# Splitting Field of a Polynomial in $\mathbb{Q}$

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.

• 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 $\mathbb{Q}(\sqrt{2}, \sqrt{1-i})$ – Wojowu Jan 16 at 9:59
• Do you mean $X^4-2X^2+2$? – Lord Shark the Unknown Jan 16 at 10:01
• Guys... I'm sorry. I edited both my mistakes. Thank you. – KingDingeling Jan 16 at 10:02

## 1 Answer

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 $$\sqrt{1+i}\in \mathbb{Q}(\sqrt{2}, \sqrt{1-i})$$.