What is the isomorphism map between $\mathbb Z$ and $\mathbb Z[x] / (x).$?
I am trying to prove that the ideal $(x)$ is a prime ideal and to show this, I want to show that $\mathbb Z \cong \mathbb Z[x] / (x),$ I do not know how to write this isomorphism, could someone help me, please?
Also, I do not know how to show that it is not a maximal ideal, could someone show me this please?
An ideal $\mathscr I\subset \mathcal R$ is prime iff $\mathcal R/\mathscr I$ is an integral domain. Apparently you knew that.
Consider the evaluation homomorphism: $\rm{ev_0}:\Bbb Z[x]\to\Bbb Z$ by $\rm{ev_0}(p)=p(0)$.
$(x)=\rm{ker}(\rm{ev_0})$, because $x$ is the minimal polynomial of $0$. Also $\rm{ev_0}$ is surjective.
Now by the first isomorphism theorem we get $$\Bbb Z[x]/(x)\cong\Bbb Z$$.
Two things: since $\Bbb Z$ is an integral domain, $(x)$ is a prime ideal.
Second, $\Bbb Z$ an integral domain $\implies \Bbb Z[x]$ is an integral domain. Then since the quotient of $\Bbb Z[x]$ by $(x)$ is not a field, $(x)$ is not maximal.
Alternatively on the second part, $\Bbb Z[x]\supsetneq(2,x)\supsetneq (x)$. This may seem confusing, since $2$ and $x$ are relatively prime. But we don't have Bezout, because $\Bbb Z[x]$ isn't a PID.