All unit orders are at most two, whereas the unit group has four elements. There is no primitive root modulo eight.

Let $L$ be the splitting field of the separable polynomial. Its Galois group is $G={\mathrm{Aut}}_K(L)$, acting faithfully on its $n$ distinct roots, hence as a subgroup of $S_n$. An automorphism preserves each root's minimal polynomial. Conversely an isomorphism $K(\alpha )\to K(\beta )$ for roots of the same irreducible factor extends to an embedding of $L$ in an algebraic closure, and normality of the splitting field makes the image $L$ itself. Thus the root orbits are precisely the irreducible factors' root sets. The action is transitive exactly when the polynomial is irreducible.

If $n=p$ is prime, a transitive action has $p\mid |G|$ by orbit-stabilizer, so Cauchy theorem for groups supplies an element of order $p$. Its permutation must be a $p$-cycle. Conversely a $p$-cycle already acts transitively on all $p$ roots. This proves the stated equivalence for prime degree.

The factorization theorem modulo a good prime says that if an integer polynomial retains its degree and has square-free reduction modulo $p$, and that reduction's irreducible factor degrees are $d_1,\dots ,d_r$, its Galois group over $\mathbb{Q}$ contains a permutation with disjoint cycle lengths $d_1,\dots ,d_r$. Equivalently the finite-field Galois group, generated by Frobenius on the roots, occurs as an unramified decomposition subgroup. The degree-retention qualification matters for nonmonic polynomials.

For the given quintic, Eisenstein criterion at three proves irreducibility. Modulo two it factors asNeither factor has a root in ${\mathbb{F}}_2$, so both are irreducible, and they are distinct; the reduction is separable. Thus $G$ contains an element of cycle type $(2)(3)$, whose cube is a transposition. It is also transitive of degree five.

To prove that this forces the full symmetric group here, form the graph on the five roots with an edge for each conjugate in $G$ of that transposition. The graph is $G$-invariant, and transitivity makes its connected components all the same size. That size divides five; there is an edge, so the size cannot be one. Hence the graph is connected. Transpositions along the edges of a connected graph generate all transpositions (move along a path and conjugate successive swaps), and hence generate $S_5$. They all belong to $G$, giving

For a finite connected plane graph, with $v$ vertices, $e$ edges and $f$ faces including the unbounded face, Euler formula for a connected planar graph isIf the graph is a tree, $e=v-1$ and $f=1$. Otherwise remove an edge on a cycle. This preserves connectivity and joins the two faces on its sides, so both $e$ and $f$ decrease by one. Repeating reaches a tree, preserving $v-e+f$ throughout and proving the formula.

For a simple connected graph with $v\ge 3$, each face boundary walk has length at least three, with bridges counted twice in the walk; the sum of lengths is $2e$. Hence $3f\le 2e$, which with Euler gives $e\le 3v-6$. If it has no triangles in a graph, no face boundary walk can have length three, and every boundary length is at least four. Thus $4f\le 2e$, givingFor a disconnected graph, add noncrossing bridges between its components; this preserves simplicity and creates no triangles, reducing to the connected bound.

Every triangle-free planar graph with at least three vertices has average degree $2e/v<4$, and therefore a vertex of degree at most three. Remove one such vertex. Its remaining graph is again triangle-free and planar. Induction, with graphs on at most two vertices as the base cases, provides a graph colouring with four colours of the remainder. At most three colours appear on the removed vertex's neighbours, so a fourth colour extends the graph colouring. This proves four-colourability without using the four-colour theorem.

A chain-complete partially ordered set has a supremum for each chain (and a least element if the empty chain is included). The weaker condition needed for Zorn lemma is that every chain have an upper bound; some terminology also calls this chain-complete. Zorn's lemma says that a nonempty partially ordered set with this upper-bound property has a maximal element. We verify that exact hypothesis below.

Consider the lattice ideals containing $a$ but not $b$, ordered by inclusion. The principal ideal $\downarrow a$ belongs to this family because $b\nleqq a$. The union of a nonempty chain of such ideals is an ideal: two of its elements lie together in one member of the chain, which contains their join and everything below it. The union still avoids $b$. The empty chain has the initial principal ideal as an upper bound. Zorn supplies a maximal such ideal $I$.

We prove that $I$ is a prime lattice ideal. If $x,y\notin I$, maximality says that the ideals generated by adjoining $x$ and by adjoining $y$ both contain $b$. Thus $b\le i\vee x$ and $b\le j\vee y$ for some $i,j\in I$: the first generated ideal consists of elements below $i\vee x$, and similarly for $y$. If $x\wedge y\in I$, distributivity givesa contradiction. Consequently $x\wedge y\in I$ implies $x\in I$ or $y\in I$.

Define $f(x)=0$ on $I$ and one elsewhere. An ideal contains $x\vee y$ exactly when it contains both $x$ and $y$, and its primeness says that it contains $x\wedge y$ exactly when it contains at least one of them. Thus $f$ preserves both joins and meets, and $f(a)=0,f(b)=1$ as required.

For completeness, take the Lindenbaum algebra of classical propositional formulae modulo provable equivalence, including any given premises as axioms. It is a Boolean, hence distributive, lattice; provable implication is its order. If a formula $P$ is not provable from the premises, then $1\nleqq [P]$. The separation just proved, with $a=[P],b=1$, gives a homomorphism of lattices sending $P$ to zero and one to one. It also sends zero to zero, since $0\le [P]$, and preserves complements because their meet is zero and join is one. Its values on atomic propositions form a Boolean valuation satisfying all premises and falsifying $P$. Thus semantic consequence implies provability, the completeness theorem for propositional logic.