Ambiguity regarding $\mathbb Q(\sqrt[4]{-2})$

Question

I am studying field theory.I found a question in Artin's algebra which asks if $i\in\mathbb Q(\sqrt[4]{-2})$.Now I am confused with the meaning of $\sqrt[4]{-2}$ here because there are $4$ of them.It is not specified which one I should take.So,I am having problem.Can someone please help me?

Answer 1

It doesn’t matter which root you choose. $i$ is in one $\mathbb Q[\sqrt[4]{-2}]$ if and only if it is in all of them.

If $p(x)$ is an irreducible polynomial in $\mathbb Q[x],$ and $\alpha$ is a root, then $\mathbb Q[\alpha]$ is isomorphic to $\mathbb Q[x]/\langle p(x)\rangle.$ So any other root $\beta$ of $p(x)$ has $\mathbb Q[\beta]$ isomorphic to $\mathbb Q[\alpha],$ since both fields are isomorphic t9 the quotient field.

This means, given $q(x)\in \mathbb Q[x]$ with a root in $\mathbb Q[\alpha],$ there is also a root of $q(x)$ in any $\mathbb Q[\beta]$ where $\beta$ is another root of $p(x).$

Now, $i$ exists in a subfield $K\subseteq\mathbb C$ if an only of $q(x)=x^2+1$ has a root in the field. This is because a root is either $i$ or $-i,$ and if $-i\in K,$ the. $i\in K.$

So what this means is that it doesn’t matter which $\sqrt[4]{-2}$ you use. $i$ is in one if and only if it is in all.

This isn’t always true - it depends on properties of $q.$ But it is always true for $q(x)$ rational and quadratic - that if one root of $q(x)$ is in $K,$ then all roots of $q(x)$ are in $K.$

You are right to ask, though.

An example where this doesn’t work is $p(x)=x^8+9$ and $q(x)=x^4+9.$ We know that $\mathbb Q[\sqrt[8]{-9}]$ will contain some complex root of $q(x),$ but which one will depend on which complex value $\sqrt[8]{-9}$ we choose.

Answer 2

The four distinct roots of $x^{4}+2$ are given by:-

$$\sqrt[4]{-2}=\sqrt[4]{2}\exp(\dfrac{(2k+1)i\pi}{4})\,,k=0,1,2,3$$.

For $k=0$ we have

$\sqrt[4]{2}(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))=\sqrt[4]{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})$

For $k=1$

$\sqrt[4]{2}(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}))=\sqrt[4]{2}(\frac{-1}{\sqrt{2}}+\frac{i}{\sqrt{2}})$

For $k=2$.

$\sqrt[4]{2}(\cos(\frac{5\pi}{4})+i\sin(\frac{5\pi}{4}))=\sqrt[4]{2}(\frac{-1}{\sqrt{2}}+\frac{-i}{\sqrt{2}})$

For $k=3$

$\sqrt[4]{2}(\cos(\frac{7\pi}{4})+i\sin(\frac{7\pi}{4}))=\sqrt[4]{2}(\frac{1}{\sqrt{2}}+\frac{-i}{\sqrt{2}})$

Say I pick $k=0$ and denote $\sqrt[4]{2}(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))=\sqrt[4]{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})=\sqrt[4]{-2}$ for the rest of the answer.

Suppose $i\in \mathbb{Q(\sqrt[4]{-2})}$.

Then $\sqrt[4]{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}),i\in\mathbb{Q(\sqrt[4]{-2})}$

So taking their product we have

So $\sqrt[4]{2}(\frac{i}{\sqrt{2}}-\frac{1}{\sqrt{2}})\in\mathbb{Q(\sqrt[4]{-2})} $

Then their $\frac{\text{difference}}{2} =\sqrt[4]{2}\frac{1}{\sqrt{2}}\in \mathbb{Q(\sqrt[4]{-2})}\implies \frac{1}{\sqrt[4]{2}}\in\mathbb{Q(\sqrt[4]{-2})}$.

Thus $\mathbb{Q}(\sqrt[4]{2},i)\subset\mathbb{Q}(\sqrt[4]{-2})$ and it is evident from the expression $\sqrt[4]{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})$ that $\mathbb{Q}(\sqrt[4]{-2})\subset \mathbb{Q}(\sqrt[4]{2},i)$.

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

But this gives a contradiction since $[\mathbb{Q(\sqrt[4]{2},i):Q]}=8$ but $[\mathbb{Q(\sqrt[4]{-2}):Q]}=4$ as the minimal polynomial for $\sqrt[4]{-2}$ is given by $x^{4}+2$. So $i$ does not belong to the $\Bbb{Q}(\sqrt[4]{-2})$

So in the proof you can clearly see that it does not depend on the $k$ chosen as you will end up picking a $-$ or a $+$ sign in the real or imaginary parts in the expression $\sqrt[4]{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})$ . The crux of the argument is that the moment $i$ lies in the field it becomes the splitting field of $x^{4}+2$ and also the splitting field of $x^{4}-2$ which should be a degree $8$ extension and not degree $4$.

Answer 3

This is indeed a confusing case; choosing different fourth roots of $-2$ will in fact give different extensions of $\mathbb Q$. Of course the extensions would be the same if the primitive ratio of the roots, $i$, were in the field ... so I've given the answer away!

In fact there is a simple proof that $i$ will not belong in the field. Since the generator is a fourth-degree algebraic number, any element in $\mathbb Q[r]$ where $r$ is any fourth root of $-2$ must have the form

$a+br+cr^2+dr^3; a,b,c,d\in\mathbb R.$

We then seek a solution to the equation

$(a+br+cr^2+dr^3)^2=-1.$

We expand the square and equate the linearly independent (over the rationals) components:

$r^0: a^2-4bd-2c^2=-1$

$r^1: 2ab-4cd=0$

$r^2: 2ac+b^2-4bd-2d^2=0$

$r^3: 2ad+2bc=0.$

Now solve the $r^1$ and $r^3$ equations for $d$. This gives

$d=ab/2c=-bc/a,$

from which either $b=0$ or $a/2c=-c/a$. The latter possibility fails by inconsistent signs for all nonzero rational numbers $a$ and $c$, so $b$ is forced to zero. Similarly, solving the $r^1$ and $r^3$ equations for $b$ leads to $d=0$.

Then with $b=d=0$ the $r^2$ equation forces either $a=0$ or $c=0$, neither of which then admits a rational solution for the $r^0$ equation. So $\mathbb Q[r]$ with $r^4=-2$ cannot contain a square root of $-1$.

The reader should explore what happens if $-1$ is replaced by a general rational number $q$. Intuitively we might expect the above contradiction to kill the proposition unless $q$ is either a rational square or $-2$ times a rational square. Is that result actually obtained?

Yes, the only rational numbers whose square roots lie in $\mathbb Q[r]$ are as given in the last paragraph.

Answer 4

This answer has many similarities to that of Oscar Lanzi, but I thought a bit more detail might be helpful.

---

It is true that for each root of $\mu^4+2=0$, $\mathbb{Q}[\mu]$ is a different ring, but all four of these rings are isomorphic.

Suppose $\mu^4+2=0$ and $i\in\mathbb{Q}[\mu]$. That is, we have $a,b,c,d\in\mathbb{Q}$ so that $$ \begin{align} 0 &=\left(a\mu^3+b\mu^2+c\mu+d\right)^2+1\\ &=\underbrace{2(ad{+}bc)\vphantom{\left(a^2\right)}}_0\,\mu^3-\underbrace{\left(2a^2{-}2bd{-}c^2\right)}_0\,\mu^2-\underbrace{2(2ab{-}cd)\vphantom{\left(a^2\right)}}_0\,\mu+\underbrace{\left(d^2{-}2b^2{-}4ac{+}1\right)}_0\tag1 \end{align} $$ Since $x^4+2$ is irreducible over $\mathbb{Q}$ (via Eisenstein), $\left\{1,\mu,\mu^2,\mu^3\right\}$ are independent over $\mathbb{Q}$. Thus, each of the coefficients in $(1)$ must be $0$.

Since $ad+bc=0$ and $2ab-cd=0$ we get $$ \begin{align} 0 &=2(ad+bc)^2+(2ab-cd)^2\\ &=\left(2a^2+c^2\right)\left(2b^2+d^2\right)\tag2 \end{align} $$ That is, $a=c=0$ or $b=d=0$.

If $a=c=0$, then $2a^2-2bd-c^2=0\implies b=0$ or $d=0$.
$\quad$if $b=0$, then $d^2-2b^2-4ac+1=0\implies d^2+1=0\quad\Rightarrow\Leftarrow\tag3$
$\quad$if $d=0$, then $d^2-2b^2-4ac+1=0\implies2b^2=1\quad\Rightarrow\Leftarrow\tag4$

If $b=d=0$, then $2a^2-2bd-c^2=0\implies2a^2=c^2\quad\Rightarrow\Leftarrow\tag5$

Cases $(4)$ and $(5)$ rely on the irrationality of $\sqrt2$.

Cases $(3)$-$(5)$ are exhaustive; thus, $i\not\in\mathbb{Q}[\mu]$.

Since we only used that $\mu^4+2=0$, this works for any root: $\mu=\frac{\pm1\pm i}{\sqrt[4]{2}}$.