First, some context. I am working with Dummit and Foote's Abstract Algebra, 2nd edition. I stumbled upon this while working on Section 8.3 Exercise 11, which is to prove that all Bezout UFDs are PIDs. While reasoning through this problem, I noticed something which is quite odd, yet I can't figure out what.
Suppose $R$ is a UFD. Then for every $n \in R$, we can write $n=p_1^{i_1}p_2^{i_2}...p_k^{i_k}$ where this factorization is essentially unique (i.e. unique up to multiplication by units.) Now, it is a fact that $R$ is Bezout if and only if every pair of elements in $R$ has a gcd in $R$ by a past exercise (I think it's exercise six or seven in the previous section). However, in every UFD, we can decompose any pair of elements into primes as above, i.e. for every $b \in R, b = p_1^{j_1}p_2^{j_2}...p_k^{j_k}$, with the appropriate $p$'s, and each $i,j \geq 0$. Then it is relatively obvious that gcd $(a,b)$= $p_1^{ \text {min} (i_1, j_1)}p_2^{ \text {min} (i_2, j_2)}...p_k^{\text {min} (i_k, j_k)}$. Since each $p$ was in the UFD, the product is in the UFD, so gcd $(a,b)$ belongs to the UFD, so in any UFD, any pair of elements has a gcd. So every UFD is Bezout.
Now I know this is wrong, Since $\mathbb Z [x]$ is a UFD but is not Bezout, as $(2,x) \subset \mathbb Z [x]$ is an ideal which is not principal, but I don't understand why.
My hope is that by understanding what is wrong with this proof, I can solve the larger problem of showing that all UFDs which are Bezout are PIDs.
Your proof that a gcd exists is correct. However, it is not a fact that $R$ is Bezout if and only if every pair of elements in $R$ has a gcd in $R$ -- the gcd must also be a linear combination of the two elements. That fails in $\mathbb{Z}[x]$.