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I see the statement that if R is UFD then R[x] is a UFD, but is the converse of the statement, which is that if R[x] is a UFD, then R is a UFD, true?

I have written a full proof at the bottom (in case the hint isn't enough).

Hint: Since $Rlbrack xrbrack$ is a UFD, it is (in particular) an integral domain. Hence degree considerations work -- i.e. we always have equality $deg (fg)=deg f+deg g$, unlike in the general case where
$deg (fg) leq deg f + deg g$.

It follows that whenever we factorize elements of $R$ in $Rlbrack xrbrack$, we actually get (by degree considerations) a factorization in $R$.

Also, elements of $R$ which are irreducible in $Rlbrack xrbrack$ must in particular be irreducible in $R$. This is a fairly straightforward argument, just remember that the units of $Rlbrack xrbrack$ are simply the units of $R$ viewed as constant polynomials.

Addendum. In general, a subring of a UFD need not be a UFD. As mentioned in an answer to this question, the easiest counterexamples are constructed by choosing an integral domain which is not a UFD, then noting that this ring is a subring of its field of fractions (which is trivially a UFD). This is why we need arguments like the one sketched above, which in one way or another use the fact that $Rlbrack xrbrack$ consists of polynomials over $R$.

So that you don't have to take my word for it: There are, of course, many examples of non-UFD integral domains. Trotter observed (in an article cited by Lang) that the well-known identity $cos^2x + sin^2x = 1$ implies the non-unique irreducible factorizations
$$sin^2(x)=(1+cos x)(1-cos x)$$
in the ring of trigonometric polynomials $mathbb R lbrack sin x,cos xrbrack$. Meanwhile, $mathbb R lbrack sin x,cos xrbrack$ is easily seen to be an integral domain.

Here is the promised proof that if $Rlbrack xrbrack$ is a UFD, then $R$ is a UFD.

Proof. This (simple) proof is divided into three steps. Recall that the units of $Rlbrack xrbrack$ are precisely the units of $R$. Also note that since $Rlbrack xrbrack$ is an integral domain and $Rsubset Rlbrack xrbrack$, then $R$ is an integral domain. Therefore, we just need to show existence and uniqueness of irreducible factorizations in $R$.

Step 1) We claim that $pin R$ is $R$-irreducible (i.e. irreducible as an element of $R$) iff $p$ is $Rlbrack xrbrack$ irreducible. We may assume $pneq 0$. The "if" part is trivial. Indeed, if $p = ab$ for $a,bin R$, then $p = ab$ is also a factorization in $Rlbrack xrbrack$. Hence wlog $a$ must be a unit in $Rlbrack xrbrack$, so also a unit in $R$. As for "only if", suppose $p$ is $R$-irreducible and $p = fg$ is a factorization in $Rlbrack xrbrack$. Then
beginalign*
deg f + deg g = deg p = 0 &Rightarrow deg f = deg g = 0 \
&Rightarrow f,gin R.
endalign*
Hence $p = fg$ is a factorization in $R$, so wlog $f$ is a unit in $R$, thus also a unit in $Rlbrack xrbrack$.

Step 2) If $ain R$, then $a$ has a factorization
beginalignlabelfactors
a = p_1dots p_n tag1
endalign
where $p_i$ is an irreducible element of $Rlbrack xrbrack$ for $i = 1,dots ,n$. As in step 1,
$$
deg p_1 + dots + deg p_n = deg a = 0 Rightarrow p_i in Rquad i = 1,dots ,n.
$$
By step 1, each $p_i$ is therefore $R$-irreducible. Then eqreffactors is an irreducible factorization in $R$.

Step 3) It remains to show uniqueness. Suppose $ain R$ and $p_1,dots ,p_n$ and $q_1,dots ,q_m$ are irreducible elements in $R$ with
beginalignlabeltwofactors
p_1dots p_n = a = q_1dots q_m.tag2
endalign
By step 1), eqreftwofactors gives two $Rlbrack xrbrack$-irreducible factorizations of $a$. But $Rlbrack xrbrack$ is a UFD, so $m = n$ and there is a permuation $sigma colon lbrace 1,dots ,nrbracerightarrow lbrace 1,dots ,nrbrace$ and units $u_i$ ($i=1,dots ,n$) in $Rlbrack xrbrack$ so that
beginalignlabelperm
p_i = u_iq_sigma (i)quad i=1,dots ,n. tag3
endalign
But $u_i$ is also a unit in $R$, so eqrefperm shows that $p_1dots p_n$ and $q_1dots q_m$ are the same factorization (up to permutation and unit) in $R$. This completes the proof. QED.

Hint: Suppose otherwise; i.e. there exist an element of $R$ which has distinct prime factorisations ("really" distinct, they're not reorderings/unit multiplications of each other). Then this element is also an element of $R[x]$. All we need to show is that primes in $R$ remain primes when viewed as an element of $R[x]$, and that non-units remain non-units, in order to contradict that $R[x]$ is a UFD.