What implies that $D[X]$ is an UFD?

Question

If $D[X]$ is an UFD, then $D$ is at least another UFD and here is my thought: if $x \in D$, I define $f(X) \in D[X]$ by $f(X) = x$. As $D[X]$ is an UFD, then there exists ${\{f_i(X)\}}_{i = 1}^n \subset D[X]$ such that $$ x = f(X) = \prod_{i = 1}^n f_i(X)\mbox{.} $$ As $D[X]$ is an UFD, then particularly it is an integral domain, so $$ 0 = \deg f = \sum_{i = 1}^n \deg f_i $$ and it implies that $\deg f_i = 0$ for all $i = 1 , \ldots , n$, so exist ${\{x_i\}}_{i = 1}^n \subset D$ such that $f_i(X) = x_i$ for all $i = 1 , \ldots , n$ and thus $$ x = \prod_{i = 1}^n x_i $$ and it implies automatically that $D$ is an UFD. But my question is: can I state that $D$ is a field or a PID (supposing that $D[X]$ is only an UFD)? I have wondered this question because the next statement is (as my thought) true: $R$ is a field if $R[X]$ is a PID, and is a stronger than to state: if $R[X]$ is a PID, then $R$ is another PID.