# A question on very ample line bundle and closed immersion

Let $X$ be a projective scheme, $i:X \hookrightarrow \mathbb{P}^n$ a closed immersion, $\mathcal{L}:= i^*\mathcal{O}_{\mathbb{P}^n}(1)$ a very ample line bundle. Let $j:{\mathbb{P}^n} \hookrightarrow \mathbb{P}^N$ be a closed immersion and $\mathcal{L}':= (j \circ i)^* \mathcal{O}_{\mathbb{P}^N}(1)$. Is it true that $\mathcal{L}\cong \mathcal{L}'$?

It need not be true in general. For example, take $X = \mathbb{P}^1$, $n=1$, $i$ the identity map, $N = 2$ and $j$ the degree 2 Veronese map into $\mathbb{P}^2$. In this situation $\mathcal{L} \cong \mathcal{O}_{\mathbb{P}^1}(1)$ but $\mathcal{L}' \cong \mathcal{O}_{\mathbb{P}^1}(2)$.
• Thank you. So, is there any condition under which $\mathcal{L}$ will be isomorphic to $\mathcal{L}'$? – user54369 Jul 18 '15 at 12:04
• Yes, I would say so. To cause them to be non-isomorphic in my example I needed a specific choice of map $j$. If instead I had chosen $j$ to embed $\mathbb{P}^1$ as a line in $\mathbb{P}^2$ then the line bundles would be isomorphic, and I believe this is true more generally, if you embed $\mathbb{P}^n$ as a linear subspace of $\mathbb{P}^N$. Perhaps there are weaker conditions than that, but I don't know them :) – john Jul 18 '15 at 12:14