Birational but non-isomorphic projective/affine spaces

[Mathmos6]

Homework Statement

How would I go about showing that if $X = \{(x,\,y) \in \mathbb{C}^2 | x^2 = y^3\}$ then X is birational but not isomorphic to the affine space [itex] \mathbb{A}^1[/tex]? I have found the obvious birational map, sending $(x,\,y) \to \frac{x}{y}$, so I have shown the spaces are birational, but how do I show they are not isomorphic? (Isomorphic here means there exists a morphism in the algebraic geometry sense with a two-sided inverse.) It is obvious that X is a sort of line with a 'cusp' at the origin, but I can't see how to show explicitly that they are non-isomorphic: presumably the behaviour at the origin is the reason why this occurs, but I don't think I can get away with saying "it's not smooth around 0 in one case and it's smooth around 0 in the other".

I am only an undergraduate so please do keep the algebraic geometry as simple as possible! Many thanks.

 

[micromass]

Hi mathmos6!

Can you give me the book you're using, so I know exactly what algebraic geometry I can use?

In fact, it is indeed the behavior at (0,0) that causes the problem. The curve $x^2=y^3$ is singular there. Did you see the notions singular/regular and how did you define it?

What I suggest is showing that the affine line $\mathbb{A}^1$ has no singular points, and that singular points are being preserved by isomorphisms.

 

[Mathmos6]

Hi Micromass! Thanks for the quick response! I have primarily been using any online lecture notes I could find, but Hartshorne has come in handy once or twice too.

I have seen the notion of singularity, yes: so essentially the tangent space has dimension 2 at the origin whereas the curve itself clearly has dimension 1, so the origin is a singular point. It is also obvious that the affine line has no singular points, though I think maybe the definition of the affine line is so basic it is confusing me in this particular example! Usually I can work with tangent space of something with functions $f_j$ such that $\sum_i v_i \frac{\partial{f_j}}{\partial{x_i}}=0 \, \forall j$, but in this case I'm confused as to how we find the functions $f_j$: I suspect maybe there are none if we have a 1-dimensional space with no restrictions like the affine line, so maybe it is a degenerate case.

In general if our affine space is the zero set of some polynomials $g_j$ then it is our $g_j$ that we take the partial derivatives of in the definition of the tangent space, right? Anyway, so once happy with the fact the affine line has no singular points, I need to show the second part, that isomorphisms preserve singular points. Intuitively, this feels very obvious to me, but in terms of a proof quite the opposite: I find the definition of the tangent space is a bit cumbersome to work with, is there a relatively easy proof somewhere I could look up that isomorphisms preserve these points? I shall go and check Hartshorne now.

 

[micromass]

Hi Micromass! Thanks for the quick response! I have primarily been using any online lecture notes I could find, but Hartshorne has come in handy once or twice too.

Just for the record, I find Hartshorne a horrible, horrible book (check my blog for an entire rant).
Although it's first chapter is decent.

I have seen the notion of singularity, yes: so essentially the tangent space has dimension 2 at the origin whereas the curve itself clearly has dimension 1, so the origin is a singular point. It is also obvious that the affine line has no singular points, though I think maybe the definition of the affine line is so basic it is confusing me in this particular example! Usually I can work with tangent space of something with functions $$f_j$$ such that $$\sum_i v_i \frac{\partial{f_j}}{\partial{x_i}}=0 \, \forall j$$, but in this case I'm confused as to how we find the functions $$f_j$$: I suspect maybe there are none if we have a 1-dimensional space with no restrictions like the affine line, so maybe it is a degenerate case.


In general if our affine space is the zero set of some polynomials $$g_j$$ then it is our $$g_j$$ that we take the partial derivatives of in the definition of the tangent space, right?

Right. Note that $\mathbb{A}^2$ is the zero set of the polynomial 0. So 0 is the $g_j$ you should take here.

Also, the modern definition of tangent space is quite different. Basically, given a point x in a variety X, then we can consider $O_{X,x}$, this is a local ring with maximal ideal m. The vector space [itex]m/m^2[/tex] is the modern definition of the tangent space.

Anyway, so once happy with the fact the affine line has no singular points, I need to show the second part, that isomorphisms preserve singular points. Intuitively, this feels very obvious to me, but in terms of a proof quite the opposite: I find the definition of the tangent space is a bit cumbersome to work with, is there a relatively easy proof somewhere I could look up that isomorphisms preserve these points? I shall go and check Hartshorne now.

OK, can you first give me a precise definition of how you defined the tangent space? There are many definitions in the running, and I want to know exactly with what I'm working here.

Also, if you check Hartshorne, chapter 1.5, then you'll see he gives another definition of singlular/regular. What you do is correct of course, but you'll need to tell me exactly with what you're working!

p.s. my "[ tex ]" and "[/ tex ]" tags seem to be making everything look funny, what am I doing wrong? How silly.

You need to use [ itex ] ... [ /itex]. If you use [ /tex] then you'll start a new line...

 

[Mathmos6]

Wonderful: that last post is fixed, I don't remember having to use itex the last time I used PF but perhaps that's just how long I've been away for! I concur with your blog, Hartshorne is boring and unintelligible, I don't like it at all but sadly it seems to be a bit of an industry standard, if you have any other good algebraic geometry books which are readable as an undergraduate then I'm open to suggestions!

My definition of tangent space is as follows: let $X \subseteq \mathbb{A}^n$ be an affine variety, $k[X] = k[x_1,\ldots,x_n]/\langle f_1, \ldots,f_r\rangle$ and let $p \in X$ be a point. The tangent space is $T_p X = \{(v_1,\ldots,v_n) \in \mathbb{A}^n: \sum_{i=1}^n v_i \frac{\partial{f_j}}{\partial{x_i}}(p) = 0,\,j = 1,\ldots,r\}$, and for an abstract variety we define the tangent space $T_p X = T_p U$ where U is any open nbhd of p.

I hope that makes sufficient sense, I detest all the higher level differential/algebraic geometry and notation, I am sure it is all obvious to professionals working with the subject, but I find it incredibly off-putting compared to something like analysis where to me everything feels lovely and intuitive. Then again, perhaps it's just my undergraduate perspective! Though I gather Grothendieck didn't have a very easy time making sense of the subject either so maybe I can take some comfort in that...

 

[micromass]

Wonderful: that last post is fixed, I don't remember having to use itex the last time I used PF but perhaps that's just how long I've been away for! I concur with your blog, Hartshorne is boring and unintelligible, I don't like it at all but sadly it seems to be a bit of an industry standard, if you have any other good algebraic geometry books which are readable as an undergraduate then I'm open to suggestions!

Uh, I would consider Hartshorne to be the last book on Earth I would give an undergraduate.
There are a lot of good books on schemes out there, but as an undergraduate, it's better to deal with varieties first. Here are some good books recommendations: http://mathoverflow.net/questions/2446/best-algebraic-geometry-text-book-other-than-hartshorne

See especially post 16. I also like Milne's online lecture notes.

My definition of tangent space is as follows: let $X \subseteq \mathbb{A}^n$ be an affine variety, $k[X] = k[x_1,\ldots,x_n]/\langle f_1, \ldots,f_r\rangle$ and let $p \in X$ be a point. The tangent space is $T_p X = \{(v_1,\ldots,v_n) \in \mathbb{A}^n: \sum_{i=1}^n v_i \frac{\partial{f_j}}{\partial{x_i}}(p) = 0,\,j = 1,\ldots,r\}$, and for an abstract variety we define the tangent space $T_p X = T_p U$ where U is any open nbhd of p.

Yes, that makes sense. So, now you have to prove that if F is an isomorphism, then the tangent space at p is isomorphic as the tangent space of F(p) (or at least has the same dimension).
So, first I would find g₁,...,g_r such that

$$F(X)=V(g_1,...,g_r)$$

then you can calculate the tangents space.

I hope that makes sufficient sense, I detest all the higher level differential/algebraic geometry and notation, I am sure it is all obvious to professionals working with the subject, but I find it incredibly off-putting compared to something like analysis where to me everything feels lovely and intuitive. Then again, perhaps it's just my undergraduate perspective! Though I gather Grothendieck didn't have a very easy time making sense of the subject either so maybe I can take some comfort in that...

I feel your pain Algebraic geometry is a very hard topic to learn and I still struggle with it every day. I guess you just need to persist, and you'll get there. After a while some things become intuitive. But when you first encounter it, it can be very hard and discouraging...

 

[micromass]

If you need a good reference for finding a suitable isomorphism between the tangents spaces, check out Milne page 94,95 http://jmilne.org/math/CourseNotes/AG.pdf

 

[Mathmos6]

If you need a good reference for finding a suitable isomorphism between the tangents spaces, check out Milne page 94,95 http://jmilne.org/math/CourseNotes/AG.pdf

You've been such a brilliant help Micromass, thank you so much :)

I do have one final question which has come up, if you don't mind. The final part of a problem I've been doing says: "let $V \subset \mathbb{P}^2$ be the plane curve given by the vanishing of the polynomial $X_0^2X_1^3 + X_1^2 X_2^3 + X_2^2X_0^3$ over a field of characteristic zero. Show that V is irreducible, and that φ determines a birational equivalence between V and a nonsingular plane quartic."

I have shown V is irreducible, but I am not entirely sure what the "nonsingular plane quartic" in question is, or how I'm meant to show the birational equivalence. Could you give me a nudge in the right direction? Thank you again :)

 

[micromass]

Show that V is irreducible, and that φ determines a birational equivalence between V and a nonsingular plane quartic."

May I first know what φ is? Or is that something you need to find.

Anyway, the way to go is first determine the singlular points of V, these are the points you don't want in your birational equivalence.

 

[Mathmos6]

May I first know what φ is? Or is that something you need to find.

Anyway, the way to go is first determine the singlular points of V, these are the points you don't want in your birational equivalence.

Oh, of course, I'm sorry! I thought I wrote it down, obviously not :)

$\varphi: \mathbb{P}^2 - \to \mathbb{P}^2$ is the rational map given by $(X_0:X_1:X_2) \to (X_1 X_2: X_0 X_2: X_0 X_1) = (1/X_0: 1/X_1 : 1/X_2)$: I have shown already that this map is not regular at the points (1:0:0), (0:1:0), (0:0:1) and I believe it follows immediately that it is regular elsewhere: it is clearly a birational map since $(X_0:X_1:X_2) \to (1/X_0: 1/X_1 : 1/X_2)$ is obviously self-inverse. I guess the first thing to do is find the image of the curve under the map φ?

I am a bit confused about what a 'plane quartic' is in this case: in normal Euclidean space it's going to be a curve defined by an equation of degree 4 which lies in a plane, but in projective space my notion of a 'plane' goes out the window, since lines aren't really lines in 3D space, they're all projected onto the unit sphere (if you want to look at it that way) and I find it much harder to picture. What would you suggest?

Thank you again!

 

[micromass]

OK, what we need to do is to find the image of the curve (sorry for working with X,Y,Z instead of your variables, but somehow I find it easier...)

$$X^2Y^3+Y^2Z^3+Z^2X^3$$

So we know that the image of $(X:Y:Z)$ is $(YZ:XZ:XY)$, the question is to find an equation which this image satisfies. So we need to rewrite the equation

$$X^2Y^3+Y^2Z^3+Z^2X^3$$

Into a polynomial of the form $f(YZ,XZ,XY)$.

Now, one obvious thing to do is

$$(XY)^2Y+(YZ)^2Z+(ZX)^2X$$

So now we only need to do something with the Y, Z and X. I suggest multiplying the equation with XYZ...

 

[Mathmos6]

OK, what we need to do is to find the image of the curve (sorry for working with X,Y,Z instead of your variables, but somehow I find it easier...)

$$X^2Y^3+Y^2Z^3+Z^2X^3$$

So we know that the image of $(X:Y:Z)$ is $(YZ:XZ:XY)$, the question is to find an equation which this image satisfies. So we need to rewrite the equation

$$X^2Y^3+Y^2Z^3+Z^2X^3$$

Into a polynomial of the form $f(YZ,XZ,XY)$.

Now, one obvious thing to do is

$$(XY)^2Y+(YZ)^2Z+(ZX)^2X$$

So now we only need to do something with the Y, Z and X. I suggest multiplying the equation with XYZ...

I'm a little confused about why you're doing this: rather than "rewriting" the old equation in terms of the images of X, Y, Z, wouldn't we take the map $\varphi$ to "substitute in" YZ, XZ, XY for each instance of X, Y, Z respectively? For instance the first term of the equation would become $(YZ)^2 (XZ)^3$?

 

[micromass]

Well, we need to find the image of V(g) through some map f. So, I take x in V(g), that means that g(x)=0. Now, I need to find a polynomial h such that f(x) is in V(h), that is h(f(x))=0.

In our example, we have to find a polynomial h such that

$$h(yz:xz:xy)=0$$

while knowing that

$$x^2y^2+y^2z^3+z^2x^3=0$$

What you suggest is taking h=g(f(X,Y,Z)), but as you can see h(f(x)) isn't necessarily zero. (also, if we follow through your suggestion, we won't end up with a quartic polynomial)

Hope that clears things up.

 

[Mathmos6]

Well, we need to find the image of V(g) through some map f. So, I take x in V(g), that means that g(x)=0. Now, I need to find a polynomial h such that f(x) is in V(h), that is h(f(x))=0.

In our example, we have to find a polynomial h such that

$$h(yz:xz:xy)=0$$

while knowing that

$$x^2y^2+y^2z^3+z^2x^3=0$$

What you suggest is taking h=g(f(X,Y,Z)), but as you can see h(f(x)) isn't necessarily zero. (also, if we follow through your suggestion, we won't end up with a quartic polynomial)

Hope that clears things up.

Ah yes of course, I was being stupid

So then as you suggested we get out a quartic in the required variables, I'm just not sure what a "plane" quartic is? Thanks!

 

[micromass]

Ah yes of course, I was being stupid

So then as you suggested we get out a quartic in the required variables, I'm just not sure what a "plane" quartic is? Thanks!

I guess it's just a quartic in the "plane" $\mathbb{P}^2$.