Prove$(2,x^2+x+1)$ is a maximal ideal of $\mathbb{Z}[x]$

Question

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.

Answer 1

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 $A/I \cong B/J$ .

Answer 2

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 $(2,x^2+x+1)/(2) \cong (x^2+x+1) \mod 2$.

Answer 3

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 $\Bbb Z_[x]$.