# Tag Archives: ring

## Simple cases of Jacobson’s theorem

A celebrated theorem of Jacobson states that

Theorem. Let $R$ be a ring, not necessarily containing $1$. If, for each $a\in R$ there exists a positive integer $n$ such that $a^n=a$, then $R$ is commutative.

This is a very strong and difficult result (although not very useful in practice). However, we can obtain some special cases via elementary means.

Proposition 1. Let $R$ be a ring such that for each $a\in R$ we have $a^2=a$. Then $R$ is commutative.

Proof. Let $a,b\in R$. Then $a+b=(a+b)^2=a^2+ab+ba+b^2=a+ab+ba+b$, i.e., $ab=-ba$. Again, $a-b=(a-b)^2=a^2-ab-ba+b^2=a-ab-ba+b$, i.e., $ab=-ba+2b$. Thus $2b=0$, i.e., $b=-b$ for each $b\in R$. Thus $ab=-ba=ba$, as desired. $\square$

The next case is already considerably harder.

Proposition 2. Let $R$ be a ring such that for each $a\in R$ we have $a^3=a$. Then $R$ is commutative.

Proof. Let $a,b\in R$. Then $a+b=(a+b)^3$ shows that

$(*)\qquad\qquad\qquad a^2b+aba+ba^2+ab^2+bab+b^2a=0$,

and $a-b=(a-b)^3$ shows that

$a^2b+aba+ba^2=ab^2+bab+b^2a$.

Hence

$(**)\qquad\qquad\qquad\qquad 2(a^2b+aba+ba^2)=0$

for all $a,b\in R$.

Plugging $a=b$ into $(**)$ gives $6a=0$, i.e., $3a=-3a$ for each $a\in R$.

Plugging $b=a^2$ into $(*)$ gives $3(a^2+a)=0$, i.e., $3a^2=3a$ for each $a\in R$. Replacing $a$ by $a+b$ gives $3(ab+ba)=0$, i.e., $3(ab-ba)=0$.

Also, multiplying $(**)$ by $a$ first on the left and then on the right and then subtracting the two gives $2(ab-ba)=0$.

From the last two paragraphs we conclude that $ab-ba=0$ for all $a,b\in R$. $\square$

Corollary. Let $R$ be a ring such that for each $a\in R$ we have $a^n=a$ for some $n\le 3$. Then $R$ is commutative.

Proof. Note that if $a^n=a$ for some $n\le 3$ then $a^3=a$. Hence the result follows by Proposition 2. $\square$

Filed under Algebra

## GCD and LCM via groups and rings

Let $a$ and $b$ be integers. Consider the ring

$G=\{ax+by:x,y\in\mathbb Z\}$.

This is a well-ordered group. So by a result in this post it is infinite cyclic. We call the positive generator $(a,b)$ the greatest common divisor (GCD) of $a$ and $b$.

Consider now the ring

$L=\{m\in\mathbb Z:m$ is divisible by both $a$ and $b\}$.

It is also a well-ordered group. Hence it is generated by a single positive element $[a,b]$, called the least common multiple (LCM) of $a$ and $b$.

Let

$M_a=\{ax:x\in\mathbb Z\}$ and $M_b=\{by:y\in\mathbb Z\}$.

$M_a$, $M_b$ and $L$ are subings of $G$. Moreover, $G=M_a+M_b$ and $L=M_a\cap M_b$. So by the Chinese remainder theorem,

$G/L\cong G/M_a\times G/M_b$,

which can be written as

$(\mathbb Z/L)/(\mathbb Z/G)\cong(\mathbb Z/M_a)/(\mathbb Z/G)\times(\mathbb Z/M_b)/(\mathbb Z/G)$

by the third isomorphism theorem. The groups in brackets are all finite groups of orders $[a,b]$, $(a,b)$, $a$ and $b$. Hence $[a,b]/(a,b)=ab/(a,b)^2$, i.e.,

$\boxed{ab=(a,b)[a,b]}$.

In general, let $a_1,\dots,a_n\in\mathbb Z$. As before, we can define

$G=\{a_1x_1+\cdots+a_nx_n:x_i\in\mathbb Z\ \forall i\}$

$L=\{m\in\mathbb Z:m$ is divisible by $a_i\ \forall i\}$

$M_i=\{a_ix:x\in\mathbb Z\}$ for $i=1,\dots,n$.

Then $G=M_1+\cdots+M_n$ and $L=M_1\cap\cdots\cap M_n$. As before,

$G/L\cong\displaystyle\prod_{i=1}^n(G/M_i)$,

i.e.,

$(*)\qquad\qquad\qquad (\mathbb Z/L)/(\mathbb Z/G)\cong\displaystyle\prod_{i=1}^n((\mathbb Z/M_i)/(\mathbb Z/G))$.

So $\displaystyle [a_1,\dots,a_n]/(a_1,\dots,a_n)=a_1\cdots a_n/(a_1,\dots,a_n)^n$, i.e.

$\boxed{a_1\cdots a_n=(a_1,\dots,a_n)^{n-1}[a_1,\dots,a_n]}$.

If we replace $\mathbb Z$ by $\mathcal O_K$ for any number field $K$, then $(*)$ takes the form

$(\mathcal O_K/L)/(\mathcal O_K/G)\cong\displaystyle\prod_{i=1}^n((\mathcal O_K/M_i)/(\mathcal O_K/G))$.

Taking cardinalities gives the following equation in terms of ideal norms

$N(L)/N(G)=\displaystyle\prod_{i=1}^n(N(M_i)/N(G))$.

Thus

$\boxed{N_{K/\mathbb Q}(a_1\cdots a_n)=N(G)^{n-1}N(L)}$.

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Filed under Algebra, Number theory

## A countability argument

Here is a countability argument that I like because it relies on almost nothing. Let $A$ be a ring.

Theorem. If $A$ is countable, then the polynomial ring $A[X]$ is countable.

Proof. Since $A$ is countable, there is an injection $f:A\to\mathbb N$. Let $p_0 be prime numbers and consider the map

\begin{aligned}g:A[X]&\to\mathbb N\\ a_0+a_1X+\cdots+a_nX^n&\mapsto p_0^{f(a_0)}p_1^{f(a_1)}\cdots p_n^{f(a_n)}.\end{aligned}

By unique factorisation in $\mathbb N$ it follows that $g$ is an injection. Thus $A[X]$ is countable. $\square$

We can use this to prove in a rather simple manner that

Corollary 1. The set $\mathbb A$ of all algebraic numbers is countable.

Proof. It follows from the above that $\mathbb Z[X]$ is countable. Let $\alpha\in\mathbb A$ be a root of some minimal polynomial $f_\alpha\in\mathbb Z[X]$. We can assign to each $\alpha\in\mathbb A$ a unique element $f\in\mathbb Z[X]$ as follows: if $\alpha_1(=\alpha),\dots,\alpha_n$ are the zeros of $f_\alpha\in\mathbb Z[X]$, assign $jf_\alpha$ to $\alpha_j$. This gives an injection from $\mathbb A$ to $\mathbb Z[X]$, as desired. $\square$

More generally,

Corollary 2. A countable union of countable sets is countable.

Proof. Let $A_0,A_1,\dots$ be countable sets. Then there are injections $A_i\to X^i\mathbb Z$ for $i=0,1,\dots$. Hence we have an injection

$\displaystyle\bigcup_{i=0}^\infty A_i\to \bigcup_{i=0}^\infty X^i\mathbb Z\subseteq\mathbb Z[X]$,

showing that $\bigcup_{i=0}^\infty A_i$ is countable. $\square$

Filed under Set theory

## Symmetric polynomials in two variables

Let $R$ be a commutative ring. The fundamental theorem of symmetric polynomials says that any symmetric polynomial in $R[X,Y]$ can be expressed uniquely as a polynomial in $X+Y$ and $XY$.

Recently I was thinking about this along the following lines. Let $R[X,Y]^{S_2}$ denote the set of all symmetric polynomials in $R[X,Y]$. Then the theorem above is saying that $R[X,Y]^{S_2}$ is generated by $\{X+Y,XY\}$ as an $R$algebra, i.e., $R[X,Y]^{S_2}=R[X+Y,XY]$. Being unsuccessful at utilising this I ended up with the following. (I can’t see the best way of showing that a set generates an algebra.)

Let $f(X,Y)=\sum_{i,j}a_{i,j}X^iY^j\in R[X,Y]^{S_2}$. Since $f(X,Y)=f(Y,X)$, we have $a_{i,j}=a_{j,i}$ for all $i,j$. Hence $f(X,Y)=\sum_{i\le j}a_{i,j} (XY)^i(X^{j-i}+Y^{j-i})$. So it suffices to show that $X^n+Y^n$ can be expressed as a polynomial in $X+Y$ and $XY$ for each $n\ge 0$. But this follows easily by induction and the following identity

$X^n+Y^n=(X+Y)(X^{n-1}+Y^{n-1})-XY(X^{n-2}+Y^{n-2})$.

However, this argument doesn’t generalise immediately to more variables, and I don’t particularly like any of the proofs that I’ve found so far.

Filed under Algebra

## 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