Basic ring isomorphism question Let $R=\mathbb{Z}[x], I=(x^2+1,x+1).$ Prove that $R/I \cong \mathbb{Z}[i]/(i+1) \cong \mathbb{Z}/2\mathbb{Z}$.
I am confused with the rather messy looking of $R/I$. My first step is to define the homorphism $f: R/I \rightarrow \mathbb{Z}[i]$ but then I don't know what to do, please help!
 A: Remember the third ismorphism theorem for rings: $\,R\,$ is a ring with ideals $\,I\le J\le R\, $ ,then
$$\left(R/I\right)/\left(J/I\right)\cong R/J$$
In your problem  we have 
$$\langle x^2+1\,\rangle\le\langle\,x^2+1\,,\,x+1\,\rangle\le\Bbb Z[x]\implies$$
$$\left(\Bbb Z[x]/\langle\,x^2+1\,\rangle\right)/\left(\langle\,x^2+1\,,\,x+1\,\rangle/\langle\,x^2+1\,\rangle\right)\cong\Bbb Z[x]/\langle\,x^2+1\,,\,x+1\,\rangle$$
Now just apply that under the same isomorphism , we have
$$\Bbb Z[x]/\langle\,x^2+1\,\rangle\cong\Bbb Z[i]\;,\;\;\langle\,x^2+1\,,\,x+1\,\rangle/\langle\,x^2+1\,\rangle\cong\langle\,x+1\,\rangle$$
the isomorphism being the one we get from the first isomorphism theorem and the homomorphism
$$\phi:\Bbb Z[x]\to\Bbb Z[i]\;,\;\;\phi(f(x)):=f(i)\;\;\;(\text{i.e., determined by}\;\;x\mapsto i\,)$$
soo that under the above, $\,\Bbb Z[x]\ge\langle\,x+1\,\rangle\mapsto\langle i+1\,\rangle\le\Bbb Z[i]\;$
A: Alternatively, one may use the $\color{blue}{\rm FIT}$ = First (vs. Third) Isomorphism Theorem  as follows.
$\quad  \smash[t]{\Bbb Z\stackrel{h}{\to}}\, \Bbb Z[i\,]/(1\!+\!i)\ \ {\rm is}\ \ \color{#0b0}{\bf onto,}\ \ {\rm by\ \ mod}\,\ 1\!+\!i\!:\ i\,\equiv -1\Rightarrow\:a\!+\!b\,i\,\equiv a\!-\!b\in \Bbb Z\ \\[0.1em] 
\quad  n\in \ker\ h\iff 1\!+\!i\,\mid n\iff \dfrac{n}{1\!+\!i}\, =\, \dfrac{n\,(1\!-\!i\,)}2\,\in\, \Bbb Z[i\,] \iff \color{#c00}2\mid n\\[1.2em]  
\quad {\rm So} \ \ \ \Bbb Z[i\,]/(1\!+\!i\,)\, \color{#0b0}{\bf =\ Im\:h}\!\smash[t]{\stackrel{\color{blue}{\rm\ \ \ FIT_{\phantom{I^2}}}}\cong} \Bbb Z/\ker h \,=\, \Bbb Z/\color{#c00}2\,\Bbb Z$ 
Similarly for $\,\Bbb Z[x]/I,\,\ I = (x^2\!+\!1,x\!+\!1).\,$  $\, x\equiv -1\ $ so $\ h\,$ is onto.  $\  \ker h = \color{#c00}2\,\Bbb Z\,\ $ by
$\quad \color{#c00}2 = x^2\!+\!1-(x\!-\!1)(x\!+\!1)\in I,\ \ \ n\in I\,\Rightarrow\, n = (x^2\!+\!1)f+(x\!+\!1)g \smash[t]{\stackrel{\large\,\ x\,=\,-1\,\ }\Rightarrow} n = \color{#c00}2f(-1)$ 
Similarly $\ \Bbb Z[x]/(f(x),x\!-\!a)\,\cong\,\Bbb Z/\color{#c00}{f(a)}\Bbb Z\ \ $ by $\, \ f(x)\equiv \color{#c00}{f(a)}\,\pmod{\!x\!-\!a}$
