How to find the value of $[\mathbb{Q}(\sqrt[4]{-2},i):\mathbb{Q}(\sqrt[4]{-2})]?$

Question

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 $\mathbb{Q}(\sqrt[4]{-2},i)$

So the splitting field of $f$ has degree $$[\mathbb{Q}(\sqrt[4]{-2},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{-2},i):\mathbb{Q}(\sqrt[4]{-2})] \cdot [\mathbb{Q}(\sqrt[4]{-2}):\mathbb{Q}]$$

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

Here im unable to find the value of $[\mathbb{Q}(\sqrt[4]{-2},i):\mathbb{Q}(\sqrt[4]{-2})]$.

My confusion: How to find the value of $[\mathbb{Q}(\sqrt[4]{-2},i):\mathbb{Q}(\sqrt[4]{-2})]?$

Answer 1

To compute this degree, we need to find a basis for $\mathbb{Q}(\sqrt[4]{-2}, i)$ over $\mathbb{Q}(\sqrt[4]{-2}).$ 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.

Answer 2

Note that $i \notin \mathbb{Q}(\sqrt[4]{-2})$. Since $i^2 = -1 \in \mathbb{Q} \subseteq \mathbb{Q}(\sqrt[4]{-2})$, it follows that $$[\mathbb{Q}(\sqrt[4]{-2},i):\mathbb{Q}(\sqrt[4]{-2})] = 2$$

Answer 3

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} $.

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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 $[L:\mathbb {Q}] =8$.