# Does $\mathbb{F}_9$ contain a 4th root of unity?

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?

• The multiplicative part of a finite field is always a cyclic group: since $\mathbb{F}_9^*$ has $8$ elements, it clearly contains a fourth root of unity. – Jack D'Aurizio Feb 18 at 17:12

## 6 Answers

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?

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 $$-\omega$$.

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.

$$\mathbb F_9\cong\mathbb F_3[X]/(X^2+1)\cong \mathbb F_3[\alpha ],$$ where $$\alpha ^2=-1$$. In particular, $$\alpha ^4=1$$.

Just for information, $$\mathbb F_9=\{0,1,2,\alpha ,2\alpha ,1+\alpha ,2+\alpha ,1+2\alpha ,2+2\alpha \}.$$

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 $$\Bbb F_3[X]/\left$$.

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 $$\,n.\,$$