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What is meant when one says that a finite field contains an $n$th root of $a$? Is it an exquisite way to say that the multiplicative group of the field has an element $x$ such that $x^n=a$? If $a=-1$, is it equivalent to saying that the group contains an element of order $2n$?

What if the field is infinite?

user21820
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user557
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3 Answers3

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Yes. An $n$-th root of $a$ is some solution of $x^n = a$. The field could be finite or infinite.

An $n$-th root of $1$ is some solution of $x^n=1$. For example, it could be $1$. There is no implicit meaning that an $n$-th root of $1$ actually has order $n$. An $n$-th root of $-1$ need not have order $2n$. For example, a $3$rd root of $-1$ is the name for a solution of $x^3 = -1$. One such solution is $-1$, which in characteristic not $2$ has order $2$, not $2n = 6$.

KCd
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  • Then would it be correct to say that if $x$ is an $n$th root of $-1$ in a field, then the field contains an element that has order $2n$ in the multiplicative group? – user557 Jun 16 '18 at 01:08
  • No. Reread what I wrote. The number $-1$ is a 3rd root of $1$, so $\mathbf Q$ has a 3rd root of $1$, but the number $-1$ does not have order $6$. – KCd Jun 16 '18 at 01:29
  • Thanks. Hm.. My questions stem from the problem which I'm working on. It assumes that a finite field has a root of $-1$, and the solution I was told begins by saying that its multiplicative group "has an element of order 4 (a root of $-1$)". Seems that in view of your replies, this claim is wrong... – user557 Jun 16 '18 at 02:12
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    No, that claim is right! That solution was much more specific, and you chose to ask a far too general question here. The "root of $-1$" meant a square root of $-1$. If you are in a field of characteristic not $2$, so $-1 \not= 1$ in the field, and there is an element $x$ in the field such that $x^2 = -1$, then $x$ has order $4$. Check that yourself. – KCd Jun 16 '18 at 05:20
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Supplementing the other answers somewhat because the OP's comment to KCd's answer brought up the need for this.

We can say a bit more!

Undoubtedly you have seen the following result when studying cyclic groups.

Fact 1. If $c$ has order $n$, then the order of $c^k$ is given by the formula $$\operatorname{ord}(c^k)=\frac{n}{\gcd(n,k)}.$$

This leads to the following (I have used it many times when answering a question about finite fields).

Fact 2. Assume that $p$ is a prime, $x^p=a$, and that $k=\operatorname{ord}(a)$ is divisible by $p$. Then $$\operatorname{ord}(x)=pk.$$

Proof. Let $m=\operatorname{ord}(x)$. The first fact tells us that $$ k=\frac{m}{\gcd(m,p)}. $$ Because $p$ is a prime that $\gcd$ is either $1$ or $p$, so we have either $k=m$ or $k=m/p$. Or, equivalently $m=k$ or $m=pk$. But, $p\mid k$, so $$ x^k=(x^p)^{k/p}=a^{k/p}\neq1. $$ Therefore $m\neq k$, so $m=pk$. QED.

Example. If $K$ is a field of characteristic $\neq2$ (so that $-1\neq1$), and $a^2=-1$, then $a$ has order $4$.

So the first fact always constrains the orders of roots of an element of a known order. We similarly see that a cubic root of an element of order $9$ has order $3\cdot9=27$. But, a cubic root of an element of order $8$ has order either $8$ or $24$. In fact, both possibilities occur as you can easily verify in the field of complex numbers. Cubing permutes the solutions of $z^8=1$, so one cubic root of an element of order eight also has order eight.

Jyrki Lahtonen
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Let $F$ be a field. Let $a \in F$. Let $n \in \Bbb N$.

We say that $F$ has an nth root of $a$ if there is $x \in F$ such that $x^n = a$ (we do not require it to be in the multiplicative group).

If $a = -1$ and $F$ is not of characteristic $2$, then if $x^n = -1$ then we know that the order of $x$ divides $2n$. It does not mean that $x$ has order $2n$ (e.g. $x=-1$, $n=3$).

It does not matter if the field is finite or infinite.

Kenny Lau
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