## 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.