Problem with units in number field Edit:There were several major mistakes by my side this post, most of which have been accounted for.Now, after editing these out, the post seems to have no purpose at all.Nevertheless, it feels wrong to delete it, so I am going to leave it as it is.
Consider the splitting field of $$x^3-2$$ which is $$K=Q(\alpha,\omega),$$ where alpha is the real cube root of 2 and omega is a primitive third root of unity.One can check that in fact $$K=Q(\alpha+\omega).$$ The galois group of K over Q is isomorphic to $$Z_2\times Z_3$$ and is abelian, so , by the Kronecker-Weber theorem, lies in a cyclotomic field, the smallest of which is dependent on the conductor of K.EDITThis in fact is wrong, the galois group of this extension is nonabelian, so K doesn't lie in a cyclotomic field.
I was trying to solve $$a^3+2b^3+4c^3=1$$ in integers, which is the norm of an element of the form $$a+b\alpha+c\alpha^2.$$EDIT The norm is actually $$a^3+2b^3+4c^3-6abc$$Let $$O_K$$ be the ring of integers, so $$Z(\alpha,\omega)$$ is contained in it.By Dirichlet's unit theorem, the ring of integers is generated by 2 elements.EDIT the fundamental units can be found here:http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/unittheorem.pdf
The diophantine equation seems to have a lot of solutions:(1,0,0),(5,-4,1),(-1,1,0) etc.So to solve this, we have to see when an element of the previous form is a product of powers of the two units.But the fundamental units look terrifying, so maybe this won't be a very fruitful process.
Also, if someone can tell me what the actual ring of integers is, that would be really helpful.
 A: The Galois group of your extension is not $Z_2\times Z_3$. This group is cyclic if order $6$, but as the Galois group of a cubic polynomial, it must be a subgroup of $S_3$, which it isn't. 
In fact, the Galois group is $S_3$, which is not abelian. This is immediate from the fact that $G$ must be a subgroup of $S_3$ of order $6$.   To see it explicitly, observe that it is generated by the automorphisms 
$$\sigma :\sqrt[3]2\mapsto \omega \sqrt[3]2\\\tau:\omega\mapsto \overline \omega=\omega^2$$
and routine checking shows that this group is indeed isomorphic to $S_3$. 
Moreover, Dirichlets unit theorem tells you that the free part of the group of units $\mathcal O_K^*$ has one generator as a multiplicative group. This does not tell you that $\mathcal O_K$ has one generator as a polynomial ring over $\mathbb Z$. 
A: Hardly an answer, but an easily-seen unit in $\Bbb Q(\sqrt[3]2\,)$ is $\alpha-1$ in your notation, and it’s hard to imagine another unit closer to the identity (in the unique real embedding of that field), though I’m not skilled enough to say that it’s a fundamental unit.
For the full field $K$, which has three inequivalent complex embeddings, two obviously independent units are $\alpha-1$ and $\omega\alpha-1$. Again, I’m not going to be so rash as to say that I know that they form a basis. The truth in these matters can be looked up easily, though.
