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?
On
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.
On
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.
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.