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In this post, we shall see some results about algebraic curves. For the past week, I have been studying algebraic curves and the great Riemann-Roch theorem. I won’t go in the details but will only sketch the basic theory behind their study.

Algebraic curves

By a curve we will mean a projective variety of dimension 1. For the sake of simplicity let us assume the base field {k} to be algebraically closed. (This is far from sufficient; indeed the most interesting applications in number theory will have {k} as the rational numbers or finite fields or {p}-adic fields). One can think of the curve as the locus of the zero set of an irreducible polynomial {F(X,Y) \in k[X,Y]}. This as it is, is an affine curve and we attach points of infinity as necessary by homogenizing {F}. Note that it is not necessary that the curve be generated by just one polynomial. It may be possible that it is the intersection of higher dimensional varieties in projective space of higher dimension. The only requirement is that the dimension of this variety be 1.

Function field of a curve

Given a curve {C}, we can associate a field to it, namely the function field of {C}. As a concrete example, let {C} be the projective circle given by the homogeneous equation {F(X,Y,Z) = X^2 + Y^2 - Z^2}. (If you are not habituated to using projective coordinates, just substitute 1 for {Z} and everything should work fine). Then the function field of {C} is

\displaystyle k(C) = \text{Field of fractions of } \displaystyle\frac{k[X,Y,Z]}{(X^2+Y^2-Z^2)}.

Note that asking the dimension of {C} to be 1 is the same as {k9C)} having a transcendence degree 1 over {k}.

Corresponding to any point {P=[a:b:c]} on the curve, there is a discrete valuation ring {k[C]_P} which is the localization of the domain {\displaystyle\frac{k[X,Y,Z]}{(X^2+Y^2-Z^2)}} at the maximal ideal {(X-a, Y-b, Z-c)}. A fundamental theorem in algebraic geometry says that the point {P} on {C} (in fact any variety) is non-singular if and only if {k[C]_P} is a regular local ring.

Maps between curves

By a map between curves {C_1} and {C_2} we will mean a morphism of the corresponding projective varieties. If

\displaystyle \phi : C_1 \rightarrow C_2

is a morphism of curves, then {\phi} is either constant or surjective! Also, {\phi} induces a map between the function fields viz.

\displaystyle \phi^* : k(C_2) \hookrightarrow k(C_1) \qquad \phi^*(f) = f \circ \phi.

If {\phi} is nonconstant this gives a finite extension of fields, {[k(C_1) : \phi^*(k(C_2))]} and we define the degree {deg(\phi)} to be the degree of this extension. A map of degree 1 is an isomorphism.

Categorical equivalence between curves and function fields

We saw that a curve {C} gives a function field {k(C)} of transcendence degree 1 over {k}. Morphisms of curves give an inclusion of function fields. Indeed, this functoriality goes beyond, it’s a categorical equivalence. Given a field {\mathbb K/k} of transcendence degree 1, one proves that the collection of local rings {R} such that {k \subset R \subset \mathbb K} actually define a non-singular projective curve. (All the {R}‘s will be dvrs since tr. deg{(\mathbb K/k)=1}). The morphisms in this category are inclusion maps of fields and they give morphisms of curves.

If you have read this far, then you are very close to understanding the connection between algebraic curves and algebraic number fields. This is explained in the next post here.


  • R. Hartshorne, Algebraic Geometry (Chapter 1)
  • J. Silverman, Arithmetic of Elliptic Curves (Chapter 2)

This post follows the previous two posts on Categories. (First and second). It explains how one uses the universal properties to prove nontrivial results. Since \LaTeX rendering in WordPress is not so efficient, I decided to upload the notes in my Notes page, and can also be accessed via this link.

Proposition: The polynomial (x^4+1) is reducible in the ring (\mathbb Z/p \mathbb Z)[x] for every prime p.


If p=2 then x^4+1 = (x+1)^4 is clearly reducible. Otherwise, let p be an odd prime. Then, p^2\equiv 1 (\mod 8), so

(x^4+1) | (x^8-1) | (x^{{p^2}-1}-1) | (x^{p^2}-x)

Hence the roots of the given polynomial lie in the field \mathbb F_{p^2}. Had the quartic been irreducible, the roots would have been found in an extension of degree 4 over \mathbb F_p.

Proposition: The polynomial (x^2-2)(x^2-3)(x^2-6) has a root modulo every prime p, that is, in (\mathbb Z/p \mathbb Z)[x].


If \displaystyle\left(\frac{2}{p}\right) = \left(\frac{3}{p}\right) = -1, then by multiplicativity of the Legendre symbol, \displaystyle\left(\frac{6}{p}\right) = 1.

I had my TIFR PhD interview today. Inspired by the Princeton Generals, where students who get through their Qualifying exam write about the questions asked, even I am writing here describing my interview experience. But I am not very enthusiastic about writing this, since the questions were very basic and not from all subjects, a very unlikely event during a Maths PhD interview. Nevertheless, here is the interview–

Chair: Prof. A. Sankaranarayanan

Other interviewers: Prof. Indranil Biswas, Prof. Raja Sridharan, Prof. Ravi Rao

I do not remember who asked which question, so in much of what follows, I am referring to all as “interviewer”. Most questions were asked by RR.

AS: So you work at Larsen & Toubro?

Me: No sir, I was working at L&T for an year from 2007 to 08. Then I joined IIT Bombay in 2008 till now.

RR: Under whom have you taken courses at IIT?

(I name Limaye sir and other teachers)

IB: What are your interests?

Me: Sir I do not have a particular interest as such, but I do like Algebra.

RR: K is a field. Is K[X^2,X^3] a UFD when considered as a subring of K[x]?

Me: Yes sir, zerodivisors in K[X^2,X^3] would mean zerodivisors in K[X] which is a domain. (Mistake.)

RR: So it is a domain. But is it a UFD?

Me: Ohh.. Yes.. I mean no. Consider X^6, it can be factored in two ways. (I proved that the two were irreducibles and not unit-times-each-other.)

RS: f:\mathbb R \to \mathbb R is smooth and \lim_{x\to\infty} f(x)=0 Then can you say if \lim_{x\to\infty} f'(x)=0?

(I tried for a long time to prove, then tried to give a counter-example. \displaystyle\frac{\sin x}{x} didn’t work. He suggested \displaystyle\frac{\sin x^3}{x} and it worked, though I goofed up in differentiating and taking limits)

(In the meanwhile, there was an interesting discussion among them if they should finish off Algebra before switching to Analysis.)

RR: Is \displaystyle\frac{\mathbb C[X,Y]}{(X^4+X^3Y+Y^4)} a domain?

Me: If it is to be a domain, then the polynomial should be prime, hence irreducible. (thinking) Eisenstein.. won’t work!

Interviewer: Hint — The polynomial is homogeneous.

(Solved it.)

Interviewer: A linear transformation T:\mathbb R^n \to \mathbb R^n sends straight lines to themselves. What can you say about T?

(I wrote, wrongly, a diagonal matrix with different diagonal entries. Later corrected the mistake after they asked me to repeat the question.)

Interviewer: V is a n-dimensional vector space and W is its subspace such that for every isomorphism T:V\to V, we have T(W)\cap W \neq (0). What can you say about W?

(Proved that the dimension must be \geq n/2.)

Interviewer: Consider S_p, the symmetric group on p-symbols and an element of order p in it.

(I wrote the element as a p-cycle. There followed a long discussion as to why it should be a p-cycle, and orders of commuting elements, their lcm’s etc. It finally ended with me group-acting S_p on those p-symbols and ‘proving’ that order of disjoint cycle types is the lcm of their orders.)

RR: Find a p-subgroup of \text{GL}_3(\mathbb F_p).

(This took the longest time of all. The question was interesting, involving some fundoo linear algebra. In spite of many hints from them, I was not able to solve it. Finally, RS told the answer.)

Me: (excitedly) Ohk, not just the elements of order p-power, I have found the  group of order p^3!!

RS: No, I have found it!

Me: (embarassed) yes sir..

(There followed a discussion if they should ask more questions. Finally, RS asked me to prove that the additive group of a field of characteristic zero is not cyclic. I was halfway done when I was told to go.)

Note: The interview lasted for an hour. The interviewers were pleasant but and did give subtle hints when I got stuck up. But they stopped me whenever they thought I was not giving the right justification or hand-waving proofs.

PS: I was not offered the fruit juice the others before me were offered. 😛

This was proved by me during the VSRP with a hint from Prof Rajan.

Theorem: Prove that the set of algebraic integers in an algebraic number field forms a ring.

A different Proof:

Denote by K, the given algebraic number field and the set of algebraic numbers in K, by \mathcal {O}_K.

Suppose \alpha \in \mathcal{O}_K satisfies
f(x)=x^n+a_{n-1} x^{n-1} + \cdots + a_0 \in \mathbb{Z}[x] .
Then, ( -\alpha ) satisfies
f'(x)= (-1)^n a_n x^n + \cdots + (-1)a_1 x + a_0. Clearly, f' \in \mathbb Z so no problem there.

Now, given \alpha and \beta in \mathcal{O}_K, we need to show that \alpha\beta and \alpha+\beta are in \mathcal{O}_K.

So assume that \alpha and \beta satisfy the monic polynomials f(x) and g(x) in \mathbb{Z}[x].

Since we can factor f and g in \mathbb{C} using the Fundamental Theorem of Algebra, we can factor f and g as
f(x) = x^m+s_{m-1} x^{m-1} + \cdots + s_0 = (x-\alpha_1) (x-\alpha_2) \cdots (x-\alpha_m)

g(x) = x^n+t_{n-1} x^{n-1} + \cdots + t_0 = (x-\beta_1) (x-\beta_2) \cdots (x-\beta_n)

where \alpha=\alpha_1 and \beta=\beta_1 for notational convenience.

Note: The s_i‘s and t_j‘s are elementary symmetric functions of \alpha_i‘s and \beta_j‘s respectively.

Now, consider the polynomials
h(x)= \displaystyle\prod_{i \in \{1,2,\cdots ,m\}} \prod_{j\in \{1,2,\cdots ,n\}} \left(x-\alpha_i \beta_j\right)


h'(x)=\displaystyle\prod_{i \in \{1,2,\cdots ,m\}} \prod_{j \in \{1,2,\cdots , n\}} (x-(\alpha_i+\beta_j) )

The polynomials h and h' are invariant under any permutation \sigma \in S_m. So, h and h' can be viewed as symmetric polynomials in the variables \alpha_1, \alpha_2, \cdots, \alpha_m. Hence, by the Fundamental Theorem on Symmetric Functions, (not proved!) we have,

h(x) = h(s_1, s_2, \cdots , s_m) (x)

Now, since the equation on the right is invariant under any permutation on n characters applied to \beta_1, \beta_2, \cdots, \beta_n, it is symmetric in \beta_j‘s. So, invoking the theorem on symmetric functions again, we have,

h(x) =  h(s_1, s_2, \cdots , s_m) (x) =  h(s_1, s_2, \cdots , s_m) (t_1, t_2, \cdots , t_n) (x)

Of course, a similar argument holds for h'.

Since the coefficients s_i‘s and t_j‘s are integers, hence h (x)\in \mathbb{Z}[x] and h' (x) \in \mathbb{Z}[x].

Clearly, \alpha\beta and \alpha+\beta are roots of h and h' respectively and hence, \alpha, \beta \in \mathcal{O}_K. \blacksquare

Remark: With a little addition, the above proof can be extended to show that the set of all algebraic numbers forms a field.

Problem: R is a commutative Noetherian domain. F is its field of quotients. If F is finitely generated as an R-module, then prove that R is a field. (So R = F.

About me

Abhishek Parab

I? An Indian. A mathematics student. A former engineer. A rubik's cube addict. A nature photographer. A Pink Floyd fan. An ardent lover of Chess & Counter-Strike.

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