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A nonempty topological space is irreducible when it is not the union of two proper closed subsets, equivalently when any two nonempty open subsets intersect.

The Zariski topology on ${\mathbb{A}}^n$ is Noetherian because $k[x_1,\dots ,x_n]$ satisfies the ascending chain condition on ideals. Suppose a closed set $X$ that is not a finite union of irreducible closed sets were minimal among such counterexamples. It is reducible, say $X=X_1\cup X_2$ with both $X_i$ proper closed subsets. Minimality expresses each $X_i$ as a finite union of irreducible closed sets, and combining those decompositions contradicts the choice of $X$. Thus every closed $X$ has a finite irreducible decomposition.

Using $x_3=x_2^2$, the second defining polynomial reduces toConsequentlyEach component is the image of $t\mapsto (\pm t,t,t^2)$ and is isomorphic to ${\mathbb{A}}^1$, hence irreducible. In characteristic two the two displayed components coincide.

Suppose a polynomial $P(x,y)={\sum }_{j=0}^m{p}_j(x)y^j$ vanishes on every $(x,e^x)$. Thenas an entire function. These exponential-polynomial functions are linearly independent: applying $(D-m{)}^{\mathrm{deg}{p}_m+1}$ kills the term with largest exponent, while it acts injectively on $p_j(x)e^{jx}$ for $j<m$; induction on the number of terms finishes the proof. Hence every $p_j$ is zero, so no nonzero polynomial vanishes on the graph. Its Zariski closure is all of ${\mathbb{A}}^2$.

Finally refine an open cover of an affine variety $X$ by distinguished opens $D(f_j)$, choosing one inside an original member around each point. Since these distinguished opens cover,The Hilbert Nullstellensatz implies that the $f_j$ generate the unit ideal. Thusfor finitely many of them, so $D(f_{j_1}),\dots ,D(f_{j_r})$ cover $X$. The corresponding original open sets form a finite subcover. This is quasi-compactness of an affine variety.

A covering map is locally a disjoint union of homeomorphisms onto the base. A space is simply connected when it is path-connected and every loop is null-homotopic. By the uniformization theorem, the simply connected Riemann surfaces areup to analytic isomorphism.

A lattice is a discrete subgroupwith ${\omega }_1/{\omega }_2\notin \mathbb{R}$. The Weierstrass elliptic functionconverges normally on compact subsets of $\mathbb{C}\setminus L$, defines a meromorphic map $\mathbb{C}\to {\mathbb{C}}_{\infty }$, and is nonconstant because it has a double pole at every lattice point. Reindexing the normally convergent derivative series shows that ${\wp }_L^{\prime }$ is $L$-periodic; evenness fixes the integration constants, so ${\wp }_L(z+\omega )={\wp }_L(z)$.

The invariance makeswell-defined and unique on $\mathbb{C}/L$. It is analytic because the quotient projection is locally biholomorphic.

Neither map is a covering. For any nonzero $\omega \in L$, periodicity and oddness of ${\wp }_L^{\prime }$ giveso ${\wp }_L^{\prime }(\omega /2)=0$. The map is not a local homeomorphism there; since the quotient projection is locally biholomorphic, the descended map fails for the same reason.

There is no covering $\mathbb{C}/L\to {\mathbb{C}}_{\infty }$. The torus is connected, while every connected covering of the simply connected sphere is a homeomorphism. A torus is not homeomorphic to a sphere, for example because their fundamental groups are ${\mathbb{Z}}^2$ and $0$.

The closest point theorem in a Hilbert space states that every nonempty closed convex subset $C$ of a Hilbert space has a unique point nearest to any given $x$.

Let $d={\inf }_{y\in C}\Vert x-y\Vert$ and choose $y_n\in C$ with $\Vert x-y_n\Vert \to d$. The parallelogram identity and convexity giveso $(y_n)$ is Cauchy. Completeness and closedness give a limit $y\in C$ with $\Vert x-y\Vert =d$. If $y,z$ were both minimizers, the same identity with their midpoint would force $\Vert y-z\Vert =0$.

Apply the theorem to a closed subspace $F$. For $x\in H$, let $y\in F$ be closest and put $z=x-y$. For every $u\in F$, minimality of $\Vert z-tu{\Vert }^2$ at $t=0$ for real and imaginary $t$ gives $⟨z,u⟩=0$. HenceThe intersection is zero, so $H=F\oplus F^{\perp }$.

If $T$ is a shift with orthonormal basis $(e_n)$, it is an isometry,Conversely, suppose these three properties hold. Choose a unit vector $e_1$ spanning $(\mathrm{Im}T{)}^{\perp }$ and set $e_n=T^{n-1}{e}_1$. Isometry makes this sequence orthonormal. Iterating the orthogonal decompositiongivesA vector orthogonal to every $e_n$ lies in every $\mathrm{Im}(T^n)$ and is therefore zero. Thus $(e_n)$ is an orthonormal basis and $T{e}_n=e_{n+1}$, so $T$ is a shift. This is the wandering-vector characterization of a unilateral shift.

A covering space is a map $p:\tilde{X}\to X$ such that every $x\in X$ has an open neighbourhood $U$ for whichand every restriction $p:U_{\alpha }\to U$ is a homeomorphism.

For path lifting, cover the compact image of $\gamma$ by evenly covered sets. The Lebesgue number lemma supplies a subdivisionfor which each $\gamma ([t_{j-1},t_j])$ lies in one such set. Starting in the sheet containing ${\tilde{x}}_0$, lift the first segment with the inverse of that sheet. Its endpoint selects the sheet for the next segment, and induction constructs a continuous lift. Two lifts starting at the same point agree on the first segment because the relevant sheet map is injective, and the same argument at successive endpoints proves uniqueness.

Write $q:Y\to Y/\varphi$ for the quotient map. The identity $p\circ Q=q\circ {\mathrm{pr}}_Y$ on $Y\times \mathbb{Z}$, where $Q$ is the quotient defining $\widetilde{Y/\varphi }$, proves continuity of $p$ by the quotient property. Away from the two gluing regions, a small open set lifts to one copy in every level. Near a glued point choose $U⋐A$ and use the matched pairAfter the prescribed identifications, each such pair maps homeomorphically to $q(U\cup \varphi (U))$, and the pairs are disjoint for different $i$. Thus $p$ is a covering map.

Each copy of $Y$ is path-connected, and adjacent copies meet through the identified copies of the nonempty set $A$, so $\widetilde{Y/\varphi }$ is path-connected. Translationis a deck transformation. Its powers act freely and transitively on every fibre, so the covering is regular. The covering-space subgroup and deck-group quotient then gives

An algebraic number is an algebraic integer when it is a root of a monic polynomial in $\mathbb{Z}[X]$.

If $\alpha$ and $\beta$ are algebraic integers, the ring $\mathbb{Z}[\alpha ,\beta ]$ is a finitely generated $\mathbb{Z}$-module. Multiplication by $\alpha \beta$ preserves this module, so the algebraic integer module criterion shows that $\alpha \beta$ is an algebraic integer.

The ring of integers of a quadratic field givessince $2\equiv 2(\mathrm{mod}4)$. For a direct verification, if $x=a+b\sqrt{2}$ with $a,b\in \mathbb{Q}$ is integral, then its trace $2a$ and norm $a^2-2b^2$ are integers. Writing $2a=m\in \mathbb{Z}$ and reducing the norm condition in lowest terms shows first that $b$ can have denominator at most $2$; a denominator $2$ with odd numerator would require $m^2\equiv 2(\mathrm{mod}4)$, which is impossible. Thus $b\in \mathbb{Z}$, and then the norm condition forces $m$ even, so $a\in \mathbb{Z}$. Conversely, every $a+b\sqrt{2}$ with $a,b\in \mathbb{Z}$ is integral because $\sqrt{2}$ is integral and algebraic integers form a ring.

Now let $f$ have degree $d$ and roots ${\alpha }_1,\dots ,{\alpha }_d$. Sinceand $|{\alpha }_1|>1$, the integer $|a_d|$ cannot be $2$; hence $|a_d|=1$. Thus $f$ is monic up to sign and ${\alpha }_1$ is an algebraic integer.

Let $B$ be the product of all roots with modulus greater than one, counting multiplicity. Complex roots occur in conjugate pairs, so $B$ is real up to the signs of the real roots andAs a product of algebraic integers, $B$ is an algebraic integer; the equality forces $B=\pm 2$. Thereforeis the absolute value of the product of the remaining conjugates, is rational, and is an algebraic integer. It is consequently a nonzero rational integer. Every remaining factor has modulus at most one, so this integer is at most one. It must equal one, and henceThis is the mahler-measure norm argument at measure two.

For an odd composite $N$ and a unit $b$ modulo $N$, $N$ is a Fermat pseudoprime to base $b$ whenIt is an Euler pseudoprime to base $b$ whenwhere the right side is the Jacobi symbol.

By the Chinese remainder theorem,and $\phi (105)=2\cdot 4\cdot 6=48$. The condition $b^{104}=1$ is automatic modulo $3$ and $5$. Modulo $7$ it hassolutions. Thus there are $2\cdot 4\cdot 2=16$ Fermat bases, giving proportion

For the Euler condition, $b^{52}=1$ modulo $3$, so the Jacobi symbol must be $+1$. The power condition is automatic modulo $5$, while modulo $7$ it again restricts $b$ to the two solutions of $b^2=1$. For each of those two residues, exactly half of the $2\cdot 4$ choices modulo $3$ and $5$ have Jacobi symbol $+1$. Hence there are $2\cdot 4=8$ Euler bases and proportion

A complex representation of $G$ is a complex vector space $V$ together with a homomorphismIt is faithful when $\ker \rho =\{e\}$, equivalently when $\rho$ is injective. Every finite group has a faithful complex representation: in the regular representation, $G$ permutes the basis $\{e_h:h\in G\}$ byand a group element fixing every basis vector must be the identity.

If $g$ is conjugate to $g^k$, then $\rho (g)$ is similar to $\rho (g^k)=\rho (g{)}^k$. Thus the two matrices have the same spectrum, while an eigenvalue $\lambda$ of $\rho (g)$ gives the eigenvalue ${\lambda }^k$ of $\rho (g{)}^k$. Hence

Now take a $p$-cycle $g\in S_p$. It is conjugate to $g^k$ for every $1\le k<p$. Faithfulness makes $\rho (g)$ have order $p$, so its spectrum contains a nontrivial $p$th root of unity $\lambda$. The preceding implication puts all the $p-1$ distinct valuesin the spectrum. Therefore $\dim V\ge p-1$. This is the spectrum orbit bound for a faithful symmetric-group representation.

Finally let $G$ be the group of all permutations of $\mathbb{N}$. It has no faithful finite-dimensional complex representation. For every $n$, $G$ contains a subgroup $(C_2{)}^n$ generated by $n$ disjoint transpositions. In a $d$-dimensional complex representation, commuting involutions are simultaneously diagonalizable and hence map into the group of diagonal sign matrices, which has order $2^d$. A faithful restriction would require $2^n\le 2^d$ for every $n$, an impossibility.

The soundness theorem for propositional logic says that if $\Gamma ⊢t$, then every valuation satisfying every member of $\Gamma$ also satisfies $t$.

The proposed function need not respect the connectives. For example, if a primitive proposition $p$ is independent of $S$, then neither $p$ nor $\neg p$ is provable from $S$, so the definition gives

Order the consistent supersets of $S$ by inclusion. The union of a chain is consistent, since a finite proof of a contradiction would already use assumptions from one member of the chain. Zorn lemma therefore gives a maximal consistent extension $T$. It is deductively closed: if $T⊢t$, adjoining $t$ preserves consistency, so maximality forces $t\in T$. Moreover, for every $t$, exactly one of $t$ and $\neg t$ belongs to $T$. They cannot both belong by consistency; if neither belonged, the inconsistency of both proper extensions would give $T⊢\neg t$ and $T⊢\neg \neg t$, again a contradiction. The usual induction on formulae now shows thatdefines a valuation satisfying $T$, and hence $S$.

Now suppose every finite subset of $S$ has a model. By soundness every finite subset is consistent. Any proof of a contradiction from $S$ uses only finitely many assumptions, so $S$ itself is consistent. Applying the preceding maximal-consistent-extension construction gives a model of $S$. This proves the propositional compactness theorem.