Suppose is a column vector and is an invertible matrix. Set , then

This formula has more general forms.

when time goes to infinity

Suppose is a column vector and is an invertible matrix. Set , then

This formula has more general forms.

An ideal of an integral domain is a free -module if and only if it is generated by one element. Indeed, given any two non-zero elements , . So any two elements are linear dependent. Hence, if were free module, it must generated by one element, or is a principle ideal. Our strategy is to find an ideal which is not principle and show that is projective.

Let . Obviously is an integral domain. Let be the ideal generated by 5 and . .

is not a principal ideal.

Define by , then it is straightforward to verify , for all . If is a unit, then , which means and . So the only units in are .

Now suppose is a principle ideal, then such that . Then such that , which means . Since has no solution in , or can not be 5. So we must have and or and . The second case is impossible, because means . Then . We can obtain , this is impossible.

The only case we left is , which means . Then . We have , corresponds to the ideal . If , then . This fact leads to , for some , . If we set , then in , this is impossible.

In conclusion, can not be principle.

is a projective module.

Define by . If we can find such that , then is a direct summand of , which is a free -module. Consequently, is a projective module.

Let , . There exist unique such that

A straightforward calculation shows that, , . Define . It is easy to verify that is an -module homomorphism. Moreover,

This means . Thus is a projective module.

As discussed in the beginning of this section, since is not principle, we get a non-free projective module.

Projective modules were introduced by Henri Cartan and Samuel Eilenberg in 1956.

In general one can define a -module to be projective by any one of the following:

(1) Any short exact sequence splits;

(2) is a direct summand of a free module, which such that is a free -module;

(3) Given an epimorphism and , then such that , that is

(4) hom is an exact functor.

These four definitions show that projective module has very rich properties and is related to other types of modules. (2) means projective module is a generalization of free modules. (3) is called the lifting property of projective module, it is also a generalization of free module, because if is a free module, there exists a unique such that the following diagram commutes

In (3) we don’t require to be unique, so does not have the universal property, consequently not necessarily a free object in category sense. Dual to definition (3), injective module is introduced

1)Given a monomorphism and a module homomorphism , then such that the following diagram commutes

2) hom is an exact (contravariant) functor;

3) Any short exact sequence splits.

Since we know free modules are all projective ones, two natural questions emerge naturally.

(1)whether there is a projective module but not free;

(2)when the projective module is free.

For the first question, there is an easy answer. Let regarded as a module for -module, then is a free module. By the Chinese remainder theorem, . Thus is a direct summand of free -module. is projective from the definition, but not free -module because actually it is a torsion -module. Carefully investigating this example, the ring is decomposable, , then it is easy to find a non-free projective module, in particular . What if we require the is not decomposable, say is an integral domain in particular? Is it still possible to find a non-free projective module? Our main result of this paper is devoted to this purpose. It turns out to be much more difficult to give such examples. Before finding an example, we should have a rough understanding on which kind of ring projective modules are always free, which is related to the answer of question (2).

The study of (2) can be summarized as following:

If is a PID (the commutative principal ideal domains), all projective modules are free.

This is because any submodule of free module over is free. Projective is direct summand of free module, hence a submodule of free module.

is a local ring, any finitely generated projective module over is free.

This result is proved by Irving Kaplansky in 1958. It is easy to prove this result for finitely generated projective modules, but the general case is difficult.

The following theorem is a well known result of H. Bass.

If is a connected (i.e.without nontrivial idempotents) Noetherian ring. Then any infinitely generated projective -module is free.

The following theorem is known as Serre’s Conjecture, which was solved by Quillen, Daniel and Suslin, Andrei A. independently in 1976.

If , where is a PID, then every projective module over is free.

Note that Bass’s theorem is a special case of Serre’s Conjecture.

Let be a cyclic extension of dimension over and let be a generator of . Let , and suppose is a non-zero element of such that for some . Show that there exists a in the (unique) subfield of of dimensionality such that .

is a cyclic group, then it has a unique subgroup of index . By the Galois corresponding theorem, there exists a unique subfield such that . . has dimensionality over .

Consider , then . We also have

,

by Hilbert’s Satz 90, such that .

Let , then , because

Surprisingly we have

.

Jacabson p300. This problem puzzled me for three weeks. Finally it turns out to be very easy.

Suppose is finite field with elements, an Galois extension of such that . Prove that is surjective.

From theorem 4.26 we know that is cyclic over generated by .

.

is a homomorphism. It is well known that any finite subgroup of the multiplicative group of field is cyclic, which means and is cyclic.

Since and , and , where , then and . Thus is surjective.

Jacobson p300.

Assume has distinct pth roots of 1. a prime, and is cyclic of dimensional . Let be a primitive th root of 1. Show that if can be imbedded in a cyclic field of dimension , then for some .

Suppose is the generating isomorphism of cyclic Galois group . Then . , so by Hilbert satz 90, there exists such that .

.

So , then

.

Jacobson p300.

Let be an extension of such that where is transcedental over and satisfies the equation over . Find the Galois group of over for any Show that

is contained in . Use this to prove that the function cos is expressible rationally with complex coefficients in terms of and . Does this hold for sin .

, we must have

(1)

(2)

(3)

This means . So there exist , a n-th root of 1, such that . Similarly such that . (3) becomes

. Combing with , we get . Since , we know that when is even and when is odd and .

In conclusion, the Galois group is trivial when is odd and Klein group when is even.

No matter is even and odd, is invariant, so is in .

If is odd, is in . If is even, is not in the ground field, because can not fix .

At first I am surprised by the result I got. I can not believe that the galois is trivial when is odd. So I conculted to Zhuohui. Thank him for his kind help. I drew the following picture after discussion with him.

Use the fact that any finite group is isomorphic to a subgroup of to prove that given any finite group there exists fields and such that

Any field , are indeterminate. Let . is the field of symmetric rational functions in Or if we let are basic symmetric polynomials, namely

, , ,

then .

As we all know, . Since , by the Galois corresponding theorem, there exists an intermediate field such that .

This is the easiest inverse Galois problem since one does not specify the ground field. If you restrict , things are much trickier.

is the group of transformations of of the form , .

is all the translations .

and , .

is a group of transformation in containing the group of translations as normal subgroup. Show that is a subgroup of , this means is the normalizer of .

Suppose and let , has the form . Hence . If , then .

Any solvable transitive subgroup of , a prime, is equivalent to a subgroup of containing the group of translations.

Suppose is this group. Then has a composition series

whose factors are all cyclic groups of prime order.

By lemma 4, we know that is also a solvable transitive subgroup of . Since is cyclic of prime order, then it must be generated by some p-cycle, otherwise it can not be transitive.

Apply some conjugation to in , we can assume is generated by . So is a translation group. We can identify the element of as the transformation of . By lemma 2, we know that is equivalent to some subgroup of .

Let be a normal nontrivial subgroup of a transitive subgroup of of transformations of . Show that all orbits have the same cardinality. Hence show that if is a prime, then is transitive.

Let be irreducible of prime degree over of characteristic 0, a splitting field over of . Show that is solvable by radicals over if and only if for any two roots of .

Since has characteristic 0, is solvable by radicals if and only if is a solvable subgroup of symmetric group . is irreducible if and only if is transitive.

(1) if is solvable by radicals, then is a solvable transitive subgroup of .

By lemma 3, we know is equivalent to a subgroup of containing . WLOG, assume . The action of on roots looks like , .

Consider , . , unique such that and . Then . Since , we get and , which means . So .

(2) If , then . This Galois group contains a normal Sylow p-subgorup which is isomorphic to . By lemma 2, we know is equivalent to a subgroup of which is solvable. So is solvable, is solvable by radicals.

Jacobson p262.

Let be a perfect field and an algebraic field extension, such that every non-constant polynomial has a root in . Show that is algebraically closed.

Suppose is a non-constant polynomial. Let be a splitting field of .

Since is perfect, is a separable extension of . Primitive element theorem implies that any finite separable extension is simple extension. That is such that . Suppose has a minimal polynomial over . By assumption, has a root . Since there exists an isomorphism by , then contains a splitting field of over .

, adjoin a root of , we get . Since and are algebraic. is also algebraic. Hence, is a root of some non-constant polynomial . By the previous proof, splits over . We conclude that and .