Complexity of factoring in number fields
The following answer was originally posted as a comment on Gil's blog

(1) Let $K=\mathbb{Q}(\alpha)$ be a number field, where we assume $\alpha$ has a monic minimal polynomial $f\in\mathbb{Z}[x]$.  One can then represent elements of the ring of integers $\mathcal{O}_K$ as polynomials in $\alpha$ or in terms of an integral basis -- the two are equivalent.

Now fixing $K$ as in (1) there's a polynomial-time reduction from the problem over $K$ to the problem in $\mathbb{Q}$.  To verify that the computations (e.g. intersecting an ideal with $\mathbb{Z}$ or factoring a polynomial mod $p$) can be done in polynomial time see Cohen's book referred to in the previous answer.

As a precomputation for each rational prime $p$ dividing the discriminant of $\alpha$ (that is the discriminant of $f$) find all primes of $\mathcal{O}_K$ lying above $p$.

(2) For primality testing, given an ideal $\mathfrak{a}\triangleleft\mathcal{O}_K$ let $p\in\mathbb{Z}$ be such that $\mathfrak{a}\cap\mathbb{Z} = p\mathbb{Z}$ (this can be computed in polynomial time and the number of bits of $p$ is polynomial in the input).  Check in polynomial time whether $p$ is prime. If not then $\mathfrak{a}$ is not prime.  If yes then find the primes of $\mathcal{O}_K$ lying above $p$ either from the precomputation or by factoring $f$ mod $p$.  In any case if $\mathfrak{a}$ is prime it must be one of those primes.

(3a),(6a) For factoring into primes, given an ideal $\mathfrak{a}\triangleleft\mathcal{O}_K$ find its norm $y = N^K_\mathbb{Q}(\mathfrak{a}) = [\mathcal{O}_K:\mathfrak{a}]$. Again this can be found in polynomial time and consequently is not too large. Factor $y$ in $\mathbb{Z}$ (either classically or using Shor's algorithm, depending on the reduction you want).  This gives a list of rational primes dividing $y$, and hence as in 2 we can find the list of primes of $\mathcal{O}_K$ dividing $y$.  Since $\mathfrak{a} | y\mathcal{O}_K$ this gives the list of primes dividing $\mathfrak{a}$.  Finally it is easy to determine the exponent to which a prime divides a given ideal.

(3b),(6b) But Gil wants factorization into irreducibles, not into primes. It turns out that given the prime factorization of $x\mathcal{O}_K$ it is possible to efficiently construct one factorization of $x$ into irreducible elements of $\mathcal{O}_K$.  For this let $h_K$ be the class number, and note that it is possible to efficiently compute the ideal class of a given ideal. Now to find an irreducible divisor of $x$ select $h_K$ prime ideals (possibly with repetition) from the factorization of $x$.  By the pigeon-hole principle some subset of those multiplies to the identity in the class group; find a minimal such subset.  Its product is then a principal ideal generated by an irreducible element.  Divide $x$ by this element, remove the relevant ideals from the factorization and repeat.  If the factorization has less than $h_K$ elements then just take a minimal subset of all the factors.

(4) I think it's possible to count the factorizations into irreducibles, but this is a bit of extra combinatorics -- please give me time to work it out.  One the other hand, determining all of them is not interesting in the context of sub-exponential factorization algorithms since there are in general exponentially many such factorizations.

(5) I have no idea.