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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?

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    $\begingroup$ What do you mean with "topological properties"? $\endgroup$ – Git Gud Apr 12 '15 at 21:25
  • $\begingroup$ @GitGud probably things like its ordering. The first proof Asvin notes depends on the fact that $x\in\Bbb R\implies x^2\ge 0$ so that $x^2+1$ cannot be a real number, the second is most explicitly a dependence upon the ordering present in $\Bbb R$. $\endgroup$ – Adam Hughes Apr 12 '15 at 21:40
  • $\begingroup$ To the op: no matter which answer you like best, you should definitely read at least the opening to mine. No proof will be able to get away with not using topology on some level. In all of the current answers (including my own) I can point out where a topological property is implicitly assumed to complete the proof. $\endgroup$ – Adam Hughes Apr 12 '15 at 22:12
  • $\begingroup$ I can't see how order is topology. Order is a concept in its own right, a fact supported by the existence of a field called Order Theory. $\endgroup$ – Git Gud Apr 12 '15 at 22:14
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    $\begingroup$ @Asvin he claims $\pm 2$ is not a square in $\Bbb Q(i)$, how would you prove that? If you go deeper you'll see this means you need to solve $a^2-b^2+2abi=\pm 2$ with $a,b\in\Bbb Q$. Any reduction to $2abi=0$ requires you to say $i\not\in\Bbb R$, which you already noted requires topology, so it needs it implicitly. The less you say in the way of details, the less apparent the topology is, but it's still there. $\endgroup$ – Adam Hughes Apr 12 '15 at 22:28
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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.


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.


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.


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!

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  • $\begingroup$ "But then this is a cyclotomic extension, hence it has no real embeddings." How do you know this? $\endgroup$ – quid Apr 12 '15 at 21:43
  • $\begingroup$ @quid I updated the answer. Ultimately anything is going to weaselly need the topology somewhere: in the spirit of the op's question I've attempted to push it as far under the rug as possible. :-) $\endgroup$ – Adam Hughes Apr 12 '15 at 22:13
  • $\begingroup$ Thanks I like it much better know. $\endgroup$ – quid Apr 12 '15 at 22:19
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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.

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  • $\begingroup$ In the comments on the original post I noted that this still requires you to use $i\not\in\Bbb R$ (or at least not in $\Bbb Q$). But in case you didn't see that, I figured you'd want to know as well since MSE doesn't notify you if you didn't comment on the original thread. $\endgroup$ – Adam Hughes Apr 12 '15 at 23:04
  • $\begingroup$ Showing that $i\not\in\Bbb Q$ requires knowing about the order (and thus the topology) on $\Bbb Q$, which doesn't require knowing anything about $\Bbb R$, for what it's worth. $\endgroup$ – Rolf Hoyer Apr 12 '15 at 23:23
  • $\begingroup$ Yeah, I agree. The original way the question was posed was doomed from the start. Cheers! $\endgroup$ – Adam Hughes Apr 12 '15 at 23:30

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