Algebra Qual Study Problems: 7

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 xm, x\in I, m \in M.
(i) Show that IM is a left submodule of M
(ii) Describe the natural left R-module structure on R/I \bigotimes_R M
(iii) Show that R/I \bigotimes_R M \cong M/IM as left R-modules.
Proof- i) We are given that IM \leq M by definition. Let r\in R, \sum_{i=1}^n x_{i}m_{i}\in IM. Then r(\sum_{i=1}^n x_{i}m_{i} ) = \sum_{i=1}^n r (x_{i}m_{i} ) = \sum_{i=1}^n (r x_{i})m_{i}. Since I is an ideal, rx_{i} \in I so that \sum_{i=1}^n (r x_{i})m_{i} \in IM. Thus IM is a left submodule of M.

ii) The natural left R-module structure on R/I \bigotimes_R M is given by r(s + I\bigotimes M) = r(s+I) \bigotimes M = rs+I \bigotimes M.

iii) Consider the map \phi M \rightarrow R/I \bigotimes M by \phi (m) = (1+I) \otimes m Then we can check that \phi is an R-module homomorphism with IM \subset Ker(\phi ). Thus by the first isomorphism theorem, there exists a unique \bar{\phi }: M/IM \rightarrow R/I \bigotimes M. Then consider the map \psi R/I \times M \rightarrow M/IM by (r+I,m) \mapsto rm + IM. Then it can be checked that \psi is a middle linear map and well defined. Therefore it induces the unique map \bar{\psi } : R/I \bigotimes M \rightarrow M/ IM. Lastly it can be easily checked that \bar{\phi } \bar{\psi } = id_{R/I \bigotimes M} and that \bar{\psi } \bar{\phi } = id_{M/IM}

Facts and counter examples to know:
1) Free module implies projective module.
2) Projective does not imply free. \mathbb{Z}_2 is a projective \mathbb{Z}_6 module but not a free \mathbb{Z}_6 module.
3) Torsion-free does not imply projective. \mathbb{Q} is torsion free as a \mathbb{Z} module but not projective as a \mathbb{Z} module.
4) Projective does not imply torsion free. \mathbb{Z}_4 is free over itself but not torsion free since 2 \cdot 2 = 0 but 2 \neq 0.
5) Torsion free does not imply free. \mathbb{Q} is torsion free as a \mathbb{Z} module but is not free.
6)Free does not imply torsion free. See number 4.

\mathbb{Q} is not a projective \mathbb{Z} module
Proof- Since \mathbb{Q} is a module there exists a free \mathbb{Z} module A and an epimorphism g: A \rightarrow \mathbb{Q} Say that A has basis X. Consider also the identity map 1_{\mathbb{Q} } \mathbb{Q} \rightarrow \mathbb{Q}. Suppose that the rationals are projective, then by definition there exists a map h:\mathbb{Q} \rightarrow A sucht that gh = 1_{\mathbb{Q} }. Then h (1) = \sum_{j=1}^n m_ix_i with x_i\in X and m_i\in \mathbb{Z}\backslash \{0\}. Since this is a finite sum, there exists some nonzero integer N that does not divide some m_i, say m_k. But we have that 1 = N\frac{1}{N} which gives that Nh(\frac{1}{N} ) = h(1) = \sum_{j=1}^n m_ix_i. But since X is linearly independent we have that N must divide m_i \forall i, which is a contradiction. Thus \mathbb{Q} is not projective.

Final Exam #9: Let K be a field. We say that a K[x]-Module M is nilpotent if x^kM=0 for some k>0. Describe isomorphism classes of all nilpotent K[x]-modules M such that dim_KM = 5.
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 dim_KM=5 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 A = Ra is isomorphic to R/(r) where (r) = \theta_a = \{ r\in R | ra =0\}. So a cyclic torsion module K[x]a is of the form K[x]/(f) with (f) = \theta_a = \{ f\in K[x] | fa =0\}. The element a of M then has order f. Note the order of any element a\in M is of the form x^n with n\leq 5 because M is nilpotent with dim_KM=5. Therefore f must be of this form. So we can say that any cyclic torsion module K[x]a \cong K[x]/(x^n) with n\leq 5. Then by the structure theorem M is a direct sum of a finite number of these with the property that orders of the summands, r_1,r_2,...r_t satisfy r_1 | r_2 | ... |r_t. Since the highest order is x^5 we can have the following for M:
K[x]/(x^5)
K[x]/(x) \bigoplus K[x]/(x^4)
K[x]/(x) \bigoplus K[x]/(x)\bigoplus K[x]/(x^3)
K[x]/(x) \bigoplus K[x]/(x) \bigoplus K[x]/(x) \bigoplus K[x]/(x^2)
K[x]/(x) \bigoplus K[x]/(x) \bigoplus K[x]/(x) \bigoplus K[x]/(x) \bigoplus K[x]/(x)
K[x]/(x^2) \bigoplus K[x]/(x^3)
K[x]/(x) \bigoplus K[x]/(x^2)\bigoplus K[x]/(x^2)
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 R = \mathbb{Z}[x_1,x_2,\ldots] be a polynomial ring in infinite variables. Then R is a finitely generated module over itself because R = \langle 1 \rangle. Recall that \langle X\rangle is the set of all linear combinations of r_ia_i where r_i\in R and a_i \in X. 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 \{1\}. Now the ideal I = \langle x_1,\ldots ,x_n,\ldots \rangle 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 x_k 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 \langle A \rangle which has a term with x_k as a factor necessarily has a term with degree at least 2. Thus x_k\notin \langle A\rangle which is a contradiction since x_k \in I \forall k. Thus I is not finitely generated.

2009 Qual #4: Let R = \mathbb{R}[x] and A = Rz_1 \bigoplus Rz_2 \bigoplus Rz_3 where
ann(z_1) = ((x+1)^2(x^2+1))
ann(z_2) = ((x^2+1)^2)
ann(z_3) = ((x^4-1)) = ((x-1)(x+1)(x^2+1))
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 Rz_i \cong R/ ann(z_i). Thus A \cong R/((x+1)^2(x^2+1)) \bigoplus R/((x^2+1)^2)) \bigoplus ((x-1)(x+1)(x^2+1)). The elementary divisors are the prime power factors of the generators of ann(z_i).
(x+1)^2,(x^2+1), (x^2)^2, (x-1), (x+1), (x^2+1)
To find the invariant factors we draw the power grid as follows. There are 3 distinct factors so there are 3 columns. The factor x^2+1 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 f_i.
f_1 = (x^2+1)
f_2 = (x^2+1) (x+1)
f_3 = (x^2+1)^2 (x+1)^2 (x-1)

2003 Qual #6: Let K be a field and let M, N be finitely generated modules over the polynomial ring K[x]. Suppose M has invariant factors: (x-1), (x-1)(x-2)^2, (x-1)(x-2)^2(x-3) and that N has invariant factors: (x-2)(x-3)^2, (x-2)^2(x-3)^2(x-5).
a) Give the elementary divisors of M \bigoplus N
List all the prime powers of M and N:
(x-1), (x-1), (x-2)^2, (x-1),(x-2)^2, (x-3), (x-2),(x-3)^2, (x-2)^2, (x-3)^2, (x-5)
b) Give all the invariant factors of M\bigoplus N
f_1 = (x-2)
f_2 = (x-2)^2(x-1)(x-3)
f_2 = (x-2)^2(x-1)(x-3)^2
f_2 = (x-2)^2(x-1)(x-3)^2(x-5)

2011 Qual #4 (Greeinstein): Let K be a field. We say that a K[x] module M is nilpotent if for every non-unit p\in K[x] we have p^nM=0 for n>>0. Prove that a finitely generated nilpotent indecomposable K[x] module is isomorphic to K[x]/(x^k) for some k>0.
Proof- Since K is a field we have that K[x] is a PID. So Then M is a finitely generated nilpotent (so torsion) module over a PID. Then if x\in K[x] is irreducible we have that x is prime. In particular every element of M has order x^k for some k>0 So by the structure theorem we have M\cong K[x]/(x^{k_1}) \bigoplus \cdots \bigoplus K[x]/(x^{k_r}) But M is indecomposible so r = 1.

2010 Qual #4: Let R be the polynomial ring \mathbb{C} [t]. Let I be the ideal generated by t^2. Let M = R/I. 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 t(t-1). Prove that N = R/J is isomorphic to the sum of two proper nonzero submodules.
Proof- First consider R/(t^2). A nonzero submodule of M is A = \{\bar{at} | a \in \mathbb{C}\}. Since t^2 is the only elementary divisor it can’t factor any further. Also, R/(t(t-1)) \cong R/(t) \bigoplus R/(t-1). 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 K[x] module structures on K^5 (that is, all elements of xK[x] act nilpotently) and End_{K[x]} (K[x]/(x^5)).
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 M \cong \bigoplus_{i=1}^n K[x]/(x^{k_i}) for k_i > 0. Therefore K^5 \cong K[x]/(x) \bigoplus K^5 \cong K[x]/(x) \bigoplus K^5 \cong K[x]/(x) \bigoplus K^5 \cong K[x]/(x) \bigoplus K^5 \cong K[x]/(x) as K[x] modules and this isomorphism is the only such nilpotent K[x]-module structure of K^5$. (WHY?)
Next we consider End_{K[x]} (K[x]/(x^5)). First note that this is a K[x] module, K[x]/(x^5) has basis \{i\} (WHY?). So any endomoprhism is determined by where \{i\} is mapped. In paricular \langle id \rangle = End_{K[x]} (K[x]/(x^5)) as K[x] modules and order(id) = x^5. Thus End_{K[x]} (K[x]/(x^5)) \cong K[x]/(x^5) by the classification theorem.
Needs a lot of explanation also.

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