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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.

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    $\begingroup$ $\mathbf F_9$ is not the residue class ring $\mathbf Z/9\mathbf Z$ \ which can't be a field since $9$ is not a prime number. $\endgroup$ – Bernard Jan 1 at 12:24
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$\Bbb F_9$ 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 $x^2=2$: $$ (x+2)(2x+2)=2x^2+6x+4\\ =2x^2+1=2\cdot2+1=2 $$

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    $\begingroup$ @greedoid Better now? $\endgroup$ – Arthur Jan 1 at 10:56
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    $\begingroup$ @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 . $\endgroup$ – DonAntonio Jan 1 at 10:56
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    $\begingroup$ @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. $\endgroup$ – DonAntonio Jan 1 at 10:57
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    $\begingroup$ @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 $\Bbb F_3=\Bbb Z/(3)$. $\endgroup$ – Arthur Jan 1 at 11:01
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    $\begingroup$ @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. $\endgroup$ – DonAntonio Jan 1 at 11:05
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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.

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

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