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The polynomial is this: $x^3 -2$

Okay, so first I can create my field extension.

I can easily extend the field to $2^{1/3}$. And I know the elements of the extension of $\mathbb{Q}(2^{1/3})$ can be written in the basis of this extension which is:

$\beta = b_0 + b_1 2^{1/3} + b_2 2^{2/3}$ I believe.

I know that the next extension is also linearly independent from this one(because it is in the imaginaries) and it has it's irreducible monic polynomial as $x^2 + 3$ which is of degree 2.

Thus $\mathbb{Q}(i\sqrt{3})$ has elements in the form of $a + bi\sqrt{3}$.

In total, since this vector thingey is completely linearly independent of $\mathbb{Q}(2^{1/3})$, this entire extention should be written as this:

$ a + b2^{1/3} + c2^{2/3} + d + ei\sqrt{3}$ Is this correct?

Why do i do all this? I have a problem that asks me to factor x^3 - 2 in linear factors with Coefficients in the extension $\mathbb{Q}(2^{1/3}, i\sqrt{3})$

How much of this did I get correct? And using all this, how do I go about factoring it with coefficients in this field? Do I just plug it all in to something? I know that this polynomial has 3 roots. One is 2(

Thanks for all the help. I feel like I get my education on Math Stack exchange at this point. XD

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The roots of $x^3-2$ are $\sqrt[3]{2}$, $\sqrt[3]{2}\omega$, $\sqrt[3]{2}\omega^2$, where $\omega$ is a primitive cubic root of unity.

Write $\omega$ and $\omega^2$ in terms of $i\sqrt3$.

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  • $\begingroup$ Are you saying I don't use any of what I just did for this? :( Wow what a waste $\endgroup$ – user121615 May 15 '15 at 17:42
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    $\begingroup$ @user121615, trying is never a waste. You always learn something. $\endgroup$ – lhf May 15 '15 at 17:44
  • $\begingroup$ That is definitely true. Thanks man $\endgroup$ – user121615 May 15 '15 at 17:52
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Curiously enough, if we take:

$\omega = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}$, it turns out that:

$\Bbb Q(\omega) = \Bbb Q(i\sqrt{3})$, in other words:

$x^3 - 2 = (x - \sqrt[3]{2})\left(x - \sqrt[3]{2}\left(-\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}\right)\right)\left(x - \sqrt[3]{2}\left(-\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}\right)\right)$

$= (x - \sqrt[3]{2})(x - \sqrt[3]{2}\omega)(x - \sqrt[3]{2}\omega^2)$.

Blame trigonometry for the curious appearance of $\sqrt{3}$ in a cube root (the $i$ part is what gets us to trigonometry in the first place, the trig functions are just the real and imaginary parts of the imaginary exponential function. Those darn circles).

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