# How to find the value of $[\mathbb{Q}(\sqrt{-2},i):\mathbb{Q}(\sqrt{-2})]?$

Determine the splitting field and its degree over $$\mathbb{Q}$$ for $$x^4+2$$

My attempt :Obviously,the splitting field of the polynomial $$f(x)=x^4 +2$$ is $$\mathbb{Q}(\sqrt{-2},i)$$

So the splitting field of $$f$$ has degree $$[\mathbb{Q}(\sqrt{-2},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt{-2},i):\mathbb{Q}(\sqrt{-2})] \cdot [\mathbb{Q}(\sqrt{-2}):\mathbb{Q}]$$

since $$\sqrt{-2}$$ is a root of the irreducible polynomial $$x^4+2$$ over $$\mathbb{Q}$$, then $$[\mathbb{Q}(\sqrt{-2}):\mathbb{Q}]=4$$.

Here im unable to find the value of $$[\mathbb{Q}(\sqrt{-2},i):\mathbb{Q}(\sqrt{-2})]$$.

My confusion: How to find the value of $$[\mathbb{Q}(\sqrt{-2},i):\mathbb{Q}(\sqrt{-2})]?$$

• $[\mathbb{Q}(\sqrt{-2}):\mathbb{Q}] = 4$, not 2 Jun 30 at 19:13
• okay@cos_dm_math21 Jun 30 at 19:22
• You should also have a look at this question Jul 1 at 4:25

I think it is better to find the roots explicitly. We have $$x^2=\pm\sqrt{2}i$$ and then $$x=\pm 2^{1/4}\cdot\frac{1+ i} {\sqrt{2} }, \pm 2^{1/4}\cdot\frac{1-i}{\sqrt{2} }$$ Hence the splitting field here is $$K=\mathbb{Q} (2^{1/4},i)$$ and one can prove that it is of degree $$8$$ over $$\mathbb{Q}$$.

Here is a bit more detail to see why the splitting field is $$\mathbb{Q} (2^{1/4},i)$$. Consider the two roots $$a, b$$ given by $$(1\pm i) /2^{1/4}$$ then $$2^{1/4}=\frac{2}{a+b},i=\frac{a-b}{a+b}$$ and hence the splitting field $$L$$ must be such that $$L\supseteq \mathbb {Q}(2^{1/4},i)$$. On the other hand all the roots are contained in $$\mathbb{Q} (2^{1/4},i)$$ and hence we have $$L=\mathbb {Q}(2^{1/4},i)$$.

Expressing the field $$L$$ in this form allows us to deduce easily that $$i\notin\mathbb {Q} (2^{1/4})$$ (real vs complex) and thereby get the degree $$[L:\mathbb {Q} (2^{1/4})]=2$$ and finally $$[L:\mathbb {Q}] =8$$.

To compute this degree, we need to find a basis for $$\mathbb{Q}(\sqrt{-2}, i)$$ over $$\mathbb{Q}(\sqrt{-2}).$$ The most obvious basis is $$1, i.$$ Indeed, this list is clearly spanning, and so the dimension is at most 2. Now we just need to show that the dimension of the vector space isn't one; but this is fairly simple, since $$i$$ is not contained in the smaller field, and if the dimension was 1, then the two fields would be the same.

• As in other answer herd, it is not obvious that $i\notin\mathbb{Q} (\sqrt{-2})$. Jul 1 at 4:00

Note that $$i \notin \mathbb{Q}(\sqrt{-2})$$. Since $$i^2 = -1 \in \mathbb{Q} \subseteq \mathbb{Q}(\sqrt{-2})$$, it follows that $$[\mathbb{Q}(\sqrt{-2},i):\mathbb{Q}(\sqrt{-2})] = 2$$

• Is $i\notin {\mathbb Q}(\sqrt{-2})$ so obvious? Jun 30 at 19:55
• You're right @Pythagoras Infact, I'm having the same doubt. Jun 30 at 20:12