$\mathbb{Z}[X]/(2x+4,x^2-3) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ Someone has asked a question before regarding the ring $\mathbb{Z}[X]/(2x+4,x^2-3)$, but the answer wasn't quite what I was looking for.  I was wondering how one would show this quotient isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2 \mathbb{Z}$.  I was thinking of using the isomorphism $$\mathbb{Z}[X]/(2x+4,x^2-3) \cong \frac{\mathbb{Z}[X]/(2x+4)}{(2x+4,x^2-3)/(2x+4)}$$ I've seen problems like this before, but I don't think I even understand the structure of $\mathbb{Z}[X]/(2x+4)$.  
 A: Note that $-2 = (X - 2)(2X + 4) - 2(X^2 - 3)$, so $2 \in (2X + 4, X^2 - 3)$.
Thus, your ring is \begin{align*}\Bbb{Z}[X] / (2, X^2 - 3) & \cong (\Bbb{Z}/2\Bbb{Z})[X] / (X^2 +1)\\
&\cong (\Bbb{Z}/2\Bbb{Z})[X]/(X+1)^2\\
&\cong (\Bbb{Z}/2\Bbb{Z})[X]/X^2\end{align*}
which, as already noted, is not isomorphic as a ring to $\Bbb{Z}/2\Bbb{Z} \times \Bbb{Z}/2\Bbb{Z}$.
A: Hint $\,\ 0\equiv \color{#0a0}{2(2\!+\!x)}\,\Rightarrow\, 0\equiv \color{#0a0}{2(2\!+\!x)}(2\!-\!x) \equiv 2(4-\color{#c00}{x^2})\equiv 2,\, $ by $\,\color{#c00}{x^2}\equiv 3$
Thus $\ 2\in I=(2x\!+\!4,x^2\!-3)\,\Rightarrow\, I = (2,\, I\ {\rm mod}\ 2) = (2,0,x^2\!+1)$
Thus $\ \Bbb Z[x]/I \,\cong\, \Bbb Z[x]/(2,x^2\!+1) \,\cong\, \Bbb Z_2[x]/(x\!+\!1)^2\,\cong\,\Bbb Z_2[x]/(x^2),\,$
the ring of dual numbers over $\,\Bbb Z_2$
A: Multiply $X^2=3$ by $2$ and use the fact $2X=-4$ to prove $0=2$ in the quotient ring.
Deduce from $X^2=3$ all elements look like $aX+b$, and from $0=2$ that wlog $a,b\in\{0,1\}$.
Prove $X$ is not $0$ or $1$, i.e. no linear combination of $2X+4$ and $X^2-3$ can yield $X$ or $X+1$.
Observe $(1+X)^2=0$ so the quotient ring is $\cong\Bbb F_2[\varepsilon]/(\epsilon^2)$, not $\Bbb F_2\times\Bbb F_2$.
