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Let $R$ be commutative with $1$. If $U$ is maximal among non-principal ideals show that $U$ is prime.

Possible proof path:

Suppose $ab \in U$ and $a \notin U$ Consider $U+(a)$. I tried to prove along the lines of "Maximal ideal is prime" but I can't seem to proceed anywhere. Any hints?

P.S. Pls do not provide links to Oka or Ako families. I just got started in Algebra.

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  • $\begingroup$ R/I is a field but only if I is maximal in general. How do you extend it to the case of I being maximal among non-principal ideals? $\endgroup$
    – Jhon Doe
    Commented Mar 3, 2018 at 4:21
  • $\begingroup$ Ah, I misread the question @Jhon. I thought it said something like "If $U$ is a maximal non-principal ideal, show $U$ is prime". Sorry. $\endgroup$
    – Kaj Hansen
    Commented Mar 3, 2018 at 4:43
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    $\begingroup$ I edited your post to $\LaTeX$ify it. Cheers! $\endgroup$ Commented Mar 3, 2018 at 5:28

2 Answers 2

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This is exercise $10$ from chapter $1$, section $1$ of the text

    Kaplansky${\,-\,}$Commutative Rings, Rev Ed (1974)

Fleshing out the author's extended hint, we can argue as follows . . .

Let $U$ be a non-principal ideal of $R$ which is maximal among all non-principal ideals of $R$.

Suppose $U$ is not prime.

Our goal is to derive a contradiction.

Let $ab \in U$, with $a,b\notin U$.

Then $(U,a) = (c)$, for some $c\in R$.

Let $V=\{x\in R\mid cx \in U\}$.

Clearly we have $U \subseteq V$.

Since $(U,a)=(c)$, we have $u+ra=c$, for some $u\in U$ and some $r\in R$. \begin{align*} \text{Then}\;\;&u+ra=c \qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\; \\[4pt] \implies\;&b(u+ra)=bc\\[4pt] \implies\;&bu+r(ab)=bc\\[4pt] \implies\;&bc\in U\qquad\text{[since $ab\in U$]}\\[4pt] \implies\;&b\in V \end{align*} Since $U \subseteq V$, and $b \in V\setminus U$, we have the proper inclusion $U\subset V$, hence $V=(d)$ for some $d\in R$.

Now let $u$ be an arbitrary element of $U$.

Since $U\subset (U,a)=(c)$, we get $u=cy$, for some $y\in R$. \begin{align*} \text{Then}\;\;&u=cy\\[4pt] \implies\;&cy \in U\\[4pt] \implies\;&y\in V&&\text{[by definition of $V$]}\\[4pt] \implies\;&y = dz,\;\text{for some}\;z\in R&&\text{[since $V=(d)$]}\\[4pt] \implies\;&u = c(dz)\\[4pt] \implies\;&u \in (cd)\\[4pt] \end{align*} Thus, $U\subseteq (cd)$.

Also, since $d\in V$, then by definition of $V$, we get $cd\in U$, hence $(cd) \subseteq U$.

But then $U=(cd)$, contrary to the assumption that $U$ is not principal.

It follows that $U$ is prime.

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Suppose,ab is in U,Consider the ideal J=U+(a),if a is not in U then J properly contains U so by maximality of U ,J=R.Hence there is u in U &c in R such that 1=u+ac,multiplying both sides by b &use ab is in U conclude b is in U.

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    $\begingroup$ You need to show that $J = U + (a)$ is non-principal to use the maximality condition in this case. $\endgroup$
    – wgrenard
    Commented Mar 3, 2018 at 6:17
  • $\begingroup$ Yah that is where i got struck $\endgroup$
    – Biman Roy
    Commented Mar 4, 2018 at 2:08

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