Yes, your proof is fine. Here is another way to view it. In a UFD every prime ideal $\rm\:P\:$ may be generated by primes. Indeed, $\rm\:0\ne p_1\cdots p_n\in P\:\Rightarrow\:$ some $\rm\: p_i\in P,\:$ so we may replace each generator by one of its prime factors. Hence if prime $\rm\:(p)\supseteq Q = (p_1,p_2,\ldots)\ne 0\ne p_i\:$ then $\rm\:(p)\supseteq (p_i)\:$ $\Rightarrow$ $\rm\:(p) = (p_i)\:$ $\Rightarrow$ $\rm\: Q = (p).\,$ QED $\, $ A converse is a famous theorem of Kaplansky.
Theorem $\ $ TFAE for an integral domain D
$\rm(1)\ \ \:D\:$ is a UFD (Unique Factorization Domain)
$\rm(2)\ \ $ In $\rm\:D\:$ every prime ideal is generated by primes.
$\rm(3)\ \ $ In $\rm\:D\:$ every prime ideal $\ne 0$ contains a prime $\ne 0.$
Proof $\ (1 \Rightarrow 2)\ $ Proved above. $\rm\ (2\Rightarrow 3)\ $ Clear.
$(3 \Rightarrow 1)\ $ The set $\rm\:S\subseteq D\:$ of products of units and nonzero primes forms a saturated monoid, i.e. $\rm\:S\:$ is closed under products (clear) and under divisors, since the only nonunit divisors of a prime product are subproducts (up to associates), due to uniqueness of factorization of prime products. Since $\rm\:S\:$ is a saturated monoid, its complement $\rm\:\bar S\:$ is a union of prime ideals. So $\rm\:\bar S = \{0\}\:$ (else it contains some prime ideal $\rm\:P\ne 0\:$ which contains a prime $\rm\:0\ne p\in P\subseteq \bar S,\:$ contra $\rm\:p\in S).\:$ Hence every $\rm\:0\ne d\in D\:$ lies in $\rm S,\:$ i.e. $\rm\:d\:$ is a unit or prime product. Thus $\rm\:D\:$ is a UFD. $\ $ QED
Remark $\ $ The essence of the proof is clearer when one learns localization. Then ones sees from general principles that prime ideals in $\rm\bar S\:$ correspond to maximal ideals of the localization $\rm\:S^{-1} D.\:$
In fact one can view this as a special case of how UFDs behave under localization. Generally the localization of a UFD remains a UFD. Indeed, such localizations are characterized by the sets of primes that survive (don't become units) in the localizations.
The converse is also true for atomic domains, i.e. domains where nonzero nonunits factor into atoms (irreducibles). Namely, if $\rm\:D\:$ is an atomic domain and $\rm\:S\:$ is a saturated submonoid of $\rm\:D^*$ generated by primes, then $\rm\: D_S$ UFD $\rm\:\Rightarrow\:D$ UFD $\:\!$ (popularized by Nagata). This yields a slick proof of $\rm\:D$ UFD $\rm\Rightarrow D[x]$ UFD, viz. $\rm\:S = D^*\:$ is generated by primes, so localizing yields the UFD $\rm\:F[x],\:$ $\rm\:F =\:$ fraction field of $\rm\:D.\:$ Therefore $\rm\:D[x]\:$ is a UFD, by Nagata. This yields a more conceptual, more structural view of the essence of the matter (vs. traditional argument by Gauss' Lemma).
I assume by $<x>$ you mean the principal ideal generated by $x\in R$.
Consider $R=\mathbb{Z}$ and let $a=3$,$b=5$,then $a$ and $b$ are distinct irreducibles.
1.$(1+a)=(4)$ is not prime.
2.$(a+b)=(8)$ is not prime
3.$(1+ab)=(1+15)=(16)$ is not prime.
For 4, you are right. An example is $R=\mathbb{R}[x][y]$, and let $a=x$.
Best Answer
You're thinking about this in entirely the wrong direction. Don't think about general abstract facts about ideals; instead think very concretely in terms of the unique factorizations of elements. If $(a)$ and $(b)$ are principle ideals, then $(a)\subseteq (b)$ iff $b$ divides $a$. What does this tell you about their factorizations? Now consider what a descending chain of principal ideals means in terms of the factorizations of the generators.
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