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

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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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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<p_1<\cdots 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

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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 Ralgebra, 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.

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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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