Find all irreducible factors of

**Solution**. Let

and see that

Thus

It’s clear that is irreducible over Now, for let be the -th cyclotomic polynomial. Using well-known properties of we have

Thus is irreducible over because cyclotomic polynomials are irreducible over Hence, by has exactly irreducible factors and they are

]]>**Problem**. Let be two idempotent matrices such that is invertible and let Let be the identity matrix. Show that

i) if then is not necessarily invertible

ii) if then is invertible

iii) is invertible

iv) if then is invertible.

**Solution**. i) Choose

See that and is invertible but is not invertible.

ii) It’s clear for For suppose that for some We need to show that Well, we have and thus But since we also have and hence

because is invertible. So and therefore So

iii) Let See that are idempotents and so, since is invertible, is invertible, by ii). The result now follows because

iv) Let and suppose that for some We are done if we show that Well, we have and thus implying that because is invertible. Hence which gives because Thus and therefore

]]>**Solution**. First notice that we are done if is abelian because then every subgroup of would be normal and since is even, has a subgroup of order two, which is clearly neither the trivial subgroup nor

So suppose, to the contrary, that is non-abelian and simple and consider the group homomorphism defined by for all Then since is a normal subgroup of and is not simple, either where is the identity matrix in or But we can’t have because then would be embedded into making abelian. Hence i.e. for all Now let be an element of order two. So which implies that is diagonalizable (because its minimal polynomial divides and so it splits into distinct linear factors). So

for some Since the set of eigenvalues of which is is a subset of Since we must have and so Since has order two, and hence Since is a central element of the subgroup is a normal subgroup of order two in and because So is not simple and that contradicts our assumption that is simple. So our assumption is wrong and is not simple indeed!

]]>i)

ii)

Show that either or

**Solution**. First notice that both and satisfy i), ii). Since is the only ring with two elements, we may assume that Let First we show that is even. Suppose, to the contrary, that is odd. Then the group would have no element of order i.e. for all and thus

contradicting the property i).

Now let We show that for some prime and To see that, let be the prime factorization of If then

contradicting the property ii), because are non-zero distinct elements of So i.e. for some prime number and positive integer Now if then we can write

and again, by the minimal property of the elements and are non-zero and distinct and that contradicts the property ii). So

We now show that in fact We just proved that for some prime and Suppose that is a prime divisor of Then there exists that has order But then which gives the false result because are coprime. So is a power of But we have already showed that is even. So and thus either or If then would be a vector space over i.e. for some integer because and But then

contradicting the property ii). So and hence contains a subring

Finally, we show that Suppose, to the contrary, that and let We have because and so, since we must have either or by property ii). If then which gives by property ii), and that’s a contradiction. So i.e. we have shown that for any element But if then too, because and so which gives contradiction. So

**Solution**. I show that only satisfy the system. First see that from we get

and so, since we have which gives

Now, and together give

and so, by or

We also have from that and so, by

It follows from that and thus by Therefore because and are relatively prime, and so because

]]>Show that and is a power of

**Solution**. First, let’s put an order on the set

We write if or Now let be any non-identity element of Let be the smallest element of such that See that, for any integer the -entry of is and the -entries of where are all zero. But since is finite, there exists an integer such that is the identity matrix and so we must have Thus (because ) and hence So the -entries of where and are all zero.

Now if is not the identity matrix, we can replace with and repeat our argument to find an element such that all -entries of where are zero. Then, again, if is not the identity matrix, we repeat the argument for etc. Since has only finitely many entries, there exists some positive integer such that all the -entries of where are zero. That means is the identity matrix and hence is a power of So we have shown that the order of every non-identity element of is a power of Thus is a power of

But if, in a finite field, for some non-zero element of the field, then we can show that every element of the field is a sum of two cubes (Problem 2).

**Problem 1**. Show that every element of a finite field is a sum of two squares.

**Solution**. Let be a finite field. So we want to show that if then for some We can actually be more specific if we consider two cases. Let

Case 1: for some integer Then, since for all we get So in this case, every element of the field is a square.

Case 2: for some integer Since is finite, the multiplicative group is cyclic.

So Let and consider the sets

Clearly and Thus and hence

Therefore i.e. there exist such that and the result follows.

**Remark 1**. Regarding the second case in the solution of Problem 1, notice that, in fact, we have

and so The reason is that if for some integers then and hence must be divisible by implying that is even.

**Problem 2**. Let be a finite field and suppose that there exists such that Show that every element of is a sum of two cubes.

**Solution**. So we want to show that if then for some Let and let’s consider three cases.

Case 1: for some integer Then for all

Case 2: for some integer Then for all and clearly

So, in both cases 1 and 2, for every there exists such that

Case 3: for some integer Since is finite, the multiplicative group is cyclic. So Let and consider the sets

Clearly and So and

So at least two of the sets have non-empty intersection. If or then for some and we are done.

Now suppose that So there exist such that and so Since, as given in the problem, for some we have and Hence

**Remark 2**. Regarding the third case in the solution of Problem 2, notice that, in fact, we have

and so The reason is that if for some integers then and hence must be divisible by implying that is divisible by

]]>**Solution**. If or then gives or and we are done. Otherwise, there exist and But then contradiction!

So, as a result, if is a finite group and are two subgroups of with and then That raises this question: how large could get? The following problem answers this question.

**Problem 2**. Let be a finite group and suppose that are two subgroups of such that and Show that

**Solution**. Recall that and thus Hence and so

where and

Now, since and we have and i.e. and So if we let and then and thus

The result now follows from

**Example 1**. The upper bound in Problem 2 cannot be improved, i.e. there exists a group and subgroups of such that An example is the Klein-four group and the subgroups and Then and

**Example 2**. We showed in Problem 1 that a group can never be equal to the union of two of its proper subgroups. But there are groups that are equal to the union of three of their proper subgroups. The smallest example, again, is the Klein-four group

Let be a group and let If are conjugate in i.e. for some then clearly if and only if So have the same order. The converse however is false, i.e. if have the same order, that does not imply are conjugate. For example, in an abelian group, two elements are conjugate if and only if they are equal but you can obviously have distinct elements of the same order in the group, e.g. in both non-zero elements have the same order

We are now going to show that although two elements of the same order of a group might not be conjugate in the group, but they are certainly conjugate in some larger group.

**Problem**. Let be a group and suppose that have the same order. Show that there exists a group such that are conjugate in

**Solution**. By Cayley’s theorem, we can embed into the symmetric group using the injective group homomorphism defined by where is the permutation defined by for all So we only need to show that are conjugate in Well, let Then the cycle decomposition of are in the form

and

So have the same cycle type and hence they are conjugate in

]]>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.

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