It's a question from Artin's Algebra. The ordinary question is to verify $\mathbb{Z}[x]/(x^2-3,2x+4)$, my answer is $\mathbb{F}_2[i]$.
My way is: Firstly, figure out that the ideal is equal to $(x^2+2x+1,2)$, then we can compute $\mathbb{Z}[x]/(2)$ and it's isomorphic to $\mathbb{F}_2[x]$ then we can compute $\mathbb{F}_2[x]/(x^2+2x+1) = \mathbb{F}_2[x]/(x^2+1)$, it's isomorphic to $\mathbb{F}_2[i]$.
Am I wrong or right? If I am wrong, then what's the right answer and where did I make mistake?
There is no $i$ in $\mathbb{F}_2$.
Since $(x^2-3)+(2x+4)=x^2+2x+1$ and $2(x^2-3)-(x-2)(2x+4)=2$, the inclusion $$ (x^2+2x+1,2)\subseteq(x^2-3,2x+4) $$ holds. The reverse inclusion can be proved similarly.
By the homomorphism theorems, $$ \mathbb{Z}[x]/(x^2+2x+1,2)\cong \frac{\mathbb{Z}[x]/(2)}{(x^2+2x+1,2)/(2)} \cong\mathbb{F}_2[x]/(x^2+2x+1) $$ which is not a field, because $x^2+2x+1=(x+1)^2$ is not irreducible.
One can say that this ring is $\mathbb{F}_2[u]$, where $(u+1)^2=0$. This is the same as requiring $u^2=1$. The ring consists of the elements $$ a+bu $$ with $a,b\in\mathbb{F}_2$, componentwise addition and multiplication $$ (a+bu)(c+du)=(ac+bd)+(ad+bc)u $$