Isomorphism between two quotient ring I am self studying Artin's algebra and stuck with this exercise which asks 'Are $\mathbb{Z}[x] /\langle x^2 +7 \rangle$ and $\mathbb{Z}[x] /\langle 2x^2 + 7\rangle$ isomorphic?'
My sense tells answer is no but to
show non-isomorphism , we should execute one property that one ring has and other hasn't ... I couldn't figure out the property... Thanks in advance for any help/ hints ..
 A: I think you're right.   The first ring $\Bbb Z[X]/(x^2+7)\cong \Bbb Z[\alpha] $ where $\alpha^2+7=0$.  As abelian groups,  we have $\Bbb Z[X]/(x^2+7)\cong\{a+b\alpha \mid a,b\in\Bbb Z\}\cong \Bbb Z^2$, since we can divide any polynomial by $x^2+7$ and get a remainder of degree less than $2$.
This is not true in the case of $\Bbb Z[X]/(2x^2+7)$.  We can't perform division of any polynomial in $\Bbb Z[X]$ by $2x^2+7$, because of its not being monic.  In particular,  we can  only divide by $2x^2+7$, inside $\Bbb Z[X]$, when the dividend is a polynomial whose leading coefficient is a multiple of $2$.  (Apparently you learn more about this sort of thing when you study localization.)  But the upshot is that you get some higher degree polynomials.
In fact,  we can conclude that $\Bbb Z[x]/(f)$ is not a finitely generated abelian group if $f$ is not monic.  See this.   So it can't be isomorphic as an abelian group to $\Bbb Z^2$.
A: Hint $ $  Easy way: only one ring has a sense of parity (i.e. an image $\cong \Bbb Z/2).\,$

Or $\ \underbrace{\frac{1}2\in \Bbb Z[x]/(2x^2\!+\!7)}_{\small\textstyle 2(x^2\!+\!4)=1}\,$ but $\frac{1}2\not\in \Bbb Z[x]/(x^2\!+\!7)\,$ else $\,2w = 1\underset{\rm Norm}\Longrightarrow 4w\bar w = 1\in \Bbb Z.\,$
Remark $ $ Above a discriminating property is integrality. A crucial property of integral ring extensions is that they don't alter unit properties in the base ring, i.e. a nonunit $\,r\in R\,$ remains a nonunit in any integral extension ring $S\,$ (so they cannot introduce fractions like $1/2\,$ above). For example, this may allow us to deduce that Diophantine equations in $\Bbb Z$ are unsolvable by deducing a parity contradiction $\,2\mid 1\,$ in some convenient extension ring of algebraic integers. See here for further discussion.
