$\Bbb Z_6[x] / (x^2+5)$ quotient ring I am trying to understand quotient rings and it would be really helpful if someone could show me a general way of mapping quotient rings to simpler rings.
Consider $\Bbb Z_6[x] / (x^2+5)$. I know with coefficients in $\Bbb Z_6$ we can factor
$(x^2+5) = (x-1)(x-5)$. 
My guess is the answer is $\Bbb Z_6 \times \Bbb Z_6$ if we define the homomorphism $\varphi: f(x) \rightarrow (f(1),f(5)) \in \Bbb Z_6 \times \Bbb Z_6$. Am I correctly applying the first isomorphism theorem?
What about $\Bbb Z_4[x] / (x^2+1)$ which isn't reducible in $\Bbb Z_4$. I know the cosets are of the form $ax+b$. I think the quotient is a field since $(x^2+1)$ is irreducible. But I read somewhere that all finite fields have set order that is prime, and so I'm confused.
I do not have a lot of intuition here.
 A: We can think of $\mathbb{Z}_6[x]/(x^2 + 5)$ as a simple overring of $\mathbb{Z}_6$ which introduces a "new" root of $x^2 + 5$, and we may represent this as $\mathbb{Z}_6[r]$.  As you point out $\pm 1 \in \mathbb{Z}_6$ are already roots, as would be evident if we wrote instead $x^2 + 5 = x^2 - 1$.  Ring element $r$ does not belong to $\mathbb{Z}_6$.
Let $r$ be this new root of $x^2 - 1$, and work out the details of the addition and multiplication for this finite ring (with zero divisors, obviously).  The addition is somewhat like your idea, but involves two copies of $\mathbb{Z}_6$, i.e $a + br$ where $a,b \in \mathbb{Z}_6$ is a typical element; addition is done componentwise.
For multiplication we need only apply that $r^2 = 1$ (since $r^2 + 5 = 0$ as a result of the quotient ring homomorphism).  As a result of expressing $r^2$ in lower order terms, everything in $\mathbb{Z}_6[r]$ is of the form $a + br$.
Here's a way to think about this generally.  If we have a ring $R$ and ideal $I$ of $R[x]$ for indeterminate $x$ such that $R \cap I = \{0\}$, then the quotient ring $R[x]/I$ is isomorphic to simple overring $R[r]$ of $R$, where $r$ is the image of $x$ in an "evaluation homomorphism" mapping $R[x]$ to $R[r]$ by $f(x) \mapsto f(r)$.  The kernel of this homomorphism is $I$.
A: This is not correct. The map $\mathbb{Z}/6\mathbb{Z}[x]/[x^{2}+5]$ gives you a ring with 36 elements, $ax+b,a,b\in \mathbb{Z}/6\mathbb{Z}$. Using Chinese remainder's theorem, this only gives you $$
\mathbb{Z}/6\mathbb{Z}[x]/[x^{2}+5]=\mathbb{Z}/6\mathbb{Z}[x]/[x-1]\oplus \mathbb{Z}/6\mathbb{Z}[x]/[x-5]\cong \mathbb{Z}/6\mathbb{Z}\oplus \mathbb{Z}/6\mathbb{Z}
$$
In the second case $(x^{2}+1)$ is not a maximal ideal(thanks for the comment to point it out), and the quotient is just $ax+b$, $a,b\in \mathbb{Z}/4\mathbb{Z}$. But this is not necessarily a field partly because the base ring is not a domain. For example $2\times 2=0$ at here. 
