Say I have 2 Hypersurfaces $S,S'\subset \mathbb{P}_k^3$ with different degrees and $k$ a field. Can it happen that the two Hypersurfaces are isomorphic or can they be fully distinguished by their degree? I have the feeling that the degree should be an invariant even for non smooth and non integral Hypersurfaces and I think there is some proof using Serre-Duality or Riemann-Roch, but I can't get the details right.
I appreciate any help.
Here is a (fairly) elementary argument for the nonsingular case: let a degree $d$ hypersurface $i:S\hookrightarrow \Bbb{P}^3$ be given. There is a short exact sequence: $$ 0 \to \mathcal{O}_{\Bbb{P}^3}(-d)\to \mathcal{O}_{\Bbb{P}^3} \to i_*\mathcal{O}_S\to 0. $$ By the adjunction formula, $\omega_S = \mathcal{O}_S(d-4)$ and so twisting by $d-4$ gives $$ 0\to \mathcal{O}_{\Bbb{P}^3}(-4)\to \mathcal{O}_{\Bbb{P}^3}(d-4)\to i_*\omega_S\to 0. $$ The long exact sequence of cohomology yields $$ 0\to H^0(\Bbb{P}^3,\omega_{\Bbb{P}^3})\to H^0(\Bbb{P}^3,\mathcal{O}_{\Bbb{P}^3}(d-4))\to H^0(S,\omega_S)\to H^1(\Bbb{P}^3,\omega_{\Bbb{P}^3}). $$ In particular, $h^0(\Bbb{P}^3,\mathcal{O}_{\Bbb{P}^3}(d-4)) = g(S)$. This completely distinguishes hypersurfaces of degree $d\ge 4$. The remaining case is when $1\le d\le 3$ for both $S$ and $S'$. Now, the $d=1$ case gives $\Bbb{P}^2$, the $d=2$ case gives a quadric which is isomorphic to $\Bbb{P}^1\times \Bbb{P}^1$. For $d=3$, any cubic in $\Bbb{P}^3$ is a blowup of $\Bbb{P}^2$ at exactly $6$ points. This distinguishes it from the other low degree cases (for instance, consider the rank of the Picard group).
As for the singular case, I'm not completely sure. Probably something can be said with more attention.