Let, $Z_1 =Z(f)$ and $Z_2 =Z(g)$ be two irreducible hypersurfaces in $\mathbb{P}^{3}$of degree $m$,$n$, where $g.c.d(m,n)=1$ and $1 <n< m$.
Is it true that $Z_1 \cap Z_2$ is always an irreducible curve?(we assume that the intersection is nonempty)
My guess is no because if we compute its ideal, then we have, $I(Z_1 \cap Z_2) = I(Z_1) \cup I(Z_2) = \sqrt {(f)} \cup \sqrt{(g)}$, which need not be necessarily even an ideal(so therefore not a prime), though we know $\sqrt {(f)}$ ,$\sqrt {(g)}$ are prime ideals.
So is it true that if we impose some condition on one of the hypersurfaces then it's irreducible? for instance if we assume one of the $Z_1$ or $Z_2$ is smooth.
In order to show that it's a curve if we assume dim ($Z_1 \cap Z_2) \neq 1$,then if it's $2$,then by Hartshorne exercise $1.10(d)$(chapter -$1$), we have $Z_1 \cap Z_2 = Z_1 \Rightarrow Z_1 \subset Z_2 \Rightarrow \sqrt{(g)} \subset \sqrt{(f)}$ , from where I don't see any contradiction even after using $g.c.d(m,n)=1$ and degree comparison.When dim $Z_1 \cap Z_2 =0$ it's even more unclear.
Any help from anyone is welcome.