Assume the field $\mathbb{K}$ we're working over is algebraically closed. Let $V\subset\mathbb{A}^n$ be an irreducible affine variety such that the coordinate ring $\mathbb{K}[V]$ is NOT an U.F.D. Let $f\in\mathbb{K}[V]$ be an irreducible element. Further, assume that the vanishing set $\mathcal{V}(f)$ is irreducible. Is it true that the ideal $(f)$ is a prime ideal?
Another question I have is that- if $\mathcal{V}(\mathfrak{a})$ is irreducible for some principal ideal $\mathfrak{a}\subset\mathbb{K}[V]$, then is it true that $\mathfrak{a}$ is a prime ideal?
I know that the first statement is true when the coordinate ring is an U.F.D.
Edit: I don't think the link that is provided answers my question. The only information I get using the link is that $\sqrt{(f)}$ is prime ideal, but I'm asking more than that. I'm asking is the ideal $(f)$ itself prime? Notice that the ring is not an UFD, so irreducible element might not necessarily be prime.
Best Answer
We'll deal with the additional question first because it's easily set down: $V(\mathfrak{a})$ is irreducible iff $\sqrt{\mathfrak{a}}$ is prime, and this is true in complete generality. But there are plenty of examples where $V(\mathfrak{a})$ is irreducible yet $\mathfrak{a}$ is not prime: take $(x^2)\subset k[x]$, for instance.
The first statement is a little more involved. It is not true that $V(f)$ irreducible with $f$ irreducible implies $(f)$ prime. Consider $V(xy-z^2)\subset\Bbb A^3$ with coordinate algebra $k[x,y,z]/(xy-z^2)$ and take $f=x$. Then $V(x)\subset V$ is just the $y$-axis, which is clearly irreducible, and $(x)$ is not prime because $(k[x,y,z]/(xy-z^2))/(x)\cong k[x,y,z]/(x,xy-z^2)\cong k[y,z]/(z^2)$ is not a domain. All that's left is to verify that $x$ is irreducible.
To show that $x$ is irreducible, consider the norm map $N:k[x,y,z]/(xy-z^2)\to k[x,y]$ which sends $p(x,y)+q(x,y)\cdot z$ to $(p+qz)(p-qz)=p^2-xyq^2$. $N$ is multiplicative, and I claim that $N(a)$ is a unit iff $a$ is a unit. One direction is easy: if $ab=1$, then $1=N(1)=N(ab)=N(a)N(b)$. On the other hand, suppose $q\neq 0$. Then for $p^2-xyq^2=u$ to be true, $\deg q=\deg p-1$ and looking at highest homogeneous parts, we must have $(p)_{\deg p}^2=xy(q)_{\deg q}^2$, where $(-)_d$ denotes the degree $d$ part of that polynomial. But this cannot be true: if $x\mid (p)_{\deg p}^2$, then $x^2\mid (p)_{\deg p}^2$ and similarly with $y$, and we reach a contraction by descent. So $q=0$ and therefore $p=\sqrt{u}$.
Therefore to prove that $x$ is irreducible it suffices to show that there is no element $p(x,y)+q(x,y)z$ so that $p^2-xyq^2=ux$. Evaluating both sides at $y=0$, we find that $p(x,0)^2=ux$ which is impossible by looking at the top-degree term of $p(x,0)$. So $x$ is irreducible and we're done.