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I'm trying to prove that $K$= $\frac{\mathbb{Z}_{3}[T]}{(T^{2}+2T+2)}$ is a commutative field. So I thought i needed to prove that $K$ is commutative and $K$ is a ring with unit element so that for every $x \in (K_{0})$ there is an $y \in K$ so that $xy=e=yx$.

I'm really sorry but I have no clue how to start this. I can't find an similair execrices to base me on. Can someone maybe just give me an regular way to prove this so that I can make this exercise by myself en edit it here how I did it?

I found this online How to show that a finite commutative ring without zero divisors is a field?

But i can't do it with my exercise... I hope I can make the exercise with you're hints. Thanks a lot.

questmath
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    The important bit is the fieldness, not the commutativity. Is $T^2+2T+2$ irreducible over $\Bbb Z_3$? – Angina Seng Aug 25 '20 at 09:34
  • yes because it's not constant and has no points that it become zero – questmath Aug 25 '20 at 09:38
  • If $k$ is a field then $k[X]/(p)$ is a field iff $p$ is irreducible. – Noel Lundström Aug 25 '20 at 12:41
  • so i need to prove that $\mathbb{Z_{3}}$ is a field? – questmath Aug 25 '20 at 12:47
  • That's right! You can just check it manually if you want or use Bezouts identity. – Noel Lundström Aug 25 '20 at 13:22
  • @mathmath Or you can use the theorem you linked in your question, If $\Sigma a_i T^i $ is a polynomial of degree $n > 2$ it is equivalent mod $T^2 + 2T + 2$ to a polynomial of degree $n-1$, $(\Sigma a_i T^i) - (a_n T^{n-2})(T^2 + 2T^{1} + 2)$ and then by induction every polynomial in $\mathbb Z_3 [T]$ is equal to a polynomial of degree $2$ or less which there are finitely many of. Thus your ring $\mathbb Z_3 [T]/(T^2 + 2T +2)$ is finite and is an integral domain by the fact that $T^2 + 2T +2$ is irreducible. – Noel Lundström Aug 25 '20 at 16:03

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As people have suggested in the comments, first we prove that $\mathbb{Z}_3$ is a field. Since it only has three elements, we can just manually compute: $\overline{2}\cdot \overline{2} = \overline{1}$ and $\overline{1} \cdot \overline{1} = \overline{1}$. Since those are the only non-zero elements in $\mathbb{Z}_3$, and $\mathbb{Z}_n$ is a commutative ring with unity for all $n \in \mathbb{N}$, we get the desired result.

Note now that $p(x) = x^2 + 2x + 2$ has no roots in the field $\mathbb{Z}_3$ (you can just test all of the possibilities). Since it has degree $2$, this means it must be irreducible.

Using the first result we proved, $\mathbb{Z}_3[T]$ is an integral domain, which means it is a commutative ring with unity. And, since $p(x)$ was irreducible, $(p)$ is a maximal ideal. We can now apply the following theorem:

Theorem: If $A$ is a commutative ring with unity and $J$ is a maximal ideal of $A$, then $A/J$ is a field

Gauss
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    you have an error: even though $p(x)$ has no integer roots, it is still possible for it to be reducible over a finite field. For example, take $p(x)=x^2 +1$ over $\mathbb{Z}/5$. – Brian Shin Aug 27 '20 at 13:26
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    I'm sorry - I meant "integer in the finite field". I just didn't really have a term for such an object... I will edit my answer forthwith. Thank you for pointing it out! – Gauss Aug 27 '20 at 13:40
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    it looks perfect now :) – Brian Shin Aug 27 '20 at 13:54