Simple extension of $\mathbb{Q} (\sqrt[4]{2},i)$

Solution 1:

Discussed in the comments above, one method would be to use the field operations and known elements in $\mathbb{Q}( \sqrt[4]{2} + i)$ to deduce that $\sqrt[4]{2}$ and $i$ are both elements of this field per additive/multiplicative closure, giving us the reverse containment and thus equality. This works, but the trial-and-error can be frustrating in general for problems like these. Fortunately, other methods exist to remove any guesswork:

As you've noted, we already have $\mathbb{Q}(\sqrt[4]{2} + i) \subset \mathbb{Q}(\sqrt[4]{2}, i)$. Therefore, to show equality, it suffices to show that the degrees of these two field extensions over $\mathbb{Q}$ are equal. Finding the degree of $\mathbb{Q}( \sqrt[4]{2}, i)$ over $\mathbb{Q}$ is made relatively easy by constructing a tower of inclusions $\mathbb{Q} \subset \mathbb{Q}( \sqrt[4]{2}) \subset \mathbb{Q}(\sqrt[4]{2}, i)$ and then applying the multiplicativity formula for degrees. How can we find $[ \mathbb{Q}( \sqrt[4]{2} + i): \mathbb{Q}]$?

One approach would be to find the degree of the minimal polynomial of $\sqrt[4]{2} + i$ over $\mathbb{Q}$, since that will be equal to the degree of the extension. The following lemma will help us find the degree of this polynomial:

Lemma: If $K/F$ is a Galois extension, then the minimal polynomial in $F[x]$ for any $a \in K$ has as its roots the elements in the orbit of $a$ under the action of $\text{Gal}(K/F)$. That is, if $S = \{ \phi(a) \ | \ \phi \in \text{Gal}(K/F) \}$, then $\displaystyle \min_a(x) = \prod_{u_k \in S} (x-u_k)$.

Proof: This is a consequence of the fact that a separable polynomial is irreducible $\iff$ its Galois group acts transitively on its roots. For a proof of this fact, see Theorem 2.9(b) here. Notice that $\displaystyle \min_a(x)$ will be an irreducible polynomial (by definition), and its Galois group will be a subgroup of $\text{Gal}(K/F)$. This latter fact is because, if $L \subseteq K$ is the splitting field of $\displaystyle \min_a(x)$, every $F$-automorphism of $L$ extends to an $F$-automorphism of $K$. This is to say: for any $\varphi \in \text{Gal}(K/F)$, either $\varphi \in \text{Gal}(K/L)$, in which case the restriction of $\varphi$ to $L$ is merely the identity, or $\varphi \notin \text{Gal}(K/L)$, in which case its restriction to $L$ corresponds to an element of $\text{Gal}(L/F)$. Hence, the action of $\text{Gal}(K/F)$ on $\alpha$ is identical to the action of $\text{Gal}\Big(\min_{\alpha}(x)\Big) = \text{Gal}(L/F)$ on $\alpha$, and we know per the aforementioned fact that this action is transitive with respect to the roots of $\min_{\alpha}(x)$. $ \qquad \blacksquare$

Note that $\mathbb{Q}( \sqrt[4]{2}, i)/\mathbb{Q}$ is a Galois extension: finite extensions of $\mathbb{Q}$ are separable and this is the splitting field for $f(x) = x^4 - 2$. First, compute the Galois group of this field. Next, find the number of elements in the orbit of $\sqrt[4]{2} + i$ under the action of $ \text{Gal}( \mathbb{Q}(\sqrt[4]{2},i)/ \mathbb{Q})$; this will be the degree of the minimal polynomial of $\sqrt[4]{2} + i$ per the above lemma. If that number is equal to $[ \mathbb{Q}( \sqrt[4]{2}, i): \mathbb{Q}]$, then the degrees of the two extensions in question are equal, and we'll indeed have $\mathbb{Q}(\sqrt[4]{2} + i) = \mathbb{Q}( \sqrt[4]{2}, i)$.

As a side note, the primitive element theorem tells us that every finite extension $K/ \mathbb{Q}$ can be written as $K \cong \mathbb{Q}(\alpha)$ for some $\alpha \in K$. In the above, of course, we had $K = \mathbb{Q}( \sqrt[4]{2}, i)$ with $\alpha = \sqrt[4]{2} + i$. If one has sufficient understanding of $\text{Gal}(K/\mathbb{Q})$, notice, per the above discussion, that it's possible to find a suitable $\alpha$ from scratch by constructing it so as to have $[K: \mathbb{Q}]$ distinct Galois conjugates.

Footnote: If you have access to Artin's Algebra (second edition), there is (if I recall correctly) a paragraph or two in the Galois theory chapter discussing the above technique of using Galois groups to find minimal polynomials. It should fill in any missing details or context in this post.

Solution 2:

This one is easy like most similar problems. Let $a=2^{1/4}+i$ so that $(a-i) ^{4}=2$ or $$a^{4}-4a^{3}i-6a^{2}+4ai+1=2$$ or $$i=\frac{a^{4}-6a^{2}-1}{4a^{3}-4a}$$ so that $i\in\mathbb{Q} (a) $ and hence $2^{1/4}=a-i\in\mathbb{Q}(a)$. It should now be obvious that $\mathbb{Q} (2^{1/4},i)=\mathbb{Q}(2^{1/4}+i)$.

Solution 3:

Without using the general method suggested by @Kaj Hansen, it is not so difficult to get hold directly of the extension $L=\mathbf Q(\sqrt [4]2+ i)$. First note that $K= \mathbf Q(\sqrt [4]2, i)$ is the splitting field of the irreducible polynomial $X^4 - 2$ (Eisenstein), hence is Galois of degree $8$. Denoting by $s$ and $t$ the automorphisms of $K$ fixing respectively $\sqrt [4]2$ and $i$, one can readily check that $s^2=t^4=1$ and $sts^{-1}=t^{-1}$, i.e. $Gal(K/\mathbf Q)=D_8$, the dihedral group of order 8. Since the degree over $L$ is obviously $\neq 2$,we only need to rule out the degree 4. But the previous relations imply that $D_8$ has exactly 4 subgroups of order 2, which are the fixators of $\pm \sqrt [4]2$ and $\pm i\sqrt [4]2$. Hence $[L:\mathbf Q]=8$ and $L=K$ .

Solution 4:

You might find it easier, in evaluating $[K:Q]$ where $K=\Bbb Q(\sqrt[4]2+i)$, to look at the tower $\Bbb Q\subset k\subset K$, where $k$ is the Gaussian numbers $\Bbb Q(i)$. Since $K=k(\sqrt[4]2+i)=k(\sqrt[4]2\,)$, you only need to find $[k(\sqrt[4]2\,):k]$, and this is easy. Using the fact that $\sqrt2\notin k$, we get $k$-irreducibility not only of $X^2-2$ but also of $X^4-2$.