# 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.

Filed under Algebra, Number theory

## 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!

Filed under Uncategorized

## 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 $K=LM$Firstly, $LM$ 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 $\gcd(\lambda,\mu)=1$This 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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Filed under Algebra, Number theory