How many solutions can $x^4=-1$ have on a field of characteristic $0$?

Question

I am looking at the equation $x^4=-1$ over a generic field $k$ of characteristic $0$. If $k$ is algebraically closed, it will certainly have $4$ roots. I think that they should all be different, but I am not completly sure about this. Could someone tell me if this is the case, and explain why? If we are looking at the complex numbers for example, it is clear to me why they are all different. But since $k$ is unknown, I don't really know how to deal with this.

Now let's look at the case where $k$ is not algebraically closed. In this case, it could be that the equation had no solutions in the ground field (for example if $k=\mathbb{Q}$). But could it be that it had only two solutions? Again, I think that the answer is no, but wouldn't be able to argue it.

In both cases, my intuition comes from this answer, and also from the fact that I know that every field of characteristic $0$ must contain a copy of $\mathbb{Q}$. So it cannot do very weird things with the solutions of this equation.

Edit: I am aware that I may not have tagged the question properly. Feel free to add/remove tags.

Answer 1

Let $F$ be a field of characteristic not $2$.

All the roots of $x^4+1=0$ are distinct, and if $\alpha$ is one such root, the set of the $4$ distinct roots are $\{\alpha, -\alpha, \alpha^3, -\alpha^3\}$.

We next establish this, by first noting that if $\alpha$ is a root of $x^4+1$, then so is both $-\alpha$ and $\alpha^3$. Indeed, that $-\alpha$ is also a root is easy to check. So to see that $\alpha^3$ is also a root, note that $$\alpha^4=-1 \implies (\alpha^4)^3 = (-1)^3$$ $$\implies (\alpha^3)^4 = (-1)^3 = -1.$$ Next note that $\alpha^3$ is not in the set $\{-\alpha, \alpha\}$. Indeed, if on the one hand the equation $\alpha^3=-\alpha$ is true, then [multiplying both sides by $\alpha$ and noting $\alpha^4=-1$] the string of equations $-1=\alpha^4 = -\alpha^2$ must also be true, giving $\alpha^2=1$ which [in a field of characteristic not $2$] implies $\alpha^4=1 \not = -1$, a contradiction. If on the other hand the equation $\alpha^3=\alpha$ is true, then [multiplying both sides by $\alpha^{-1}$] this would give $\alpha^2 = 1$ , which gives $\alpha^4=1 \not = -1$, which is also a contradiction. Likewise, $-\alpha^3$ is not in $\{-\alpha, \alpha\}$ either.

As $\alpha, \alpha^3$ are nonzero [as $0$ is not a root of $x^4+1$] and $\alpha \not = -\alpha$ and $\alpha^3 \not = -\alpha^3$ [due to $F$ not being characteristic $2$], this gives $\{\alpha^3,-\alpha^3\}$ and $\{\alpha, -\alpha\}$ nonintersecting [by the above paragraph] sets with $2$ distinct elements each, which gives $\alpha, -\alpha, \alpha^3, -\alpha^3$ the $4$ distinct roots of $x^4+1$.

The field $F$ has either all $4$ roots or none.

So now it remains to show that if one roots $\alpha$ is in $F$ then all $4$ roots are. However, if $F$ contains one root $\alpha$ of $x^4+1=0$, then $F$ must also contain $-\alpha$ and $\alpha^3$ [by definition of a field] and thus $-\alpha^3$. So as $\alpha, -\alpha, \alpha^3, -\alpha^3$ are the $4$ distinct roots of $x^4+1$, indeed $F$ contains all $4$ roots of $x^4+1$ if it has at least one, and we are done.

Finally, as noted in the OP, we note that $\mathbb{Q}$ contains no roots of $x^4+1$, whereas $\mathbb{C}$ contains all $4$ roots. So $F$ containing either all $4$ roots of $x^4+1$ or no roots is possible, as the OP already noted above.

Answer 2

For the first question, you know that if $\operatorname{char} k = 0$ any polynomial $p$ has no double root iff $\gcd (p,p') = 1$, so in your case by applying Euclides algorithm you get $$\begin{align*} x^4 + 1 = \frac{x}{4} \cdot 4x^3 + 1/4 \\ 4x^3 = 16x^3 \cdot \frac{1}{4} +0 \end{align*}$$ So $\gcd (p,p') = 1$ up to multiplication by a scalar (I wrote $p = x^4 + 1$). Hence all possible roots are distinct because the gcd is invariant with field extension. So if $(x-r)| x^4$ then it does not divide $4x^3$ hence it is not root of $4x^3$ so every root is singular. Also you don’t need this since we explicitely prove later that we can find 4 distinct roots if we have one.

Now suppose you have a root $j \in k$, $i=j^2$ is a root of $x^2+1$ and $(j^3)^4 = (-1)^3 = -1$ so $j^3$ is also a root. Then $(-j)^4 = j^4$ so $-j$ is also a root as well as $-j^3$.

Let us check that they are different:

• $j-(-j) = 2j \neq 0$ since $\operatorname{char} k =0$ and $j \neq 0$ since $j$ is not a root of $x$, hence (j \neq -j).
• $(j-j^3) = j(1-i)$ and $1$ is not a root of $x^2+1$ so $1 \neq i$ hence $j \neq j^3$
• So on the other are the same, very direct.

Hence either you have $0$ roots or you have every $4$ distinct roots.

Answer 3

Let $k$ be a field of characteristic $0$. Then $x^4=-1$ has at most $4$ roots. If $\alpha\in k$ is a root of $x^4=-1$ then $\alpha^8=1$. Then also \begin{eqnarray} (\alpha^3)^4&=&\alpha^{12}&=&\alpha^8\alpha^4&=&\alpha^4=-1,\\ (\alpha^5)^4&=&\alpha^{20}&=&(\alpha^8)^2\alpha^4&=&\alpha^4=-1,\\ (\alpha^7)^4&=&\alpha^{28}&=&(\alpha^8)^3\alpha^4&=&\alpha^4=-1,\\ \end{eqnarray} and so $\alpha^3$, $\alpha^5$ and $\alpha^7$ are also roots of $x^4=-1$. To see that all four are distinct, first note that $$\alpha^5=\alpha^4\alpha=-\alpha\qquad\text{ and }\qquad \alpha^7=\alpha^4\alpha^3=-\alpha^3,$$ and so $\alpha^5\neq\alpha$ and $\alpha^7\neq\alpha^3$, because of course $\alpha\neq0$. Similarly, if either $$\alpha=\alpha^3\qquad\text{ or }\qquad \alpha=\alpha^7\qquad\text{ or }\qquad \alpha^5=\alpha^3\qquad\text{ or }\qquad\alpha^5=\alpha^7,$$ then it follows that $\alpha^2=1$, but then also $\alpha^4=1$, a contradiction. So $\alpha,\alpha^3,\alpha^5,\alpha^7\in k$ are the four distinct roots of $x^4=-1$ in $k$. This also immediately shows that $k$ either contains four roots of $x^4=-1$ or none at all, because of course $\alpha^3,\alpha^5,\alpha^7\in k$ if $\alpha\in k$.