Are the rings $R=\mathbb{Z}[x]/(x^2+7)$ and $R'=\mathbb{Z}[x]/(2x^2+7)$ isomorphic? 
Are the rings $R=\mathbb{Z}[x]/(x^2+7)$ and $R'=\mathbb{Z}[x]/(2x^2+7)$ isomorphic?

I tried to use this method:
Suppose there exists an isomorphism $\phi$ such that $\phi$ sends $2|_R$ to $2|_{R'}$. Then
set $$A=\dfrac{\mathbb{Z}[x]/(x^2+7)}{(2)},\ B=\dfrac{\mathbb{Z}[x]/(2x^2+7)}{(2)}.$$
It's easy to see that $A$ and $B$ aren't isomorphic. So $R$ and $R'$ aren't isomorphic. Is it right?
 A: Yes, that works. Slightly simpler, note that the method essentially shows that they are not isomorphic because $\,2\,$ is a unit in $\,R',\,$ since $2(x^2+3) = -1,\,$ but $\,2\,$ is a nonunit in $\,R,\,$ else $\,2f + (x^2\!+7)g = 1\,$ in $\,\Bbb Z[x],\,$ so $\,(x^2\!+7)g = 1\,$ in $\,\Bbb Z/2[x],\,$ so $\,8g(1) = 1\,$ in $\,\Bbb Z/2,\,$ contradiction.
Remark $\ $ Conceptually, this is an instance of an important fact, namely integral extensions like $\,\Bbb Z\subset R\,$ cannot change a nonunit into a unit, but this can happen for a nonintegral extension like $\,\Bbb Z\subset R',\,$ which contains the proper fraction $\,x^2 = -7/2,\,$ so also $\,x^2+4 = 1/2,\,$ so the nonunit $\,2\in\Bbb Z\,$ becomes a unit in the extension ring $\,R'.\,$ This property of integral extensions is a generalization of the well-known property that rationals that are roots of monic polynomials $\in\Bbb Z[x]\,$ must be integers, by the monic case of the Rational Root Test. These properties are clarified when one studies algebraic integers in commutative algebra or algebraic number theory. 
A: Short answer: yes.
In general, if $\phi: R \rightarrow S$ is a ring isomorphism and $I \subset R$ an ideal, then $\phi': R/I \rightarrow S/\phi(I)$ is also an isomorphism. Now suppose that $\phi: R \rightarrow R'$ is an isomorphism. Clearly, $\phi(2) = 2$, so if we let $I = (2)$, then $\phi(I) = (2)$. But $R/(2)$ is not isomorphic to $R'/(2)$, as stated in your question. This contradicts the fact that $\phi$ is an isomorphism, so no isomorphism $R \rightarrow R'$ exists.
A: It looks like the actual question was: why is  $\phi(2)=2?$ (given that $\phi$ is an isomorphism). Equivalently, one needs  to prove that $\phi(1)=1$.
Since both $x^2+7$ and $2x^2+7$ are irreducible in $\mathbb{Z}[x]$, the corresponding ideals 
are prime. Therefore, the quotient rings $R$ and $R'$ are domains. Since
$$\phi(1_R)=\phi(1_R\cdot 1_R)=\phi(1_R)\cdot \phi(1_R)$$
We get  $\phi(1_R)=1_{R'}$, as desired.
