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?

  • 2
    $\begingroup$ 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). $\endgroup$
    – Aphelli
    Commented Mar 30, 2023 at 21:10

1 Answer 1


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!


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