Prove that $\mathbb{Z}[x]/(2,x^3+1)≅\mathbb{Z}_2[x]/(x^3+1)$.

Question

The Problem: Prove that $\mathbb{Z}[x]/(2,x^3+1)\cong\mathbb{Z}_2[x]/(x^3+1)$.

The same question has been asked before here; but the answers are more confusing than helpful.

My Attempt: Let $I=(2, x^3+1)$. By the Third Isomorphism Theorem, we have $(\mathbb{Z}[x]/(2))/(I/(2))\cong\mathbb{Z}[x]/I$. Now, if we can show that $\mathbb{Z}[x]/(2)\cong\mathbb{Z}_2[x]$ and $I/(2)\cong(x^3+1)$ (the last "$(x^3+1)$" is an ideal of $\mathbb{Z}_2[x]$), then we'd be done; but these two isomorphisms aren't actually all that easy to show. My question is: is there a better way? Any help would be greatly appreciated.

Answer 1

Consider $f:\mathbb{Z}[x] \to \frac{\mathbb{Z_2}[x]}{(x^3+1)}$ such that $f(p) =\bar{p}+(x^3+1) $ where $\bar{p}$ denotes p in $\mathbb{Z_2}[x]$

Clearly, $f$ is a ring homomorphism. Now, We have to observe kernel of $f$.

Let, $g \in kerf$ Then, $\bar{g} \in (x^3+1) $ hence, $\bar{g} =\bar{h}(x^3+1)$

Now, $g=\bar{g} + 2q$, $ g= \bar{h}(x^3+1) +2q$, $ g \in (2,x^3+1). $ Hence, $kerf \subseteq (2,x^3+1) $. Conversely , If $q \in (2,x^3+1) $ Then, $f(q) =(x^3+1)$ which is additive identity of codomain ring.

The map is onto by definition .

By 1st isomorphism theorem, $\frac{\mathbb{Z}[x]}{(2,x^3+1)} \cong $ $\frac{\mathbb{Z_2}[x]}{(x^3+1)}$

Answer 2

One alternative proof.

Consider a coset-element $p(x)+(2,x^3+1)$ of the quotient ring $\Bbb{Z}[x]/(2,x^3+1)$, where $p(x)∈\Bbb{Z}[x]$. If $p(x)$ has a term with even coefficient, then this term belongs to the ideal $(2,x^3+1)$, so that it vanishes from $p(x)+(2,x^3+1)$. If $p(x)$ has a term with odd coefficient, then this odd coefficient can be written as $2k+1$ with $k∈\Bbb{Z}$, hence the term can be written as a sum of two terms, one of which has even coefficient hence belongs to the ideal $(2,x^3+1)$ hence vanishes from $p(x)+(2,x^3+1)$, and the other has coefficient $1$. Thus, all nonzero terms of a coset-element $p(x)+(2,x^3+1)$ of the quotient ring $\Bbb{Z}[x]/(2,x^3+1)$ have coefficient $1$, hence $p(x)+(2,x^3+1)$ is actually a polynomial with coefficients from $\Bbb{Z}_2$, so that it can be bijectively mapped with a copy of itself in the quotient ring $\Bbb{Z}_2[x]/(2,x^3+1)$; note here that the ring $\Bbb{Z}_2[x]$ indeed has an ideal $(2,x^3+1)=(0,x^3+1)=(x^3+1)$. This bijection $\Bbb{Z}[x]/(2,x^3+1)↔\Bbb{Z}_2[x]/(2,x^3+1)$ or, which is the same, $\Bbb{Z}[x]/(2,x^3+1)↔\Bbb{Z}_2[x]/(x^3+1)$ defines a natural isomorphism between these quotient rings because in fact it maps each element to itself.
Thus, $\Bbb{Z}[x]/(2,x^3+1)≅\Bbb{Z}_2[x]/(x^3+1)$.