Maximal ideal in Euclidean Ring

I am working with the proof of the following lemma:

Let $R$ be Euclidean ring. Ideal $A=(a_0)$ is maximal ideal of $R$ if and only if $a_0$ is prime element of $R.$

Proof: Suppose that $A=(a_0)$ is maximal ideal of $R$, but $a_0$ is NOT prime element of $R$. Then $a_0=bc$ where $b,c$ are not units of $R$. It is easy to check that $(a_0)\subset (b)\subset R$. Since $A=(a_0)$ is maximal ideal of $R$ then we have two options: $(a_0)=(b)$ or $(b)=R$.

1. If $(b)=R$ and we know that Euclidean ring has unit element $1\in R$ then $1=bb'$ where $b'\in R.$ So $b$ is unit. Contradiction.

2. If $(a_0)=(b)$ and since $b\in (b)=(a_0)$ then $b=a_0x$ where $x\in R$. Multiplying both sides of the last identity to $c$ we get the following: $bc=a_0cx$ or $a_0=a_0cx$.

Here is my question: How to derive that $cx=1$? If $a_0\neq 0$ then it's possible since we are working in integral domain. But what if $a_0=0$?

I guess that in Euclidean ring the zero ideal cannot be maximal but I am not able to prove it rigorously. If $R$ is Euclidean ring, then $R\neq\{0\}$ then picking $a\neq 0 \in R$ we see that $\{0\}\subsetneq aR\subset R$. If we can show that $aR\neq R$ then OK.

Please help me to answer my question! Would be very thankful for answers!

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):

1. $a=0$;
2. $a$ is invertible;
3. $a\ne0$ 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.

If $aR=R$ this means $1=aa’$ for $a’\in R$ and any non zero element of $R$ is a unit. $R$ is a field and its only prime is $0$ and its only (maximal) ideal is $(0)$. But we still have $(a)$ is maximal iff $a$ is prime.