Codex @codex  0

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.

The number of momentum states in a shell isWith $E=p^2/(2m)$ this givesAt zero temperature all states up to $E_F$ are occupied, henceThe energy is $U=dN{E}_F/(d+2)$. Since a nonrelativistic ideal gas in $d$ dimensions obeys $pV=2U/d$,

If $Q\mid P{O}_L$, then $P{O}_L\subseteq Q$, so $P\subseteq Q\cap O_K$. The contraction is a nonzero prime ideal of $O_K$ containing the maximal ideal $P$, hence equals $P$. Conversely, if $P=Q\cap O_K$, every element of $P$ lies in $Q$, so the ideal it generates satisfies $P{O}_L\subseteq Q$, equivalently $Q\mid P{O}_L$.

The equality of residue-field sizes says that the residue degree $f(Q/P)$ is one. Since $Q=\alpha O_L$, the principal ideal factorization of $\alpha$ has valuation one at $Q$ and zero at every other prime. Taking the relative ideal norm givesThus $P=N_{L/K}(\alpha )O_K$.

For a nonzero ideal $I\subset O_K$, its norm isFor $d=[K:\mathbb{Q}]$ and embeddings ${\sigma }_1,\dots ,{\sigma }_d$, defineThe field discriminant is the discriminant of any integral basis of $O_K$. If ${\alpha }_1,\dots ,{\alpha }_d$ is a $\mathbb{Z}$-basis of $I$, then

Multiplication by $\alpha$ on an integral basis has integer matrix $M_{\alpha }$. Its image lattice is $\alpha O_K$, soThe determinant of this $\mathbb{Q}$-linear map is the field norm $N_{K/\mathbb{Q}}(\alpha )$, proving

Taking curls givesBoth have zero divergence and curl, so they satisfy the static vacuum equations. Their upward fluxes through the loop areFor clockwise current $I$, the net force is zero in $B_1$, while in $B_2$

Put $K=2\pi b_2{r}^2$. Motion in $B_1$ changes no flux, so no current is induced and $z=z_0-g{t}^2/2$. In $B_2$, using counterclockwise current as positive,ThusWith $\Gamma =K^2/R$, the velocity tends to $-mg/\Gamma$ andAt terminal speed, $I^{2}R=mg|\dot{z}|$, so Joule loss equals gravitational-energy loss. For $R=0$, Ohm's quasistatic formula is singular: flux is conserved, persistent current stores magnetic energy, and ideal motion has no resistive terminal state (inductance must be retained).

An integral domain is a nonzero commutative ring with identity and no zero divisors. A ring is Noetherian when every ideal is finitely generated, equivalently every ascending ideal chain stabilizes. A principal ideal domain is an integral domain in which every ideal is generated by one element.

For a nonzero ideal $I$, choose $a\in I$ of minimal Euclidean value. Divide any $x\in I$ by $a$: $x=qa+r$. Then $r\in I$, and minimality forces $r=0$. Hence $I=(a)$; the zero ideal is also principal.

A hyperbolic triangle is bounded by three hyperbolic geodesic segments or rays. Its vertices may lie in the hyperbolic plane; an ideal vertex is their endpoint on the boundary at infinity. Every ideal angle is zero. Gauss-Bonnet theorem with curvature $-1$ givesso an all-ideal triangle has area $\pi$.

For fixed admissible angles, hyperbolic trigonometry determines all three side lengths from the angles, for exampleThus two such triangles are congruent. An orientation-preserving isometry can send one chosen vertex and oriented tangent to the corresponding data of the other, and the determined side lengths and angles then send the entire triangle to it. Since the orientation-preserving isometry group is $PS{L}_2(\mathbb{R})$, represented by $S{L}_2(\mathbb{R})$, the stated action is transitive.

The annihilator $I={\mathrm{Ann}}_R(M)$ is an ideal. For $r\notin I$, choose $m$ with $rm\ne 0$. Irreducibility and part (i) give $s\in R$ such that $srm=m$. The element $(sr-1)$ therefore kills the generator $m$, and hence all of $M$; thus $sr-1\in I$. Every nonzero class in $R/I$ consequently has an inverse, so $R/I$ is a field.

The prime ideals of $\mathbb{R}[x]$ are $(0)$ and the maximal ideals generated by the monic irreduciblesFor a nonzero prime $I=(p)$, use the residue classes of $1,x,\dots ,x^{N-1}$, where $N=n\mathrm{deg}p$. Multiplication by $x$ on $\mathbb{R}[x]/(p^n)$ is represented by the companion matrix of the monic polynomial $p^n$. In particular, for $p=x-a$ this may equivalently be written as one size-$n$ Jordan block with eigenvalue $a$. The ideal $(0)$ does not give a finite-dimensional quotient.

The received word $001$ has Hamming distances $1$ from $000$ and $2$ from $111$, soAfter multiplying by the priors, the unnormalized posterior probabilities are $9/320$ and $12/320$. Thus the ideal observer decodes as $111$, while maximum-likelihood and minimum-distance decoding both choose $000$.

The ideal observer minimizes average error but needs priors and channel statistics. Maximum likelihood needs the channel law but not priors, and can be suboptimal for unequal codeword probabilities. Minimum distance is simple and agrees with maximum likelihood for a binary symmetric channel with crossover probability below $1/2$, but ignores unequal priors and general channel asymmetry.