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We are studying splitting fields and I just wanted to make sure that I really understood them, so we needed to do the following:

Show that $\mathbb{Q}(\sqrt{2}, \sqrt{1-i})$ is a splitting field of the Polynomial $f=X^4 -2X^2 + 2$.

My attempt:

Substituting $z:=X^2$ we get $f'=z^2 - 2z +2$. Computing the roots we get $x_{1,2} = 1\pm i$ so $f$ has the roots $y_{1,2,3,4}=\pm \sqrt{1\pm i}$. The splitting field of f is therefore $\mathbb{Q}(\sqrt{1+i}, -\sqrt{1+i}, \sqrt{1-i}, -\sqrt{1-i})$. Now we show that $\mathbb{Q}(\sqrt{1+i}, -\sqrt{1+i}, \sqrt{1-i}, -\sqrt{1-i})$ = $\mathbb{Q}(\sqrt{2}, \sqrt{1-i})$.

i) $\mathbb{Q}(\sqrt{2}, \sqrt{1-i})\subset \mathbb{Q}( \sqrt{1+i}, -\sqrt{1+i}, \sqrt{1-i}, -\sqrt{1-i})$ is true since $\sqrt{1+i}\cdot \sqrt{1-i} = \sqrt{(1+i)\cdot(1-i)}=\sqrt{2}$

ii) $ \mathbb{Q}( \sqrt{1+i}, -\sqrt{1+i}, \sqrt{1-i}, -\sqrt{1-i}) \subset \mathbb{Q}(\sqrt{2}, \sqrt{1-i})$ Now for $\sqrt{1+i}$ and $-\sqrt{1+i}$ this is trivial. But I am not 100% sure about $\sqrt{1-i}$. Am I allowed to assume that because $\sqrt{2}= \sqrt{(1+i)\cdot(1-i)}= \sqrt{1+i}\cdot \sqrt{1-i}$ then $\sqrt{1-i}$ has to be in $\mathbb{Q}(\sqrt{2}, \sqrt{1+i})$ because $\sqrt{1+i}$ is and if $\sqrt{1-i}$ isn't then $\sqrt{2}$ also wouldn't be? I am not sure my argumentation is 100% correct at this point.

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  • $\begingroup$ The notation $(\mathbb{Q},\sqrt{2}, \sqrt{1-i})$ is nonstandard, I presume you mean the smallest field containing $\mathbb{Q},\sqrt{2}$ and $\sqrt{1-i}$? That's usually denoted by $\mathbb{Q}(\sqrt{2}, \sqrt{1-i})$ $\endgroup$ – Wojowu Jan 16 at 9:59
  • $\begingroup$ Do you mean $X^4-2X^2+2$? $\endgroup$ – Lord Shark the Unknown Jan 16 at 10:01
  • $\begingroup$ Guys... I'm sorry. I edited both my mistakes. Thank you. $\endgroup$ – KingDingeling Jan 16 at 10:02
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Now for $\sqrt{1+i}$ and $-\sqrt{1+i}$ this is trivial. But I am not 100% sure about $\sqrt{1-i}$.

Actually, the cases $\sqrt{1-i}$ and $-\sqrt{1-i}$ are trivial since we are considering $\mathbb{Q}(\sqrt{2}, \sqrt{1-i}) $.

Now for $\sqrt{1+i}$, your argument is almost correct. Note that $\sqrt{1-i}\neq 0\in \mathbb{Q}(\sqrt{2}, \sqrt{1-i})$. Since $\mathbb{Q}(\sqrt{2}, \sqrt{1-i})$ is a field, the multiplicative inverse of $\sqrt{1-i} $ is in $\mathbb{Q}(\sqrt{2}, \sqrt{1-i})$. So now we can express $\sqrt{1+i}$ as $\frac{\sqrt{2}}{\sqrt{1-i}}$(obtained from the equation you wrote). This implies that $\sqrt{1+i}\in \mathbb{Q}(\sqrt{2}, \sqrt{1-i})$.

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