Q($\sqrt[3]{2}$) - Unique Factorisation Domain? I am considering the set of "integers" of the from
$$ a+b\sqrt[3]{2} + c\sqrt[3]{4} $$
where $a,b,c$ are integers. It is easy to show this field is closed under addition and multiplication. I then wish to determine whether Unique Factorisation holds for these integers. Motivated by other examples such as the Gaussian Integers and the Eisenstein Integers, I attempt to define a norm for these integers such that $ N(ab) = N(a)*N(b) $ and the norm always an integer, to then show the field is Euclidean and hence a Unique Factorisation Domain. However, here I get stuck. Unlike the Gaussian and Eisenstein integers, I cannot see how to define the Norm for this field.
Thanks in advance.
 A: make this CW; 
the result mentioned by Robert Soupe is page 327, Example 12.6.9,  Introductory Algebraic Number Theory by Alaca and Williams, http://www.amazon.com/Introductory-Algebraic-Number-Theory-Saban/dp/0521540119
See primes represented integrally by a homogeneous cubic form
The part that was done entirely was class number one, (rational) primes $q \equiv 2 \pmod 3,$ also $p = u^2 + 27 v^2.$
My belief, with calculation done for a different norm form, is that when $q = 4 u^2 +2uv+7v^2,$ with integers $u,v$ (not necessarily positive), whenever we have
$$ a^3 + 2 b^3 + 4 c^3 - 6abc \equiv 0 \pmod q,  $$
THEN
$$ a,b,c \equiv 0 \pmod q.  $$
The first few such primes are
$$      7,     13,     19,     37,     61,     67,     73,     79,     97,    103,
    139,    151,    163,    181,    193,    199, $$
NOTE: This has been confirmed by Noam Elkies. Also note that these are precisely the primes for which there is no cube root of $2.$
I actually have the IANT book somewhere, I will check. Meanwhile, a complete proof for everything for a different norm form is at  https://mathoverflow.net/questions/127160/numbers-integrally-represented-by-a-ternary-cubic-form
