Two nonassociated functions defining the same hypersurface? Let $X\subseteq\mathbb P^n$ be a complex, irreducible projective variety. Let $R$ be the projective coordinate ring of $X$, i.e. $R=\mathbb C[x_0,\ldots,x_n]/I$ for some homogeneous prime ideal $I$. Can there be homogeneous elements $f,g\in R$ which are irreducible, not associated and which cut out the same codimension one subvariety of $X$? In other words, can the ideals generated by $f$ and $g$ have the same radical ideal? It feels unintuitive, so I am having a hard time coming up with an example - but I see no formal reason why it can't happen.
 A: Some foolish thoughts that can't fit in the comments box:
If $f$ and $g$ are prime, then $(f)$ and $(g)$ are prime ideals and hence their own radicals. Now $(f)=(g)$ implies $f=ag$ and $g=bf$ for some $a,b\in R.$ Then $f=ag=abf \implies ab=1,$ $a$ and $b$ are units and so $f$ & $g$ are associates. 
This is the case if $R$ is a a UFD (in a UFD, irreducible implies prime). Consider the following:

Let $R$ be a noetherian domain. Then $R$ is a UFD if and only if every prime ideal
  of height one is principal.

This is Theorem 20.1 (on page 161) of Matsumura's Commutative ring theory, or you can also see its proof here. Geometrically, this means your $R$ is a UFD if and only if every codimension one subvariety of $V(I)$ is the set of zeros of some $f\in R.$
A: Here's an example with a smooth X. Consider $X = \mathbb{P}^1$ embedded in $\mathbb{P}^3$ in the Veronese (twisted cubic) embedding. The coordinate ring is
$$k[X,Y,Z,W] / (XW - YZ, Y^2 - XZ, Z^2 - YW)$$
(which we can think of as the ring $k[S^3,S^2T,ST^2,T^3]$ if we want, where $S,T$ are the coordinates on $\mathbb{P}^1$ from $\mathcal{O}(1)$. So $S=0$ is the point "$0$" and $T=0$ is "$\infty$".)
Then the linear polynomials $Y$ and $Z$ are evidently irreducible (being linear), and they generate different ideals, since the relations are all quadratic. (Note that $Y=S^2T$ has divisor $2[0] + [\infty]$, and $Z=ST^2$ has divisor $[0] + 2[\infty]$. So they have the same support but are not multiples of each other.)
But the quadratic relations directly state that $Y^2$ is divisible by $Z$ and $Z^2$ is divisible by $Y$. So they have the same radical (which I think is just $(Y,Z)$.)
Edit. Was not thinking clearly when I first posted this, and did it with the divisors $2[0]+[1]$ and $[0]+2[1]$ by mistake. Fixed now.
