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What are interesting and important examples of morphisms of schemes (especially varieties) to keep in mind when trying to understand a new concept or looking for a counterexamples?

Examples of what I'm looking for:

  • The projection from the hyperbola to the affine line has finite fibers but it is not a finite morphism
  • The Froebnius map is a morphism of varieties which is a bijection nevertheless not an isomorphism
  • The map $(x,y) \mapsto (xy, y)$ shows that the image of an affine variety need not be affine

Thanks for the excellent answers! To be a little more specific, I am especially interested in reasonable examples (so no line with infinitely many origins or the product of infinitely many fields). If the examples already make sense in the classical settings then all the better.

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This question isn't really focussed and is very subjective. –  Martin Brandenburg Feb 6 at 16:22
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Astonishingly I think that this is a very good question! I have noticed that an amazing lot of questions in algebraic geometry and commutative algebra can be answered by resorting to a very small set of judicious examples. I'll try to give a small list of examples I found strongly recyclable. –  Georges Elencwajg Feb 6 at 18:57
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But Martin has a point: the required examples might be reminiscent of Borges's classification from a fictional Chinese encyclopedia: "Animals are divided into (a) those that belong to the Emperor, (b) embalmed ones, (c) those that are trained, (d) suckling pigs, (e) mermaids, (f) fabulous ones, (g) stray dogs, (h) those that are included in this classification, (i) those that tremble as if they were mad, (j) innumerable ones, (k) those drawn with a very fine camel's hair brush, (l) others, (m) those that have just broken a flower vase, and (n) those that resemble flies from a distance." –  Georges Elencwajg Feb 6 at 20:09
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I also wonder why the two answers gather random general facts about morphisms and not as much specific examples of morphisms. –  Martin Brandenburg Feb 7 at 9:55
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@MartinBrandenburg Feel free to add your own :) I would have personally loved to have these lists when I was starting to learn AG! –  Alex Youcis Feb 7 at 10:45

4 Answers 4

up vote 13 down vote accepted

At the request of Mr. Elencwajg, I'll make my comments an answer.

I will include the "correct" (read most general) hypothesis below. If they are unfamiliar to you, most can be done away by assuming we are working in variety land (reduced and separated):

1) Chevalley's theorem: finite type morphisms between Noetherian schemes send constructible sets to constructible sets. Constructible just means a finite union of locally closed (locally closed=intersection of an open and a closed).

For example, take Georges's nice example of the map $\mathbb{A}^2\to\mathbb{A^2}:(x,y)\mapsto (xy,y)$. The image is not open, or closed, but it is constructible. Indeed, its image is $\mathbb{A}^2$ minus the $y$-axis, but with the origin replaced. This is the union of the $x$-axis (closed) and $\mathbb{A}^2-\{y\text{-axis}\}$ (open).

Note that this seems contravariant to intuition from complex geometry since there are all maps are open (the open mapping theorem). But, the difference is that there we are in the "smooth category". It turns out that all "smooth morphisms" are flat and thus, by 4) of Georges's response, open.

This is also important because it helps you relate schemes and varieties. In particular, it follows pretty quickly from Chevalley that a map of varieties (thought of scheme theoretically) is surjective if and only if it's surjective on closed (classical) points.

Lastly, just because I can't resist saying it, this gives a very nice proof of Zariski's lemma (a form of the Nullstellensatz) that if $L$ is a field finitely generated over a field $K$, then it's actually a finite extension of $K$. Indeed, since $L/K$ is a finitely generated (as an algebra) extension, it suffices to show it's algebraic. But, if $L/K$ weren't algebraic there would exist some injection $K[x]\to L$, which would induce a dominant finite type map $\text{Spec}(L)\to\mathbb{A}^1_K$ of Noetherian schemes. In particular, the image of $\text{Spec}(L)$ in $\mathbb{A}^1_K$ would be constructible by Chevalley. But, since $\text{Spec}(L)\to\mathbb{A}^1_K$ is dominant, the image is just the generic point of $\mathbb{A}^1_K$. But a quick check shows that the generic point of $\mathbb{A}^1_K$ isn't constructible--contradiction.

2) If $X$ and $Y$ are $Z$-schemes, $X$ reduced, and $Y$ separated over $Z$, then if $f,g:X\to Y$ are morphisms which agree on an open set (i.e. they are equalized by $U\hookrightarrow X$ for some open subscheme $U$ of $X$) then they are equal. Or, if we stick to the category of separated (over $Z$) reduced $Z$-schemes, then open embeddings are epimorphisms. In particular, in the category of varieties open embeddings are epimorphisms.

This is very obvious in classical language, say of affine schemes over an algebraically closed field. Indeed, there, since the difference of morphisms is a morphism, the locus of agreement is closed. In scheme theory land this is much stickier, and is handled by the above theorem.

Since each of thse should probably include an example, you can consider the map $X\to\text{Spec}(k)$ where $X$ is the line with doubled origin (i.e. glue two copies of $\mathbb{A}^1$ together along $\mathbb{G}_m$). This map is NOT separated. Consider the theorem I stated above and the two morphisms $\mathbb{A}^1_k\to X$ which are inclusions, but with different choices of which origin in $X$ to send the origin of $\mathbb{A}^1_k$. They agree on $\mathbb{G}_m$, but aren't equal.

3) The maps $\mathbb{A}^1\to\mathbb{A}^1:z\mapsto z^2$ and the projection of the cuspidal cubic $y^2=x^3$ onto the $x$-axis

These are nice maps to understand ramification. In particular, both are ramified at the origin. The first seems like it's ramified because the actual map is bad, the second seems like it's bad because the cuspidal cubic is bad (although, if you think birationally it's because the curve you started with was obtained by doing something bad--twisting/pinching $\mathbb{P}^1$).

The second example also shows how the cotangent sheaf can keep track of infinitesimal data. In particular, you can compute that $(\Omega_{C/\mathbb{A}^1}^1)_{(x,y)}$ (where $C$ is the cuspidal cupic) is $k[x]/(x^3)$.

4) If $X$ and $Y$ are reduced irreducible curves over a field $k$, with $X$ proper over $k$, then any $k$-morphism $X\to Y$ is either constant or surjective. In the latter case $Y$ is proper over $k$, and the morphism is finite.

The first fact follows since the image of $X$ in $Y$ must be irreducible and closed, and thus by dimension considerations must be a point or all of $Y$. If it's all of $Y$, since $Y$ is reduced, it is the scheme theoretic image of $X$ must be proper. Finally, it's finite because it's proper and quasifinite (finite type and finite fibers)--this equivalence follows from ZMT. The latter is true because only the generic point of $X$ can map to the generic point of $Y$, and closed points can't have infinite fibers else they'd be dense (since $X$ carries the cofinite topology).

This is particularly useful when one wants to study elliptic curves.

5) In classical language, given a complete variety $X$ over a field $k$, and a nonsingular curver $C/k$ then one can always extend morphisms $C/\{P\}\to X$ to morphisms $C\to X$ (where $P$ is a point of $C$). In fact, this turns out to be equivalent to completeness. This is believable, at least for projective space. Indeed, you can just multiply by an appropriate power of the uniformizer at $P$ to cancel out any issues for the morphism at $P$. For example, consider the curve $\mathbb{A}^1-\{0\}\to\mathbb{P}^1$ defined by $x\mapsto [\frac{1}{x}:1]$. This can obviously be extended to $\mathbb{A}^1\to\mathbb{P}^1$ by rewriting our map as $x\mapsto [x:1]$. This trivial example is actually quite indicative of the general case.

If you're interested, this is a primitive form of the valuative criterion for properness.

6) Projection of a projective variety onto a hyperplane. This is a very classical construction in projective geometry which is not only a pivotal example of a morphism from a projective variety, but is part of many theorem statements.

7) The morphism $\text{Spec}(\mathbb{Z}[i])\to\text{Spec}(\mathbb{Z}[3i])$.

This is very much an arithmetic analogue of Geroges's 1) (the normalization of the cuspidal cubic). Indeed, this is a normalization map which is finite and bijective, but of course not an isomorphism. You can think about it as patching up the singularity at $(3,3i)$ (i.e. there the localization isn't regular).

Having number theoretic analogues is important, if only to convince yourself of the power of scheme theory to connect the geometric and arithmetic. It will also enhance your understanding of number theory to be able to see these things as inherently geometric constructions.

8) An extension of fields $\text{Spec}(L)\to\text{Spec}(K)$.

At the risk of sounding melodramatic, this is one of the most important examples I can think of. Structurally it is very simple, but it doesn't take long to realize that, in fact, maps between points encompass a HUGE amount of complex data. For example, the automorphism group of the point $\text{Spec}(\mathbb{Q})$ is nothing more than the absolute Galois group $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, the (in some sense) subject of all of modern number theory. Understanding how complex maps between points are will give you a real appreciation of how intricate the construct of schemes really is.

Moreover, the group $\text{Gal}(\mathbb{C}/\mathbb{Q})$ gives uncountably many examples of isomorphisms between two schemes that are equal on the underlying space.

9) Let $X/\mathbb{C}$ be finite type, and let $\sigma\in\text{Gal}(\mathbb{C}/\mathbb{Q})$. We can then define $X^\sigma$, usually called "$X$ twisted by $\sigma$", to be the scheme making the following diagram fibered:

$$\begin{matrix}X^\sigma & \xrightarrow{p} & X\\ \downarrow & & \downarrow\\ \text{Spec}(\mathbb{C}) & \xrightarrow{\text{spec}(\sigma)} & \text{Spec}(\mathbb{C})\end{matrix}$$

Or, in other words, if $X$ is affine just act on the coefficients of the defining polynomials of $X$ by $\sigma$. The important morphism is then $p$. It is an isomorphism of abstract schemes, but is patently not an isomorphism of $\mathbb{C}$-schemes.

While the fact that $p$ is an isomorphism of schemes over $\mathbb{Z}$, which is not an isomorphism of schemes over $\mathbb{C}$, isn't that special, it's a really good example to keep in mind when learning things like the statement of GAGA, or just analytification in general. In particular, let's suppose that $X$ was projective (proper), then the analytification $X^{\text{an}}$ of $X$ is a projective (resp. compact) analytic variety. Similarly, we have a projective (resp. proper) analytic variety $(X^\sigma)^{\text{an}}$.

Now, the isomorphism $p:X^\sigma\to X$ does not respect $\mathbb{C}$, and thus really does not induce an isomorphism $X^{\text{an}}\cong (X^\sigma)^{\text{an}}$. But, you might expect that the two analytic schemes can't be "that different" since we have an isomorphism $p$ as abstract schemes between their underlying schemes. That said, Serre gave an example where $X^\text{an}$ and $(X^\sigma)^{\text{an}}$ aren't even homotopy equivalent. This helps you keep in mind how important it is to keep track of relative data of maps, especially when passing between algebraic geometry and other subjects.

10) The open embedding $\iota:\mathbb{A}_k^2-\{(0,0)\}\hookrightarrow \mathbb{A}_k^2$ and the map $\varphi:\mathbb{A}^2_k\to X$, where $X$ is the affine plane with doubled origin, and the map just embeds in the natural way, choosing one of the origins.

Why these maps are important relies on a commonly stated fact--the plane minus the origin is not affine. There are several ways of proving this, with varying levels of sophistication. Probably the least sophisticated, but totally rigorous way, is to use what Vakil calls "Algebraic Hartog's Theorem" and our first map $\iota$. Algebraic Hartog's says that if $X$ is a locally Noetherian integral normal scheme, and if $U\subseteq X$ is open, with $\text{codim}_X(X-U)\geqslant 2$, then the restriction map $\mathcal{O}_X(X)\to\mathcal{O}_X(U)$ is an isomorphism (of rings). In particular, since $\mathbb{A}_k^2-\{(0,0)\}$ is the complement of a closed subset of $\mathbb{A}_m^2$ of codimension $2$ (namely the point $(0,0)$) Hartog's lemma says that the induced map on global sections is a ring isomorphism. But, the restriction map is precisely the induced map on global sections coming from our morphism $\iota$. If $\mathbb{A}^2_k-\{(0,0)\}$ were affine, then the antiequivalence of categories between rings and affine schemes would say that since the induced map on global sections $\iota^\sharp$ is an isomorphism, then $\iota$ must be an isomorhpism. Since this is patently false, we see that $\mathbb{A}^2_k-\{(0,0)\}$.

Now we can say why these two maps are good examples. It's true that any map between affine schemes is an affine map (this follows because affineness is an affine local condition on the target). That said, the two maps given above are both non-affine, one with affine source and the other with affine target. We've explained why $\iota$ is non-affine. For the other, consider the copy $U$ of $\mathbb{A}^2_k$ sitting inside of $X$ NOT containing the origin $\varphi$ picks out. Then, $U$ is affine, but $\varphi^{-1}(U)=\mathbb{A}^2_k-\{(0,0)\}$ and thus is non-affine.

Both have these have come up in other scenarios, when trying to drop affineness conditions from a theorem.

11) The closed embedding $X_\text{red}\to X$.

This is a good example of a kind of degenerate closed embedding, and often forces you to to appreciate the difference between closed embeddings in the classical setting, and closed embeddings in the modern setting. For one, very obvious, wacky thing, consider that $X_\text{red}\to X$ defines a closed embedding whose image is everything, but which is not just the identity embedding (when $X$ is non-reduced).

It also gives a counterexample to the statement that a locally closed embedding with open image need not be an open embedding. While this may seem obviously not true, consider that it IS true if one replaces the words "open" with "closed".

12) The morphism $\text{Spec}(\mathbb{C}[x,y,t]/(y^2-x^3+t))\to \text{Spec}(\mathbb{C}[t])$.

This is an excellent morphism to keep in mind. It really nicely allows you to see how algebraic geometry is very well suited to discussing degenerations of smooth objects into singular objects. In particular, you see that the fiber over every closed point of the base, except $(t)$, corresponds to a non-singular curve (consider the discriminant), but the fiber over $(t)$ is singular--it corresponds to the cuspidal cubic that has been mentioned many times before.

13) The map $\mathbb{A}^1_\mathbb{C}\to\mathbb{A}^1_\mathbb{R}$.

Perhaps you have heard the way to picture schemes like $\mathbb{A}^1_\mathbb{R}$. Namely, we first move to $\mathbb{A}^1_\mathbb{C}$, where we can see all the points (this is just the complex plane). We then see what points we have "pulled apart" by extending our field, and the reidentify them. A little thinking shows that what has happened in somewhat simple. We already had all of the real points $(x-a)$ for $a\in\mathbb{R}$, and each conjugate pair of non-real points $(x-(a\pm bi))$ were obtain by pulling apart the point $(x^2-2ax+(a^2+b^2))$.

Thus, we can think about $\mathbb{A}^1_\mathbb{R}$ as taking the complex plane, and folding it in half, where the crease is at $\mathbb{R}$. So, right now, the picture is something like this where, as we have already said, the fold is at $\mathbb{R}$. But, now, the on each of the two "folds" there are one of the two conjugate points corresponding to an irreducible quadratic over $\mathbb{R}$. Thus, finally, to get the picture of $\mathbb{A}^1_\mathbb{R}$ we identify these points.

Now, we can think of our map $\mathbb{A}^1_\mathbb{C}\to\mathbb{A}^1_\mathbb{R}$ as this last step--going from folded plane to identifying the two flaps of the folded plane along conjugate points.

Why is this map important? Because, it's étale, and, in fact, a theorem of Chevalley says (interpreted correctly) that all étale maps look like this very close up. They all look like projections of folded sheets where sheets can come together, but not "transversely" (not in a ramified manner).

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I hadn't thought of giving arithmetic examples: great idea to mention them! –  Georges Elencwajg Feb 7 at 11:06
    
@GeorgesElencwajg Will do ;) –  Alex Youcis Feb 7 at 11:18
    
On point (2): you only need $Y \to Z$ to be separated. This is because the equaliser (= locus of equality) is the pullback of the diagonal $Y \to Y \times_Z Y$ along the morphism $\langle f, g \rangle : X \to Y \times_Z Y$. –  Zhen Lin Feb 7 at 14:23
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@ZhenLin And not reduced? That's not true. From what you're saying we can conclude that the locus of agreement is a closed subscheme of X whose underlying set is all of X. But, if X isn't reduced that doesn't imply that we are actually looking at X. –  Alex Youcis Feb 7 at 18:30
    
@ZhenLin There are schemes whose underlying topological spaces are homeomorphic but are not the same scheme: Consider $X$ and $X_{red}$. –  user38268 Feb 9 at 15:24

Here are some (random) useful facts/examples about morphisms:

1) The normalization of the cusp $C\subset \mathbb A^2$ given by $y^2=x^3$ is the morphism $\nu:\mathbb A^1\to C:t\mapsto (x=t^2,y=t^3)$
That morphism is finite, bijective but is not an isomorphism and is not flat. More generally:

2) A non trivial normalization is never flat

3) Any proper morphism between smooth varieties all of whose fibers have the same dimension is flat: "Miracle Flatness".

4) Flat morphisms between varieties are open.

5) The inclusion $\text {Spec} (k)\hookrightarrow \text {Spec}( k[\epsilon]/(\epsilon^2))$ of the simple point into the double point is open but not flat.

6) Any morphism $\mathbb A^1 \to \mathbb A^n $ has closed image but the image of the morphism $\mathbb A^2\to \mathbb A^2:(x,y)\mapsto (x,xy)$ is not closed. [This is an extension of an example in the question]

7) Smooth, or even étale, morphisms needn't have local sections: think of the squaring map $\mathbb G_m\to \mathbb G_m:z\mapsto z^2$ from the line with origin deleted to itself.

8) An affine variety can map surjectively onto a projective variety: $\mathbb A^1\to \mathbb P^1:u\mapsto (u:1-u^2)$

9) Morphisms $\mathbb P^N\to \mathbb P^n$ with $N\gt n$ are constant.

10) Morphisms $X\to Y$ between smooth irreducible projective curves with $\text {genus} (X)\lt \text{genus}(Y)$ are constant.

11) The morphism $\mathbb A^1\to \mathbb A^n:t\mapsto (t^{n+1},\cdots,t^{2n})$ has as image a closed curve $C\subset \mathbb A^n$ [cf. 6)] .
That curve, although of dimension $1$, has a huge $n$-dimensional Zariski tangent space $T_O(C)= T_O(\mathbb A^n)$ at the origin $O$ and $C$ is not isomorphic to any closed subvariety of a smooth variety of dimension $\lt n$.

12) The Viète morphism constructed from the elementary symmetric functions $s_i(x_1,\cdots,x_n)$ over a ring $k$ : $$s:\mathbb A^n_k \to \mathbb A^n_k: x=(x_1,\cdots,x_n)\mapsto (s_1(x),\cdots, s_n(x))$$ is finite of degree $n!$ and faithfully flat.
It is the quotient morphism $\mathbb A^n_k \to (\mathbb A^n_k)^{S_n}$ of affine space by the natural group action of the symmetric group $S_n$ on that affine space.
Notice that the quotient space is non singular even though the action is far from being free.

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@Benja You are the third person to mention this--look at the above comments ;) –  Alex Youcis Feb 9 at 12:02
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@AlexYoucis Ah yes. My example shows we need the finite type hypothesis and Georges is saying flat implies open for varieties (which are finite type according to my definition, variety = integral, separated scheme of finite type over $k$) :D –  user38268 Feb 9 at 12:05
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@Benja We actually need finitely presented, but honestly, who cares. Georges is right, slogans over precision is better in these cases :) –  Alex Youcis Feb 9 at 12:05
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@AlexYoucis: Dear Georges and Alex, Another argument to see that if a normalization is flat, it is trivial: under some mild hypotheses (which are tacitly in force) the normalization is a finite morphism. If it is also flat then it is finite locally free. (Recall that for modules, finite rank locally free is equivalent to finitely presented and flat.) But we can compute this rank over the generic point (or points, if there is more than one component), where it is one (because a normalization is a birational map by its very construction). Thus our normalization is an isomorphism. Regards, –  Matt E Feb 9 at 13:46
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@MartinBrandenburg: Dear Martin, Yes, you need a mild finiteness hypothesis (locally finitely presented) to conclude that flat morphisms are open. I think that all of Georges's examples should be taken to have such tacit hypotheses in force (in the spirit of the original question, which asked for examples which fit into the classical world of varieties, where certainly all these finiteness hypotheses are in force). Regards, –  Matt E Feb 9 at 13:49

Let me add some remarks to Georges' and Alex's nice answers above.

  1. A few words about the normalization of the cuspidal cubic $k[t^2,t^3] \to k[t]$: As simple as this example may be it shows that

    • A finite surjective morphism from a non-singular variety to any variety need not be flat. We really do need the codomain to be non-singular (or at least CM, I can't remember), this is Hartshorne III Ex. 9.3(a).

    • A universally closed morphism of $k$-schemes need not be a closed immersion of $k$-schemes! The morphism $k[t^2,t^3] \to k[t]$ is finite and since a finite morphism is closed and stable under base change, $k[t^2,t^3] \to k[t]$ is universally closed. But it is not a closed immersion because this would imply $k[t^2,t^3] \cong k[t]/(f(t))$ with $f(t) \neq 0$ by Hartshorne II Ex. 3.12(b); the latter is always finite dimensional while $$\dim_k k[t^2,t^3] = \infty.$$

  2. Let me give a counterexample to Alex's point (5) above when the domain is not a curve. This shows we really, really need the domain to be dimension one! My counterexample is this: Consider the rational map $$\pi :\Bbb{P}^2_k - \{[0:0:1\} \to \Bbb{P}^1_k$$ that wipes out the last coordinate. I claim this cannot be extended to a map on all of $\Bbb{P}^2$. If we could extend it to a map on all of $\Bbb{P}^2$ then $\pi$ would send the connected set $L_1 \cup L_2$ to a union of at least two points, contradiction. Here $L_1 = [t:0:1]$ and $L_2 = [0:u:1]$.

  3. When Georges says that a flat morphism is open, this is of course under the assumption that the morphism is of finite type: Take $k[t] \hookrightarrow k(t)$ which is flat but obviously $\operatorname{Spec} k(t) \to \operatorname{Spec} k[t]$ is not open. The reason why $k(t)$ is flat as a $k[t]$-module is because a module over a PID is flat iff it is torsion free.

  4. Let me give a quick proof as to why there is no morphism from $\Bbb{P}^2 \to \Bbb{P}^1$: modulo all the business involving sections of line bundles, any morphism from $\Bbb{P}^2 \to \Bbb{P}^1$ on $k$-valued points is given by $[x_0:x_1:x_2] \mapsto [f_1(x_0,x_1,x_2): f_2(x_0,x_1,x_2)]$ where $f_1,f_2$ are homogeneous polynomials. If this is defined everywhere, the polynomials $f_1,f_2$ cannot be simultaneously zero on all of $\Bbb{P}^2$. But this contradicts Bezout's theorem (here I assume $k$ is algebraically closed) and so no such morphism can exist.

  5. Here's one fact I find very useful. Let $Y$ be an affine scheme and $X \to Y$ a closed immersion. Then $X$ is affine. The proof is really easy in the special case that $X$ is Noetherian: Let $\mathscr{F}$ be a quasi-coherent sheaf on $X$. For any $i > 0$, $$H^i(X,\mathscr{F}) = H^i(Y,f_\ast \mathscr{F}) = 0$$ since $Y$ is affine and $X$ Noetherian implies $f_\ast \mathscr{F} \in \operatorname{QCoh}(Y)$. By Serre's Theorem, $X$ is affine. Done.

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I think there are much more elementary ways to prove your number 5 :) I also think that some people (myself included) would define a closed embedding to be an affine morphism for which, if $f^{-1}(\text{Spec}(A))=\text{Spec}(B)$ then $A\to B$ is surjective. –  Alex Youcis Feb 11 at 8:45
    
Dear @AlexYoucis, the notion of closed immersion I have studied is a closed immersion of schemes is topological closed immersion + surjective on stalks. This is the one in Hartshorne. –  user38268 Feb 11 at 9:30
    
And they are equivalent. Or its the natural scheme associated to a quasicoherent sheaf of ideals. All the same. I learned mostly from Vakil, who defines it the way I did in my original comment :) –  Alex Youcis Feb 11 at 10:52

Since my list of examples is becoming too long to be easily legible, let me start a new one.

Let $k$ be an algebraically closed field of characteristic $2$ and $C\subset \mathbb P^2_k$ be the conic $x^2-yz=0$ .
Identify the $y$-axis $V(x)$ with $\mathbb P^1_k$ through $(0:y:z)=(y:z)$, project $C$ onto $V(x)$ from the point $A=(1:0:0)$ and consider the projection morphism $$p:C\to \mathbb P^1_k: (x:y:z)\mapsto (y:z)$$ That morphism is bijective but ramified at every point of $C$ i.e. its tangent map is zero everywhere!
Geometrically all lines joining $A$ to a point $P\in C$ are tangent to $C$ at $P$.
The conic $C$ certainly deserves its technical denomination of "strange curve" !

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Dear Georges, I thought the technical term (at least in Hartshorne) was "funny curve"? In the high-stakes world of algebraic geometry, these distinctions are crucial. (By the way, I meant to add this example myself!) –  Asal Beag Dubh Feb 11 at 12:34
    
Dear @Asal: Here is an article with "Strange Curves" in the title. But I agree that they are funny too -and that the terminology is of galactic importance :-) . –  Georges Elencwajg Feb 11 at 12:43
    
Dear Georges: ah, in fact I see that Hartshorne's "funny curve" is something different --- a nonsingular quartic in characteristic 3 whose map to its dual is purely inseparable (Exercise IV.2.4). And indeed he talks about strange curves in Section IV.3. –  Asal Beag Dubh Feb 11 at 13:02

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