Identify $\mathbb{Z}[x]/(x^2-3,2x+4)$. I need help to identify $\mathbb{Z}[x]/(x^2-3,2x+4)$.
I've been solving such problems in an approach like:
$$
2(x^2-3)=2x^2-6, x(2x+4)=2x^2+4x \\
(2x^2+4x)-(2x^2-6)=4x+6, 2(2x+4)=4x+8 \\
(4x+8)-(4x+6)=2
$$
What shall I do next, please? Thank you.
Simon
 A: we have $2=2(x^2-3)+(2-x)(2x+4)$ and then $(x^2-3,2x+4,2)=(2,x^2-3)$
$\mathbb Z[x]/(x^2-3,2x+4) \cong\mathbb Z[x]/(2,x^2-3)\cong \mathbb Z_2[x]/(x^2-1) \cong  \mathbb Z_2[x]/(x-1)^2 \cong \mathbb Z_2[x]/(x)^2 $
A: First divide $x^2-3$ by $x+2$, dividing by a linear polynomial is the same as evaluating at the root of the linear term, so $x^2-3=4-3=1$. Thus $x^2-3=q(x)(x+2)+1$ for some $q(x)$. Multiplying by 2, and rearranging, we get $2(x^2-3)-q(x)(2x+4)=2$. Hence $2\in (x^2-3,2x+4)$, so since $2\mid 2x+4$, we have 
$$(x^2-3,2x+4)=(x^2-3,2x+4,2)=(x^2-3,2).$$
Then $\newcommand{\ZZ}{\mathbb{Z}}\ZZ[x]/(x^2-3,2)\cong \ZZ_2[x]/(x^2-3),$
since the kernel of the natural map $\ZZ[x]\to \ZZ_2[x]/(x^2-3)$ is $(2,x^2-3)$.
Then $x^2-3\equiv x^2+1=x^2+1^2\equiv (x+1)^2\pmod{2}$, so
$$\ZZ_2[x]/(x^2-3)\cong \ZZ_2[x]/((x+1)^2).$$ 
Then by a change of variables, $$\ZZ_2[x]/((x+1)^2)\cong \ZZ_2[x]/(x^2).$$
I'm pretty sure it's impossible to simplify this further. The ring has four elements, $0,1,x,x+1$, only 0 and 1 are idempotents, since $x$ is nilpotent, so $x+1$ is a unit. Thus the ring is not a product of smaller rings.
