Prime ideals below a principal prime ($0\ne P\subseteq (p)⟹P=(p)$)

Let $R$ be a unique factorization domain and let $p$ be a prime element. The only prime ideals contained in the principal ideal $(p)$ are $(0)$ and $(p)$. To prove this, suppose $0\ne P⊊(p)$. Then $p\notin P$. For $0\ne f\in P$, unique factorization allows us to write $f=p^{k}g$ with $g\notin (p)$. Since $P$ is a prime ideal and $p\notin P$, repeated primality gives $g\in P\subseteq (p)$, a contradiction.

In the particular polynomial ring $\mathbb{Z}[t]$, when $p$ is a prime number, the same proof uses only the integer coefficients: factor out the largest common power of $p$ from the coefficients of a nonzero polynomial $f$. This gives $f=p^{k}g$ with $g\notin (p)$ directly, without requiring the general unique factorization domain theorem for polynomial rings.