I myself want to prove the fundamental fact that is:
If $R$ is UFD then so is $R[X]$.
Here is my approach:
Let $f$ is a polynomial of $R[X]$ and assume $\deg(f)=d$, $d$ is non-negative integer.
We just consider the case $f$ is not irreducible then we have the factorization:
$f$= $d_1d_2...d_m.f_1.f_2...f_n...$.
Where $d_i $ is an irreducible element of $R$ and $f_i$ is a polynomial which have degree at least $1$.
Because the degree of $f$ is $d$ is finite and each $f_i$ has degree at least $1$ so we can not have infinite those elements $f_i$.
Now, if we have two factorizations $f$=$f_1.f_2...f_n$ = $g_1.g_2...g_m$ $(1)$ so:
$g_1| f_1.f_2...f_n$ we must have some $f_i$, say $f_1$ such that: $g_1|f_1$ by the irreducibility of $f_1$ we conclude $f_1$= $a_1$.$g_1$, where $a_1$ is unit in $R$. Cancel both $f_1$ and $g_1$ from $(1)$ and continue on this fashion we have $m \le n$.
Conversely, $f_1|g_1...g_n$, use the same argument we also have $n\le m$ then $n=m$.
This proof is the first pop in to my head, I also red the popular proof for the theorem in textbook. I would say my process is wrong at certain point, but I cant see it. Please illustrate the point to me. Thank you.