Gauss lemma for polynomials states that the product of two primitive polynomials over a unique factorization domain is primitive. In particular, a primitive polynomial in $\mathbb{Z}[X]$ is reducible over $\mathbb{Q}$ exactly when it is reducible over $\mathbb{Z}$.

Eisenstein criterion says that a primitive polynomialis irreducible over $\mathbb{Q}$ if there is a prime $p$ such that $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$. To prove it, suppose $F=GH$ with both factors of positive degree. After removing contents, Gauss's lemma lets us take $G,H\in \mathbb{Z}[X]$ primitive. Reduction modulo $p$ givesso both reduced factors are monomials of positive degree. Their constant terms are therefore divisible by $p$, making $p^2$ divide $a_0=G(0)H(0)$, a contradiction.

An algebraic integer is a complex number satisfying a monic polynomial in $\mathbb{Z}[X]$. Let $\alpha$ be one and let $m_{\alpha }\in \mathbb{Q}[X]$ be its monic minimal polynomial. The usual proof using polynomial content shows that $m_{\alpha }\in \mathbb{Z}[X]$. The setis the kernel of the polynomial-evaluation ring homomorphism, hence an ideal. Every $f\in I_{\alpha }$ is divisible by $m_{\alpha }$ in $\mathbb{Q}[X]$. Because $m_{\alpha }$ is monic, polynomial division of $f$ by $m_{\alpha }$ takes place in $\mathbb{Z}[X]$, and the remainder must vanish by minimality. Thereforewhere $m_{\alpha }$ is monic and irreducible.

Fortranslation givesThis is Eisenstein at $p=3$, so $f(X+1)$ and therefore $f(X)$ are irreducible over $\mathbb{Q}$. HenceIrreducibility and Gauss's lemma also show that $(f)$ is a prime ideal of $\mathbb{Z}[X]$, soIt is not a field. The class of $2$ is nonzero but not a unit: reducing further modulo $2$ gives a nonzero quotientin which the image of $2$ is zero, whereas a unit must remain a unit under every unital quotient map. Thus

SetConditioning on $X_n$ gives the affine recurrenceIts fixed point is the first component of the stationary distribution,Thus $p_n-{\pi }_1=(1-\alpha -\beta {)}^n(p_0-{\pi }_1)$, and $p_0=1$ gives

Starting instead from state two gives the same recurrence with initial value zero. Therefore

Let the upper plate in flow A move at speed $U$, with the lower plate fixed. The Couette flow profile isso its mid-plane speed is $U/2$ and its total viscous dissipation per unit plate area is

Write $G=-dp/dx>0$ for the constant pressure-gradient magnitude in flow B. The plane Poiseuille flow profile isEquality of the mid-plane speeds givesSince $u_B^{\prime }(y)=G(h-2y)/(2\mu )$,Hence

Let $r$ be distance from the common centre. By Gauss's law, the electric field is radial and equalsTaking the electric potential to vanish at infinity and requiring continuity across each shell gives

The electrostatic energy is the integral of the electric-field energy density:Only the two nonzero-field regions contribute, soTherefore

Use the angular-frequency Fourier transformThenwith the continuous value $\hat{f}(0)=2A$.

The convolution $(f\ast f)(x)$ is $A^2$ times the length of the overlap of the intervals $[-1,1]$ and $[x-1,x+1]$. Thus

The convolution theorem states, for this convention, thatSince $g=(B/A^2)(f\ast f)$, it follows thatagain interpreted continuously at $k=0$, where $\hat{g}(0)=4B$.

Because the constraint determinesthe constrained function isIts stationary points satisfyWhen $yz=0$, these equations giveWhen $yz\ne 0$, subtracting the two bracketed equations gives $z^2=a^2{y}^2$, and substitution then giveswith the two signs chosen independently. These are exactly the constrained stationary points; equivalently they follow from the Lagrange multiplier equations.

Under the additional restriction $x\ge 0$, the constraint gives the compact elliptical disc $a^2{y}^2+z^2\le b^2$. On its boundary $x=0$, the objective is zero. At the four nonzero interior stationary points,with the sign determined by $yz$. Therefore

Write $D(f)=\mathbb{C}\setminus f^{-1}(0)$. The zero and constant-one polynomials give $D(0)=\varnothing$ and $D(1)=\mathbb{C}$. Also,so $\tau$ is closed under finite intersections. Every nonzero one-variable complex polynomial has only finitely many roots, and every finite subset $\{a_1,\dots ,a_m\}$ is the zero set of ${\prod }_j(z-a_j)$. Consequently $\tau$ is exactly the cofinite topology: its open sets are the empty set and the complements of finite sets. An arbitrary union of such sets is again empty or has finite complement, so $\tau$ is a topology.

The product topology on $X\times Y$ is the topology with basisThe proposed complement need not be open. TakeThen $g^{-1}(0)$ is the diagonal. If its complement were open, a point such as $(0,1)$ would have a basic neighbourhood $U\times V$ contained in that complement. But $U$ and $V$ are nonempty cofinite subsets of $\mathbb{C}$, so $U\cap V\ne \varnothing$. For $t\in U\cap V$, the point $(t,t)$ belongs both to $U\times V$ and to the diagonal, a contradiction. Hence

Fix $\alpha \in \Omega$. If $\beta =g\alpha$, then normality of $H$ givesThus the transitive group action of $G$ permutes the orbits of a group action of $H$, so all $H$-orbits have the same cardinality $d$. They partition the prime-sized set $\Omega$, hence $d$ divides $|\Omega |$. The only possibilities are $d=1$ and $d=|\Omega |$. The first would make every element of $H$ fix every point, contrary to the assumption that $H$ acts nontrivially. Therefore $d=|\Omega |$ and

Now suppose $H\cap G_{\alpha }=\{1\}$. Define the orbit mapIt is surjective by transitivity of $H$. If $\theta (h_1)=\theta (h_2)$, then $h_2^{-1}{h}_1\in H\cap G_{\alpha }$, so $h_1=h_2$; hence $\theta$ is bijective. For $k\in G_{\alpha }$,Thus $\theta$ is an equivariant map: conjugation by $G_{\alpha }$ on $H$ corresponds exactly to the given action of $G_{\alpha }$ on $\Omega$.

The Lagrangian sufficiency theorem for a maximization problem says the following. Suppose the constraints are $g_i(q)\ge 0$ and $h_j(q)=0$, and defineIf a feasible $q^{\ast }$ satisfies complementary slackness ${\lambda }_i{g}_i(q^{\ast })=0$ and globally maximizes $L(\cdot ,\lambda ,\mu )$, then $q^{\ast }$ globally maximizes $f$. Indeed, for every feasible $q$,This proves the theorem. In a concave maximization problem, stationarity and the boundary optimality conditions ensure the required global maximum of the Lagrangian.

For the problem at hand, retain $x,z\ge 0$ as the domain and attach a multiplier $\lambda$ to the equality:For $\lambda \ge 1$, this is concave in $(x,y,z)$ on the domain. Its separate maximizers satisfywhile the $x$ term is $(1-\lambda )x$. Thus $x$ may be positive only when $\lambda =1$; when $\lambda >1$, its maximizing value is $x=0$.

Iftake $\lambda =1$. Then $y=1$, $z=a^2-1$, and feasibility fixesThe Lagrangian sufficiency theorem gives the maximum

Ifthen $\lambda >1$ and $x=0$. The equality constraint becomessoConsequentlyThe assumptions $a\ge 1$ and $b\ge 1/2$ ensure $z\ge 0$. At the stationary point $\sqrt{1+z}=ay$, so the maximum value isThe two formulas agree at $b=a^2-1/2$.