# Indian Statistical Institute, ISI MMATH 2021 PMB Solutions & Discussions

## Problem 1.

Let be a real matrix with all diagonal entries equal to and all non-diagonal entries equal to s. Compute the determinant of .

# Solution:

Fix Let be the matrix whose diagonal elements are equal to and non-diagonal elements are equal to Denote Then is a polynomial of degree in We observe that The sum of rows of the matrix is equal to

If then this row is zero, Now we compute the derivative of

By induction, It follows that is a root of . Hence, is a root of order of and is a root of order Finally,

## Problem 2.

Let be the polynomial ring over a field . Prove that the rings and are isomorphic if and only if the characteristic of is .

# Solution:

Assume that the rings and are isomorphic, and let be an isomoprhism. Then

Since is an isomorphism, necessarily or

It follows that

That is,

Then hence and If then and can’t be an isomorphism. So, and we find that is of characteristic

Conversely, assume that is of characteristic Consider the mapping

The mapping is an isomorphism. Indeed,

as

## Problem 3.

Let be a subset of endowed with the subspace topology. If every continuous real-valued function on is bounded, then prove that is compact.

# Solution:

The function is continuous. Hence it is bounded on

The set is bounded.

Let (i.e. is in the closure of ). If then the function

is continuous on For every there exists a point such that Then As is arbitrary, this shows that is unbounded on . The obtained contradiction proves that is bounded and closed, hence compact.

## Problem 4.

Let be a nonzero real matrix such that for . If for some , then prove that . Here is the -th power of .

# Solution:

Let be the first number such that for some (such exists as and for ) Denote We check by induction that for all

For the statement is true by the choice of Assume the statement is proved for Let Then

the sum is taken over non-empty set of indices when i.e. when The statement is proved.

Let be such that For all we have hence We find that

( because it is the off-diagonal element of the identity matrix). Also, for all we have hence and

## Problem 5.

Let be the function given by

Prove that has a local maximum and a local minimum in the interval for any positive integer .

# Solution:

On each interval the function is strictly increasing. Indeed, its derivative is

Let At we have and Then there is a unique such that

Consider the derivative of the function for

Let and take Then For we have

Further,

and

Function attains a local maximum at

For we have

Further,

and

Function attains a local minimum at

On each interval function attains a local maximum and a local minimum.

## Problem 6.

Fix an integer . Suppose that is divisible by distinct natural numbers such that

Pick a random natural number uniformly from the set . Let be the event that is divisible by . Prove that the events are mutually independent.

# Solution:

Let divides There are numbers in that are divisible by So,

Since and are co-prime when a number is divisible by if and only if it is divisible by and by

Similarly, and

The equality

holds for all and all Events are mutually independent.

## Problem 7.

Let be a function. Assume that there exists
such that for all and for all
Show that the set is countable.

# Solution:

Remark: I assume that in the statement of the problem are distinct. Otherwise, must be identically zero.

Denote Then We will show that each set is finite. Indeed, if are distinct, then

So, and the set has at most elements.

## Problem 8.

Let be a group having exactly three subgroups. Prove that is cyclic of order for some prime .

# Solution:

has three subgroups: and some other subgroup . The group has an element such that Consider a cyclic subgroup generated by If this subgroup coincides with , then is cyclic. Otherwise, There is an element Let be a cyclic subgroup generated by Since and then and is cyclic. The subgroup has no proper subgroups, so its order is prime: Correspondingly, for some integer By the fundamental theorem of cyclic groups, if divides then has a subgroup of order It follows that the only possible divisor of is .