A question on ring isomorphisms Is the ring $\mathbb{Z[x]}/(2x^2 + 5)$ isomorphic to any subring of $\mathbb{Z[x]}/(x^3+x+1)$? I believe that the answer is No, but I am not 100 % sure. Here is my reasoning: 
$\mathbb{Z[x]}/(2x^2 + 5) = \{b + ax + \sum_{n = 2}^{N}x^n| 2x^2 = -5\}$ and can have cubic terms. However, the other ring cannot have cubic terms because $x^3 = -x - 1$, so none of its subgroups can either. Is this correct, is there a better explanation, or is there a way to show an isomorphism? Any help will be appreciated
 A: The argument that "the other ring cannot have cubic terms" is not clear to me. An isomorphic subring could have "cubic terms".
Perhaps the following remark is useful:
Let $R=\Bbb Z[x]/(2x^2+5)$ and $S=\Bbb Z[x]/(x^3+x+1)$. 
Then $2$ is an invertible element in $R$ since we have $2(x^2+3)=1$. But is $2$ also invertible in $S$? If not, you are done.
A: Maybe the following argument is along the lines you meant in your original post...
(Though the argument outlined in the other answer is obviously much simpler!)

Let $S=\mathbb{Z}[X]/(X^3+X+1)$ and let $x=X+(X^3+X+1) \in S.$
Every element of $S$ is equal to an element of the form $ax^2+bx+c$ for some $a,b,c \in \mathbb{Z}.$
[Can you show this? 
As you've already pointed out $x^3=-x-1,$ so we can always swap $x^3$ for lower order terms.
What about $x^4$, can you do something with that? And so on...]
Hence $S$ is generated as an abelian group by $\{1,x,x^2\};$ in particular, it is finitely generated.

Now let $R=\mathbb{Z}[Y]/(2Y^2+5)$ and let $y=Y+(2Y^2+5) \in R$
Since $2(y^2+3)=1$ in $R,$ we see that $2$ is a unit in $R.$
It follows that we can view $\mathbb{Z}[1/2]$ as a subring of $R.$
You should check that $\mathbb{Z}[1/2]$ is not finitely generated as an abelian group.
[Hint: Denominators in $\mathbb{Z}[1/2]$ can be arbitrarily large.]

Suppose now, for contradiction, that there is an injective ring homomorphism $R \hookrightarrow S.$
Then we could view $\mathbb{Z}[1/2]$ as an additive subgroup of $S.$
But it is a fact that subgroups of finitely generated abelian groups are also finitely generated.
We therefore have a contradiction!
