Your proof does indeed seem to be correct. It is written in a very convoluted way, however. Perhaps you should start with:
Let $J$ be an ideal containing $f^{-1}(M)$. As $f$ is surjective, $f(J)$ is an ideal, and it contains $M$. As $M$ is maximal, either $f(J)=M$ or $f(J)=S$. If $f(J)=M$, then...
Notice that this avoids your imprecision with the "contrary with our assumption", which actually is not contrary to any assumption you made. Also, it cleans up the proof, by making the chain of deduction clearer.
If I may suggest, a cleaner way of proving this is by an altogether different method, bypassing elements. Since $M$ is maximal, $k:=S/M$ is a field. Since $f$ is surjective, its composition with the canonical projection $\bar{f}:R\to k$ is a surjection. This means that $\ker \bar{f}$ is a maximal ideal. Can you compute it?
The most common definition of a prime element in a domain $R$ is
$a\in R$, $a\ne0$ is prime when it is not invertible and, for all $b,c\in R$, if $a\mid bc$, then $a\mid b$ or $a\mid c$.
What you're using is irreducibility:
$a\in R$, $a\ne 0$, is irreducible when it is not invertible and, for all $b,c\in R$, if $a=bc$, then either $b$ or $c$ is invertible.
Every prime element is irreducible: suppose $a$ is prime and $a=bc$; in particular $a\mid bc$ so, by assumption, either $a\mid b$ or $a\mid c$. If $a\mid b$, then $b=ax$ for some $x\in R$ and therefore $a=axc$ and, from $a\ne0$, we deduce $1=xc$, so $c$ is invertible. Similarly, if $a\mid c$ we conclude that $b$ is invertible.
In a Euclidean domain, however, every irreducible element is prime. This is a consequence of existence of the Euclidean algorithm that provides a greatest common divisor and Bézout's identity: if $a$ is irreducible, $a\mid bc$ and $a\nmid b$, then $1$ is a greatest common divisor of $a$ and $b$ (by irreducibility of $a$. Hence $1=ax+by$, which implies $c=axc+bcy$, so $a\mid c$.
Now let's look at your argument. Suppose $(a)$ is maximal, but $a$ is not irreducible (or prime, as the concepts are equivalent here). Then three cases are possible (see the definition of irreducible above):
- $a=0$;
- $a$ is invertible;
- $a\ne0$ and $a=bc$ where neither $b$ nor $c$ is invertible.
Case 2 is not possible, because if $a$ is invertible, then $(a)=R$ and so $(a)$ is not maximal.
Suppose we are in case 3. Then $(a)\subseteq(b)$ and, by maximality, $(b)=(a)$ or $(b)=R$. If $(b)=R$, then $b$ is invertible: contradiction. Therefore $(b)=(a)$, implying $b=ay$ for some $y$. Then $a=bc=ayc$. Since $a\ne0$, we conclude $yc=1$ and so $c$ is invertible: contradiction.
It only remains case 1. And you're right! The ideal $(0)$ can be maximal, precisely when $R$ is a field. And every field is trivially a Euclidean domain.
The correct statement would be:
Suppose $R$ is a Euclidean domain that is not a field. Then an ideal $(a)$ is maximal if and only if $a$ is prime (irreducible).
In the argument above, case 1 cannot happen, because $(0)$ is not a maximal ideal in a ring that is not a field.
Best Answer
Since $2$ is in $J$ and $J$ is an ideal , $2 + 2 + \ldots + 2 = 2n \in J$ by closure for every integer $n$. This means that