Do prime elements generate a prime ideal?

Question

If $p_1, p_2, p_3 \in R$ are prime elements, does it follow that $\left<p_1,p_2,p_3\right>$ will be a prime ideal?

Answer 1

No: this is not true even for $n \ge 2$ prime elements. A counterexample follows:

let $R= \Bbb{C}[X,Y]$. By Hilbert's Nullstellensatz we know that radical ideals of $R$ are in bijection with algebraic subsets of $\Bbb{C}^2$. Moreover, prime ideals correspond to irreducible algebraic sets. Now, if you consider the family of "complex parabolas" with equations $$y= ax^2-a \ \ \ \ \ a \in \Bbb{C}$$ all of these correspond to prime elements of $\Bbb{C}[X,Y]$, namely $\{ Y- aX^2+a\}_{a \in \Bbb{C}}$. Denote these elements by $\{ p_a \}_{a \in \Bbb{C}}$. All these belong to $m_1 \cap m_2$, where $$m_1 = \langle X-1, Y\rangle\ \ \ , \ \ \ m_2 = \langle X+1, Y\rangle$$ which are maximal ideals of $R$: indeed all the parabolas have in common exactly two points: $(1,0)$ and $(0,1)$.

Fix now finitely many distinct complex numbers $a_k$ with $k = 1, \dots , n$, and consider the ideal $\langle p_{a_1}, \dots , p_{a_n} \rangle$ : this is a sum of prime ideals, hence it corresponds to the intersection of the corresponding parabolas. But this intersection is given by the union of the two points above, which is not irreducible: in other words $$\langle p_{a_1}, \dots , p_{a_n} \rangle = m_1 \cap m_2$$ and this is not a prime ideal of $\Bbb{C}[X,Y]$.