Quotient rings examples/intuition I'm confused with some ideas about quotient rings and wanted to check if my intuition was flawed in any ways:
1:
$$\mathbb{Z}[\sqrt{-5}]/(2,1+\sqrt{-5})\cong \mathbb{Z}/2\mathbb{Z}$$
The way I approached this was by using the homomorphism
$f(a+b\sqrt{-5})=a+5b\: \; mod(2)$ and showing that the kernel included all multiples of 2 in the ring, as well as for $(1+\sqrt{-5})$. Is this a proper approach to proving the isomorphism? I'm sorry if this is blatantly obvious, I felt like it was a little too easy and am skeptical if whether this is a valid way of showing the isomorphism
2: $\mathbb{Z}[X]/(X^{2}+3,3)\cong (\mathbb{Z}/3\mathbb{Z})[X]/(X^2)$
For this one, I found this answer somewhere and I am confused because I ended up with something else. Like the previous example, I set up a homomorphism with a proper kernel to arrive at the isomorphism: $f(a+bx)=a+b\sqrt{-3}\; mod(3)$. This clearly has the ideal of 3 (unless I'm mistaken) and $(x^{2}+3)$ as its kernel, so would the above  also be isomorphic to $(\mathbb{Z}/3\mathbb{Z})[\sqrt{-3}]$? Apart from this, how would the original isomorphism be proved? 
I don't know what I'm not understanding/doing wrong. I apologize for the naivety, I'm trying to teach myself algebra using only online resources and it hasn't been very easy. Thanks for any help .
 A: The intuition for quotient rings is that the generators of the ideal give you elements that are zero in the quotient ring.  So in your first example, we know that $2=0$ and $1+\sqrt{-5}=0 \Rightarrow \sqrt{-5}=-1\equiv 1 \pmod{2}$.  That gives us a pretty good idea how we should try to define our homomorphism.  Define $\phi(a+b\sqrt{-5})=a+b~(\mod 2).$
Show that $\phi$ is a ring homomorphism (not an isomorphism) from $\Bbb Z[\sqrt{-5}]$ to $\Bbb Z/2\Bbb Z$ and that $\ker(\phi)=(2, 1+\sqrt{-5})$ and you've proved your isomorphism, because for any ring homomorphism $\phi:R \to S, R/\ker(\phi) \cong \operatorname{image}(\phi)$.
Similarly, for problem $2$, you know $3=0$ and $X^2+3=X^2=0$.  So for any polynomial $p(X)=a_0+a_1X+X^2q(X) \in \Bbb Z[X]$, define $\phi(p)=a_0 (\mod 3)+ a_1 (\mod 3)X$.  (You're ignoring the terms of higher degree in $p$ because they are all multiplied by $X^2$, which will be $0$ in the quotient ring.)  As above, show that this really is a ring homomorphism and show that it has the correct kernel, and you're done.
