Jtdylktuy

I found this observation on my book, but I did not understand it well.

Let $R$ a commutative ring without identity, but having a single generator, then $R$ contains nonprime maximal ideal.

We suppose that $R=(a)$. First observe that the principal ideal $(a^2)$ is a proper ideal of $R$, since the generator $anotin (a^2)$. Indeed, were $a$ in $(a^2)$, we could write $a=ra^2+na^2$ for same $rin R$ and $nin mathbb{Z}$;

Why, in this case, the element $e=ra+na$ is a multiplicative identity for $R$? Therefore, since $e$ is a identity of $R$, they are violated the hypothesis.

Now, since $Rne (a^2)$, exist a maximal ideal $M$ of $R$ with $(a^2)subseteq M$. However $M$ is not a prime ideal, as can be seen by considering the product of elements in te complement of $M$ ($r,snotin M$), the product $rsin (a^2)subseteq M$. (Why this is the negation of a definition of prime ideal?)

Anyone could explain to me how things are?

Thanks!

Since every element of $R$ can be written as $ra+na$ for some $rin R$ and $ninmathbb{Z}$, if we find $ein R$ such that $ae=a$, then $e$ will be the identity of $R$.

If $ain(a^2)$, then $a=ra^2+na^2$, for some $rin R$ and $ninmathbb{Z}$. Set $e=ra+na$; then $ae=a$.

Now take an ideal $M$ of $R$ maximal with respect to the property that

1. 
   $(a^2)subseteq M$;


2. 
   $anotin M$.

Such a maximal element exists by Zorn's lemma: the union of a chain of ideals containing $(a^2)$ and not containing $a$ is again an ideal satisfying the same property. Clearly such an ideal is not prime, because $anotin M$, but $aa=a^2in M$.

Let's prove that $M$ is maximal. If $I$ is an ideal of $R$ properly containing $M$, then $I$ contains $(a^2)$, so by the choice of $M$ we must have $ain I$. Hence $I=R$.

Note: the fact that $R=(a)$ is crucial. A ring without an identity can fail to have maximal ideals at all.

Suppose that $a=ra^2+na^2$. Then, $a(ra+na)=a$, so $ae=a$. Then, for any $xin R$, $x=ta+ma$ for some $tin R$, $min mathbb{Z}$. So $xe= (ta+ma)e= tae+mae= ta+ma=x$. Therefore, $e$ is the identity element.

To prove that $M$ is not a prime ideal, it is enough to check that there are two ideals $A$, $B$ with $ABsubseteq M$, and $A$ and $B$ are not contained in $M$. By taking $A=B=(a)$, $AB=(a)(a)subseteq (a^2)subseteq M$, but $(a)=R$ is not contained in $M$ because it is a maximal ideal, so by definition $Mneq R$.

$4mathbb{Z}$ in $2mathbb{Z}$. Note that $4 = 2 cdot 2 in 4mathbb{Z}$, but $2 not in 4 mathbb{Z}$. There is no ideal between $4mathbb{Z}$ and $2 mathbb{Z}$.

(Note that this is a concrete example of your more abstract example.)

For the first part, you have the idea, but if $ain (a^2)$, you can write $a=P(a^2) $ where $P$ is a polynomial. Let $M$ be a maximal ideal containing $(a^2)$, $a.a=a^2$ is in $M$ but $a$ is not in $ M$. This implies that $M$ is not prime.

For the second question, the definition of a prime ideal is that $I$ is prime if for any $a, b in R$ such that $ab in I$ then either $a in I$ or $b in I$. So if we can find $a, b in R setminus I$ with $ab in I$ then this would show that $I$ was not prime.

Now in the situation of the question we have a maximal ideal $M$ of $R = (a)$ with $(a^2) subseteq M$. Let $r, s in R setminus M$. Then $r = ta + ma$ and $s = ua + na$ for some $t, u in R$, $m, n in Bbb Z$, so $rs = (tu + tn + mu + mn)a^2 in (a^2) subset M$. Thus $rs in M$ but $r notin M$, $s notin M$, as required to show that $M$ was not prime.