Why is $\mathbb{F}_{9}^*$ a multiplicative group

Question

Since $\mathbb{F}_9$ is a field, its units $\mathbb{F}_{9}^* = (1,2,3,4,5,6,7,8)$ should form a multiplicative group. However in this group $3 \times 3 = 0 \notin \mathbb{F}_{9}^*$. I'm trying to understand how this is possible. Don't rush on me since I'm new to the literature.

Answer 1

$\Bbb F_9$ is a quotient ring of the polynomial ring $\Bbb F_3[X]$. As such, the elements of $\Bbb F_9$ are written as $a+bX +(f)$ where $a,b\in\Bbb F_3$ and $f$ is an irreducible quadratic polynomial over $\Bbb F_3$. Usually we shorten this to $a+bx$, where $x$ is thought of one of the two roots of $f$.

Addition is done the regular way, and multiplication is done as with regular polynomials, then reduced through $f$ to be on the above form again. Exactly which $f$ you choose is up to you, but be consistent.

The elements of $\Bbb F_9^\times$ are $$1,2,\\x,x+1,x+2,\\2x,2x+1,2x+2$$ An example of multiplication, using $f(X)=X^2-2$, meaning $x^2-2=0$, or $x^2=2$: $$ (x+2)(2x+2)=2x^2+6x+4\\ =2x^2+1=2\cdot2+1=2 $$

Answer 2

The error is that $\Bbb{F}_9$ is not $\Bbb{Z}/9$. For any field $K$ we have that $K^{\times}$ is a multiplicative group because $K$ is a field. But $\Bbb{Z}/9$ is not a field, as $3\cdot 3=0$ and $3\neq 0$.

Reference: This duplicate.