Constructive decision of the dyadic cyclotomic norm equation

For any $t\in D$, both $t^{†}t=\mathrm{\mid }t{\mathrm{\mid }}^2$ and $(t^{†}t{)}^{\bullet }=\mathrm{\mid }{t}^{\bullet }{\mathrm{\mid }}^2$ are nonnegative. If either real embedding of $\xi$ is negative, reject. The signs are decidable by rational arithmetic and squaring the rational coefficients of $a+b\sqrt{2}$ after checking their signs. If either embedding is zero, injectivity of that embedding implies $\xi =0$, and the output $t=0$ suffices. Thus the remaining case has both embeddings strictly positive. Choose $r$ by clearing the powers of two in the two coefficients of $\xi$; multiplication by the positive rational square $4^r$ preserves their signs.

Put $\delta =1+\omega$ and $\lambda =1+\sqrt{2}$. Direct multiplication gives $N_O(\delta )=2$, ${\delta }^{†}={\omega }^{-1}\delta$, and ${\delta }^4=2i{\lambda }^2$. Norm two makes $\delta$ irreducible, hence prime by [F1]. As $i$ and $\lambda$ are units, the last identity shows that $2$ has just the prime associate class of $\delta$, with multiplicity four. Every element $w\in D$ has a unique-factorization valuation $v_{\pi }(w)\in \mathbb{Z}$ at each prime associate class of $O$, obtained from its integral numerator minus the valuations of its power-of-two denominator; only $v_{\delta }(w)$ can be negative. Conjugation acts on these valuations by permuting the prime classes, since it is a ring automorphism. In particular, for a fixed class, $v_{\pi }(w^{†}w)=2v_{\pi }(w)$.

If $w^{†}w=\eta \in O$ for $w\in D$, then $2v_{\delta }(w)=v_{\delta }(\eta )\ge 0$, so all valuations of $w$ are nonnegative and $w\in O$. To see the last implication without any assumption on units of $D$, write $w=a\mathrm{/}{2}^m$ with $a\in O$; its only possible denominator class is $\delta$, and the valuation inequality says ${\delta }^{4m}\mid a$, hence $2^m\mid a$. Consequently a solution to $t^{†}t=\xi$ gives an integral solution $w=2^{r}t$ to $w^{†}w=\eta$. Every conjugation-fixed prime exponent of $\eta$ must therefore be even. Conversely an integral solution for $\eta$ gives a solution for $\xi$ by division by $2^r$. This proves both necessity of the parity test and completeness of the denominator-clearing reduction.

Factor nonzero $\eta$ by the finite algorithm of [F1]. Determine conjugation orbits by conjugating each factor and testing associates; ${\eta }^{†}=\eta$ and uniqueness of prime factorization imply that both classes in every two-element orbit occur with the same exponent. This is also an effective prime splitting test: each factor is assigned either to a fixed class or to a pair of distinct conjugate classes; no unproved classification by rational primes modulo eight is required. If a fixed exponent is odd, reject as justified in step 2.1. Otherwise, for each fixed class of exponent $2h$ choose its displayed prime to power $h$, and for each pair of exponent $m$ choose one displayed prime to power $m$. Let $s$ be the product of these choices, with empty product one. Then $s^{†}s$ and $\eta$ have exactly the same prime-class exponents.

Their quotient $e=\eta \mathrm{/}(s^{†}s)$ is a unit of $O$ by unique factorization, and is real. Its inverse is real too, so it is a unit of $R=\mathbb{Z}[\sqrt{2}]$ by [F1]. Both embeddings of $s^{†}s$ are strictly positive, as are those of $\eta$, so both embeddings of $e$ are positive. The unit-square construction of [F1] therefore returns $v\in R$ with $v^2=e$. Because $v^{†}=v$, $(vs{)}^{†}(vs)=e{s}^{†}s=\eta$, and division by $2^r$ gives the claimed $t$. All loops terminate: denominator clearing is finite, factorization terminates, associate tests are exact, the factor list is finite, and square-root recovery terminates for a positive real unit. Together with the rejection cases and necessity this proves the entire decision and construction claim.