Monthly Archives: April 2015

Binomial sum modulo prime power: Part 2

Let p be a prime number and \omega=\exp(2\pi i/p). Recall that we conjectured in this post that

(*)\qquad\qquad\qquad\displaystyle\sum_{j=0}^{p-1}(1-\omega^j)^n=p\sum_{k=0}^{\lfloor n/p\rfloor}(-1)^{kp}\binom{n}{kp}

is divisible by p^{\lceil n/(p-1)\rceil}. From exercise 2 of this post we know that (p) factors as (p,\omega-1)^{p-1} into prime ideals in the ring \mathbb Z[\omega]. So (\omega-1)^{p-1}\in (p). Therefore the right-hand side of (*) is divisible by p^{\lfloor n/(p-1)\rfloor}. So we are off by at most one factor of p! I believe this approach can be improved upon to account for the extra factor, since we haven’t used any property of the sum.

Furthermore,

\displaystyle\sum_{j=0}^{p-1}\omega^{-rj}(1-\omega^j)^n=p\sum_{m\equiv r\pmod p}(-1)^m\binom{n}{m}

for any r. So our best result thus far is the following:

Theorem (weaker version of Fleck’s).

\displaystyle\sum_{m\equiv r\pmod p}(-1)^m\binom{n}{m}\equiv 0\pmod{p^{\lfloor\frac{n}{p-1}\rfloor-1}}.

Goal: Improve the floors to ceilings.

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Emmy Noether

I rarely pay tribute to people, but this will be an exception.

I first came to know about Emmy Noether back in 2009 when, like many others at my level in formal education, I became very interested in physics. (Although as time went by I soon realised that mathematics, more precisely pure mathematics, was in fact where the “complete picture” lay; but that’s a different story.) I was trying to solve a problem using conservation of energy. Eventually I stumbled upon her wikipedia page from articles on conservation laws (cf. Noether’s theorem).

Later on, in my second undergraduate year I learned about rings and modules, when the term “noetherian” came up. It was merely a brief introduction. Recently I took an introductory course on commutative algebra, and her name was so ubiquitous throughout the course that there probably wasn’t a single lecture in which she wasn’t referred to. So I grew an urge to know more about her. As I read more and more about her I developed an immense respect toward this woman.

Some quotes by famous mathematicians about Noether: (excerpted from wikipedia)

  • “Absolute beyond comparison.” – Van der Waerden, regarding her mathematical abilities
  • “Changed the face of algebra by her work.” – Weyl
  • “Revolutionary.” – Kaplansky, regarding her idea of the ascending chain condition
  • “The greatest woman mathematician of all time.” – Alexandrov
  • “The development of abstract algebra, which is one of the most distinctive innovations of twentieth century mathematics, is largely due to her – in published papers, in lectures, and in personal influence on her contemporaries.” – Jacobson
  • “Miss Noether is… the greatest woman mathematician who has ever lived; and the greatest woman scientist of any sort now living, and a scholar at least on the plane of Madame Curie.” – Norbert Wiener

If you love to read and have some time, I suggest going through her wikipedia page. It’s such a powerful story that I believe you won’t want to stop until you’ve read the entire article.

A belated happy birthday, Emmy Noether!

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Ring of integers of cyclotomic field

Let \zeta=\zeta_n=e^\frac{2\pi i}{n} be a primitive n-th root of unity, and let K=\mathbb Q(\zeta). I am going to outline a proof that \mathcal O_K=\mathbb Z[\zeta], based on several homework problems from one of my recent courses. There are probably many other proofs of this, but I particularly like this one because it’s easy to follow, touches on a wide range of topics, and I worked hard through it!

First, let n=p^m be a prime power. The minimal polynomial of \zeta is the n-th cyclotomic polynomial

\displaystyle\Phi_n(X)=\Phi_{p^m}(X)=\sum_{j=0}^{p-1}X^{jp^{m-1}}.

Let f(X)=\Phi_n(X+1).

Exercise 1. Following the above notation, show that f satisfies the conditions of Eisenstein’s criterion for the prime p.

Consider the discriminant \Delta(f)=\Delta(R), where R=\mathbb Z[\zeta-1]. If q is a prime factor of \Delta(f), then f must have a multiple root modulo q. Hence X^{p^m}-1 will also have a multiple root modulo q. But \gcd(X^{p^{m}}-1, p^{m}X^{p^{m}-1})=1 in \mathbb Z/q\mathbb Z unless q=p. Thus \Delta(f)=\Delta(R)=[\mathcal O_K:R]^2\Delta(\mathcal O_K) is a power of p. In particular, [\mathcal O_K:R] is a power of p.

Exercise 2. Suppose that f\in\mathbb Z[X] satisfies the conditions of Eisenstein’s criterion for a prime number p. Let \alpha be a root of f and let R=\mathbb Z[\alpha]. Prove that there is exactly one prime ideal P\subseteq R that contains p, and that the local ring R_P is a DVR with uniformiser \alpha.

Now if Q\subseteq R is another prime ideal, then p\not\in Q. So [\mathcal O_K:R]\not\in Q.

Exercise 3. Let K be a number field and R\subseteq\mathcal O_K a subring of finite index d. If Q\subseteq R is a prime ideal not containing d, show that R_Q is a DVR.

I’ll include this solution because I love it!

Solution. Let D=\mathcal O_K. By going up, there is a prime ideal \tilde Q\subseteq D with \tilde Q\cap R=Q. We’ll show that D_{\tilde Q}=R_Q, so the result will follow.

Let a/b\in R_Q, so that a\in R, b\in R\setminus Q. If b\in \tilde Q, then b\in R\cap \tilde Q=Q, a contradiction. Hence b\not\in\tilde Q, i.e. a/b\in D_{\tilde Q}. Thus R_Q\subseteq D_{\tilde Q}.

Now let a/b\in D_{\tilde Q}, so that a\in D, b\in D\setminus\tilde Q. Then da\in R, db\in R, and a/b=(da)/(db). Note that b\not\in\tilde Q\supseteq Q, and d\not\in Q by assumption. So db\not\in Q since Q is a prime ideal. Thus a/b=(da)/(db)\in R_Q, i.e. D_{\tilde Q}=R_Q. \square

So R localised at any prime ideal is a DVR, implying that R=\mathbb Z[\zeta] is a Dedekind domain. Thus R=\mathcal O_K. This completes the proof for the prime power case.

Now we proceed by induction on the number of distinct prime factors of n. The base case has already been taken care of. So suppose that n=ab, where a,b>1 are coprime integers. Let L=\mathbb Q(\zeta_a) and M=\mathbb Q(\zeta_b). By the induction hypothesis, \mathcal O_L=\mathbb Z[\zeta_a] and \mathcal O_M=\mathbb Z[\zeta_b].

Exercise 4. Let L and M be number fields with discriminants \lambda and \mu respectively. Let \{a_1,\dots,a_m\} and \{b_1,\dots,b_n\} be integral bases for L and M respectively. Let K=LM=\mathbb Q(a_1,\dots,a_m,b_1,\dots,b_n) be the composite field of L and M. Suppose that [K:\mathbb Q]=[L:\mathbb Q][M:\mathbb Q] and that \gcd(\lambda,\mu)=1. Show that \{a_ib_j:1\le i\le m; 1\le j\le n\} is an integral basis for K.

So our main result will follow from exercise 4 once we have checked that all the hypotheses are satisfied.

(i) Checking K=LMFirstly, LM contains \zeta_a\zeta_b=e^{\frac{2\pi i(a+b)}{n}}=\zeta^{a+b}.

Exercise 5. If \gcd(a,b)=1, show that \gcd(a+b,ab)=1.

So writing j=a+b shows that \gcd(j,n)=1 and \zeta^j\in LM. If j^{-1}\in\{1,\dots,n\} is the multiplicative inverse of j\pmod n, then \zeta=(\zeta^j)^{j^{-1}}\in LM. Thus K\subseteq LM.

Again, since \zeta^a=\zeta_b and \zeta^b=\zeta_a, we have LM\subseteq K. Thus K=LM.

(ii) Checking [K:\mathbb Q]=[L:\mathbb Q][M:\mathbb Q]. We have

[K:\mathbb Q]=\varphi(n)=\varphi(ab)=\varphi(a)\varphi(b)=[L:\mathbb Q][M:\mathbb Q].

(iii) Checking \gcd(\lambda,\mu)=1This is slightly harder. We need a few more facts.

Exercise 6. Let E=\mathbb Q(\alpha) be a number field, and let f be the minimal polynomial of \alpha. Show that

\Delta(\alpha)=(-1)^{\binom{\deg(f)}{2}}N_{E/\mathbb Q}(f'(\alpha)).

Solution. Let f(x)=(x-\alpha_1)\cdots(x-\alpha_n). Then

\displaystyle\Delta(\alpha)=(-1)^{\binom n2}\prod_{i\neq j}(\alpha_i-\alpha_j)=(-1)^{\binom n2}\prod_i\prod_{j\neq i}(\alpha_i-\alpha_j).

Using Leibniz’s rule,

\displaystyle f'(x)=\sum_i\prod_{j\neq i}(x-\alpha_i)\Rightarrow f'(\alpha_i)=\prod_{j\neq i}(\alpha_i-\alpha_j).

Hence

\displaystyle\Delta(\alpha)=(-1)^{\binom n2}\prod_if'(\alpha_i)=(-1)^{\binom n2}N_{E/\mathbb Q}(f'(\alpha)). \square

Exercise 7. Using the previous exercise, show that \Delta(\zeta_n)\mid n^{\varphi(n)}.

Hint. Write X^n-1=\Phi_n(X)g(X), and use Leibniz’s rule to get n\zeta_n^{n-1}=\Phi_n'(\zeta_n)g(\zeta_n).

So \lambda=\Delta(\zeta_a) divides some power of a, and \mu=\Delta(\zeta_b) divides some power of b. Since \gcd(a,b)=1, we conclude that \gcd(\lambda,\mu)=1.

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