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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.

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How does (b) follow from (a)? (And why is it necessary? You don't need to prove this to show that it has unique factorization.) What is your definition of ord_3? (It would be better to say ord_{1-zeta} since this will actually give you a discrete valuation: see en.wikipedia.org/wiki/Discrete_valuation_ring ). –  Qiaochu Yuan Feb 28 '11 at 23:12
    
$O_{Q(\sqrt{-3})}=O_{Q(\zeta)}$, since by (a), $Q(\sqrt{-3})=Q(\zeta)$. –  Jason Smith Feb 28 '11 at 23:15
    
I agree that it is not necessary, but Im trying to get the best sense of how all of this fits together. As for ord_3, I am using it in the proof. Is my use of ord_3 standard? There is a line in the proof that I am trying to understand that seems to use ord_3 in this way. –  Jason Smith Feb 28 '11 at 23:18
    
This isn't really Eulers proof, is it? –  Myself Feb 28 '11 at 23:33
    
Not exactly, but I think it is essentially the same. To clarify, my question is not the proof. It is only questions about some facts that I want to be sure of before I get up in front of aome people and present the proof. –  Jason Smith Feb 28 '11 at 23:36

1 Answer 1

up vote 3 down vote accepted

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."

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Awesome! Thank you. What do you mean for (d)? As for (c) I know a way around using it, so will do so. –  Jason Smith Mar 1 '11 at 18:30
    
My definition of conjugate is: For $a+b\zeta$, the conjugate is $a-b\zeta$. –  Jason Smith Mar 1 '11 at 19:58
    
@Jason: that is an incorrect definition. If $K$ is Galois, you can define the conjugates of an element of $\mathcal{O}_K$ to be the images under the action of the Galois group, and in this case the conjugate of $a + b \zeta$ is $a + b \zeta^2$. –  Qiaochu Yuan Mar 1 '11 at 20:40
    
So, $(a-\zeta b)'=a-\zeta^2 b=a-zeta' b$ since $\zeta^2=\zeta'$. –  Jason Smith Mar 1 '11 at 21:30
    
Can I say this? Taking conjugates of $z-\zeta y=\zeta'\alpha^3$, we have $z-\zeta' y=\zeta (\alpha')^3$. –  Jason Smith Mar 1 '11 at 21:54

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