Prove$(2,x^2+x+1)$ is a maximal ideal of $\mathbb{Z}[x]$ I tried : From the third isomorphism theorem we have $\frac{\mathbb{Z}[x]}{(2,x^2+x+1)} \cong \frac{\mathbb{Z}[x]/(2)}{(2,x^2+x+1)/(2)}$, if we can prove 1: $(2,x^2+x+1)/(2) \cong ( \overline {x^2+x+1}) $ and 2: $\frac{\mathbb{Z}[x]/(2)}{(2,x^2+x+1)/(2)} \cong  \frac{\mathbb{Z}_{2}[x]}{(\overline {x^2+x+1})}$, then because $\overline {x^2+x+1} $ has no root in $\mathbb{Z}_{2}[x] $ so it is irreducible and they are fields, then we are done. I can prove the second assertion, but I can’t prove the first assertion. I don’t know if this is a right proof direction. Could someone help me, thank you.
 A: You can use this. 
$\mathbb Z[x]/(2) \cong (\mathbb Z/2) [x]$ 
The same isomorphism takes $\frac {(2, x²+x+1) }{(2)}$ to $(x²+x+1) $ in $\mathbb Z/2[x]$  so $\mathbb Z[x]/(2, x²+x+1) \cong \mathbb Z/2[x]/(x²+x+1) $ . 
Here I am using if $A\cong B$ and the isomorphism takes $I$ to $J$ then it induces an isomorphism $A/I \cong B/J$ . 
A: You asked the first assertion i.e., $ (2,x^2+x+1)/(2) \cong (x^2+x+1) \mod 2$. 
It is clear that $(2,x^2+x+1)/(2)=2c+(x^2+x+1)d+(2), \ c,d \in \mathbb{Z}$.
As $2c \in (2)$, we get $(2,x^2+x+1)/(2) \cong (x^2+x+1) \mod 2$. 
A: It is well-known that the maximal ideals of $\Bbb Z[z]$ are of the form $(p, f(x))$ where $p \in \Bbb Z$ is a prime and $f(x) \in \Bbb Z[x]$ is irreducible mod $p$.  See for example
https://web.ma.utexas.edu/users/voloch/Homework/zx.pdf.  We apply this result to the present problem, with $p = 2$ and
$f(x) = x^2 + x + 1; \tag 1$
since
$\deg f(x) = 2, \tag 2$
$f(x)$ is reducible modulo $p = 2$ if and only if it has a zero in
$\Bbb Z \mod 2 = \Bbb Z_2, \tag 3$
but it is easily seen that this is not so, via direct evaluation of $f(x)$ in $\Bbb Z_2$:
$f(0) = 0^2 + 0 + 1 = 1 \mod 2,  \tag 4$
$f(1) = 1^2 + 1 + 1 = 1 + 1 + 1 = 1 \mod 2; \tag 5$
therefore $f(x)$ is irreducible modulo $2$, and the cited theorem may then be invoked to conclude that $(2, x^2 + x + 1)$ is maximal in $\Bbb Z_[x]$.
