TITLE: An application of Keith Conrad's " Splitting field of $X^3-2$ over $\mathbb{Q}$ "
QUESTION [8 upvotes]: After reading This Paper, I thought in the case of FLT for n = 3: 
Let  $x^3+y^3+z^3=0$ ,  for  $x,y,z$  relatively coprime integer solutions.  
If  $3$  doesn't divide  $x,y,z$  then  $x^3+y^3+z^3\not\equiv{0}$ mod $9$ .  Because  $(\mathbb{Z}/9\mathbb{Z})^3=\{0,1,-1\}$ . So, let's assume that  $3$  divides  $z$ . 
The pure cubic field $\mathbb{Q}(\sqrt[3]{2})$,  as Professor Keith Conrad explains,  has ring of integers  $\mathbb{Z}(\sqrt[3]{2})$  and an integer basis is:  $\left\lbrace 1,\sqrt[3]{2},\sqrt[3]{4}\right\rbrace$ .  The prime $3$  is ramified:  $\pi^3\nu$ ,  for:  $\pi=1+\sqrt[3]{2}$  and  $\nu=\sqrt[3]{2}-1$ ,  such that  $\nu$  is a unit in  $\mathbb{Q}(\sqrt[3]{2})$ .
We have:  $x^3+y^3+z^3=0$   and   $-z^3=x^3+y^3$ $\Rightarrow$ $-z^3+3y^3=x^3+4y^3$ .  The expression $x^3+4y^3$   factors in  $\mathbb{Q}(\sqrt[3]{2})$  as:  $(x+y\sqrt[3]{4})(x+\omega y\sqrt[3]{4})(x+\omega^2 y\sqrt[3]{4})\,=\,(x+y\sqrt[3]{4})(x^2+2y^2\sqrt[3]{2}-xy\sqrt[3]{4})$ .  So:  $-z^3+3y^3=(x+y\sqrt[3]{4})(x^2+2y^2\sqrt[3]{2}-xy\sqrt[3]{4})$ (1). (Where  $\omega=(-1+\sqrt{-3})/2$  is a primitive cube root of unity).
I will consider here two lemmas as proved. An impossibility of being rational numbers in  $\mathbb{Z}(\sqrt[3]{2})$  implies the impossibility of being it in  $\mathbb{Z}$  ($\mathbb{Z}\subseteq\mathbb{Z}(\sqrt[3]{2})$) .  And the extension of arithmetic rules (for example: to be relatively coprime) of  $\mathbb{Z}$  to  $\mathbb{Z}(\sqrt[3]{2})$ .       
A)  If  $x^2+2y^2\sqrt[3]{2}-xy\sqrt[3]{4}$  is coprime to  $x+y\sqrt[3]{4}$ ,  it will be coprime to  $(x+y\sqrt[3]{4})^2=x^2+2y^2\sqrt[3]{2}+2xy\sqrt[3]{4}$ .  Let's see it. The sum is:  $2x^2+4y^2\sqrt[3]{2}+xy\sqrt[3]{4}$  and the difference:  $-3xy\sqrt[3]{4}$ .  And only by  $3$  we see  $-3xy$  doesn't divide  $2x^2$  either  $4y^2$  or  $xy$ . Because  $3$  only divides  $z$ . 
B)  Let's see it in  $\mathbb{Q}(\sqrt[3]{2},\omega)$ .  If  $x+y\sqrt[3]{4}$  is coprime to  $x+\omega y\sqrt[3]{4}$ ,  its addition and its substraction will neither have common factors. The sum is:  $2x+y\sqrt[3]{4}(1+\omega)$  and the difference:  $y\sqrt[3]{4}(1-\omega)$ .  The expression $1+\omega$  is a unit in  $\mathbb{Q}(\omega)$  and  $1-\omega=-(\omega-1)$  divides  $3=-\omega^2(\omega-1)^2$ .  Then  $1-\omega$  doesn't divide  $2x$  or  $y\sqrt[3]{4}(1+\omega)$ .  It is only left to see with  $x+\omega^2 y\sqrt[3]{4}$ .  Its sum is:  $2x+y\sqrt[3]{4}(1+\omega^2)$  and its difference:  $y\sqrt[3]{4}(1-\omega^2)$ .  But  $1+\omega^2$  is also a unit in  $\mathbb{Q}(\omega)$  (the conjugate of  $1+\omega$ )  and  $1-\omega^2=(1+\omega)(1-\omega)$ .  Then, for the same reason than before  $x+y\sqrt[3]{4}$  is coprime to  $x+\omega^2 y\sqrt[3]{4}$  and therefore to  $x^2+2y^2\sqrt[3]{2}-xy\sqrt[3]{4}$ .         
As  $3$  divides  $-z^3$ (1) ,  then  $\pi=1+\sqrt[3]{2}$ ,  where is prime in  $\mathbb{Z}(\sqrt[3]{2})$ ,  must divide  $x+y\sqrt[3]{4}$  or  $x^2+2y^2\sqrt[3]{2}-xy\sqrt[3]{4}$  and only one. Let's check it out: 
$\dfrac{(x+y\sqrt[3]{4})(\pi^2\nu)}{\pi\cdot\pi^2\nu}\,=\,\dfrac{(x+y\sqrt[3]{4})(1-\sqrt[3]{2}+\sqrt[3]{4})}{3}\,=\,\dfrac{x-2y+(2y-x)\sqrt[3]{2}+(x+y)\sqrt[3]{4}}{3}$ 
Without loss of generality we assume  $x^3+y^3+z^3=0$ $\Rightarrow$ $1-1+0\equiv{0}$ mod $3$ .  So:  $x\equiv{1}$ mod $3$  and  $y\equiv{-1}$ mod $3$ .  In this way, we observe that:  $x-2y\equiv{0}$ mod $3$   ,   $2y-x\equiv{0}$ mod $3$   and   $x+y\equiv{0}$ mod $3$ .  Therefore  $\pi$  divides  $x+y\sqrt[3]{4}$ .
Now let's check this:  $\pi$  doesn't divide  $x^2+2y^2\sqrt[3]{2}-xy\sqrt[3]{4}$ .
$\dfrac{(x^2+2y^2\sqrt[3]{2}-xy\sqrt[3]{4})(\pi^2\nu)}{\pi\cdot\pi^2\nu}\,=\,\dfrac{(x^2+2y^2\sqrt[3]{2}-xy\sqrt[3]{4})(1-\sqrt[3]{2}+\sqrt[3]{4})}{3}$
$=\,\dfrac{x^2+2y^2\sqrt[3]{2}-xy\sqrt[3]{4}-x^2\sqrt[3]{2}-2y^2\sqrt[3]{4}+2xy+x^2\sqrt[3]{4}+4y^2-2xy\sqrt[3]{2}}{3}$ 
$=\,\dfrac{x^2+2xy+4y^2+(2y^2-x^2-2xy)\sqrt[3]{2}+(x^2-xy-2y^2)\sqrt[3]{4}}{3}$ 
But:  $x^2+2xy+4y^2\equiv{0}$ mod $3$   ,   $2y^2-x^2-2xy\equiv{0}$ mod $3$   and   $x^2-xy-2y^2\equiv{0}$ mod $3$ .  This is a contradiction and tell us if  $x^3+y^3+z^3=0$ ,  then  $3$  divides  $x,y$  and  $z$  in  $\mathbb{Z}(\sqrt[3]{2})$ . In particular, if  $3$  divides  $x+y\sqrt[3]{4}$  and  $x^2+2y^2\sqrt[3]{2}-xy\sqrt[3]{4}$ ,  at A) we have seen that  $3$  should also have divided  $x,y$ ,  not just  $z$ ;  and at B), in  $\mathbb{Z}(\sqrt[3]{2},\omega)$ ,  that  $1-\omega$  should have divided  $x,y$  in addition to  $z$ ,  ($1-\omega$  is a factor of  $3$) .    
But if  $3$  divides  $x,y,z$ in $\mathbb{Z}(\sqrt[3]{2})$  and consequently in  $\mathbb{Z}$ ; then:  $x^3=3^3x'\,^3\,,\,y^3=3^3y'\,^3$  and  $z^3=3^3z'\,^3$ . And it there will be a sum  $x'\,^3+y'\,^3+z'\,^3=0$ , for  $x',y',z'$  integral relatively prime coefficients, less than  $x^3+y^3+z^3=0$ .  And so on. Beginning an infinite decreasing. Hence $x,y,z$  cannot be rational.
Is it correct? 
Have I used the concepts of the Paper well? 
Thank you in advance¡

REPLY [2 votes]: The above proof has a mistake.
$2x^2+4y^2\sqrt[3]{2}+xy\sqrt[3]{4}$  and  $-3xy\sqrt[3]{4}$  aren't relatively coprime. Because  $1+\sqrt[3]{2}$  divides both.
$\dfrac{(2x^2+4y^2\sqrt[3]{2}+xy\sqrt[3]{4})(1-\sqrt[3]{2}+\sqrt[3]{4})}{(1+\sqrt[3]{2})(1-\sqrt[3]{2}+\sqrt[3]{4})}$
$=\,\dfrac{2x^2-2xy+8y^2+(4y^2-2x^2+2xy)\sqrt[3]{2}+(xy-4y^2+2x^2)\sqrt[3]{4}}{3}$
$2x^2-2xy+8y^2\equiv{0}$ mod $3$  ,  $4y^2-2x^2+2xy\equiv{0}$ mod $3$  and  $xy-4y^2+2x^2\equiv{0}$ mod $3$ .
And certainly  $1+\sqrt[3]{2}$  divides  $-3xy\sqrt[3]{4}$ .   
I'm sorry