TITLE: An injective morphism between varieties that is not an immersion
QUESTION [5 upvotes]: I believe this is relatively elementary, but I'm struggling to think of an example of a morphism $f: X \rightarrow Y$ between varieties which isn't an immersion in the sense of algebraic geometry.  
I'd prefer to see something "geometric", say of varieties over $\mathbb{C}$, to rule out "algebraic" counterexamples involving weird characteristic $p$ things (I believe the Frobenius map over a non-perfect field could possibly be an example?) or possibly different base fields. 
Bonus points if $X$ is projective (or proper, I suppose) so the map is closed and thus a topological embedding.
Context: I'm reading - in Qing Liu's Algebraic Geometry and Arithmetic Curves, Chapter 7.4 -  about under what conditions a divisor $D$ on a projective curve $X$ is very ample. The argument proceeds by first showing the map is injective, then showing that the differential is injective. Intuitively, in the analogy with the complex analytic case, this latter step is verifying that the map "preserves the complex structure" in a differential-geometric sense.

REPLY [5 votes]: Let $X = \{ (x, y) \in \mathbb{C}^2 : x^2 - y^3 = 0 \}$, let $Y = \mathbb{C}$ and let $X \to Y$ be the first projection. This is a homeomorphism (!) but it is not even an immersion: indeed, the corresponding ring homomorphism $\mathbb{C} [t] \to \mathbb{C} [x, y] / (x^2 - y^3)$ is the one that sends $t$ to $x$, and it is clear that $y$ is not in the image of the homomorphism.

There is also a projective version of the above. Let $X = \{ (x : y : z) \in \mathbb{C P}^2 : z x^2 - y^3 = 0 \}$ and let $Y = \mathbb{C P}^1$. Then $(x : y : z) \mapsto (x : z)$ is a regular map $X \to Y$ and is again a homeomorphism but not an immersion.