Are these two quotient rings of $\Bbb Z[x]$ isomorphic? Are the rings $\mathbb{Z}[x]/(x^2+7)$ and $\mathbb{Z}[x]/(2x^2+7)$ isomorphic?
Attempted Solution:
My guess is that they are not isomorphic. I am having trouble demonstrating this. Any hints, as to how i should approach this?
 A: Let $A = \mathbb{Z}[x]/(x^2+7)$
Let $B = \mathbb{Z}[x]/(2x^2+7)$
$A$ and $B$ are clearly integral domains.
Let $\alpha$ be the image of $x$ by the canonical homomorphism $\mathbb{Z}[x] \rightarrow A$.
Let $\beta$ be the image of $x$ by the canonical homomorphism $\mathbb{Z}[x] \rightarrow B$.
Since $\alpha$ is integral over $\mathbb{Z}$, $A$ is integral over $\mathbb{Z}$.
Suppose $A$ and $B$ are isomorphic.
Then $\beta$ is integral over $\mathbb{Z}$.
Let $K$ be the field of fractions of $B$.
Since $2\beta^2+7 = 0$, $\beta^2 = -\frac{7}{2}$ in $K$.
Hence $-\frac{7}{2}$ must be integral over $\mathbb{Z}$.
This is a contradiction.
A: Hint $\ \ 2\:$ is invertible in $\rm\ \Bbb{Z}[x]/(2x^2\!+7),\:$ but  not in $\rm\, \Bbb{Z}[x]/(x^2\!+7)\,\cong\, \Bbb Z[\sqrt{-7}].\:$  Indeed, in the first ring $\rm\:2(x^2\!+4) = 1.\:$ In the second,  $\rm\:2\alpha = 1\:\Rightarrow\:2\alpha'=1\:$ $\Rightarrow$ $\rm\:4\alpha\alpha' = 1,\:$ $\rm\: \alpha \alpha'\in \Bbb Z,\:$ contradiction, where $\,\rm (a+b\sqrt{-7})' = a-b\sqrt{-7}\,$ is the conjugation automorphism. 
Remark $\ $ The proof is accessible at high-school level by eliminating use of the conjugation automorphism in $\rm\,R \cong \Bbb Z[\sqrt{-7}].\:$ If $\,2\,$ is invertible in $\rm\,R\,$ then $\rm\:2\,(a\!+\!b\sqrt{-7})= 1,\,$ for $\rm\, a,b\in \Bbb Z.\:$ Therefore $\rm\:b\ne 0\ $ (else $\rm\:2a=1,\ a\in\Bbb Z)\ $ hence $\rm\,\sqrt{-7}\, =\, (1\!-\!2a)/(2b)\in \Bbb Q,\,$ contradiction.   
This elementary approach may be helpful for readers not familiar with the more advanced techniques applied in the other answers. 
A: a) The ring $\mathbb{Z}[x]/(x^2+7)=\mathbb Z[\xi]$ is finitely generated  as a $\mathbb Z$-module, since  $\mathbb Z[\xi]=\mathbb Z\cdot1\oplus \mathbb Z\cdot \xi$ 
Thus the ring $\mathbb Z[\xi]$ integral over $\mathbb{Z}$.    
b) On the other hand the ring $\mathbb{Z}[x]/(2x^2+7)=\mathbb{Z}[\eta]$ contains the element $\eta^2=-\frac {7}{2}$ which is not integral over $\mathbb Z$, since the the only  numbers in $\mathbb Q$ integral over $\mathbb Z$ are the elements of $\mathbb Z$ .
So the ring  $\mathbb Z[\eta]$ is not integral over $\mathbb Z$.
Hence our two rings are not isomorphic.
A: Another approach, let $f:\mathbb{Z}[x]/(2x^2+7)\to \mathbb{Z}[x]/(x^2+7)$ be an isomorphism. 
Now as per condition $x^2+7|f(2x^2+7)$. Let $f(x)=ax+b+(x^2+7)$. 
So, $f(2x^2+7)=2(ax+b)^2+7+(x^2+7)$. 
Now that implies $x^2+7|−14a^2+2b^2+7+4abx \implies 7+2b^2=14a^2$. Absurd.
A: Suppose they are isomorphic. Then
$\left(\mathbb{Z}[x]/(x^2+7)\right)/(2) \cong \left(\mathbb{Z}[x]/(2x^2+7)\right)/(2)$. Ravi helpfully pointed out that considering ideals in either ring in terms of $x$ will often give us different ideals, but we do not suffer from this problem when using ideals generated by integers such as $(2)$.
Letting overlines represent the canonical homomorphism from $\mathbb{Z}[x]$ to $\mathbb{F}_2[x]$, applying the correspondence theorem for rings on the left side tells us that $$\left(\mathbb{Z}[x]/(x^2+7)\right)/(2) \cong \left(\mathbb{Z}[x]/(2)\right)/(\overline{x^2 + 7}) \cong \mathbb{F}_2[x]/(x^2 + 1)\,.$$
Similarly, on the right, we get
$$\left(\mathbb{Z}[x]/(2x^2+7)\right)/(2) \cong \left(\mathbb{Z}[x]/(2)\right)/(\overline{2x^2 + 7}) \cong \mathbb{F}_2[x]/(1) = \{0\}\,.$$
One can verify the former ring contains four distinct elements. The latter is trivial, which is absurd.
EDIT: To further bolster intuition for an approach like this, see Bill's mention of it in his first comment on his own post.
