Exploring the Spectrum of Rings: Analytic & Differentiable Functions

[Hurkyl]

Spectrum of a Rings

So I think I'm finally getting the definition of a scheme, but the texts I'm using aren't exactly ripe with examples. So, I wanted to condier some ring spectra to construct examples...

I played around with the simple examples Z and Z[x], and my next thought was to look at manifolds and stuff...

So I wondered what the spectrum is of A, the ring of continuous functions over Rⁿ... in order for a manifold to be a scheme, the spectrum would have to be homeomorphic to Rⁿ... but I'm almost certain now that it is not.

I can discern some of the prime ideals of A; each of the ideals I(P) = {f in A| f(P) = 0} is prime, which is good. However, so is the ideal Z = {f | f is zero on a set with nonempty interior}, which is annoying because it doesn't even correspond to a subset of Rⁿ. Furthermore, I can't prove to myself that, for instance, there is no subideal of I(P) that is not prime, and I'm having similar troubles when you replace P with some closed connected set of measure zero.

Is there a good way, at all, to describe the spectrum of the ring of continuous functions over Rⁿ? What about differentable, twice differentiable, ..., infinitely differentiable functions? And analytic functions (which I suspect might be nicer)?

 

[matt grime]

Let's make it simple and just consider R. There is a paper by Neeman, though i can't seem to find it at the moment, that demonstrates the derived category of sheaves (of abelian groups) over R is not compactly generated. This kind of result implies some pathological misbehaviour of non-compact manifolds.

The ring of continuous functions seems too big (is it even noetharian?). Obviously it isn't even finitely generated over R, so it's not going to be nice.

 

[styler]

The ring of continuous functions seems too big (is it even noetharian?).

Yeah seems too big.
Somebody suggested that the following ideal contains an infinite ascending chain:
Update: found a mention of it on the webpage of somebody's alg course:
http://www.math.gatech.edu/~singh/sol3.pdf

 

[Hurkyl]

Well, I have gotten an answer to my question; I feel silly not noticing it earlier. The spectrum of a ring is always compact, thus cannot be isomorphic to Euclidean n-space.


Considering for the moment the case of infinitely differentiable functions on Euclidean n-space, I can produce two types of "counterexamples"; that is, ideals that are not simply the set of functions zero on a point.

(a) The set of functions zero at the origin and whose derivatives are all zero, form a prime ideal.

(b) Given any ultrafilter (right term?) on closed sets, the set of functions whose zero set is in the ultrafilter form a prime ideal. One can generate a "counterexample" by extending the filter of closed sets with bounded complement.


Now, I haven't been able to come up with a variant of (a) for any of the other rings in question (rings of n-th differentiable functions, or simply continuous functions) and if I work in a compact space, (b) doesn't seem to generate counterexamples because the only ultrafilters seem to be collections of sets containing a given point.


So my amended question is that if I have a compact manifold (compact metrizable space? compact T1 space?) and take the ring of continuous functions (n-th differentiable functions?) on the manifold, is the spectrum isomorphic to the manifold? That is, is the collection of prime ideals of this ring simply those ideals consisting of all functions zero at a given point?

 

[matt grime]

I think you need to say what you mean by isomorphic. Spectra arise from the Zariski topology which is very badly separated, and the zariski topology on R for instance is compact, but obviously not hausdorff, indeed every open set is dense.

I still can't see what you're after. Take the unit circle, T, then whilst its set of continuous functions might have a countable analytic basis, algebraically it is disgusting, and there's little chance of its spectrum being P^1.

Its almost as though you wish to take the spectrum of the spectrum of something, which seems to me at least to be the wrong thing to examine.

 

[Hurkyl]

and there's little chance of its spectrum being P^1.

This is the part I'd like to see shown; I can't figure out why it wouldn't be so. The only prime ideals I can find in this ring are those that are the set of all functions zero at a given point. If, indeed, these are all the prime ideals, then we can identify each prime ideal with its point of the circle, and then the closed subsets of the spectrum will be precisely the closed subsets of the circle.

 

[matt grime]

Well, which topology on T do you want? (given the example of the unit circle)

 

[Hurkyl]

Ok, take T with the topology given by its embedding in R^2. Take the ring of continuous functions on T. Is there a prime ideal that, for all points P of T, is not equal to the set of all continuous functions zero at P?



(actually, I had another related question; given a continuous map f from Spec A to Spec B, is there an associated homomorphism from B to A? I know the other way around is true)

 

[matt grime]

This is proving really tricky to figure out since I have no idea if I should be proving it's true or finding a counter example. Does anyone have references for spectra of (uncountably) infinitely generated rings over R? Perhaps even thinking aobut C-star algebras mght help, though we ought to do this algebraically. Of course, even if we've got the set of prime ideals as those vanishing at some point, you would then need some sort of inclusion reversing correspondence withe radical ideals etc.

 

[Hurkyl]

Suppose the prime ideals of A, the ring of continuous functions over a compact space X, has only prime ideals of the form of all functions zero at a given point.


Suppose I is any ideal of A, and consider the zero set of this ideal. Each element of f is zero on a closed set in the Euclidean sense (since f is continuous), and thus Z(I) is closed in the Euclidean sense.

Furthermore, if F is closed in the euclidean sense, then I(F) = {f | f is zero on F} is an ideal with Z(I(F)) = F. So, we have a 1-1 correspondence between the closed sets in the Euclidean sense and in the scheme-theoretic sense.


(Incidentally, another question I had is if, in an arbitrary Spec A, the closed sets really are in 1-1 correspondence with radical ideals)

 

[mathwonk]

Take a closed bounded interval, like [0,1] and the ring of continuous functions on that. For each point p of the interval, there is an evaluation map from the ring of continuous functions to the reals, taking f to f(p). This is obviously an algebra surjection onto a field, so the kernel, or set of functions mapping to zero, i.e. the ideal of f such that f(p) = 0, is a maximal ideal of the ring R.

What about in the opposite direction? take any maximal ideal M in R. I claim it is the set of functions vanishing at some point of the interval. The proof is by contradiction.

Suppose for each point p of the interval there is a function in the ideal not vanishing at p. Then it also does not vanish on some open nbhd of p. Since this gives an open cover of the interval, which is compact, we obtain a finite cover and a corresponding finite number of functions. Then the sum of the squares of these functions does not vanish anywhere on the interval, hence is an unit, so the ideal contains a unit and hence is not maximal, but is the unit ideal.

Thus in fact every proper ideal of the ring R of continuous functions on a compact interval has a common zero, hence the ideal is contained in the maximal ideal corresponding to that point, and by maximality must equal it.

The same proof seems to work for almost any compact space like the circle.

I seem to recall from topology class, that in general the space of maximal ideals of the ring of continuous functions, on a space on which the topology can be defined by continuous functions, is a compactification of the original space called the Stone Cech compactification.

In fact I believe I recall (after 35 years), there is a one one correspondence betwen all constant - containing, point - separating, subalgebras of this ring, and all compactifications of the space. So the full set of maximal ideals defines the largest possible compactification.



Thus for a non compact space like R^n, the space of maximal ideals defines a larger compact space, indeed much larger.

for example, the set of functions "vanishing at infinity" is a maximal ideal, where f vanishes at infinity if for every epsilon, there is a compact subset K in R^n such that f is less than epsilon outside K.

If you take as your subalgebra the set of functions having limits at infinity, defined in an analogous way, you get the famous one point compactification of R^n, i.e. you get a sphere S^n.

Note that in a space like R^n, the Zariski topology defined by continuous functions agrees with the usual topology.

 

[Hurkyl]

There are two sorts of ways my hypothesis could fail:

There exists a prime ideal M such that, for each p, there's a function f in M with f(p) != 0

There exists a prime ideal M that is a proper subset of some maximal ideal.


I was fairly convinced the first way couldn't happen on a compact set; I didn't work out all the details, but it's good to see proof that the maximal ideals are precisely what they "should" be. Do you have any idea about how to tackle the other problem?

 

[mathwonk]

I am not sure I am understanding, so let me just blather on.

In general, in small rings like polynomials in two variables over a field, there are a lot of non maximal prime ideals. Indeed the definition of dimension of a spectrum is (one less than) the length of a strictly nested chain of prime ideals, such as (0), (x), (x,y), in K[x,y], where K is a field.

Now in the more flexible ring of continuous functions on an interval like [0,1], there may possibly be no prime ideals that are not maximal, but I do not see a proof immediately.

In general algebraic geometry the prime ideals correspond to "irreducible" algebraic sets, such as a line in the plane. I.e. no product of two polynomials in x,y can vanish on a whole line without one of them doing so.

But in the world of continuous functions, for any subinterval say, we can find a function vanishing on part of it and another vanishing on the rest, so their product is in the ideal of functions vanishing on that set, yet neither factor is. so that ideal is not prime.

So I do not see an analogous family of "irreducible" subsets of an interval that are not just one point sets.

There is a famous book from the 60's called Rings of Continuous functions by Gillman and Jerison, but I do not have a copy, that might discuss this stuff.

 

[mathwonk]

In an arbitrary commutative ring A, there is a 1-1 correspondence between the clsoed sets of the spectrum and radical ideals of the ring.

By definition a set S is closed if for some ideal J , S equals the set of primes which contain J. Then rad(J) determiens the same clsoed set, and conversely, the intersection of all the prime idelas in S equals rad(J).

see a basic introd to schemes such as Mumford's redbook.

 

[Hurkyl]

Here are two examples of the reasons why I still have some doubt:

If, instead of continuous functions, you take the ring of analytic functions, then you get all sorts of irreducible closed sets. (if you're in more than 1 variable)

In the ring of infinitely differentiable functions, an example of a non-maximal prime ideal is {f | f⁽ⁱ⁾(0) = 0 for n in N}.

 

[mathwonk]

well, analytic functions are not so different from polynomials, but I like the example from C infinity.

I guess I see the proof. If the lowest order derivative of f that is not zero is r, and the lowest order non zero derivative of g has order s, then fg has non zero (r+s)th derivative, by the producr rule.

 

[mathwonk]

does every continuous map from specB to specA arise from a homomorphism a to B? answer: no.

Easiest case: take C[X] = A = B. then the Zariski topology on the set of prime ideals, which equal {0} and the maximal ideals, is such that the closed sets are the finite subsets, excluding {0}, and the whole space.

The closure of {0} is the whole space. so let's see, i guess any function at all which takes {0} to {0} and is other wise at most finite to one, should be continuous, but hardly likely to be obtained from a homomorphism.

2) why is there a 1-1 correspondence betwen closed sets and radical ideals?

key point: show that if A is a radical ideal and f does not belong to A, then there is a prime ideal containing A but not f. Construction: look at the localization of R/A at the element f. This is a ring and contains maximal hence prime ideals by the zorn lemma, hence the pullback of anyone such under the map R/A goes to the localization, is a prime ideal of R not containing A but not f. QED. (R is any ring)

 

[mathwonk]

On page 29 of gillman and jerison there is a theorem that says that a prime ideal of C(X) is contained in a unique maximal ideal. Thus there is, on a compact (completely regular, i.e. "good") space, only one common zero of all the functions in the prime ideal.

Thus your example of functions with all derivatives zero at 0, is the typical type of example.

Gillman and Jerison is alittle tedious tor ead as they sue their own peculair notation for everything so you cannot begin except at the beginning, but they had some pair of ideals,

associated to each point p of a compact space called say O(p) and M(p), where M(p) is the amximal ideal of all functions avnishing at p.

Then O(p) was something like all functions f whose zero set contained a whole nbhd of p.

Then they proved that O(p) is a radical ideal, hence an interscetion of prime ideals, hence there exist other prime ideals that are not maximal.

Then they also proved I believe that every prime ideal is caught between two such ideals O(p) and M(p) for some unique point p.

But they did not give, or at least I did not notice it, your very beautiful example, which illustrates the spirit of their other results perfectly.

I only spent a few minutes perusing the book.

 

[Hurkyl]

Well, I get the impression that the continuous case is somewhat uglier than the infinitely differentiable case. An "explicit" example will likely require the axiom of choice.

This sounds like an attack on the problem I might be able to carry out. There are three major points:

(a) Prove that all maximal ideals are of the form M(p)
(b) Prove that O(p) is a radical ideal.
(c) Prove that a radical ideal is the intersection of prime ideals.
(d) Show that O(p) cannot be the intersection of maximal ideals.

(c) is the only one that seems difficult at first glance, though it's probably a well-known result. (I admit that part of my motivation for looking at Algebraic Geometry is to strengthen my algebra. )

Merci beaucoup!

 

[mathwonk]

I sketched the proof of (c) in post 17 but I was in a hurry.

Let J be a radical ideal in a (commutative) ring R (with identity) and f an element of the ring R but f is not in J. Then since J is radical, no non negative power of f is in J. To show J is an intersection of prime ideals, we must show there is a prime ideal containing J but not containing f.

Consider first the "quotient ring" R/J, whose elements are elements of R but considered equal if they differ by an element of J. For example, R is the ring of polynomials on the plane and say J is the ideal of polynomials vanishing on some curve in the plane, and then R/J is the ring of restrictions to that curve of polynomials on the plane. I.e. two polynomials on the plane are considered equal in R/J if their difference is zero on the curve, i.e. if they have the same values on the curve.

OK, now since J is a proper ideal, the ring R/J is not the zero ring, i.e. it the elements 1 and 0 are different in there. Now moreover the element f is not zero in the ring R/J because it does not lie in J. Moreover no positive power of f is zero in the ring R/J because J is radical.

That means we can "localize" the ring R/J at positive powers of f. I.e. we can enlarge the ring R/J = S by introducing denominators which are non negative powers of f.

I.e. the localization of S at the multiplicative set {1,f,f^2,f^3,...} consists of all elements formally of form s/f^n, with n >=0, and s in S. Of course as usual with fractions we equate s/f^n and t/f^m if sf^m = tf^n.

However, because we are not necessarily in a domain, we must also equate them if there is some power of s, which kills the difference, i.e. also s/f^n = t/f^m, if for some k we have f^k( sf^m - tf^n ) = 0 in S.

This is needed since we are dividing by f in the new ring, so f has become an invertible element. So anything a power of f kills must become zero.

Notice however 1 still does not become zero, since for 1 to equal zero, we would have to have f^k (1-0) = 0, in S, and we chose J so that J is radical, i.e. since f is not in J also no positive power of f is in J, so f^k is not zero in S = R/J.


OK. Now all this implies (by Zorn's lemma) that the localization of S does contain maximal ideals.

Now consider the map from S to the localization of S, taking s to s/1, or to (fs)/f, if you prefer. The inverse image of a prime ideal is always a prime ideal, and any maximal ideal is prime. So pullback any maximal ideal from the localization of S, to get a prime ideal in S. Now a prime ideal in S pulls back further to a prime ideal in R that contains J.

On the other hand, since we have forced f to become an invertible element of the localization of S, f cannot belong to any proper ideal there. Hence the maximal ideal we pulled back did not contain f, and so neither did any of its pullbacks.

So the pullback to R of any maximal proper ideal of the localization of S at powers of f, is a prime ideal of R containing J but not f. QED.

To someone familiar with localizations, this is actually the easier proof, but a direct argument which in essence is exactly the same, would be to check directly that in R any ideal maximal wrt the property that it contains J but not f, is prime.

Actually the construction above is a standard general way of checking that.

 

[mathwonk]

to actually prove the pullback of a maximal ideal M does not contain f, proceed as follows: if f became equal to an element s/f^n of M in the localization of S, then there would be a power f^k such that f^k ( f^n+1 - s) = 0 in S, where s is in the pullback of M. But this says that f^(n+k+1) = s, which says that since s belongs to M, so does that power of f. But f is invertible in the localization, so no power of it can belong to a proper ideal.

 

[Hurkyl]

I did it without peeking. I have roughly the same proof you gave, but I didn't bother modding by J first... I simply considered the ideal J' = {j / f^n | j in J, n in N} and extended that to a maximal ideal.

a direct argument which in essence is exactly the same, would be to check directly that in R any ideal maximal wrt the property that it contains J but not f, is prime.

I actually started out this way, but this afternoon, this use of localization finally clicked in the sesne that I could recognize that it could be applied usefully.

 

[mathwonk]

sorry. I felt kind of foolish giving the proof in detail, since you clearly didn't need it, but I was trying to show off. It seems to have done no harm, except to confirm my reputation as a pedant.

 

[Hurkyl]

No, detail is good! I'm not yet an expert on the subject, so seeing full detail can be helpful, bringing to light things I've overlooked, or didn't even know.

For instance, I had missed this fact:

However, because we are not necessarily in a domain, we must also equate them if there is some power of s, which kills the difference, i.e. also s/f^n = t/f^m, if for some k we have f^k( sf^m - tf^n ) = 0 in S.

Even worse, until the ride home today, I had never even considered the possibility of localizing at zero divisors; if it hadn't struck me then, this quote would have been fairly enlightening.


Anyways...


BLARG

I finally discovered my explicit example of a nonmaximal prime ideal... and it's very annoying since I had been dancing around the idea for quite a while and is so annoyingly similar to the C^inf example.

Let I be the set of all functions with a zero of infinite degree at P. More precisely,

$$I := \{ f | \forall n \in \mathbb{N}: \lim_{x \rightarrow P} \frac{f(x)}{|x - P|^n} = 0 \}$$

*sigh* At least I learned something new today, so I don't feel so silly about missing this!

 

[mathwonk]

Well I didn't know those examples, and they certainly enlightened me on the topic of primes in rings of functions more general than polynomials or analytic functions. I like them too.

 

[mathwonk]

In the world of algebraic geometry, my little project for the last few weeks was learning how to prove the hirzebruch riemann roch theorem for curves and for smooth surfaces in three space, and at least what it says for arbitrary dimensional smooth projective varieties, if you are ever interested in that type of stuff.

also what riemann's approach to the classical result was for compact riemann surfaces. this is all very clasical stuff of course between 50 and 150 years old.

 

[Hurkyl]

Well, I'm usually interested in just about anything I can kinda understand! I've seen the theorem in the guise that div f = [p] - [q] --> p = q in the case of elliptic curve functions. It was suggested that it's a fairly difficult theorem, though.

 

[mathwonk]

I'm a little tanked right now, but RR is not so hard especially for elliptic functions, it just needs the theory of the weierstrass p function.

 

[mathwonk]

oh yes, the fact you stated is very elementary, i.e. if div(f) = p-q, where p is different from q, it means there is a holomorphic map from the elliptic curve E to the complex projective line P^1, such that the inverse image of 0 is p and the inverse image of infinity is q. That means the map is of degree one from E onto P^1. But E is a torus, i.e. topologically a compact surface of genus one, and since P^1 is a sphere that is a contradiction.

 

[mathwonk]

the more general theorem that follows from the theory of the P function, is that if D = p1+...pn is any effective divisor, then the space of meromorphic functions f such that div(f) + D >= 0, has dimension equal to n.

In your case above, is p is different from q, then the space of meromorphic functions f such that div(f) +q >= 0, has dimension one, hence equals the constants so f is constant so div(f) = 0, so p = q, contradiction.


the general theorem is that if D is any divisor, and K is the divisor of a differential form, then the dimension of the space of meromorphic functions f such that div(f) + D >= 0, equals 1-g + deg(D) + dim L(K-D),

where L(K-D) is the space of meromorphic functions f such that
div(f) +K-D >= 0.

The general HRR theorem gives a computation of the alternating sum chi(D), (analogous to the euler characteristic V-E+F of a polyhedron), of dimensions of cohomology groups associated to a divisor D, such that chi(D) = Todd(D), where Todd(D) is some number computed from the topological invariants of the divisor D such as itself - intersection numbers etc...

I'm fading...

 

[mathwonk]

Hurkyl, We have been getting into using schemes to define differential one forms on singular spaces, in the thread "what is a tensor?" if you are interested.

the "elementary" argument above was supposed to be essentially that a meromorphic function f with divisor p-q where p,q are different, defines an isomorphism between a torus and a sphere, which is not even topologically possible, much less analytically so. did that make sense?

i.e. you just send the point x on the torus to the value of f at x. That defines a holomorphic map from the torus to the complex number plane, and it extends to a continuous map from the torus to the one point compactification of the plane sending the "pole" q to infinity.

that gives a map from the torus to the sphere, but since the divisor of f is p-q that says only one point goes to zero (i.e. p), and only one point goes to infinity (namely q), counting multiplicities.

The general theory of complex maps then implies that every point of the sphere has exactly one preimage, so the map is a homeomorphism, and even an isomorphism.

I.e. for complex maps of compact connected complex one manifolds, each point has the same number of preimages, counting multiplicities, if that number is finite.

 

[Hurkyl]

I've been distracted by a side problem.

Algebraically, I'm afraid I might be using what I'm trying to prove, I have to go back and review. I can see why it's a degree 1 map because if div(f) = [P] - [Q], then f = g (X - P)/(X - Q) where div(g) = 0. I know this implies g is a constant, but I don't remember off hand what was used to prove that... and from here it's obvious that f is an isomorphism. (Of course, I also have not seen the algebraic proof that the EC is a torus, though I believe it)

I've seen the p-functions, so I can follow the argument over the complexes. Passing to the complex plane as the covering of the EC, we have that g is bounded and entire, thus constant.

 

[mathwonk]

Let X,Y be compact Riemann surfaces, i.e. compact real 2 dimensional manifolds,with a complex structure.

And let f:X-->Y be a holomorphic map. Then f is a "branched covering". I.e. away from the points of X where the derivative of f is zero, the map is a covering map. In particular, the preimage of every point of Y which does not contain a zero of the derivative, contains the same number of points.

If f is a meromorphic function on X, to say that div(f) = p-q means that the inverse image of 0 is p, counting multiplicites, and the inverse image of infinity is q, also counting multiplicities.

This impies that the dergee of f is one, i.e. that the number of inverse images of every point is one, counting multiplicities.

Thus the map X-->P^1 is one to one everywhere, hence bijective.

Now the structure of holomorphic mappings implies that this map is thus an isomorphism, hence also a homeomorphism from X to P^1, where P^1 is a sphere.

Hence X is homeomorphic to a sphere.

Now assume that X is an elliptic curve, e.g. a smooth cubic curve in the plane P^2.

We claim that then X is actually homeomorphic to a torus. The easiest way to see this is to note that if we degenerate X to a union of three lines, i.e. a triangle, then X looks like a union of three spheres joined pairwise at points. This is obviously a degenerate torus.

Another way is to note that a smooth cubic curve can be put in the form
y^2 = x^3 + g1x + g2. where g1 and g2 are complex numbers, by some change of coordinates.

Then there exists a meromorphic function called a P function, such that the map from X to P^2 given by P and P' (the derivative of P), takes X isomorphically onto the given cubic curve. Then the theory of (doubly periodic) P functions shows that X is isomorphic to the quotient of C^2 by a lattice, hence to a torus.

Thus every elliptic curve X is topologically a torus, hence X is not homeomorphic to a sphere, hence X cannot have a meromorphic function f with divisor div(f) = p-q, with p different from q.

 

[mathwonk]

oh you want the algebraic proof, over a field other than the complexes where we cannot use the usual topology and homeomorphism argument.

That is more work. Then we have to use some algebraic way to distinguish betwen P^1 and an EC. There are basically two versions of the "genus" of a curve.

One is the number of handles in the usual topology, which we cannot use, and the other is the number of linearly independent differential forms. Those two numbers are equal in the classical case so we can use either of them as the genus. In the algebraic case then we use the number of linearly independent regular algebraic differentials on the curve, defined of course as you konw, by globalizing the I/I^2 definition of differentials.

Here is one argument that an EC and P^1 cannot be isomorphic. First of all, the quotient of two differential forms is a rational function (analog in arbitrary algebraic setting of a meromorphic functrion). Second a rational function defines a map from a smooth projective curve to P^1, which is a branched covering, and again every point of p^1 has the same number of preimages counted properly. This number, the algebraic degree of the mapping, equals the dergree of the field extension defined by the map, i.e. the degree of the field of rational functions on the curve as an extension of the field of rational functions on P^1. [This is why a function with divisor p-q has degree one, because the divisor exhibits the unique inverse image of zero as p. I.e. since the divisor p has degree one, by the theorem the map has degree one. I did not understand your short argument about the degree being one.]

Hence the divisor of any non constant rational function has degree zero. This is not trivial to prove in the algebraic case and uses the theory of finite extensions of rings and the fact that a torsion free finite module over a principal ideal domain (such as the local ring at a smooth point on a curve) is free, of well determined rank. There is a nice proof in Shafarevich's book Basic Algebraic geometry.

Anyway, from the fact that the quotient of two differentials is a rational function and the fact that the divisor of a rational function is zero, it follws that all differentials on a curve have divisors of the same degree.

So all we have to do is compute the degree of any divisor of a differential on an EC and the degree of any divisor of a differential on P^1 and show they do not agree. On P^1 just write down dz and note it has a pole of order 2 at infinity and no other zeroes aor poles, so has degree -2. (This is 2g-2 where g=0.)

For an EC with equation y^2 = (x-a)(x-b)(x-c), dx/y is a regular differential everywhere with no zeroes or poles (the zeroes of y at the points (a,0), (b,0), (c,0) are canceled by the zeroes of dx at those points). The degree of the divisor of dx/y is thus zero = 2g-2 where g = 1. Hence we have a contradiction to assuming that EC and P^1 are algebraically isomorphic isomorphic.


There is a more elementary proof that does not develop the theory of the genus, just finding a numerical contradiction, but it is harder to motivate.

I.e. assume there is an isomorphism from P^1 to an EC, say the fermat curve with equation (1) x^3 +y^3 = z^3 . I.e. assume we can express each variable x,y,z as a rational function of t, not all constant.

Then differentiate to obtain (2): x^2x' + y^2y' = z^2z', where ' denotes differentiation w.r.t. t. Now we want to eliminate the z terms. multiply (1) by z' and (2) by z, and subtract, to obtain z'x^3 + z'y^3 = x'x^2z + y'y^2z. Collecting terms, and factoring gives x^2( xz'-zx') = y^2 ( y'z - yz'). If either ( y'z - yz')=0, or ( xz'-zx') = 0, then both do, and hence by the quotient rule for derivatives (which make sense algebraically), we would have x/z and y/z constant. Since x,y are rel prime, x^2 divides (yz'-zy'), in k[t], and thus 2degree(x) <= deg(y)+deg(z)-1. Repeating the argument for each of the other two variables, i.e. eliminating the x and y terms, leads to the same inequality with the variables permuted. Adding the 3 inequalities gives 2[deg(x)+deg(y)+deg(z)] <= 2[deg(x)+deg(y)+deg(z)] - 3, a contradiction. QED.

This argument can be generalized to any EC with some work, but I prefer the more general conceptual approach above.

In algebra, the whole (smooth) curve X with all its geometry is determined just by the algebra of the field of rational functions k(X), as a finitely generated field over k the algebraically closed base field. I.e. if two curves have isomorphic fields k(X) isom to k(Y) as k algebras, then X is isomorphic to Y as curves.

One can define the regular differentials just from this field, hence one can show purely algebraically that the field of an EC is not isomorphic to the field of P^1.

Of course that is what our elementarya rgument above did. I.e. it showed that there is no embedding of the field of the fermat curve as a subfield of the field k[t] of rational functions on P^1. In particular they are not isomorphic.

It also shows further than there is no non constant mapping from P^1 to an EC, even of degree higher than one. In topology thisa could be done by showing that such a map of curves cannot raise genus. And that resault two can be proved using the differential form version of genus. I.e. a surjective algebraic mapping induces in the other direction an injection of differential forms. (In the spirit of the threads in differential geometry and tensors, covector fields are distinguished from vector fields, in that covector fields pull back, although vector fields do not even push forward.)

Anyway it follows from this that the vector space of differential forms on the target curve (or variety of any dimension) must be no larger in dimension that that of the source curve. Thus surjective morphisms, cannot raise the algebraic genus.

 

[mathwonk]

Another proof, by Riemann Roch.

the genus is the most important invariant of a smooth projective curve.

It figures in the famous Riemann Roch theorem as well as follows:

let a divisor D = p1+p2+...+pr -q1-q2...-qs be given on a projective smooth curve X over the algebraically closed field k.

Define L(D) as the vector space of rational functions f on X such that div(f)+D >= 0. (i.e. f has poles at most among the p's, and zeroes at least on the q's.)

Then the dimension of L(D) is finite, and if K is the divisor of a differential form on X, then w e have:

dimL(D) - dimL(K-D) = deg(D) 1-g, where g is the algebraic genus of X.

Let's apply this to X = an EC of genus one, and D = q. Then the constants provide one dimension of functions f with div(f)+q = q>=0, so L(q) has dimension at least one.

However a function f with div(f) = p-q, for some p different from q, would be non constant so we would have dimL(q) >= 2.

Now we know a differential form on an EC has divisor equal to zero, se we get from RRT:

dimL(q) -dimL(-q) = 1-g +deg(D) = 1-1+1 = 1.

Now L(-q) is the zero vector space since no function can have a zero on q without also having a pole somewhere, as noted above.

So RRT gives dimL(q) = 1, a contradiction to the existence of a non constant f in there.

This kind of easy numerical argument shows why people love the RRT, and why it is the most powerful, most useful theorem in the theory of curves.