0
$\begingroup$

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

$\endgroup$
3
  • 2
    $\begingroup$ $[\mathbb{Q}(\sqrt[4]{-2}):\mathbb{Q}] = 4$, not 2 $\endgroup$ Jun 30 at 19:13
  • $\begingroup$ okay@cos_dm_math21 $\endgroup$
    – jasmine
    Jun 30 at 19:22
  • 1
    $\begingroup$ You should also have a look at this question $\endgroup$ Jul 1 at 4:25
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

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

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

$\endgroup$
2
  • 2
    $\begingroup$ Is $i\notin {\mathbb Q}(\sqrt[4]{-2})$ so obvious? $\endgroup$
    – Pythagoras
    Jun 30 at 19:55
  • $\begingroup$ You're right @Pythagoras Infact, I'm having the same doubt. $\endgroup$
    – jasmine
    Jun 30 at 20:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.