# Thread: Factor x^3 - 1 in Q[√3]

1. ## Factor x^3 - 1 in Q[√3]

How can you factor $x^3 - 1$ completely into linear factors in $Q[sqrt(3)]$?

2. Originally Posted by elemental
How can you factor $x^3 - 1$ completely into linear factors in $Q[sqrt(3)]$?
Dear elemental,

I dont understand what you mean by $"Q[\sqrt 3]"$. Can you please elaborate? However if you factor this,

$x^3-1=(x-1)(x^2+x+1)=(x-1)((x+\frac{1}{2})^2+\frac{3}{4})=(x-1)((x+\frac{1}{2})^2-\frac{3i^2}{4})~where~i=\sqrt -1$

$x^3-1=(x-1)(x+\frac{1}{2}-\frac{i\sqrt 3}{2})(x-\frac{1}{2}+\frac{i\sqrt 3}{2})$

3. Originally Posted by Sudharaka
Dear elemental,
I dont understand what you mean by $"Q[\sqrt 3]"$. Can you please elaborate?
It is the algebraic extension

$\mathbb{Q}[\sqrt{3}]=\mathbb{Q}[x]/(x^2-3)$

$x^3-1=(x-1)(x^2+x+1)=(x-1)((x+\frac{1}{2})^2+1)$
It should be:

$x^3-1=(x-1)(x^2+x+1)=(x-1)\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\r ight)=\ldots$

Fernando Revilla

4. Originally Posted by elemental
How can you factor $x^3 - 1$ completely into linear factors in $Q[sqrt(3)]$?
Possibly you meant $\mathbb{Q}[\;\sqrt{3}\;i\;]$ instead of $\mathbb{Q}[\;\sqrt{3}\;]$

Fernando Revilla

5. " $Q[\sqrt{3}]$" can also be expressed as "the smallest field containing all rational numbers and $\sqrt{3}$". It can be shown that all such numbers are of the form $p+ q\sqrt{3}$ for some rational numbers p and q.

$x^3- 1= (x- 1)(x^2+ x+ 1)$ and that is as far as it can be factored in Q or, for that matter, $Q[\sqrt{3}]$.

However, $x^2+ x+ 1= (x^2+ x+ 1/4- 1/4)+ 1= x^2+ x+ 1/4+ 3/4= (x+ 1/2)^2+ 3/4$
$= (x+ 1/2)^2-(-3/4)= (x+1/2- i\sqrt{3}{2})(x+ 1/2+ i\sqrt{3}{2}$
which are in $Q[i\sqrt{3}]$

So: $x^3- 1= (x-1)(x^2+ x+ 1)$ in $Q[\sqrt{3}]$ but $x^3- 1= (x- 1)(x+ 1/2- i\sqrt{3}{2})(x+ 1/2+ i\sqrt{3}/2)$ in $Q[i\sqrt{3}]$.

6. Sorry! I did mean -3, not 3. Thank you very much.