**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.

]]>**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.

**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

**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!

]]>**Problem**. Let be any map which satisfies the following two conditions: and for all Let be a set with Prove that if is an abelian subgroup of then

**Solution**. The proof is by induction on If then Let be a set with and suppose that the claim is true for any set of size Let be an abelian subgroup of Clearly defines an action of on Let be the orbits corresponding to this action and consider two cases.

*Case 1*. : Fix an element Then Suppose that for some and let Then for some Thus, since is abelian, we have

Hence for all and thus So the stabilizer of is trivial and therefore, by the orbit-stabilizer theorem,

*Case 2*. : Let Clearly and, since we have for all For every and let the restriction of to and put

Then and is an abelian subgroup of Thus, by the induction hypothesis

for all Now, define by for all It is obvious that is one-to-one and so

**Remark**. The map defined by for all satisfies both conditions in the above Problem. So if and if is an abelian subgroup of then

**Problem**. Prove that if for all then is commutative.

**Solution**. Clearly is reduced, i.e. has no nonzero nilpotent element. Note that for all because Hence is an idempotent for every because

Thus is central for all by Remark 3 in this post. Therefore is central for all But

and hence is central. Therefore which gives us

]]>**Theorem**.

*Proof*. Suppose first that is finitely generated and let be a frame of Let Since we have

Let Clearly is a frame of and

for all because every element of commutes with every element of Therefore, since and for all we have and Thus and hence

Therefore by and we are done.

For the general case, let be any finitely generated – subalgebra of Then, by what we just proved,

and hence Now, let be a -subalgebra of generated by a finite set Let be the -subalgebra of generated by all the coefficients of Then and so

Thus

and the proof is complete.

**Corollary**. for all In particular,

*Proof*. It follows from the theorem and the fact that

**Theorem (**Borho and Kraft**, **1976) Let be a finitely generated -algebra which is a domain of finite GK dimension. Let be a -subalgebra of and suppose that Let Then is an Ore subset of and Also, is finite dimensional as a (left or right) vector space over

*Proof*. First note that, by the corollary in this post, is an Ore domain and hence both and exist and they are division algebras. Now, suppose, to the contrary, that is not (left) Ore. Then there exist and such that This implies that the sum is direct for any integer Let be a frame of a finitely generated subalgebra of Let and suppose that is the subalgebra of which is generated by For any positive integer we have

and thus because the sum is direct. So and hence Taking supremum of both sides over all finitely generated subalgebras of will give us the contradiction A similar argument shows that is right Ore. So we have proved that is an Ore subset of Before we show that we will prove that is finite dimensional as a (left) vector space over So let be a frame of For any positive ineteger let Clearly for all and

because So we have two possibilities: either for some or the sequence is strictly increasing. If then we are done because is finite dimensional over and hence is finite dimensional over Now suppose that the sequence is strictly increasing. Then because Fix an integer and let be a -basis for Clearly we may assume that for all Let be a frame of a finitely generated subalgebra of Then

which gives us

because the sum is direct. Therefore which is a contradiction. So we have proved that the second possibility is in fact impossible and hence is finite dimensional over Finally, since, as we just proved, the domain is algebraic over and thus it is a division algebra. Hence because and is the smallest division algebra containing

]]>**Problem 1**. Let be an -module and suppose that are submodules of Prove that if and only if for all

**Solution**. We only need to solve the problem for If then and because both and contain Conversely, let be a nonzero submodule of Then because and therefore because

**Problem 2**. Prove that if is an -module, then is a submodule of and is a proper two-sided ideal of In particular, if is a simple ring, then

**Solution**. First note that because Now suppose that Then by Problem 1. Therefore and hence Now let and We need to show that Let be a nonzero left ideal of Then is also a left ideal of If then and thus If then because So there exists such that and Hence So and thus is a submodule of Now, considering as a left -module, is a left ideal of by what we have just proved. To see why is a right ideal, let and Then and so i.e. Finally, is proper because and so

**Problem 3**. Prove that if are -modules, then Conclude that if is a semisimple ring, then

**Solution**. The first part is a trivial result of Problem 1 and this fact that if where the sum is direct, then The second now follows trivially from the first part, Problem 2 and the Wedderburn-Artin theorem.

**Problem 4. **Suppose that is commutative and let be the nilradical of Prove that

1)

2) it is possible to have

3) if then as -modules or -modules.

**Solution**. 1) Let Then for some integer Now suppose that Then Let be the smallest integer such that Then and hence

2) Let and put For every let and consider It is easy to see that

3) Let Then and thus there exists such that and Hence and so Thus implying that is an essential -submodule of Now, we view as a ring and we want to prove that as an essential ideal of Again, let Then and thus there exists such that and Let Then and thus implying that is an essential ideal of

]]>**Definition 1**. Let be an -module and a nonzero submodule of We say that is an **essential** submodule of and we will write if for any nonzero submodule of Clearly, that is equivalent to saying for any nonzero element So, in particular, a nonzero left ideal of is an essential left ideal of if for any nonzero left ideal of which is equivalent to the condition for any nonzero element

**Definition 2**. Let be an -module and Recall that the (left) annihilator of in is defined by which is obviously a left ideal of Now, consider the set It is easy to see that is a submodule of (see Problem 2 in this post for the proof!) and we will call it the** singular submodule** of If then is called **singular**. If then is called nonsingular. We will not discuss nonsingular modules in this post.

**Problem 1**. Prove that if is a free -module, then i.e. a free module is never singular.

**Solution**. Let be any element of an -basis of Let Then and so Thus and so

Next problem characterizes singular modules.

**Problem 2**. Prove that an -module is singular if and only if for some -module and some submodule

**Solution**. Suppose first that where is an -module and Let and let be a nonzero left ideal of If then and so If then because So there exists such that That means So we have proved that and hence i.e. is singular. Conversely, suppose that is singular. We know that every -module is the homomorphic image of some free -module. So there exists a free -module and a submodule of such that So we only need to show that Note that by Problem 1. Let be an -basis for and suppose that We need to show that there exists such that We can write, after renaming the indices if necessarily,

where For any let Now, since is singular, and so So there exists such that and Hence gives us

Note that because Now, if for all then and we are done. Otherwise, after renaming the indices in the sum on the right hand side of if necessary, we may assume that Repeating the above process gives us some such that and Then implies

The first two terms on the right hand side of are in and because If we continue this process, we will eventually have a positive integer and such that

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