As mentioned before on this platform, I am self-studying some commutative algebra out of A Course in Commutative Algebra by Kemper. I am trying to attempt the following problem from the text.
Let $R$ be a principal ideal domain which is not a field. Show that the polynomial ring $R[x]$ has (Krull) dimension 2.
My attempt at this exercise is to show that $\dim(R[x]) \geq 2$ and $\dim(R[x]) \leq 2.$ I have been able to show the first claim, namely:
We know that $R$ has at least one prime element $p \in R$, giving rise to a chain $${0} \subset (x)_{R[x}] \subset (x, p)_{R[x]}$$ of prime ideals. So $\dim(R[x]) \geq 2.$
Its showing that $\dim(R[x]) \leq 2$ that is giving me problems. Now, I am aware that for a Noetherian ring $R \neq \{ 0 \}$ we have $$\dim (R[x]) = \dim(R) + 1.$$ This fact would prove the exercise, however this fact is not introduced until later in the text so I want to prove this exercise without it. Any help one showing why $\dim(R[x]) \leq 2$ would be appreciated!