We want to prove that in a UFD the gcd exists.
In the proof we let $R$ be a UFD and $a,b \in R$ So we suppose that $a,b$ are not units because if one of them were the unit, the proof is trivial.
The proof now let $a,b$ be both non units. Well we know that we can write $$a=T_1×...×T_m$$ and $$b=R_1×....×R_m$$ were $T_i$ and $R_j$ are irreducible elements. Now we can rewrite them as $$a=p_1^{α_1}.....p_n^{α_n}$$ And $$b=p_1^{β_1}....p_n^{β_n}$$ were $p_i$ are prime obviously. Here's my 1st question: how did we let the prime elements in both $a$ and $b$ be equal and with same index $n$?? Shouldn't they be different? Shouldn't we say that $b=q_1^{β_1}...q_m^{β_m}$??
Also: it defines $s_i=min(α_i,β_i)$ and let $d=p_1^{s_1}...p_n^{s_n}$. again how are the primes equals to the case of $a$ and $b$? The proof proceeded to show that $d$ is the gcd. I can see why $d$ is the common divisor. To show that d is the gcd, we let $x|a$ and $x|b$ Then if x is a unit then $x|d$ But what if x is non unit??
The proof needs details that the author didn't include or they're trivial but I don't see them. Please help! :(
The trick is that we allow the exponents $\alpha_i,\beta_i$ to be zero, and the list $p_1,\dots,p_n$ contains all primes that divide either $a$ or $b$.
The other important thing is that in an UFD, $a\,|\,b$ iff for every prime $p$ and exponent $\gamma\in\Bbb N$ we have $p^\gamma\,|\,a\implies p^\gamma\,|\,b$, that is, for every prime $p$, if the exponent of $p$ in $a$ is $\alpha$ and in $b$ is $\beta$ (and a priori they can be zero as well), then $\alpha\le\beta$.