If $u=0$, any nonzero $w$, for instance the first Cartesian basis vector $e_1$, gives $H(w)u=0$ and $\alpha =0$. Suppose $u\ne 0$. Choosewhere ${\mathrm{sgn}}_{+}(r)=1$ for $r\ge 0$ and $-1$ otherwise. This choice ensures $w\ne 0$, including when $u$ is already on the first coordinate axis. Since ${\alpha }^2=u^{T}u$,Therefore the Householder reflection satisfiesThe orthogonality of $H(w)$, used here to check length preservation, gives $|\alpha |=\Vert H(w)u{\Vert }_2=\Vert u{\Vert }_2$. The chosen sign also avoids catastrophic cancellation in $w_1=u_1-\alpha$. This is a Householder reflection sending a vector to a coordinate axis with no zero-normal exception.

A Givens rotation acts as the identity matrix outside two selected coordinates $i,j$, and has the following block on those coordinates:The convention with the opposite signs of the sines is equivalent after replacing $\theta$ by $-\theta$. A Householder reflection with nonzero normal vector $w\in {\mathbb{R}}^n$ isIt reverses the component parallel to $w$ and fixes the perpendicular hyperplane. Both are orthogonal matrices, so their transposes are their inverses and they preserve the Euclidean norm.

For a splitting field $L$ of $f$ over $K$, define $\mathrm{Gal}(f)={\mathrm{Aut}}_K(L)$. Each automorphism permutes the distinct roots, since it fixes the coefficients. An automorphism fixing every root fixes the field they generate, so this action is faithful. If there are $r\le n$ distinct roots, it gives an embedding into $S_r$, and hence into $S_n$ by fixing $n-r$ additional symbols.

Equality $(\alpha )=(\beta )$ gives $\alpha =u\beta$ and $\beta =v\alpha$ for some $u,v\in {\mathcal{O}}_K$. Substitution and cancellation of the nonzero $\alpha$ yield $uv=1$. Thus $u$ is a unit and $\alpha =u\beta$.

For the ideal-power assertion, suppose first that all ideals are nonzero. A prime ideal divides at most one $Q_i$ because the $Q_i$ are pairwise coprime. Taking its valuation in the given equality givesSince $\gcd (n_i,k)=1$, every exponent $v_P(Q_i)$ is divisible by $k$. Dividing all its prime-factorization exponents by $k$ constructs an integral ideal $I_i$ with $Q_i=I_i^k$. If a $Q_i$ is zero, pairwise coprimality forces all other $Q_j$ to be the unit ideal; taking $I_i=0$ and the other $I_j={\mathcal{O}}_K$ covers that degenerate case too.

For the first Diophantine equation, $x$ cannot be odd: then $x^2+17\equiv 2(\mathrm{mod}8)$, while a cube modulo eight is $0$ or odd. Hence $x$ is even and $y$ is odd. Also $17\nmid x$, since otherwise $v_{17}(x^2+17)=1$, impossible for a cube.

In $\mathcal{O}=\mathbb{Z}[s]$, $s=\sqrt{-17}$, factorThe two principal ideals are coprime. A common prime ideal would divide their sum $2x$ and difference $2s$, and hence lie above $2$ or $17$; odd norm excludes $2$, while $17\nmid x$ excludes $17$. Their coprime product is the cube $(y{)}^3$, so the ideal-power assertion makes $(x+s)=I^3$. The class group has order $4$, while $[I{]}^3=1$, so $[I]=1$. Write $I=(a+bs)$ for integers $a,b$.

The only units are $\pm 1$, since their norm equation is $a^2+17b^2=1$. Equality of principal ideals therefore gives $x+s=\pm (a+bs{)}^3$. Absorb the sign into $a+bs$. Comparing the coefficient of $s$ yieldsThus $b=1$ or $-1$. The first gives $a^2=6$ and the second gives $3a^2=16$, both impossible for an integer $a$. Consequently the first equation has no integral solutions.

For the product equation, odd $x$ gives $(x^2+17)(9x^2+17)\equiv 4(\mathrm{mod}8)$, again not a cube. If $17\mid x$, each factor has $17$-adic valuation one, so the product has valuation two, also impossible. Thus any hypothetical solution has $x$ even and $17\nmid x$. The two positive factors are odd, and their greatest common divisor dividesIt must therefore be one. A product of coprime positive integers is a cube only if each factor is a cube. This would make $x^2+17$ a cube, contradicting the result already proved. Hence the product equation also has no integral solutions.

The product of ideals $AB$ consists of finite sums ${\sum }_j{a}_j{b}_j$ with $a_j\in A$, $b_j\in B$. Ideal divisibility $P\mid A$ means $A=PC$ for an integral ideal $C$; in this ring it is equivalent to $A\subseteq P$. The ideal norm is $N(A)=|{\mathcal{O}}_K/A|$, equivalently its additive lattice index.

If the prime ideal $P$ divides $AB$ but does not divide $A$, choose $a\in A\setminus P$. For every $b\in B$, $ab\in AB\subseteq P$, so primality gives $b\in P$. Thus $B\subseteq P$, proving $P\mid B$.

Every nonzero proper ideal $A$ is contained in a maximal ideal $P$, which is prime. Then $P\mid A$, so $A=PB$. Norm multiplicativity gives $N(B)=N(A)/N(P)<N(A)$ because $N(P)\ge 2$. Induction on the positive integer $N(A)$ gives a prime-ideal factorization, ending at norm one. For uniqueness, a prime in one factorization divides some prime in the other. Every nonzero prime ideal is maximal, since its quotient is a finite integral domain and hence a field. The two primes therefore coincide. Cancel that common factor using the permitted cancellation law, and continue. This proves uniqueness up to order.

The finite additive group ${\mathcal{O}}_K/A$ has order $N(A)$, so that integer annihilates the class of $1$. ThereforeIn particular an ideal of norm $m$ contains $(m)$. Such ideals correspond to certain additive subgroups of the finite group ${\mathcal{O}}_K/(m)$, so there are finitely many. Finiteness of these quotients follows from an integral basis; a nonzero ideal contains a nonzero principal ideal, whose index is the absolute field norm of its generator.

For the computation put $s=\sqrt{-6}$. The Kummer-Dedekind theorem says that when $p$ does not divide the power-basis index, a factorization $\bar{f}={\prod }_i{\bar{g}}_i^{e_i}$ gives $(p)={\prod }_i(p,g_i(s){)}^{e_i}$, with each factor having norm $p^{\mathrm{deg}{g}_i}$. The given full power basis has index one. Since $X^2+6\equiv X^2(\mathrm{mod}2)$ and $X^2+6\equiv (X-1)(X+1)(\mathrm{mod}7)$,Here $N(P)=2$ and $N(Q)=N(Q^{\prime })=7$. A prime ideal dividing an ideal of norm $98=2\cdot 7^2$ lies above $2$ or $7$, and these are all the prime ideals above them. Unique factorization and norm multiplicativity therefore leave exactlyThey are distinct by their prime-ideal valuations, and each has norm $98$.