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I know that for $E=\mathbb{Z}_3[x]/\langle x^2+x+2\rangle $, since $x^2+x+2$ is irreducible over $\mathbb{Z}_3$, $E$ is a field, because $\langle x^2+x+2\rangle $ is a maximal ideal.

In addition, $E$ is a field of order $3^2$, and $\mathbb{Z}_3$ is a field of order 3.

For this reason, $E=GF(3^2)$ and $\mathbb{Z}_3=GF(3)$.

Now, considering $E$ as additive group, $E \cong \mathbb{Z}_3\oplus \mathbb{Z}_3$. We see that every element $(a,b) \in \mathbb{Z}_3\oplus \mathbb{Z}_3$ has order $|(a,b)|=\mathrm{lcm}(|a|,|b|)$. Also, for $a$ and $b$ in $\mathbb{Z}_3$, they must have order 1 or 3. Then, apart from $(0,0)$, all elements in $E$ have order 3. Therefore, there would be 8 elements of order 3, and none of order 2.

In this part I am confused because $E=(\mathbb{Z}_3[x]/\langle x^2+x+2\rangle ) \cong \mathbb{Z}_3(i)$. Then all elements of $\mathbb{Z}_3(i)$ other than zero, have order 3(?) because it does not seem to be the case.

Finally, I am also asked for how many generators does $E$ have. I suppose that here we need to consider the multiplicative group of $E$, which is isomorphic to $\mathbb{Z}_8$, right? How do I move ahead? Do I consider $\mathbb{Z}_8$ as a cyclic additive group generated by 1, and I count the coprimes with 8? I think I am very confused swapping from additive to multiplicatie groups. For example, in this last part, I considered $E$ as multiplicative group but it is isomorphic to an additive group(?).

Robin
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Camilo Diaz
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    It may be helpful to to think of the multiplicative group $E^$ as being more like the group of eighth roots of unity. That is, like the group $$\mu_8={\zeta^i\mid 0\le i<8}\le \Bbb{C}^,$$ where $\zeta=(1+i)/\sqrt2$. Observe that in characteristic three $2=-1$, so $\sqrt2$ becomes $\sqrt{-1}=i$. Indeed, modulo three $$x^2+x+2\equiv x^2-2x+2=(x-1)^2+1,$$ so the zeros of $x^2+x+2$ are a lot like $1\pm i$. Furthermore, $(1+i)/i=1-i$. It follows that in the field $E$ the element $g=x+\langle x^2+x+2\rangle$ behaves just like the complex number $(1+i)/\sqrt2$. – Jyrki Lahtonen Jun 12 '24 at 08:17
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    (cont'd) That may have felt like a lot of hot air. To get solid ground under your feet you can verify as an exercise that if in any field $K$ we have elements $u$ and $v$ such that $u^2=-1$ and $v^2=2$, then $g=(1+u)/v$ satisfies the equations $g^4=-1$ and $g^8=1$. – Jyrki Lahtonen Jun 12 '24 at 08:19

2 Answers2

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A finite field $K$ has $\mathbb{F}_p=\mathbb{Z}/\mathbb{Z}p$ as its prime field for some prime number $p>0$ which is its characteristic.

Thus $px=0$ for all $x\in K$ which means that every element of $K$ has additive order dividing $p$ and since $p$ is prime $$ {\rm ord}(x)=\left\{ \begin{array}{ll} 1 & \text{if $x=0$,}\\ p & \text{if $x\neq0$.} \end{array}\right. $$ It follows that $$ (K,+)\simeq(\mathbb{Z}/\mathbb{Z}p)^d $$ where $d=\dim_{\mathbb{F}_p}(K)$. In particular $(K,+)$ is not cyclic, i.e. no generators, as soon as $d>1$.

The multiplicative order is a different story: one knows that the multiplicative group $K^\times$ is cyclic of order $q-1=p^d-1$ ($d$ as above). So the choice of a generator defines an isomorphism $$ (K^\times,\cdot)\simeq(\mathbb{Z}_{q-1},+) $$ (which some call a discrete logarithm). From this we get at once that $K^\times$ has elements of order $r$ if and only if $r$ is a divisor of $q-1$, and there are exactly $r$ of them.

Andrea Mori
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Then all elements of $\mathbb{Z}_3(i)$ other than zero, have order 3?

Yes, that's exactly the case. After all, additively $\mathbb Z_3(i)=\mathbb Z_3\oplus\mathbb Z_3i$. Here $i^2=-1$ doesn't play any role, other than it's linearly independent with $1$ over $\mathbb Z_3$. In fact this is a more general fact: If $V$ is a vector space over $k$ where $k$ is a field of characteristic $p$, then every nonzero element of $V$ (as an abelian group) has order $p$. This is because for any $b\in V\setminus\{0\}$, $v+v+\cdots+v=nv=0$ iff $n=0$ in $k$ iff $p|n$ in $\mathbb Z$.

how many generators does $E$ have?

Do you mean the number of $\alpha\in E$ such that $E=\mathbb Z_3(\alpha)$, i.e. the number of primitive elements for the extension? If so, the answer is $|E|-|\mathbb Z_3|=6$, and we do not have to think about the multiplicative group $E^*$. The idea is simple, if $\alpha\not\in \mathbb Z_3$, then $$[E:\mathbb Z_3]\ge [\mathbb Z_3[\alpha]:\mathbb Z_3]\ge 2 = [E:\mathbb Z_3] \\ [\mathbb Z_3[\alpha]:\mathbb Z_3]=2, \mathbb Z_3[\alpha]=E$$

Similarly given any finite field extension: Let $E/F$ be a finite extension over a finite field $F$, then $\alpha\in E$ is a primitive element generator over $F$ iff $\alpha$ is not in any subfield $F'$ such that $E\subseteq F'\subsetneq F$. Therefore we can count the number of primitive elements generators by inclusion-exclusion principle.

If you are asking for the number of generators for $E^*\simeq\mathbb Z/8\mathbb Z$ (which is not isomorphic to the additive group: they don't even have the same order; and those generators are unfortunately called primitive elements, not to be confused with primitive elements for simple extension, as mentioned by Jyrki Lahtonen in the comments), then yes it is Euler $\phi$ of $8$ which is $4$. However, this is not the number of primitive elements generators, as primitive elements generators can generate the whole field by both multiplication and addition.

Just a user
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    The math here is ok, and you do cover all the interpretations! Just adding a warning about a terminological conflict (which reappears on this site frequently). In the context of finite fields a primitive element is a generator of the multiplicative group. In other words, addition is not allowed when generating. So the field $\Bbb{F}_9$ has $\phi(8)=4$ primitive elements, and $\Bbb{F}_3$ has one primitive element (namely $-1=2$). My elaboration. – Jyrki Lahtonen Jun 12 '24 at 10:22
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    @JyrkiLahtonen Oh, thank you! I'm not aware of the abuse of the terminology. I guess primitive elements for finite fields generalize primitive roots in elementary number theory, which make sense, but still an awkward coincidence. – Just a user Jun 12 '24 at 10:24
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    It is, indeed, a bit unfortunate. On this site I have witnessed several occasions when quite knowledgeable users misunderstand each other. I haven't studied the history, but my best guess is that the finite field usage is an extension of the convention of calling a generator of $\Bbb{Z}_p^$ a primitive root modulo $p$*. – Jyrki Lahtonen Jun 12 '24 at 10:28