Complete linear system induce an immersion iff the associated sheaf is very ample

Let $$X$$ be a geometrically integral projective scheme over a field $$k$$, let $$D$$ be a Cartier divisor over $$X$$, and let $$\mathcal L=\mathcal O_X(D)$$ be the associated invertible sheaf. Then $$H^0(X,\mathcal O_X)=k$$ (Liu, Corollary 3.3.21), and by defining the complete linear system $$|D|=\{E\text{ effective Cartier divisor}\mid E\sim D\}$$ we get a 1-1 correspondence $$|D|\leftrightarrow\mathbb P(H^0(X,\mathcal L))$$ (I hope: this is basically Proposition II.7.7 of Hartshorne, which has additional hypotheses on $$X$$, but I believe that everything follows through). If we additionally suppose that $$|D|$$ is base-point-free (which is equivalent to the fact that $$\mathcal L$$ is generated by its global sections: again, I believe we can follow Hartshorne, Lemma II.7.8 even if not all hypotheses are met), we can associate to $$|D|$$ a $$k$$-morphism $$\varphi\colon X\to\mathbb P_k^n$$ by taking a base $$s_0,\dots,s_n$$ of $$H^0(X,\mathcal L)$$ (which is indeed a finite-dimensional vector space over $$k$$, by Hartshorne, Theorem II.5.19), and constructing the morphism described in Hartshorne, Theorem II.7.1 or here.

We say that $$\mathcal L$$ is very ample if there exists an immersion (that is, an open immersion followed by a closed immersion) $$i\colon X\to\mathbb P_k^m$$ such that $$\mathcal L\simeq i^*(\mathcal O_{\mathbb P_k^m}(1))$$. This also means that there exist non-zero global sections $$t_0,\dots,t_m\in H^0(X,\mathcal L)$$ which generate $$\mathcal L$$ and induce the immersion $$i$$, by Hartshorne, Theorem II.7.1.

My question is: is it true that $$\mathcal L$$ is very ample if and only if $$\varphi$$ is an immersion? I ask this because I think that in the case that $$\mathcal L$$ is very ample, using the notation above the global sections $$t_0,\dots,t_m$$ are not necessarily a base of $$H^0(X,\mathcal L)$$ (or even generators).

I think that we can rephrase the question as follows: is it equivalent that there exist non-zero global sections of $$\mathcal L$$ which induce an immersion, and that a base of $$H^0(X,\mathcal L)$$ induce an immersion? What about the same but with closed immersions?

• Note that every such immersion $X \rightarrow \mathbb{P}^r$ is a closed immersion ($X$ is projective, so this map is closed; since it’s an immersion, it’s a closed immersion). Commented Mar 30, 2023 at 21:10

Thanks to @Aphelli's observation that every such immersion $$X\to\mathbb P_k^n$$ is a closed immersion, I think I may have an answer.

One direction is obvious: if the morphism $$\varphi\colon X\to\mathbb P_k^n$$ induced by $$|D|$$ is an immersion, then obviously $$\mathcal L\simeq\varphi^*(\mathcal O_{\mathbb P_k^n}(1))$$ which means that $$\mathcal L$$ is very ample.

Conversely, suppose $$\mathcal L$$ is very ample and $$s_0,\dots,s_m\in H^0(X,\mathcal L)$$ are non-zero global sections such that the induced morphism $$\psi\colon X\to\mathbb P_k^m$$ is a (closed) immersion. We can suppose that among those, the first $$l+1$$ sections $$s_0,\dots,s_l$$ are a base of the linear subspace of $$H^0(X,\mathcal L)$$ generated by $$s_0,\dots,s_m$$. The subsets $$X_{s_i}=\{x\in X\mid\mathcal L_x=(s_i)_x\mathcal O_{X,x}\}=\{x\in X\mid (s_i)_x\notin\mathfrak m_x\mathcal L_x\}$$ (where $$\mathfrak m_x$$ is the maximal ideal of $$\mathcal O_{X,x}$$) for $$i=0,\dots,m$$ form an open cover of $$X$$ (otherwise the immersion wouldn't be defined everywhere), but actually $$X_{s_0},\dots,X_{s_l}$$ are sufficient to cover the whole space: if $$s=\lambda_0 s_0+\dots+\lambda_l s_l$$ with $$\lambda_i\in k$$, then $$X_s\subseteq\bigcup_{i=0}^lX_{s_i}$$ because if $$x\notin\bigcup_{i=0}^lX_{s_i}$$, then $$(s_i)_x\in\mathfrak m_x\mathcal L_x$$ for $$i=0,\dots,l$$ and so also $$s\in\mathfrak m_x\mathcal L_x$$, hence $$x\notin X_s$$.

Now, let's complete $$s_0,\dots,s_l$$ to a base $$s_0\dots,s_l,t_{l+1},\dots,t_n$$ of $$H^0(X,\mathcal L)$$. To prove that the morphism $$\varphi\colon X\to\mathbb P_k^n$$ induced by these sections (which is the morphism induced by $$|D|$$) is an immersion, we make use of Proposition II.7.2 of Hartshorne:

Let $$\varphi\colon X\to\mathbb P_A^n$$ be a morphism of schemes over $$A$$, corresponding to an invertible sheaf $$\mathcal L$$ on $$X$$ and non-zero global sections $$s_0,\dots,s_n\in H^0(X,\mathcal L)$$ which generate $$\mathcal L$$. Then $$\varphi$$ is a closed immersion if and only if

1. each open set $$X_{s_i}$$ is affine, and
2. for each $$i$$, the map of rings $$A[y_0,\dots,y_n]\to H^0(X_i,\mathcal O_{X_i})$$ defined by $$y_j\mapsto s_j/s_i$$ is surjective.

By applying this theorem to $$\psi$$ we get that the $$X_{s_i}$$ are affine for $$i=0,\dots,l$$ $$(\star)$$ and that the morphisms $$k[y_0,\dots,y_m]\to H^0(X_i,\mathcal O_{X_i})$$ are surjective, but that implies that also the similarly defined morphisms $$k[y_0,\dots,y_n]\to H^0(X_i,\mathcal O_{X_i})$$ are surjective for $$i=0,\dots,l$$ $$(\star\star)$$. We can't apply theorem II.7.2 to $$\varphi$$ directly because we don't know a lot about $$X_{t_i}$$ for $$i=l+1,\dots,n$$, but since the $$X_{s_i}$$ cover $$X$$, the restrictions $$\varphi|_{X_{s_i}}\colon X_{s_i}\to U_i\subseteq\mathbb P_k^n$$ glued together are enough to define the whole $$\varphi$$, and we can use Hartshorne's argument for the proof of theorem II.7.2 to prove that $$\varphi$$ is a closed immersion to $$\bigcup_{i=0}^lU_i$$ (so, an immersion to $$\mathbb P_k^n$$, as that is a closed immersion followed by an open immersion, so it is an immersion by Liu, Exercise 3.2.3b): by $$(\star)$$ and $$(\star\star)$$, for each $$i=0,\dots,l$$ the restriction $$\varphi|_{X_{s_i}}\colon X_{s_i}\to U_i$$ is a closed immersions to $$U_i$$, so $$\varphi(X)\cap U_i$$ is closed in $$U_i$$. Finally, $$\varphi(X)\subseteq\bigcup_{i=0}^lU_i$$ so $$\varphi(X)$$ is closed in $$\bigcup_{i=0}^lU_i$$, the map is an homeomorphism with its image because it is on each set $$\varphi^{-1}(U_i)=X_{s_i}$$ for $$i=0,\dots,l$$, and the other requirements for $$\varphi$$ to be a closed immersion are of a local nature, so they're met by looking at the restrictions.

Please, someone tell me if this makes sense!