I realised that I don't know how to construct $\mathbb{F}_9$. I'm guessing that $\mathbb{F}_9 = \mathbb{F_3(\theta)}$, where $\theta$ is the root of some irreducible polynomial over $\mathbb{F}_3[x]$ of degree two?
Must I even construct $\mathbb{F}_9$ in order to determine whether it contains a 4th root of unity or is there some other simpler way I'm missing?
$\endgroup$ 16 Answers
$\begingroup$HINT: To make that conclusion there is no need to explicitly constuct it. Use the fact that the units in any finite field constitute a cyclic group. Now can the cyclic group in your example contain an element of order 4?
$\endgroup$ $\begingroup$A non-trivial $4$th root of unity is a square root of $-1$, since $(x^4-1)=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)$, and $x^2+1$ is irreducible over $\mathbf F_3$. So the answer is yes:$$\mathbf F_9\simeq \mathbf F_3[x]/(x^2+1),$$and if you denote $\omega$ the congruence class of $x$, the non-trivial $4$th roots of unity are $\omega$ and $-\omega$.
$\endgroup$ $\begingroup$$$\mathbb F_9\cong\mathbb F_3[X]/(X^2+1)\cong \mathbb F_3[\alpha ],$$where $\alpha ^2=-1$. In particular, $\alpha ^4=1$.
Just for information, $$\mathbb F_9=\{0,1,2,\alpha ,2\alpha ,1+\alpha ,2+\alpha ,1+2\alpha ,2+2\alpha \}.$$
$\endgroup$ $\begingroup$No construction necessary. The elements of $GF(p^n)$ are exactly the zeros (splitting field) of the polynomial $x^{p^n}-x$ over $GF(p)$. In particular, the nonzero elements of $GF(p^n)$ are exactly the roots of $X^{p^n-1}-1$ and they form a cyclic group of order $p^n-1$. E.g., the nonzero elements of $GF(9)$ are the zeros of $X^8-1$ and form a cyclic group of order 8. If $a$ is a primitive generator, then $a^2$ has order 4.
$\endgroup$ $\begingroup$The multiplicative group of a finite field is cyclic. Therefore the nonzero elements $\Bbb F_9$ form a cyclic group of order $8$ under multiplication. Therefore one of them has order $4$, that is it is a primitive $4$-th root of unity.
But that gives an easy way to construct $\Bbb F_9$. As such a fourth root of unity satisfies $\alpha^2=-1$, consider $\Bbb F_3[X]/\left<X^2+1\right>$.
$\endgroup$ $\begingroup$There are already several good answers, but here is another way. It is known from Fermat's theorem on sums of two squares that$\,x^2 \equiv -1 \pmod p\,$ has no solutions in integers if $\,p = 4n+3.\,$This applies in our case $\,p=3\,$ and so we extend $\,\mathbf F_p\,$ with an element $\,X\,$ such that $\,X^2 = -1.\,$ Now $\,X^4 = 1\,$ which implies $\,X\,$ and $\,-X\,$ are two $4$th roots of unity. The other two $4$th roots of unity are $\,1\,$ and $\,-1\,$. Historically this construction was first used to extend the real numbers into the complex numbers, which have $n$th roots of unity for every positive integer $\,n.\,$
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