Isomorphisms for rings modulo ideals Let the following ring be given: $\mathbb{Z}[\sqrt{-3}] := \{a+b\sqrt{-3}: a,b\in\mathbb{Z}\}$. I was wondering what the following quotients would look like, given the ideals.
For the ideal $(\sqrt{-3})$, we have that: $$(\sqrt{-3})=\{\sqrt{-3}(a+b\sqrt{-3}): a,b\in\mathbb{Z}\}=\{-3b+a\sqrt{-3}:a,b\in\mathbb{Z}\}=\{3x+y\sqrt{-3}:x,y\in\mathbb{Z}\} := I$$
We can take a look at the homomorphism $f:\mathbb{Z}[\sqrt{-3}]\to \mathbb{Z}/3\mathbb{Z}$; $f(a+b\sqrt{-3}) = a \mod 3$. For any $a+b\sqrt{-3}$ where $a$ is a multiple of 3, this maps to 0, so that $\ker f= I$. Clearly this map is surjective, if you just look at the mappings of $0,1,2\in\mathbb{Z}[\sqrt{-3}]$ so with the first isomorphism theorem, we have $\mathbb{Z}[\sqrt{-3}]/I \cong \mathbb{Z}/3\mathbb{Z}$.
However, during lectures and in my course, we often have used a rule that stated that for any ring $R$, its polynomial ring $R[x]$ and any ideal $(x-a)$, we have that $R[x]/(x-a) \cong R$ for any constant $a\in R$. If we use $R=\mathbb{Z}$, $x=\sqrt{-3}$ and $a=0$, wouldn't we find that  $\mathbb{Z}[\sqrt{-3}]/I \cong \mathbb{Z}$? Clearly one (or both) of these conclusions must be wrong, but for what reasons?
For another ideal, $J=(2)$, I thought the following, but it felt strange considering it could give an answer that looks like that of $I=(\sqrt{-3})$, even when these ideals don't look alike. Here, I found that $(2) = \{2(a+b\sqrt{-3}):a,b\in \mathbb{Z}\} = 2\mathbb{Z}[\sqrt{-3}].$ Bluntly, I just went ahead and said that $$\mathbb{Z}[\sqrt{-3}]/2\mathbb{Z}[\sqrt{-3}] \cong \mathbb{Z}/2\mathbb{Z}$$
I also tried the same approach as before; now with the homomorphism $g:\mathbb{Z}[\sqrt{-3}]\to \mathbb{Z}/2\mathbb{Z}$ where $g(a+b\sqrt{-3})=a+b \mod 2$; right as I typed this out I realised that this does not give us the desired $\ker g$, so this doesn't work. However the isomorphism felt natural, although at this point I have no argument why, and I can't figure it with homomorphisms and theorems so I'm not exactly sure what I could deduct from this quotient from this point on.
 A: It is indeed true that for any commutative ring $R$ and any $a\in R$, there is an isomorphism $R[x]/(x-a)\cong R$. This can be seen by applying the first isomorphism theorem to the evaluation homomorphism
$$\varepsilon_a:\ R[x]\ \longrightarrow\ R:\ f\ \longmapsto\ f(a),$$
because $\ker\varepsilon_a=(x-a)$.
However, the ring $\Bbb{Z}[\sqrt{-3}]$ is not the same as the polynomial ring $\Bbb{Z}[x]$. In fact, the ring $\Bbb{Z}[\sqrt{-3}]$ is isomorphic to a quotient of the polynomial ring; again the evaluation homomorphism
$$\varepsilon_{\sqrt{-3}}:\ \Bbb{Z}[x]\ \longrightarrow\ \Bbb{Z}[\sqrt{-3}]:\ f\ \longmapsto f(\sqrt{-3}),$$
is clearly surjective, and its kernel equals $\ker\varepsilon_{\sqrt{-3}}=(x^2+3)$. So by the first isomorphism theorem
$$\Bbb{Z}[\sqrt{-3}]\cong\Bbb{Z}[x]/(x^2+3).$$
This also gives another description of the quotient $\Bbb{Z}[\sqrt{-3}]/I$, because
$$\Bbb{Z}[\sqrt{-3}]/I\cong(\Bbb{Z}[x]/(x^2+3))/(x)\cong\Bbb{Z}[x]/(x^2+3,x).$$
Of course $(x^2+3,x)=(x,3)$, which shows that
$$\Bbb{Z}[x]/(x,x^2+3)\cong\Bbb{Z}[x]/(x,3)\cong\Bbb{Z}/(3)=\Bbb{F}_3,$$
which agrees with your finding. We can do the same for the ideal $J$; as before
$$\Bbb{Z}[\sqrt{-3}]/J\cong(\Bbb{Z}[x]/(x^2+3))/(2)\cong\Bbb{Z}[x]/(x^2+3,2),$$
where of course $(x^2+3,2)=(2,x^2+1)$. This shows that
$$\Bbb{Z}[x]/(x^2+3,2)\cong\Bbb{Z}[x]/(2,x^2+1)\cong(\Bbb{Z}[x]/2\Bbb{Z}[x])/(x^2+1)\cong\Bbb{F}_2[x]/(x^2+1),$$
which is a ring of order $4$. In particular this is not isomorphic to $\Bbb{F}_2=\Bbb{Z}/2\Bbb{Z}$. In fact, this shows that the quotient $\Bbb{Z}[\sqrt{-3}]/J$ contains a nilpotent element, i.e. there exists some nonzero $\xi\in\Bbb{Z}[\sqrt{-3}]/J$ satisfying $\xi^2=0$. Backtracking the isomorphisms above shows that one such element is 
$$\xi=1+\sqrt{-3}.$$
Indeed $\xi\notin J$ but $\xi^2=-2+2\sqrt{-3}\in J$, illustrating that the quotient is not isomorphic to $\Bbb{Z}/2\Bbb{Z}$.
