Qual 2011 #1 (Hungerford 4.5.9a): Let I be a two-sided ideal in a ring R and let IM be the abelian subgroup of an R-module M generated by all elements of the form .
(i) Show that IM is a left submodule of M
(ii) Describe the natural left R-module structure on
(iii) Show that as left R-modules.
Proof- i) We are given that by definition. Let . Then . Since I is an ideal, so that . Thus IM is a left submodule of M.
ii) The natural left R-module structure on is given by .
iii) Consider the map by Then we can check that is an R-module homomorphism with . Thus by the first isomorphism theorem, there exists a unique . Then consider the map by . Then it can be checked that is a middle linear map and well defined. Therefore it induces the unique map . Lastly it can be easily checked that and that
Facts and counter examples to know:
1) Free module implies projective module.
2) Projective does not imply free. is a projective module but not a free module.
3) Torsion-free does not imply projective. is torsion free as a module but not projective as a module.
4) Projective does not imply torsion free. is free over itself but not torsion free since but .
5) Torsion free does not imply free. is torsion free as a module but is not free.
6)Free does not imply torsion free. See number 4.
is not a projective module
Proof- Since is a module there exists a free module A and an epimorphism Say that A has basis X. Consider also the identity map . Suppose that the rationals are projective, then by definition there exists a map sucht that . Then with and . Since this is a finite sum, there exists some nonzero integer N that does not divide some , say . But we have that which gives that . But since X is linearly independent we have that N must divide , which is a contradiction. Thus is not projective.
Final Exam #9: Let K be a field. We say that a K[x]-Module M is nilpotent if for some . Describe isomorphism classes of all nilpotent K[x]-modules M such that .
Proof- Recall that if F is a field then F[x] is a PID (in fact Euclidean). Since M is nilpotent, M is a torsion module. If then M is a finitely generated torsion module. So we may apply the structure theorem of modules over a PID to classify all such modules. M must be the direct sum of a finite number of cyclic torsion modules. A cyclic torsion module is isomorphic to where . So a cyclic torsion module is of the form with . The element a of M then has order f. Note the order of any element is of the form with because M is nilpotent with . Therefore f must be of this form. So we can say that any cyclic torsion module with . Then by the structure theorem M is a direct sum of a finite number of these with the property that orders of the summands, satisfy . Since the highest order is we can have the following for M:
Note the powers are all partitions of 5.
This solution needs a ton of explanation.
1993 Qual #4: A submodule of a finitely generated module which is not finitely generated.
Proof- Let be a polynomial ring in infinite variables. Then R is a finitely generated module over itself because . Recall that is the set of all linear combinations of where and . Any polynomial in R, can be written as itself with 1 multiplied in front of each term and thus satisfies the condition to be finitely generated by . Now the ideal is not equal to R and is not finitely generated. Suppose that the set I was finitely generated by some set A. Then A is finite so every element of A must be a finite sum of terms in finitely many factors. Then there exists an index k such that does not appear as a factor of any terms in A. Also, A does not contain any elements with nonzero constant term. Therefore any element of which has a term with as a factor necessarily has a term with degree at least 2. Thus which is a contradiction since . Thus I is not finitely generated.
2009 Qual #4: Let and where
Find the elementary divisors and invariant factors of A.
Proof- Since the real numbers are a field, R is a pid. Thus A is a finitely generated module over a PID. So by the structure theorem we know that each summand is cyclic and that . Thus . The elementary divisors are the prime power factors of the generators of .
To find the invariant factors we draw the power grid as follows. There are 3 distinct factors so there are 3 columns. The factor appears the most, 3 times, so there are 3 rows. Then we arrange the factors in the following way and multiply across to get the polynomials .
2003 Qual #6: Let K be a field and let M, N be finitely generated modules over the polynomial ring . Suppose M has invariant factors: and that N has invariant factors: .
a) Give the elementary divisors of
List all the prime powers of M and N:
b) Give all the invariant factors of
2011 Qual #4 (Greeinstein): Let K be a field. We say that a module M is nilpotent if for every non-unit we have for . Prove that a finitely generated nilpotent indecomposable K[x] module is isomorphic to for some .
Proof- Since K is a field we have that is a PID. So Then M is a finitely generated nilpotent (so torsion) module over a PID. Then if is irreducible we have that x is prime. In particular every element of M has order for some So by the structure theorem we have But M is indecomposible so .
2010 Qual #4: Let R be the polynomial ring . Let I be the ideal generated by . Let . Prove that M has a proper nonzero submodule and that M cannot be written as a direct sum of proper nonzero submodules. Suppose now thtat we take J to tbe the ideal generated by . Prove that is isomorphic to the sum of two proper nonzero submodules.
Proof- First consider . A nonzero submodule of M is . Since is the only elementary divisor it can’t factor any further. Also, . Needs a ton of explanation. A submodule of M should be of the form R/(something), how is A even a submodule?
2013 Qual #6: Let K be a field. Describe all, up to an isomorphism, nilpotent module structures on (that is, all elements of act nilpotently) and .
Proof- K is a field so that K[x] is a PID. So by the classification theorem, any nilpotent K[x] module which is finitely generated, M, has the structure for . Therefore as K[x] modules and this isomorphism is the only such nilpotent K[x]-module structure of K^5$. (WHY?)
Next we consider . First note that this is a K[x] module, has basis (WHY?). So any endomoprhism is determined by where is mapped. In paricular as K[x] modules and . Thus by the classification theorem.
Needs a lot of explanation also.