But what if doesn’t have The following problem answers this question.

**Problem**. Let be a ring, which may or may not have Show that if has no proper left ideals, then either is a division ring or and for some prime number

**Solution**. Let

Then is a left ideal of because it’s clearly a subgroup of and, for and we have and so i.e. So either or

Case 1: That means for all or, equivalently, Thus every subgroup of is a left (in fact, two-sided) ideal of Hence has no proper subgroup (because has no proper left ideals) and thus for some prime

Case 2: Choose So and hence because is clearly a left ideal of Thus there exists such that Now

the left-annihilator of in , is obviously a left ideal of and we can’t have because then So Since

we have Thus Let

Clearly is a left ideal of and Thus So for all Now let be any element of Then, by what we just proved, On the other hand, by the same argument we used for we find such that Thus i.e.

So and hence i.e. and thus

So and hence for all Thus proving that is a division ring.

**Remark**. The same result given in the above problem holds if has no proper right ideals.

**Example**. Let be a prime number. The ring

is not a division ring and it has no proper left (or right) ideals.

]]>In other words, is -abelian if the map defined by is a group homomorphism.

**Definition 2**. If is both -abelian and -abelian, for some integers then is also -abelian because then for we will have

So the set

i.e., the set of those integers for which is -abelian, is a multiplicative subset of Clearly The set is called the **exponent semigroup** of

**Remark 1**. Since we have if and only if So a group is -abelian if and only if or, equivalently, for all So a group is -abelian if and only if it is -abelian. In other words, if and only if

**Example 1**. Every abelian group is obviously -abelian for all It is also clear that -abelian groups are abelian because gives As the next two examples show, there exists a non-abelian -abelian group for any

**Example 2**. Let be an odd integer and consider the Heisenberg group which is a non-abelian group (why?). We show that i.e. is both -abelian and -abelian. To see that, let

An easy induction shows that

for all integers Thus the identity matrix, and so for all

Hence and for all proving that is both -abelian and -abelian.

**Remark 2**. Notice that, in Example 2, if is even, then (as elements of and so is not always the identity element in this case. However, there’s a way to fix this, as the next example shows. But first, notice that can also be viewed as the group of triples

with multiplication defined by

In the next example, we modify the above multiplication such that we get for all

**Example 3**. Let be any integer and consider the set Define multiplication in by

It’s easy to see that is a non-abelian group. We show that Let A quick induction shows that

for all integers So for all and thus for all proving that is -abelian.

Also, since for all we have and so is -abelian.

**Problem 1**. Let be a group. Show that

i) if for some integer then for all

ii) if for some integer then is abelian.

**Solution**. i) Let We have

and so

ii) Let By i), we have and Thus

and so

**Problem 2**. Let be an -abelian group and let Show that

i)

ii)

**Solution**. i) By Remark 1, for all Thus

and the result follows.

ii) Again, using the Remark 1, we have

**Remark 3**. Let be an -abelian group for some integer An immediate result of Problem 2, ii), is that if either is torsion-free or is finite and then is abelian.

In this post, we showed that the ring has idempotents, where is the number of prime divisors of Clearly is also the number of prime (= maximal) ideals of (recall that, in general, in commutative Artinian rings, prime ideals are maximal). Let be the Euler’s totient function. The number of units of is If is the prime factorization of then If is odd, then is even for all and hence divides So if is odd, then the number of idempotents of divides the number of units of As we are going to show now, this is a property of any finite commutative ring of odd order.

Recall that a commutative ring is called **semilocal** if the number of maximal ideals of is finite and it is called **local** if it has only one maximal ideal.

**Example 1**. If is a prime and is an integer, then is a local ring with the unique maximal ideal If is an integer, then is a semilocal ring (what are its maximal ideals?).

**Example 2**. Generalizing the above example, Artinian rings are semilocal. This is easy to see; let be an Artinian ring and let be the set of all finite intersections of maximal ideals of Since is Artinian, has a minimal element Let be any maximal ideal of Since and is a minimal element of we must have So for some and hence Therefore are all the maximal ideals of

**Problem 1**. Show that if is a local ring, then are the only idempotents of

**Solution**. Let be the maximal ideal of and let be an idempotent of Then Thus either or If then because otherwise which is false. So is a unit because there’s no other maximal ideal to contain So for some and thus Similarly, if then and thus is a unit. So for some implying that and hence

**Problem 2 **Show that the number of idempotents of a commutative Arinian ring is where is the number of maximal ideals of

**Solution**. Since is Artinian, it has only finitely many maximal ideals, say (see Example 2). The Jacobson radical of is nilpotent, hence there exists an integer such that

Thus, by the Chinese remainder theorem for commutative rings, Since each is a local ring, with the unique maximal ideal it has only two idempotents, by Problem 1, and so has exactly idempotents.

**Problem 3**. Let be a finite commutative ring. Show that if the number of elements of is odd, then the number of idempotents of divides the number of units of

**Solution**. Since is finite, it is Artinian. Let be the set of maximal ideals of By problem 2, the umber of idempotents of is and for some integer

Since is odd, each is odd too because is a subgroup of and so divides Also, units in a local ring are exactly those elements of the ring which are not in the maximal ideal. So the number of units of each is which is an even number because both and are odd. So the number of units of which is the product of the number of units of is divisible by which is the number of idempotents of

**Remark 1**. The result given in Problem 3 is not always true if the number of elements of the ring is even. For example, has one unit and two idempotents. However, the result is true in which has units and two idempotents.

Can we find all even integers for which the result given in Problem 3 is true in ? Probably not because this question is equivalent to finding all integers such that where is the number of prime divisors of and that is not an easy thing to do.

**Remark 2**. The result given in Problem 3 is not necessarily true in noncommutative rings with an odd number of elements. For example, consider the ring of matrices with entries from the field of order three. Then has units (see Problem 3 in this post!) but, according to my calculations, has idempotents and does not divide

**Solution**. Suppose first that is a division ring. Then, since is a finite ring, is a finite field, by the Wedderburn’s little theorem. Let Then for all and so has no solution in

Conversely, suppose that is not a division ring (equivalently, a field because is finite). So and hence the equation has solutions in (just choose and ).

Let be the Jacobson radical of Since is finite, it is Artinian and so is nilpotent.

So if then there exists such that and so for all Therefore the equation has a solution in

If then by the Artin-Wedderburn’s theorem,

for some finite fields Since is not a field, we have either for some or for all and If for some then and hence will have a non-zero nilpotent element and we are done. If for all and then

will satisfy

]]>**Solution**. The proof is by induction over There’s nothing to prove for For since we have

and thus, by the induction hypothesis, for all So for some

If then and so Hence, by the induction hypothesis, for all

If then (so because ). Thus

and so, by the induction hypothesis, for all

If then gives and so implying, again by the induction hypothesis,that for all

**Example**. Show that for all integers

**Solution**. Let where for all Since and for all we have, by the above problem, for all Thus and we are done because obviously

It is known that is invertible and if then We are going to use these two properties of Hilbert matrices to solve the following calculus problem.

**Problem**. Let be an integer and let be a continuous function. Suppose that for all Show that

**Solution**. Since the Hilbert matrix, is invertible, there exist real numbers such that

So the polynomial satisfies the conditions

Clearly is the sum of all the entries of and so Now let be a real-valued continuous function on such that

Let be the above polynomial.Then since

integrating gives

]]>**Solution**. Since is orthogonal, its eigenvalues have absolute value and it can be be diagonalized. Let be a diagonal matrix such that for some invertible matrix Then

We claim that the eigenvalues of are for some Well, the characteristic polynomial of has degree three and so it has either three real roots or only one real root. Also, the complex conjugate of a root of a polynomial with real coefficients is also a root. So, since the eigenvalues of are either all which is the case or two of them are and one is which is the case or one is and the other two are in the form for some So

Note that given the matrix

is orthogonal, and

]]>**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 I’m not going to prove 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.

**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

]]>**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.

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