List the distinct elements in the ring $\mathbb{Z}[x]/\langle 3 , x^2+1 \rangle$. I tried to do this problem in the following way:
As, $x^2+1 + \langle 3 , x^2+1 \rangle= 0 + \langle 3 , x^2+1 \rangle \implies x^2+1 \equiv 0 \implies x^2 \equiv -1.$
Also, $3+ \langle 3 , x^2+1 \rangle=0 +\langle 3 , x^2+1 \rangle \implies 3 \equiv 0$.
Now, any element of  $\mathbb{Z}[x]/\langle 3 , x^2+1 \rangle$ is of form $p(x)+\langle 3 , x^2+1 \rangle$ where $p(x) \in \mathbb{Z}[x]$. So, divide $p(x)$ by $x^2+1$ to get a linear polynomial $ax+b$ as a remainder. So, any element of $\mathbb{Z}[x]/\langle 3 , x^2+1 \rangle$ can be written as $ax+b+ \langle 3 , x^2+1 \rangle$ where $a,b \in \mathbb{Z}/3\mathbb{Z}.$ Let $I=\langle 3, x^2+1\rangle$. So, the elements of the ring are $I, 1+I, 2+I, x+I,(x+1)+I,(x+2)+I,2x+I,(2x+1)+I,(2x+2)+I.$
Is my solution correct?
Thanks!
 A: Looks right, except for one nit-picky thing: the $a$ and $b$ in your $ax + b$ aren't elements of $\Bbb Z/3\Bbb Z$, they're elements of $\{0,1,2\}\subseteq\Bbb Z$ (you could also take $a,b\in S$ for any fixed system $S$ of representatives of $\Bbb Z/3\Bbb Z$ in $\Bbb Z$).
A: Well, note that all your polynomials have the condition that their coefficients are $0,1,2$ and are of degree $0,1$, because $x^2+1=0$. So you have that the elements without $x$ are $0,1,2$. Then you consider the monic elements with $x$, that are $x,2x$. Finally you sum all the possible combinations of the firsts with the seconds:
$$x+1,x+2,2x+1,2x+2$$
Then all the elements of your ring are $0,1,2,x,2x,x+1,x+2,2x+1,2x+2$ (they're 9), so yes, your answer is right. 
Is clear to see that all of them are different because: 
If any different of them of degree $0$ is equal, then you will have that $1$ or $2$ is equal to $0$. If any of degere $0$ is equal to another of degree $1$, then $x$ will be equal to $1$ or $2$, and the degree of all your polynomials will be $0$. In the same way, all the polynomials of degree $1$ won't be equal between them, or you will have the same case as before. 
(I didn't add the $I$ of the ideal because we can consider all the elements I put here as the classes of the elements)
