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Write $K=\mathbb{Q}(\alpha )$ with minimal polynomial of degree $n$. An embedding is determined by the image of $\alpha$, and each of the $n$ distinct complex roots gives one embedding, so there are exactly $n$.

For an integral basis ${\omega }_1,\dots ,{\omega }_n$, the field discriminant isOrder the embeddings with all real embeddings first and each nonreal embedding adjacent to its conjugate. Replacing each conjugate pair of rows by their real and imaginary parts extracts one factor $2i$ from the determinant. The remaining determinant is real and nonzero. Squaring showswhere $s$ is the number of conjugate pairs.

If $L/K$ is Galois, form the norm polynomialThe Galois group permutes the factors, so $F\in K[X]$. Since the identity factor is $f$, taking $g=F/f\in L[X]$ proves the claim.

If $L/K$ is merely finite separable, place it in a finite Galois closure $M/K$ and use the same product over $\mathrm{Gal}(M/K)$. It gives $F\in K[X]$ divisible by $f$ in $M[X]$. Polynomial division of $F$ by $f$ uses only coefficients in $L$, and the remainder is zero, so the quotient $g=F/f$ actually lies in $L[X]$. It is nonzero and satisfies $fg=F\in K[X]$.

For roots $r_1,r_2,r_3$, the discriminant isand for $f=X^3+pX+q$,If $f$ is irreducible, its Galois group is transitive in $S_3$: it is $A_3\cong C_3$ when $\Delta$ is a square in $K$, and $S_3$ otherwise. If $f$ has one root in $K$ and an irreducible quadratic factor, the group is $C_2$; if it splits, it is trivial. A repeated-root depressed cubic already has all roots in $K$, so adds no case.

For $X^3+3X+1$, the rational-root test proves irreducibility andis not a rational square. Thus $G\cong S_3$. By the fundamental theorem of Galois theory, the complete subfield list iswhere the three cubic fields fix the three order-two subgroups and the quadratic field fixes $A_3$.

The edge chromatic number ${\chi }^{\prime }(G)$ is the least number of colors in a proper coloring of edges, where incident edges receive distinct colors. Hall marriage theorem says a bipartite graph has a matching saturating one side exactly when $|N(S)|\ge |S|$ for every subset $S$ of that side.

In a 4-regular bipartite graph, the $4|S|$ edges leaving $S$ all enter $N(S)$, whose vertices can receive at most $4|N(S)|$ such edges. Hence Hall's condition holds and there is a perfect matching. Remove it and repeat in the resulting 3-, 2-, and 1-regular bipartite graphs. The four perfect matchings give a four-edge-coloring, while every vertex requires four distinct colors. Therefore

Euler's formula for a connected plane graph is $|V|-|E|+|F|=2$. If the graph is triangle-free, every face has boundary length at least four, soEuler's formula then gives $|E|\le 2|V|-4$, so the average degree is less than four. There is a vertex of degree at most three. Delete it, color the remaining graph inductively with four colors, and restore it using a color absent from its at most three neighbors. Thus

A graph is $k$-connected when it has more than $k$ vertices and remains connected after deletion of fewer than $k$ vertices. In a noncomplete 3-connected graph, choose a vertex $v$ with two nonadjacent neighbors $a,b$. Since $G-\{a,b\}$ is connected, take a spanning tree rooted at $v$ and order its vertices so every vertex other than $v$ has a later tree neighbor. Color $a$ and $b$ first with the same color, then greedily color the other vertices in reverse tree order, leaving $v$ last. Every nonfinal vertex has one uncolored neighbor and therefore sees at most $\Delta -1$ colors. At $v$, the two neighbors $a,b$ share a color, so again at most $\Delta -1$ colors occur. HenceThis is the relevant 3-connected case of Brooks' theorem.

A proper $k$-coloring assigns one of $k$ colors to each vertex so adjacent vertices receive different colors. The chromatic number $\chi (G)$ is the least such $k$. Order the vertices arbitrarily and color greedily. At most $\Delta (G)$ colors are forbidden by previously colored neighbors, soEquality occurs for every possible maximum degree: use $K_1$ for $\Delta =0$, $K_2$ for $\Delta =1$, an odd cycle for $\Delta =2$, and $K_{\Delta +1}$ for every $\Delta \ge 3$.

Let $f:X\to A\subseteq Y$ be an order isomorphism onto a down-set, and let $g:Y\to U=X\setminus B$ be an order isomorphism onto the complement of a down-set $B$. Define increasing down-setsThe order assumptions ensure inductively that each $D_n$ is a down-set. DefineThe usual Cantor-Schröder-Bernstein theorem orbit argument shows that these two pieces partition both domain and codomain bijectively: points generated from $B$ move forward through $f$, while every remaining point lies in $U$ and moves backward through $g$. Each branch preserves order. If $x\in D$ and $x^{\prime }<x$ then $x^{\prime }\in D$; hence the only mixed case has $x\notin D<x^{\prime }\in D$, and the initial/final-segment hypotheses put $h(x)<h(x^{\prime })$. Thus $h$ is an order isomorphism and

The Knaster-Tarski theorem says that the fixed points of a monotone self-map of a complete lattice form a complete lattice. In particular,For the first formula, let $a$ be the displayed meet. Monotonicity gives $f(a)\le f(x)\le x$ for every prefixed point $x$, hence $f(a)\le a$. Then $f(f(a))\le f(a)$, so $f(a)$ is itself prefixed; minimality of $a$ gives $a\le f(a)$. Thus $f(a)=a$. The dual argument gives the greatest fixed point. Applying the same construction above the join of any family of fixed points, and dually below its meet, supplies joins and meets within the fixed-point set.

A down-set contains every element below any of its members. Arbitrary unions and intersections of down-sets are down-sets. Thus the down-sets of $X$, ordered by inclusion, have joins given by unions and meets by intersections, so they form a complete lattice.

For the requested counterexample, takewith their usual orders. The set $X$ is a down-set in $Y$, while $Y$ is order-isomorphic to the down-set $[0,1/2]$ of $X$. They are not isomorphic because $Y$ has a greatest element and $X$ does not.

Neglecting the pressure and self-gravity bracket leavesDuring radiation domination, $a(t)\propto t^{1/2}$ and $H=1/(2t)$, soConsequentlyMatter perturbations therefore grow only logarithmically while radiation dominates.

Take the divergence of the linear Euler equation, use $\delta P=c_s^{2}m\bar{n}\delta$, and differentiate the continuity equation. Eliminating $\nabla \cdot \delta \mathbf{v}$ givesThe perturbed Poisson equation isThusHere $k_J$ is the comoving Jeans wavenumber.

The homogeneous Euler equation removes the background force. Products of perturbations are second order, while differentiating the peculiar velocity inside the expanding flow produces the Hubble drag term. The first-order equation is therefore

The homogeneous part of the continuity equation gives $\dot{\bar{n}}+3H\bar{n}=0$. At first order,For the density contrast $\delta =\delta n/\bar{n}$, the background equation cancels the expansion terms, leaving

For real $s>0$, collapse the Hankel contour onto the negative axis and use the jump in the argument of $t^{s-1}$. The reflection formula for the gamma function yieldsso $K(z,s,t)=z{t}^{s-1}/[\Gamma (s)(e^t-z)]$. At positive integer $s$, the original Hankel expression is interpreted by analytic continuation: the pole of $\Gamma (1-s)$ cancels the corresponding zero of the contour integral.

When $z$ crosses the slit at $x>1$, the pole of the real-integral kernel at $t=\log x$ crosses the integration path. The two boundary values differ by $2\pi i$ times its residue. Since $d(e^t-z)/dt=x$ there, that residue is $(\log x{)}^{s-1}/\Gamma (s)$, giving the jump

When $s$ is a nonpositive integer, $t^{s-1}$ is meromorphic and the Hankel contour may be shrunk to a small circle about zero. The integral is then computed by the residue at zero. This givesThese also follow by repeatedly applying $zd/dz$ to the geometric series.

For $|z|<1$ on the negative-real portions of the Hankel contour, expandLocal uniform convergence permits termwise integration. The reciprocal Hankel formula givesso $I(z,s)={\sum }_{n\ge 1}{z}^n/n^s={\mathrm{Li}}_s(z)$ initially on the disc.

For $z\notin [1,\infty )$, choose the contour around the negative axis so that it passes between zero and every pole satisfying $e^{-t}=z$. On compact subsets of the slit $z$-plane it can be chosen uniformly, and differentiation under the integral proves holomorphy. Deforming without crossing a pole gives a single-valued analytic continuation. Near the cut, the pole $t=-\log z$ lies on opposite sides of the two admissible contours; this is why the contour must pass it consistently.

For fixed $s$, the factor $n^{-s}$ grows at most polynomially in $n$, whereas $|z{|}^n$ decays exponentially when $|z|<1$. The defining series for the polylogarithm therefore converges absolutely and locally uniformly for every complex $s$ and $|z|<1$. It is holomorphic in $z$ on the unit disc and entire in $s$ there.

A standard moment estimator is the Pearson dispersion estimatorFor Bernoulli data, $V(\mu )=\mu (1-\mu )$. The deviance divided by its residual degrees of freedom is a common alternative.

The residual deviance per residual degree of freedom isreasonably close to one. The first and third quartiles of the deviance residuals, $-0.591$ and $0.621$, are also fairly close to the standard-normal quartiles $\pm 0.674$. There is mild evidence of underdispersion, but nothing in this summary makes the binomial value one plainly unreasonable.

The code fits a Bernoulli logistic-regression model. Conditionally on the covariates, the responses are independent withwhere the design includes the displayed numerical predictors and indicator columns for factor levels. It maximizesAkaike information criterion iswhere $k$ is the number of fitted parameters. Backward stepwise selection starts from the full model, tentatively removes each eligible term, chooses the removal producing the lowest AIC, and repeats while AIC decreases. It balances fit against model size rather than testing every coefficient at a fixed significance threshold.