= Prime ideals below a principal prime
{title2=$0\ne P\subseteq(p)\Longrightarrow P=(p)$}
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 $0\ne P\subsetneq(p)$. Then $p\notin P$. For $0\ne f\in P$, <unique factorization> allows us to write $f=p^kg$ 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 $f=p^kg$ with $g\notin(p)$ directly, without requiring the general <unique factorization domain> theorem for <polynomial rings>.
Back to article page
