projective variety Log Out | Topics | Search
Moderators | Register | Edit Profile

Ask NRICH » Archive 2010-2011 » Higher Dimension » projective variety « Previous Next »

Author Message
Amazigh
Poster

Post Number: 9
Posted on Sunday, 10 July, 2011 - 10:59 pm:   

I hardly know anything about algebraic geometry. But I learned a little bit from the first chapter of the famous book the arithmetic of elliptic curves (J Silverman).
My question is how do you know that
V: Y^2 Z = X^3+17 Z^3
is a projective variety. So how can I verify that I(V) is a prime ideal in k[X,Y,Z] ?
Daniel Fretwell
Veteran poster

Post Number: 1430
Posted on Monday, 11 July, 2011 - 10:13 am:   

Im currently reading this book and I must confess, I am finding it hard to get on with.

I read a few pages of the first chapter and in the end skipped to chapter 3, it just seemed like definition after definition with no explanation of why things are important. For example, why is it important that I(V) is a prime ideal?

Even at chapter 3 I am getting way too much detail and "generality for generality sake", with very little intuition and explanation of why these things are important (e.g. I have yet to read why elliptic curves are "important" from this book). Maybe it gets better towards to end.

I much prefer Silverman/Tate "Rational points on Elliptic Curves", even if it doesnt do as much as this book.
Zhen Lin
Frequent poster

Post Number: 123
Posted on Monday, 11 July, 2011 - 10:37 am:   

If the ideal of an algebraic set is prime, then its coordinate ring is an integral domain. This makes the algebra easier. Moreover, the ideal of an algebraic set is prime if and only if the set is irreducible, so this makes the geometry easier.

The easiest way to show that [inline]I(V)[/inline] is prime, I think, is as follows. First, we note that [inline]f = Y^2 Z - X^3 - 17 Z^3 \in I(V)[/inline]. Then, we show [inline]f[/inline] is irreducible. By Hilbert's Nullstellensatz, [inline]I(V) = \sqrt{(f)}[/inline], but [inline]k[X, Y, Z][/inline] is a UFD, so [inline](f)[/inline] is prime, hence [inline]\sqrt{(f)} = (f)[/inline], and in particular [inline]I(V)[/inline] is prime.

There is more geometric way, however. If [inline]V[/inline] is a hypersurface, and you can show that [inline]V[/inline] is a smooth projective variety, then a fortiori [inline]V[/inline] is irreducible. Essentially, [inline]V[/inline] would then be the union of two or more irreducible projective hypersurfaces, and Bézout's theorem tells us that any two of them have non-empty intersection, but [inline]V[/inline] cannot be smooth at the intersection.

Why are elliptic curves important? Well, I don't know. They are certainly quite nice geometrically. But number theorists seem to be fascinated by them...
Daniel Fretwell
Veteran poster

Post Number: 1431
Posted on Monday, 11 July, 2011 - 11:00 am:   

Well that was just two examples of things that arent explained in the book. For another example in your previous post, why is the "coordinate ring" of any use and why define it as such? Also why is it important that the coordinate ring is an integral domain?

I understand why elliptic curves are important but what I am saying is that the book seems to ignore the why and just focuses on the how. You have to know the answers to your own questions to get anything from it.

Like I say I havent got through the whole book yet (and tbh I doubt I will continue with it). It is written for people that know all of the stuff in the book already. Upto yet all the book seems to do is say "we can define this, then we can define this more complicated thing...then we can find this result and then define something else". I just get lost in what we are actually doing and why.

I find it very hard to learn maths when it is presented in such a dry and formal way.
Yatir Halevi
Veteran poster

Post Number: 1299
Posted on Monday, 11 July, 2011 - 05:49 pm:   

I also started reading this book, and then someone suggested that I should first learn some algebraic geometry. You are right that the first chapter of the book doesn't really teach algebraic geometry, it just reminds the reader the general definitions.
I suggest you should pick up a book about the subject and do some reading. It turns out that algebraic properties of the coordinate ring can be translated to geometric properties of the variety. One of them being, the coordinate ring is an integral domain iff the variety is irreducible.

Yatir
Billy Woods
Veteran poster

Post Number: 933
Posted on Monday, 11 July, 2011 - 05:56 pm:   


quote:

It is written for people that know all of the stuff in the book already.



Can you really judge that from the first few pages of the first chapter? The first chapter, if I remember correctly, is intended to be a reminder of the important bits of algebraic geometry for those who already know it - I really wouldn't recommend you learn from there. I believe Reid has a very good book called "Undergraduate algebraic geometry" - it's small and well written and will prepare you more than adequately for Silverman. :-)
Lloyd Smith
Veteran poster

Post Number: 421
Posted on Tuesday, 12 July, 2011 - 12:41 am:   

More generally: Algebraic geometry can get very technical indeed, and it's easy to spend a lot of effort on the algebra without understanding anything about the geometry. If you're interested in the subject (which you should be!) then it's a good idea to avoid this trap by taking some elementary courses and reading books with plenty of pictures. There are lots of good books; Billy's recommendation is good, as is Joe Harris's "Algebraic geometry - a first course"

Good luck!
Zhen Lin
Frequent poster

Post Number: 124
Posted on Tuesday, 12 July, 2011 - 01:54 am:   

From what I can remember Reid is very simplistic and doesn't talk about any of the advanced topics like differentials or divisors or Riemann–Roch, which will almost surely show up in any elliptic curves course. I don't know of any good introductory textbooks for algebraic geometry, but some people have said good things about Hulek's _Elementary Algebraic Geometry_.

Unfortunately, if you're interested in number theory, the algebraic geometry doesn't stop there. Algebraic varieties are usually defined over algebraically closed fields in introductory courses, but you're probably more interested in less pleasant fields, like [inline]\mathbb{Q}[/inline], or rings which are not even fields, like [inline]\mathbb{Z}[/inline]. This brings us into the realm of arithmetic geometry, and for that you need Grothendieck's schemes. My Number Fields supervisor recommended Mumford's _Red Book_ and Shafarevich's _Basic Algebraic Geometry_, vol. 2. There's also Hartshorne's classic textbook.

Finally, I'll mention Harris's *other* book, _Principles of Algebraic Geometry_. I've only had a brief glance at it once, but I noticed it was specialised to the case of complex varieties and always keeps the analytic picture close at hand. Those who find algebraic geometry too algebraic may find this to be a welcome change.
Daniel Fretwell
Veteran poster

Post Number: 1432
Posted on Tuesday, 12 July, 2011 - 10:15 am:   

I shall certainly have a look at all of the books mentioned above.

I might have been a little harsh in what I said above. I think my problem is that I am hugely an "intuitive mathematician", I learn via knowing the ultimate aim and how the pieces fit together.

I think other people may or may not be more appreciative of a completely formal view of things. It all depends on the way you work.
Billy Woods
Veteran poster

Post Number: 935
Posted on Wednesday, 13 July, 2011 - 04:32 pm:   


quote:

From what I can remember Reid is very simplistic and doesn't talk about any of the advanced topics like differentials or divisors or Riemann–Roch, which will almost surely show up in any elliptic curves course.



I recommended Reid because it was fairly elementary - if I remember right, Silverman introduces all the non-elementary stuff properly, including differentials and divisors and the Riemann-Roch theorem. :-)


quote:

I think my problem is that I am hugely an "intuitive mathematician", I learn via knowing the ultimate aim and how the pieces fit together.



And of course there is a lot of intuition behind algebraic geometry. Algebraic statements like "I(V) is a prime ideal" often correspond to much easier geometric statements - crudely, if I(V) is prime, V is a curve or a surface (or...), but if I(V) is not prime, V looks like several of these things intersecting. And it's often much easier to tell whether an ideal generated by a given equation is prime than it is to sketch the graph of that equation and see what it looks like.

The coordinate ring simply gives us a way of translating between algebra and geometry so that we can do one and get results in the other - very much like coordinates in plane geometry. There's a direct analogy, actually. Consider the ring [inline]\mathbb{R}[x,y][/inline] (which is the ring of normal boring Cartesian coordinates): is the ideal generated by xy prime? Of course not, and so the equation xy = 0 has a graph that looks like some stuff intersecting. Is the ideal generated by x2 - 1 prime? Yes, and so the equation x2 - 1 = 0 has a graph that looks like a single curve. :-)
Lloyd Smith
Veteran poster

Post Number: 422
Posted on Wednesday, 13 July, 2011 - 05:11 pm:   

I think Billy means x2 - y! :-)
Billy Woods
Veteran poster

Post Number: 936
Posted on Wednesday, 13 July, 2011 - 11:42 pm:   

Bah. Thank you. :-)
Amazigh
Poster

Post Number: 10
Posted on Thursday, 14 July, 2011 - 12:45 am:   

I think we all agree we should first do some AG to start with this book :-)
@ Zhen Lin: Thanks for the help. But I do not understand why (f) is prime by only using the fact that k[x,y,z] is a UFD? I mean ((x-1)˛) is not a prime.
Daniel Fretwell
Veteran poster

Post Number: 1436
Posted on Thursday, 14 July, 2011 - 09:08 am:   

But (x-1)^2 is not irreducible.

The fact that f is irreducible is important (although I have no idea how to show this without looking at it properly).

Zhen Lin
Frequent poster

Post Number: 125
Posted on Thursday, 14 July, 2011 - 02:26 pm:   

In a UFD, every irreducible is prime, and it is clear that a principal ideal is prime if and only if its generator is prime (or 0, if we are in an integral domain). (On the other hand, a prime is irreducible in any ring.)
Zhen Lin
Frequent poster

Post Number: 126
Posted on Thursday, 14 July, 2011 - 02:32 pm:   


quote:

I recommended Reid because it was fairly elementary - if I remember right, Silverman introduces all the non-elementary stuff properly, including differentials and divisors and the Riemann-Roch theorem. :-)




Ah, but that's precisely why I didn't like the book: it focused far too much on classical geometry. I couldn't care less about conic sections or the 27 lines on a cubic surface! (On the other hand, it's questionable how much algebraic geometry I actually do care about. The machinery of locally ringed spaces and sheaves is all rather interesting, and I suspect I may have lost sight of the forest for the trees...)
Amazigh
Poster

Post Number: 11
Posted on Thursday, 14 July, 2011 - 05:00 pm:   

Maybe a stupid question but why exactly is (f) prime? It is not clear for me.
Zhen Lin
Frequent poster

Post Number: 127
Posted on Thursday, 14 July, 2011 - 08:50 pm:   

It's just definition-chasing.

Let [inline]f[/inline] be a prime element of a ring [inline]A[/inline]. Then the ideal [inline](f)[/inline] of [inline]A[/inline] generated by [inline]f[/inline] is prime: suppose [inline]a b \in (f)[/inline], then [inline]a b = h f[/inline] for some [inline]h[/inline] in [inline]A[/inline]; but then either [inline]f[/inline] divides [inline]a[/inline] or [inline]f[/inline] divides [inline]b[/inline], so either [inline]a \in (f)[/inline] or [inline]b \in (f)[/inline].
Amazigh
Poster

Post Number: 12
Posted on Thursday, 14 July, 2011 - 11:05 pm:   

Yeah ofcourse, but how do you prove that f is prime (hence a irreducible polynomial) ? That was my issue :-)
Zhen Lin
Frequent poster

Post Number: 128
Posted on Friday, 15 July, 2011 - 01:52 am:   

That's the difficult part. Usually I pray that it is "obviously" radical and defines a smooth variety, because a smooth variety must be irreducible (as I argued in my post above). Sometimes simpler ad hoc methods work.
Billy Woods
Veteran poster

Post Number: 937
Posted on Friday, 15 July, 2011 - 02:18 am:   


quote:

Yeah ofcourse, but how do you prove that f is prime (hence a irreducible polynomial) ?



Because it's irreducible! Try it: can you factorise f?

Let f = X3 - (Y2Z - 17Z3) - the polynomial in question, but written slightly oddly. This is because I'm going to consider it as a polynomial in X. It's a cubic, so if it factorises at all, at least one of its factors must be linear, and in particular must be of the form (X - g), where g is secretly a polynomial in Y and Z. (Convince yourself of the second half of this sentence.) Then X = g is a root. From the way I've written it, it's obvious that the only roots of this polynomial in X are the cube roots of Y2Z - 17Z3 - but they are not polynomials, so are not in k[X,Y,Z]. In particular, f does not have a linear factor over k[X,Y,Z], so f doesn't factorise at all there.

Add Your Message Here
Posting is currently disabled in this topic. Contact your discussion moderator for more information.

Topics | Last Day | Last Week | Tree View | Search | Help/Instructions | Program Credits Administration