Some questions about $\mathbb{Z}[\zeta_3]$ I am giving a talk on Euler's proof that $X^3+Y^3=Z^3$ has no solutions in positive integers.  Some facts that I believe to be true are the following.  For some I give proof.  Please verify that my reasoning is correct and make any pertinent comments.  I use the notation $\zeta=\zeta_3$.

(a) $\mathbb{Q}(\sqrt{-3})=\mathbb{Q}(\zeta)$
Proof:   Note that $\{1, \zeta\}$ and $\{1, \frac{1+\sqrt{-3}}{2}\}$ are bases for $\mathbb{Z}[\zeta]$, $\mathbb{Z}[\sqrt{-3}]$, respectively.  If $\alpha \in \mathbb{Z}[\zeta]$, then, $\alpha=a+b\zeta=a-b+b+b\zeta=(a-b)+b(1+\zeta)=(a-b)+b(\frac{1+\sqrt{-3}}{2})\in \mathbb{Z}[\sqrt{-3}]$.  Similarly, if $\alpha \in \mathbb{Z}[\sqrt{-3}]$, then $\alpha=a+b(\frac{1+\sqrt{-3}}{2})=a+b-b+b(\frac{1+\sqrt{-3}}{2})=(a+b)+b\zeta \in \mathbb{Z}[\zeta]$.

(b) $O=\mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$.  Proof:  This follows from (a).

(c) Ord$_3(1-\zeta)=1/2$ makes sense and is well defined.
Here is my reasoning for this.  $1=$Ord$_3(3)=$Ord$_3((1-\zeta)(1-\zeta'))=$Ord$_3(1-\zeta)+$Ord$_3(1-\zeta')$.  There is a way around using this in the proof, but this is kinda cool so Im interested.

(d) The conjugate of $a-\zeta b$ is $a-\zeta'b$.  Proof:  Since the conjugate of products is the product of conjugates and the conjugate of sums is the sum of the conjugates, this follows.

b) does not follow from a). By definition, for $K$ a number field, $\mathcal{O}_K$ is the set of algebraic integers in $K$. You have shown, at best, that the ring of algebraic integers in $K = \mathbb{Q}(\zeta)$ contains $\mathbb{Z}[\zeta]$, but you have not shown that this is all of $\mathcal{O}_K$. 
Here is a complete proof of b) although, again, this is unnecessary since all you need for this proof is that $\mathbb{Z}[\zeta]$ has unique prime factorization. Let $a + b \zeta$ be an algebraic integer in $K$. Then its conjugate is $a + b \zeta^2$, hence its trace is $a - b$, which must be an integer. After multiplying by $\zeta$ we get $a \zeta + b \zeta^2 = -b + (a-b) \zeta$. Since this is also an algebraic integer, its trace $-b - (a-b) = -a$ must also be an integer, hence $a, b$ are both integers. On the other hand any element of $K$ of the form $a + b \zeta, a, b \in \mathbb{Z}$ is an algebraic integer, so $\mathcal{O}_K = \mathbb{Z}[\zeta]$ as desired.
c) is not a good idea. The correct definition is this: for a prime ideal $P$ in a Dedekind domain $D$, there is a discrete valuation $\nu_P$ defined as follows: if $x \in \mathcal{O}_K$, then let $\nu_P(x)$ be the greatest power of $P$ that divides the ideal $(x)$. So in this case the relevant prime ideal is $P = (1 - \zeta)$ since $(3)$ is not a prime ideal, and one has $\nu_P(1 - \zeta) = 1$.
Associated to any discrete valuation is an absolute value $2^{-\nu_P(x)}$ on $K$, and the reason you do not want to try defining valuations with respect to non-prime ideals is that the corresponding absolute value will not satisfy the triangle inequality. In this case you are lucky because $3$ ramifies (it is the square of a prime ideal) but, for example, $7 = (2 - \zeta)(2 - \zeta^2)$ does not ramify and the valuations $\nu_{2 - \zeta}, \nu_{2 - \zeta^2}$ do not coincide, so the naive definition of $\nu_7$ will not satisfy the triangle inequality. 
d) depends on your definition of "conjugate." 
