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

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.

• $\mathbf F_9$ is not the residue class ring $\mathbf Z/9\mathbf Z$ \ which can't be a field since $9$ is not a prime number. – Bernard Jan 1 at 12:24

$$\Bbb F_9$$ 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 $$x^2=2$$: $$(x+2)(2x+2)=2x^2+6x+4\\ =2x^2+1=2\cdot2+1=2$$

• @greedoid Better now? – Arthur Jan 1 at 10:56
• @greedoid This is Theory of Fields extensions, which many times goes together (in fact, a little before) Galois Theory in middle undergraduate studies in mathematics. Are you there? Because otherwise simply telling you that $\;x\;$ is a root of an irreducible quadratic in $\;\Bbb F_3[X]\;$ won't likely help . – DonAntonio Jan 1 at 10:56
• @Arthur Using the same symbol $\;x\;$ for the unknown of the polynomial ring $\;\Bbb F_3[x]\;$ and the elements in $\;\Bbb F_9\;$ is not a good idea, imo. – DonAntonio Jan 1 at 10:57
• @DonAntonio It's a common abuse of notation to use the same symbols for elements of a ring and the corresponding elements in some quotient ring. Just see how easily we use $2$ when talking about elements of $\Bbb F_3=\Bbb Z/(3)$. – Arthur Jan 1 at 11:01
• @Arthur I know that...but even in that case it is usual to point out that those are representatives of polynomials (and that's why that $\;x\;$ there usually) in the usual representation as quotient: $\;\Bbb F_9\cong \Bbb F_3[x]/\langle x^2+1\rangle\;$ (or modulo any other irreducible quadratic modulo $\;3\;$, of course). As it is, it could be highly confusing for greedoid and/or for the OP, I believe. – DonAntonio Jan 1 at 11:05

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.

• Can you write down explicitly this group, please. – Aqua Jan 1 at 10:48
• Which group? We have three groups, $(K,+)$, $(K^{\times},\cdot)$ and $\Bbb{Z}/9$. – Dietrich Burde Jan 1 at 10:49
• $K^x$................... – Aqua Jan 1 at 10:49
• See Arthur's answer! – Dietrich Burde Jan 1 at 10:51