# If $U$ is maximal among non-principal ideals show that $U$ is prime. (Hints?)

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.

• 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? – Jhon Doe Mar 3 '18 at 4:21
• 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. – Kaj Hansen Mar 3 '18 at 4:43
• I edited your post to $\LaTeX$ify it. Cheers! – Robert Lewis Mar 3 '18 at 5:28

## 2 Answers

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.

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.

• You need to show that $J = U + (a)$ is non-principal to use the maximality condition in this case. – wgrenard Mar 3 '18 at 6:17
• Yah that is where i got struck – Biman Roy Mar 4 '18 at 2:08