Given a ring $R$ and an ideal $I \triangleleft R$ i'm trying to make some conclusions about the quotient $I/R$.
Let for instance $R=\mathbb Q[x]$ and $I=(x^4-16)$. I'm looking for a zero divisor unequal to 0 to show that the quotient $R/I$ is not a field. Is it true to say $x=\pm 2$ since $(\pm 2)^4$=16?
Now let $R = (\mathbb Z / 8 \mathbb Z)[x]$ and $I=(x^3+x+1)$. Again I would like to find a zero divisor to show that $R/I$ is not a field, but I would also like to show that $x \in R/I$ is a unit. Could someone model this answer for me? That would be very kind. I have incredible difficulty working on any $\mathbb Z/n \mathbb Z$ and I am not sure how to do either of this.
Finally, I thought of letting $R = \mathbb Z[x]$ and $I=(x^2+2)$. In this case, I would like to find an element that does not have an inverse, and show that the quotient $R/I$ is a identity ring. Here, I was considering two somewhat related cases. First of all, say I had $(x^2+2)$, then that would be $\mathbb Z[i]$. If it were $(x^2-2)$ then it would just be $\{ a+b \sqrt 2 : a,b\in\mathbb Z\}.$ Can I use either of these to conclude anything about $R/I$?
First example. You should find a polynomial, but not substitute any number in $x$. $x^4 - 16 = (x^2 - 4)(x^2 + 4)$ so $x^2-4$ and $x^2+4$ are zero divisors.
Second one. $x(-x^2-1) = -x^3-x = 1$ so $x$ is a unit. To find a zero divisor, try to find a factorisation of $x^3+x+1$ over $\mathbb{Z}_8$. This is a key to your problem as it follows from the previous example.
The third one. Yes, $R/I = \mathbb{Z}[\sqrt{2}i]$. So the analogy with $\mathbb{Z}[i]$ can help.