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
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!