If $f$ is irreducible in a ring which is not a UFD, is the ideal $(f)$ a prime ideal? 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.
 A: 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.
