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Let $F$ = $(\mathbb{Z}/3\mathbb Z)[x]/(x^2-2)$ be the quotient ring generated by the principal ideal $(x^2-2)$. I need to find a root of $x^2+2x+2$ and find all generators of the group $F^*$.

Is $x+1$ a root for the above? I have that $(x+1)(2x) = 2x^2 + 2x = x^2 + x^2 + 2x + 2 - 2 = x^2 + 2x +2$. Is this valid?

As for finding all generators, I am a bit stuck. I could try all possibilities (like $x+1$), but aren't there $3^3 = 27$ cases?

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Is $\,\frac{\Bbb Z}{3}=\Bbb Z/3\Bbb Z=\,$ the field with three elements? –  DonAntonio Dec 16 '12 at 3:40
    
Yes, it is just elements 0, 1, and 2. –  Card Flower Dec 16 '12 at 3:42

1 Answer 1

up vote 1 down vote accepted

Put $\,\Bbb F_3:=\Bbb Z/3\Bbb Z\,$ . Since $\,x^2-2\,$ is irreducible in $\,\Bbb F_3[x]\,$ , the quotient $\,\Bbb F_9:=\Bbb F_3[x]/(x^2-2)\,$ is a field with $\,3^2=9\,$ elements , which can be expressed as

$$\Bbb F_9:=\{\,aw+b\;\;;\;\;a,b\in \Bbb F_3\,\,,\,\,w^2=2\}$$

and with operation modulo $\,3\,$

Now:

$$(w+1)^2+2(w+1)+2=w^2+2w+1+2w+2+2=2+2+2+ w +1=w+1\neq 0\Longrightarrow $$

$\,w+1\,$ is not a root or $\,x^2+2x+2\,$ , but

$$(w+2)^2+2(w+2)+2=w^2+w+1+2w+1+2=0\Longrightarrow w+2$$

is a root of that polynomial.

Now, just choose an element of $\,F_9^*\,$ of order $\,8\,$ for a generator.

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Sorry, can you explain what you mean by order 8? –  Card Flower Dec 16 '12 at 3:56
    
Well...yes, but if you don't know this then this exercise may be out of what you've covered in your studies. Are you studying in some university or just doing some self-study...? Basically, this is: if $\,F\,$ is a field, then $$p(x)\in F[x]\,\,\text{irreducible}\,\Longleftrightarrow (p(x))\leq F[x]\,\text{is a maximal ideal}\,\Longleftrightarrow\,\text{the quotient ring}\,\,F[x]/(p(x))\,\,\text{is a field}$$ and in case we have a finite field $\,|F|=q\,$ ,then the field $\,F[x]/(p(x))\,$ has $\,q^{\deg p}\,$ elements, which is our case here. Any more explanation's out of this site's scope –  DonAntonio Dec 16 '12 at 10:44

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