Let $f:X\to X'$ be isomorphism between hypersurfaces of dimension $n-1$ ( in $\mathbb P^n$). Is it true that we can always lift this $f$ to $\mathbb P^n$?
I have already seen that for very general case this is true. My approach is using the very ample line bundle $\mathcal O_X(1)=i^*\mathcal O_{\mathbb P^n}(1)$ ($i$ is the embedding in $\mathbb P^n$). If $f^*\mathcal O_{X'}(1)=\mathcal O_{X}(1)$ and $h^0(X,\mathcal O_{X}(1))=n+1$ then obviously we can lift $f$ to $\mathbb P^n$.
So we need two conditions:
$h^0(X,\mathcal O_{X}(1))=n+1$. Using the exact sequence $0\to \mathcal O(1-d)\to \mathcal O(1) \to \mathcal O_X(1)\to 0$ this is true if $h^0(\mathbb P^n,\mathcal O(1-d))=h^1(\mathbb P^n,\mathcal O(1-d))=0$ , which trivially holds for $d>1$. So this is okay.
$f^*\mathcal O_{X'}(1)=\mathcal O_{X}(1)$. I don't know if this is always true? I know for $n>3$ the Picard group of hypersurface is isomorphic to $\mathbb Z$. And the canonical bundle of $X$ is $\mathcal O_X(d-n-1)$. So if $d-n-1\neq 0$ we always have $f^*\mathcal O_{X'}(1)=\mathcal O_{X}(1)$. But I feel very uncomfortable with the condition $d-n-1\neq 0$. Is there any way to avoid it?
Or, if there is another way to show this?
Thanks in advance.