0
$\begingroup$

I am reading Silverman's The Arithmetic of Elliptic Curves. I am wondering if with the definition of morphism he gives, we can conclude that if $\phi:V_1\rightarrow V_2$ is a morphism of projective varieties then $V_1\cong \phi(V_1)$. Throughout my mathematical career, I have seen cases where a morphism induces an isomorphism to the image, and cases where it doesn't. It usually depends on how restrictive you want the morphism condition to be. I don't really know what happens in this case, and would appreciate some insight. Here are the definitions Silverman uses for (iso)morphism of projective varieties:

Let $V_1$ and $V_2 \subset \mathbb{P}^n$ be projective varieties. A rational map from $V_1$ to $V_2$ is a map of the form $$ \phi: V_1 \longrightarrow V_2, \quad \phi=\left[f_0, \ldots, f_n\right], $$ where the functions $f_0, \ldots, f_n \in \bar{K}\left(V_1\right)$ have the property that for every point $P \in V_1$ at which $f_0, \ldots, f_n$ are all defined, $$ \phi(P)=\left[f_0(P), \ldots, f_n(P)\right] \in V_2 . $$

A rational map $$ \phi=\left[f_0, \ldots, f_n\right]: V_1 \longrightarrow V_2 $$ is regular (or defined) at $P \in V_1$ if there is a function $g \in \bar{K}\left(V_1\right)$ such that (i) each $g f_i$ is regular at $P$; (ii) there is some $i$ for which $\left(g f_i\right)(P) \neq 0$.

A rational map that is regular everywhere is called a morphism. Finally:

Let $V_1$ and $V_2$ be varieties. We say that $V_1$ and $V_2$ are isomorphic, and write $V_1 \cong V_2$, if there are morphisms $\phi: V_1 \rightarrow V_2$ and $\psi: V_2 \rightarrow V_1$ such that $\psi \circ \phi$ and $\phi \circ \psi$ are the identity maps on $V_1$ and $V_2$, respectively.

$\endgroup$
1
  • $\begingroup$ No, consider $P^n\times P^m \to P^m, (x, y) \to y$ for example, with $P^n\times P^m$ projective as it can be embedded into some $P^N$. $\endgroup$
    – C. Brendel
    Commented Mar 9 at 13:50

1 Answer 1

0
$\begingroup$

Let $E\subset\Bbb P^2$ be an elliptic curve in Weierstrass form. Then the projection to the $x$-axis gives a 2-to-1 surjective map from $E\to\Bbb P^1$, and $E\not\cong\Bbb P^1$ since they have different genera.

Even if you require bijectivity, this won't hold: consider the normalization map $\Bbb P^1\to V(y^2z-x^3)\subset\Bbb P^2$, which is bijective but not an isomorphism, since the source is smooth but the target is singular.

One positive result in this direction is Zariski's main theorem - a quasi-finite separated birational map of varieties with normal target is an open immersion. Applying this to the case of a morphism between projective varieties, we have that a quasi-finite birational morphism of projective varieties with normal target is an isomorphism. So you need some assumptions.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .