$i \notin \mathbb{Q}[\sqrt[4]{2}]$ without using topological properties of $\mathbb{R}$

Question

I can think of two related ways to prove that $i \notin K = Q[\sqrt[4]{2}]$:

$K$ is a subset of the real numbers and $i$ is not a real number.

$K$ is orderable and no ordered field can contain $\sqrt{-1}$.

Both of them depend on topological properties of $\mathbb{Q}$ and might even be the same proof. Is there any way to show this purely algebraically?

Answer 1

One of the "weird" features of real vs complex algebra is that the ordering is part of what makes $\Bbb R$ what it is. Knowing that $\Bbb Q(\sqrt[4]{2})$ is even a subset of $\Bbb R$ relies on the fact that you can use continuity to prove there is such a real number as $\sqrt[4]{2}$, so it's somewhat cherry-picking when you want to say $i\not\in \Bbb R$ is topological, but $\sqrt[4]{2}\in\Bbb R$ is assumed. Nonetheless, there are proof which at least feel more algebraic, though I honestly cannot fathom one which completely evades the need for completeness on $\Bbb R$--to show there is a $4^{th}$ root of $2$ in $\Bbb R$--or the ordering--to show $i\not\in \Bbb R$. One of these facts needs to be exploited at least implicitly at some point in any proof.

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If $i\in\Bbb Q(\sqrt[4]{2})$ then $K=\Bbb Q(i,\sqrt[4]{2})=\Bbb Q(\sqrt[4]{2})$, hence the splitting field of $x^4-2$ has degree $4$. Now I note that the number

$$\zeta_8={1\over\sqrt{2}}+{i\over\sqrt{2}}\in K$$

is a primitive $8^{th}$ root of $1$ which is clearly in the field $K$. We know that it's degree is $\phi(8)=8-4=4$, so it must be that it generates $K$, i.e. $K=\Bbb Q(\zeta_8)$. But then this is a cyclotomic extension, hence it has no real embeddings *, but clearly if we present $K$ as we originally did--$K=\Bbb Q(\sqrt[4]{2})$--we have at least one real embedding, a contradiction, hence $i\not\in \Bbb Q(\sqrt[4]{2})$.

• To see that cyclotomic fields, $K=\Bbb Q(\zeta_n)$ have no real embeddings, it is sufficient to show that there are $\phi(n)$ complex embeddings, since the number of real embeddings $r$, and complex embeddings $s$ of a field, $K$ satisfy $r+2s=[K:\Bbb Q]$. But then we can explicitly provide $2s=\phi(n)$ complex embeddings, so that $r=0$. They are given by

$$\zeta_n\mapsto \zeta_n^k \quad 1\le k\le n, \gcd(k,n)=1.$$

If you still think this is too topological, an alternative proof: since $K/\Bbb Q$ has degree $4$, its Galois group is either the Klein-$4$ group or a cyclic group of order $4$, hence there are at most $3$ quadratic subfields by the FTGT. We can verify $\Bbb Q(\alpha), \alpha\in\{\sqrt 2, i,i\sqrt2\}$ are three such subfields, so the cyclic case is ruled out. But then all Klein-$4$ group extensions are given as towers of quadratic extensions, i.e. biquadratic extensions. But then this means that $K$ is the splitting field for $(x^2-a)(x^2-b)$, and by our classification of the subfields we can take $a=2, b=-1$. But then $\sqrt[4]{2}=a+b\sqrt{2}+ci+di\sqrt{2}$ for some rational $a,b,c,d$ implying $c=d=0$, and putting $\sqrt[4]{2}\in\Bbb Q(\sqrt{2})$ a contradiction to degrees.

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As a closing addendum, thanks to the comments of the op for clarifying what he's looking for, let's examine what it would mean to have a purely algebraic proof along the lines we've all been searching.

Key Ingredient: If we look at $\Bbb Q(\sqrt[4]{2})$ with $\sqrt[4]{2}$ as an actual number, we need to know it is in $\Bbb R$, which uses the completeness of $\Bbb R$ and continuity of polynomials, this is absolutely necessary and relies on the completeness. So to do it purely algebraically, we need to work with the description of the field as

$$K=\Bbb Q[x]/(x^4-2).\qquad (*)$$

But then in this case, the question is about the quotient field. For $-1\in K$ it is necessary and sufficient that there be $f(x)\in \Bbb Q[x]$ so that

$$f(x)^2=q(x)(x^4-2)-1$$

by definition of the image in the quotient field $K$ as described in $(*)$. But then there are infinitely many choices for $f(x)$ and $q(x)$, so certainly we cannot check by-hand. Instead we must write $f(x)=q_1(x)(x^2-4)+r(x)$ and then square it to get $r(x)^2=-1$. This implies--by degree considerations, which are purely algebraic--that $\deg r(x)=0$, so that all that it remains to prove is that there is no rational number ${p\over q}$ so that $p^2=-q$, which--as you have observed, needs the ordering.

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One final note: even the reductions up to the last step in the previous discussion, which I think most reasonable people would consider "purely" algebraic, has one last snag which I think should be mentioned for absolute completeness: if you go back to the very roots of the subject, the cornerstones, you will see that you need to use $\Bbb N$ is well-ordered to create the division algorithm, which you use on polynomials to even define the quotient field, so ultimately it will depend on ordering from the case of $\Bbb N$ even!

Answer 2

One way would to say that if $\Bbb Q(i) \subset \Bbb Q(\sqrt[4]{2})$, then it would necessarily be an quadratic extension. This would mean that $x^4-2$ would factor over $\Bbb Q(i)$. Any factorization $x^4-2 = (x^2+ax+b)(x^2+cx+d)$ must have $a = -c$ by consideration of the cubic term. If $a$ is nonzero we must have $b = d$ by consideration of the linear term, and if $a$ is zero we must have $b = -d$ by consideration of the quadratic term. This implies that either $2$ or $-2$ is a square in $\Bbb Q(i)$, which is easily seen to be false.