How can you factor $\displaystyle x^3 - 1$ completely into linear factors in $\displaystyle Q[sqrt(3)]$?
Dear elemental,
I dont understand what you mean by $\displaystyle "Q[\sqrt 3]"$. Can you please elaborate? However if you factor this,
$\displaystyle x^3-1=(x-1)(x^2+x+1)=(x-1)((x+\frac{1}{2})^2+\frac{3}{4})=(x-1)((x+\frac{1}{2})^2-\frac{3i^2}{4})~where~i=\sqrt -1 $
$\displaystyle x^3-1=(x-1)(x+\frac{1}{2}-\frac{i\sqrt 3}{2})(x-\frac{1}{2}+\frac{i\sqrt 3}{2})$
Hope this will help you.
It is the algebraic extension
$\displaystyle \mathbb{Q}[\sqrt{3}]=\mathbb{Q}[x]/(x^2-3)$
It should be:$\displaystyle x^3-1=(x-1)(x^2+x+1)=(x-1)((x+\frac{1}{2})^2+1)$
$\displaystyle x^3-1=(x-1)(x^2+x+1)=(x-1)\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\r ight)=\ldots$
Fernando Revilla
Possibly you meant $\displaystyle \mathbb{Q}[\;\sqrt{3}\;i\;]$ instead of $\displaystyle \mathbb{Q}[\;\sqrt{3}\;]$
Fernando Revilla
"$\displaystyle Q[\sqrt{3}]$" can also be expressed as "the smallest field containing all rational numbers and $\displaystyle \sqrt{3}$". It can be shown that all such numbers are of the form $\displaystyle p+ q\sqrt{3}$ for some rational numbers p and q.
$\displaystyle x^3- 1= (x- 1)(x^2+ x+ 1)$ and that is as far as it can be factored in Q or, for that matter, $\displaystyle Q[\sqrt{3}]$.
However, $\displaystyle x^2+ x+ 1= (x^2+ x+ 1/4- 1/4)+ 1= x^2+ x+ 1/4+ 3/4= (x+ 1/2)^2+ 3/4$
$\displaystyle = (x+ 1/2)^2-(-3/4)= (x+1/2- i\sqrt{3}{2})(x+ 1/2+ i\sqrt{3}{2}$
which are in $\displaystyle Q[i\sqrt{3}]$
So: $\displaystyle x^3- 1= (x-1)(x^2+ x+ 1)$ in $\displaystyle Q[\sqrt{3}]$ but $\displaystyle x^3- 1= (x- 1)(x+ 1/2- i\sqrt{3}{2})(x+ 1/2+ i\sqrt{3}/2)$ in $\displaystyle Q[i\sqrt{3}]$.