**Problem**. Let

i) Show that is a vector space but is not.

ii) Show that

**Solution**. Let be the standard basis of

i) Let and Then

So and thus is a vector space. To show that is not a vector space, see that, for example, because but because

ii) Since all the eigenvalues of a nilpotent matrix are zero, the trace of a nilpotent matrix zero and thus implying that because is a vector space.

So, to complete the proof of ii), we need to show that Let Let be an upper triangular element of similar to (such exists because the field of complex numbers is algebraically closed). So

for some invertible element Clearly we can write where is diagonal and is strictly upper triangular. Notice that is nilpotent because all of its eigenvalues are zero (because its eigenvalues are the entries on the main diagonal and those entries are all zero). So

and Thus, in order to complete the solution, we only need to show that Clearly because Let be the -entry of Since we have and thus

So, in order to show that we only need to show that for all and to do that, just write

and see that

So both and are in and therefore by

]]>**Solution**. Suppose first that Since are normal, is a subgroup of and

for all implying that Thus the map

defined by is an onto group homomorphism. Since is also one-to-one, hence an isomorphism.

A counter-example for is the direct product of copies of Clearly every has order two. Now choose distinct elements

(we can do that because ) and let be the subgroup of generated by Since is abelian, each is normal in and clearly for all because are distinct subgroups of order two. Now has elements and so it can’t be isomorphic to a subgroup of

]]>**Problem**. Let be a prime number and Show that if and then

**Solution**. Since the possible eigenvalues of are the roots of the polynomial which are

where the th primitive root of unity. Let be the number of eigenvalues of equal to Then, since we must have

Now consider the polynomials

in We know that is irreducible over and so, since it is the minimal polynomial of Hence divides because we also have by So is a constant multiple of and hence

If then would have at least eigenvalues, which is impossible because

Hence i.e. which means that is not an eigenvalue of Thus all the eigenvalues of are zero and therefore is nilpotent.

Let be the smallest integer such that We have for some integers with and Then, since we get

So, by the minimality of we must have which gives So and thus

]]>Let and suppose that is nilpotent, i.e. for some integer Then But what can we say about Is it equal to Not necessarily, of course! For example, consider

Then but See that in this example,

**Problem**. Let and suppose that is nilpotent. Show that if commute, i.e. then

**Solution**. since is algebraically closed and commute, are simultaneously triangularizable, i.e. there exists an invertible element such that both and are triangular. Since is both nilpotent and triangular, all its diagonal entries are zero and so the diagonal entries of are the same as the diagonal entries of Thus

because are both triangular and the determinant of a triangular matrix is the product of its diagonal entries. So

]]>Show that

**Solution**. Let be a generator of Then

Thus

for some integers and so

Thus because is the order of Hence are coprime. Now consider the map

We show that is a ring isomorphism, which is what we need.

i) is well-defined. Because if for some integer then and so

ii) is additive. That’s clear.

iii) is bijective. Suppose that for some integer Then and so because are coprime, implying that So is injective and hence bijective because and have the same number of elements.

iv) is multiplicative. Because for integers we have

**Example**. Let be a finite ring with and suppose that the number of elements of is square-free. Then That’s because is an abelian group of square-free order and every abelian group of square-free order is cyclic (why?). The result now follows from the above problem.

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

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