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?
On
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.
On
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.
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 $2n = 6$.