Explain these ring isomorphisms I need help understanding the following isomorphisms:
$$\mathbb Z[\sqrt{-5}]/(2,1+\sqrt{-5}) \cong \mathbb Z[X]/(X^2+5,2,1+X) = \mathbb Z[X]/(2,1+X) \cong \mathbb Z/2\mathbb Z$$
In general, I am also wondering whether the following are true, and if so briefly why:


*

*$\mathbb Z[\sqrt{D}] \cong \mathbb Z[x]/(x^2 - D)$ for all squarefree integers $D$

*$(R/I)/J \cong R/(I +J)$, ie why do we have things like $\mathbb Z[x]/(2,x^2+5)\simeq \mathbb Z_2[x]/(x^2+5) $ where $\mathbb Z[x] / (2) \cong \mathbb Z_2[x]$?


Also for example for something like the following:
$$\,\mathbb Z[x]/(2,x^2+5)\simeq \mathbb Z_2[x]/(x^2+1)$$
Is it true because of the third isomorphism theorem? I'm guessing we set $R = \mathbb Z[x]$, $I = (2)$, and $J = (x^2 + 5)$, but please expand more on each step of the reduction.
 A: Yes all this results from the 3rd isomorphism theorem. Here are some details:
\begin{align}\mathbf Z[\sqrt{-5}]/(2,1+\sqrt{-5})&\simeq\mathbf Z[X]/(X^2+5)/(2,1+X)\cdot Z[X]/(X^2+5)\\&\simeq\mathbf Z[X]/(X^2+5)/(2,1+X, X^2+5)/(X^2+5)\end{align}
For the second isomorphism:
\begin{align}\mathbf Z[X]/(2,X^2+5)&\simeq\mathbf Z[X]/2\mathbf Z[X]/(2,X^2+5)\mathbf Z[X]/2\mathbf Z[X]\\
&\simeq\mathbf Z/2\mathbf Z[X]/(X^2+1)\mathbf Z/2\mathbf Z[X]
\end{align}
A: For example:
$$\phi:\Bbb Z[x]\to\Bbb Z_2[x]\;,\;\;\phi(p(x)):=p(x)\pmod 2$$
is a rings epimorphism (reduction modulo $\;2\;$ of all the coefficients in the polynomial), and its kernel is clearly $\;(2)=2\Bbb Z[x]=\;$ the polynomials whose coefficients are even numbers, so by the first isomorphism theorem you have $$\Bbb Z[x]/(2)\cong\Bbb Z_2[x]$$
Then observe that $\;x^2+5=x^2+1=(x+1)^2\pmod 2\;$ and you get the last result.
For the first result I think the following is the easier approach: since using the FIT again, as above, we get
$$\Bbb Z[x]/(x^2+5)\cong\Bbb Z[\sqrt{-5}]$$
the trick used in the first line is to identify $\;\sqrt{-5}\approx x\;$ in the quotient, so
$$\Bbb Z[\sqrt{-5}]/(2,\,1+\sqrt{-5})\cong\left(\Bbb Z[x]/(x^2+5)\right)/(2,\,1+x)\cong$$
$$\cong\Bbb Z[x]/(x^2+5,\,2,\,1+x)\cong\Bbb Z[x]/(2,1+x)$$
the last step following by the fact that $\;(2,\,1+x)\;$ is a maximal ideal in $\;\Bbb Z[x]\;$ and thus $\;\Bbb Z[x]/(x^2+5,\,2,\,1+x)=\Bbb Z[x]/(2,\,1+x)\;$ , since for example
$$(x-1)(x+1)+2\cdot3=x^2+5\in(2,\,1+x)$$
A: $\mathbb Z[\sqrt{-5}]\cong  \mathbb Z[x]/(x^2+5)\Rightarrow \mathbb Z[\sqrt{-5}]/(2,1+\sqrt{-5})\cong \mathbb Z[x]/(x^2+5,2,1+x) \cong \mathbb Z_2[x]/(x^2+5,1+x)= \mathbb Z_2[x]/(x^2+1,x+1)$
we have $x^2+1=(x+1)(x+1)$ in $\mathbb Z_2$, so $x^2+1\in (x+1)$, ((or $(x^2+1,x+1)=(x^2+1(mod (x+1), (x+1))=((-1)^2+1,x+1)=(0,x+1)=(x+1)$))
and
 $\mathbb Z_2[x]/(x+1)\cong \mathbb Z_2[-1]=\mathbb Z_2[1]=\mathbb Z_2 $, so $\mathbb Z[\sqrt{-5}]/(2,1+\sqrt{-5})\cong\mathbb Z_2 $
