Happy day!

## Rings satisfying x^n = x are commutative

Posted: March 14, 2018 in Division Rings, Noncommutative Ring Theory Notes, Primitive RingsTags: central, division ring, finite field, Jacobson theorem, Kaplansky theorem, polynomial identity ring, primitive ring, reduced ring, structure theorem, Wedderburn's little theorem, x^n=x

Throughout this post, is a ring with

**Theorem** (Jacobson). If for some integer and all then is commutative.

In fact in Jacobson’s theorem, doesn’t have to be fixed and could depend on i.e. Jacobson’s theorem states that if for every there exists an integer such that then is commutative. But we are not going to discuss that here.

In this post, we’re going to prove Jacobson’s theorem. Note that we have already proved the theorem for (see here and here) and we didn’t need to have we didn’t need that much ring theory either. But to prove the theorem for any we need a little bit more ring theory.

**Lemma**. If Jacobson’s theorem holds for division rings, then it holds for all rings with

**Proof**. Let be a ring with such that for some integer and all Then clearly is reduced, i.e. has no non-zero nilpotent element. Let be the set of minimal prime ideals of

By the structure theorem for reduced rings, is a subring of the ring where is a domain. Clearly for all and all But then, since each is a domain, we get or i.e. each is a division ring. Therefore, by our hypothesis, each is commutative and hence which is a subring of is commutative too.

**Example**. Show that if for all then is commutative.

**Solution**. By the lemma, we may assume that is a division ring.

Then gives or Suppose that is not commutative and choose a non-central element Then are also non-central and so which gives contradiction!

**Remark 1**. Let be a division ring with the center If there exist an integer and such that for all then is a finite field. This is obvious because the polynomial has only a finite number of roots in and we have assumed that every element of is a root of that polynomial.

**Remark 2**. Let be a domain and suppose that is algebraic over some central subfield Then is a division ring and if then is a finite dimensional division -algebra.

**Proof**. Let So for some integer and We may assume that Then and so is invertible, i.e. is a division ring.

Since is a subring of it is a domain and algebraic over and so it is a division ring by what we just proved. Also, since for some integer we have and so

**Proof of the Theorem**. By the above lemma, we may assume that is a division ring.

Let be the center of By Remark 1, is finite. Since is a division ring, it is left primitive. Since every element of is a root of the non-zero polynomial is a polynomial identity ring.

Hence, by the Kaplansky-Amtsur theorem, and so is finite because is finite. Thus, by the Wedderburn’s little theorem, is a field.

## Limit of a sequence of 2 by 2 matrices

Posted: January 29, 2018 in Linear AlgebraTags: characteristic polynomial, diagonalizable

The following problem is from the American Mathematical Monthly. The problem only asks the reader to calculate it doesn’t give the answer; I added the answer myself.

**Problem **(Furdui, Romania). Let be real numbers with For an integer let

Let be the sign function. Show that

**Solution**. The characteristic polynomial of is

And roots of are

where

If is sufficiently large, which is the case we are interested in, then, since are either both positive or both negative (because ), and so So, in this case, are distinct real numbers and hence is diagonalizable in

Now is an eigenvector corresponding to if and only if if and only if Similarly, is an eigenvector corresponding to if and only if if and only if So if

then and hence

The rest of the solution is just Calculus and if you have trouble finding limits, see this post in my Calculus blog for details. We have

## Rank of Hermitian matrices

Posted: October 2, 2012 in Elementary Algebra; Problems & Solutions, Linear AlgebraTags: Hermitian matrix, rank, trace of a matrix

For let denote the complex conjugate of Recall that a matrix is called Hermitian if for all It is known that if is Hermitian, then is diagonalizable and every eigenvalue of is a real number. In this post, we will give a lower bound for the rank of a Hermitian matrix. To find the lower bound, we first need an easy inequality.

**Problem 1**. Prove that if then

**Solution**. We have for all and so

Adding the term to both sides of the above inequality will finish the job.

**Problem 2**. Prove that if is Hermitian, then

**Solution**. Let be the nonzero eigenvalues of Since is diagonalizable, we have We also have and Thus, by Problem 1,

and the result follows.

## Rings with an odd number of units

Posted: September 26, 2012 in Elementary Algebra; Problems & Solutions, Rings and ModulesTags: an odd number of units, Artin-Wedderburn theorem, Jacobson radical, rings, the group of units of a ring, Wedderburn's little theorem

Throughout this post, and are the group of units and the Jacobson radical of a ring Assuming that is finite and is odd, we will show that for some positive integers Let’s start with a nice little problem.

**Problem 1**. Prove that if is finite, then is finite too and

**Solution**. Let and define the map by This map is clearly a well-defined group homomorphism. To prove that is surjective, suppose that Then for some and hence implying that So is surjective and thus

Now, is a subgroup of and Thus is finite and

**Problem 2**. Let be a prime number and suppose that is finite and Prove that if then

**Solution**. Suppose that and Then, considering as an additive group, is a subgroup of and so But then by Problem 1, and that’s a contradiction!

There is also a direct, and maybe easier, way to solve Problem 2: suppose that there exists On define the relation as follows: if and only if for some integer Then is an equivalence relation and the equivalence class of is Note that because and So if is the number of equivalence classes, then contradiction!

**Problem 3**. Prove that if is a finite field, then In particular, if is odd, then and is a power of

**Solution**. The group is isomorphic to the group of invertible linear maps Also, there is a one-to-one correspondence between the set of invertible linear maps and the set of (ordered) bases of So is equal to the number of bases of Now, to construct a basis for we choose any non-zero element There are different ways to choose Now, to choose we need to make sure that are not linearly dependent, i.e. So there are possible ways to choose Again, we need to choose somehow that are not linearly dependent, i.e. So there are possible ways to choose If we continue this process, we will get the formula given in the problem.

**Problem 4**. Suppose that is finite and is odd. Prove that for some positive integers

**Solution**. If in then would be a subgroup of order 2 in and this is not possible because is odd. So Hence and Let be the ring generated by and Obviously is finite, and We also have by Problem 2. So is a finite semisimple ring and hence for some positive integers and some finite fields by the Artin-Wedderburn theorem and Wedderburn’s little theorem. Therefore The result now follows from the second part of Problem 3.

## Maximal and prime ideals of a polynomial ring over a PID (2)

Posted: September 21, 2012 in Elementary Algebra; Problems & Solutions, Rings and ModulesTags: irreducible modulo a prime, maximal ideals, PID with infinitely many primes, polynomial ring over PID, prime ideals

See part (1) here! Again, we will assume that is a PID and is a varibale over In this post, we will take a look at the maximal ideals of Let be a maximal ideal of By Problem 2, if then for some prime and some which is irreducible modulo If then for some irreducible element Before investigating maximal ideals of in more details, let’s give an example of a PID which is not a field but has a maximal ideal which is principal. We will see in Problem 3 that this situation may happen only when the number of prime elements of is finite.

**Example 1**. Let be a filed and put the formal power series in the variable over Let be a variable over Then is a maximal ideal of

*Proof*. See that and that is the field of fractions of Thus is a field and so is a maximal ideal of

**Problem 3**. Prove that if has infinitely many prime elements, then an ideal of is maximal if and only if for some prime and some which is irreducible modulo

**Solution**. We have already proved one direction of the problem in Problem 1. For the other direction, let be a maximal ideal of By the first case in the solution of Problem 2 and the second part of Problem 1, we only need to show that So suppose to the contrary that Then, by the second case in the solution of Problem 2, for some We also know that is a field because is a maximal ideal of Since has infinitely many prime elements, we can choose a prime such that does not divide the leading coefficient of Now, consider the natural ring homomorphism Since and so is invertible in Therefore for some Hence for some If then we will have which is non-sense. So for some where does not divide the leading coefficient of Now gives us and so the leading coefficient of is divisible by Hence the leading coefficient of must be divisible by contradiction!

**Example 2. **The ring of integers is a PID and it has infinitely many prime elements. So, by Problem 3, an ideal of is maximal if and only if for some prime and some which is irreducible modulo By Problem 2, the prime ideals of are the union of the following sets:

1) all maximal ideals

2) all ideals of the form where is a prime

3) all ideals of the form where is irreducible in

## Maximal and prime ideals of a polynomial ring over a PID (1)

Posted: September 21, 2012 in Elementary Algebra; Problems & Solutions, Rings and ModulesTags: Gauss's lemma, irreducible polynomial modulo p, maximal ideals of R[x], PID, prime ideals of R[x]

We know that if is a field and if is a variable over then is a PID and a non-zero ideal of is maximal if and only if is prime if and only if is generated by an irreducible element of If is a PID which is not a field, then could have prime ideals which are not maximal. For example, in the ideal is prime but not maximal. In this two-part post, we will find prime and maximal ideals of when is a PID.

**Notation**. Throughout this post, is a PID and is the polynomial ring in the variable over Given a prime element we will denote by the natural ring homomorphism

**Definition **Let be a prime element of An element is called **irreducible modulo** if is irreducible in Let be the natural ring homomorphism. Then, since an element is irreducible modulo if and only if is irreducible in Note that is a field because is a PID.

**Problem 1**. Prove that if is prime and if is irreducible modulo then is a maximal ideal of If then is a prime but not a maximal ideal of

**Solution**. Clearly So is a maximal ideal of because is irreducible in and is a field. So is a maximal ideal of If then and so is a domain which implies that is prime. Finally, is not maximal because, for example,

**Problem 2**. Prove that a non-zero ideal of is prime if and only if either for some irreducible element or for some prime and some which is either zero or irreducible modulo

**Solution**. If is irreducible, then is a prime ideal of because is a UFD. If or is irreducible modulo a prime then is a prime ideal of by Problem 1.

Conversely, suppose that is a non-zero prime ideal of We consider two cases.

*Case 1*. : Let Then is clearly not a unit because then wouldn’t be a proper ideal of So, since and is a prime ideal of there exists a prime divisor of such that So and hence is a prime ideal of Thus we have two possibilities. The first possibility is that which gives us and therefore The second possibility is that for some irreducible element which gives us because

*Case 2*. : Let be the field of fractions of and put Then is a non-zero prime ideal of because is a prime ideal of Note that So, since is a PID, for some irreducible element Obviously, we can write where and is irreducible and the gcd of the coefficients of is one. Thus and, since we have for some and But then and so because is prime and Hence We will be done if we prove that To prove this, let So for some Therefore, since the gcd of the coefficients of is one, we must have by Gauss’s lemma. Hence and the solution is complete.

See the next part here!