Algebra Qual Study Problems: 5

2011 Qual #8 (Hungerford 3.4.4): Suppose R =\mathbb{Z}_6, S = \{2,4\} \subset R. Prove that S^{-1}R is a finite field and identify the field.
Recall that S^{-1}R is the set of equivalence classes of R\times S under the relation that (r,s) \sim (r',s') \Leftrightarrow s_1(rs'-r's) = 0 for some s_1 \in S. With the sets R and S given we have that all possible elements of S^{-1}R are \{(0,2),(1,2),(2,2),(3,2),(4,2),(5,2),(0,4),(1,4),(2,4),(3,4),(4,4),(5,4)\}. Now we must see which are equivalent under the relation. We have that:
(1,2) \sim (4,2) \sim (2,4) \sim (5,4) because 8-2=4-4=10-4 = 0
(2,2) \sim (5,2) \sim (1,4) \sim (4,4) because 4-10=8-2=8-8=0
(0,2) \sim (3,2) \sim (3,4) \sim (0,4) because 8-8=12=6=0-6=0
Thus S^{-1}R = \{(0,2),(1,2),(2,2)\}. Having only 3 elements we know that S^{-1}R \cong \mathbb{Z}_3 so that it is in fact a finite field.

2013 Qual #1 (Final Exam #8): Let R be a ring and let I, J be ideals. Let 0 \rightarrow M_1 \xrightarrow{f_1} M_2 \xrightarrow{f_2} M_3 \rightarrow 0 be a short exact sequence of R modules.
(a) Prove that IM_2 = 0 \Rightarrow IM_1 = 0 = IM_3
(b) Show that IM_1 = 0 = JM_3 \Rightarrow IJM_2 = 0. Give an example of a ring R a short exact sequence of R modules as above, and an ideal I of R such that IM_1=0=IM_3 but IM_2 \neq 0.
a) Suppose that IM_2 = 0. Let i\in I, m_1 \in M_1 , m_3\in M_3. Then we have f_1 (im_1) = if_1(m_1) \in IM_2 = 0. Therefore im_1 \in Ker(f_1). But by being a short exact sequence we have that Ker(f_1) = 0. Therefore im_1 = 0. Since this was arbitrary we have that IM_1 = 0. Next, f_2 is surjective due to exactness so there exists m_2 \in M_2 such that f_2(m_2) = m_3. This gives that if_2(m_2) = im_3, but because of the R-mod homomorphism we have im_3 = if_2(m_2) = f_2(im_2), but again since IM_2 = 0 we have that im_2 = 0 so that im_3 = f_2(0) = 0. Therefore IM_3 = 0.
b) Now suppose that IM_1 = 0 = JM_3. Let elements be the same as in part a and let j\in J. Since f_2 is onto we have there exists some m_2 \in M_2 such that f_2(m_2) = m_3 \Rightarrow f_2(jm_2) = jm_3 = 0. So then jm_2 \in Ker(f_2) = Im(f_1) by exactness. This gives that there exists m_1 \in M_1 such that f_1(m_1)= jm_2. Then if_1(m_1) = f_1(im_1) = f_1(0) = 0 = ijm_2. Therefore IJM_2 = 0.
For the counterexample take the following short exact sequence:
0 \rightarrow \mathbb{Z}_2 \xrightarrow{f} \mathbb{Z}_4 \xrightarrow{g} \mathbb{Z}_2 \rightarrow 0, where the maps are defined as f(0) = 0, f(1) = 2, g(0) = 0 = g(2), g(1) = 1 = g(3). Then clearly f and g are \mathbb{Z} \text{-module} homomorphisms, with the property that Imf = Kerg. Then take I = 2\mathbb{Z} to be the ideal of R = \mathbb{Z}. We have then that (2\mathbb{Z} ) \mathbb{Z}_2 = 0 but (2\mathbb{Z} ) \mathbb{Z}_4 \neq 0. Note that this counterexample works because the sequence does not split and that the statement would be true for any short exact sequence that splits. This is because if 0 \rightarrow M_1 \xrightarrow{f_1} M_2 \xrightarrow{f_2} M_3 \rightarrow 0 splits we can write M_2 \cong M_1 \bigoplus M_3 And so if an ideal I has the property that IM_1 = IM_3 = 0, then 0 = IM_1 \bigoplus IM_3 \cong IM_2

2006 Qual #1 (Hungerford 4.1.9, Takehome midterm #4c): If f: A\rightarrow A is an R-module homomorphism such that f^2 = f then A = Kerf \bigoplus Imf.
First recall that: Kerf \bigoplus Imf = \{a_1 + \ldots +a_n | a_i\in Kerf \cup Imf\} Since the kernel and image are submodules of A we have that \{a_1 + \ldots + a_n | a_i\in Kerf \cup Imf\} \subset A. To finish we must show the opposite containment. Let a\in A. We can write a = f(a) + (a-f(a)). Then note that f(a) \in Imf by definition and that f(a-f(a)) = f(a)-ff(a) = f(a)-f(a) = 0 because of the assumption. Therefore a-f(a) \in Kerf. Thus a\in \{a_1 + \ldots + a_n | a_i\in Kerf \cup Imf\}. Next we must show that Kerf \cap Imf = \{0\}. Let a \in  Kerf \cap Imf . Then f(a) = 0 and also there exists a b\in A such that f(b) = a. But then applying f to both sides of this equality we have ff(b) = f(a) \Rightarrow a = f(b) = f(a) = 0. Therefore Kerf \cap Imf  = \{0\}. This gives that A is the internal direct sum, Kerf \bigoplus Imf.

Qual 2011 #6: Let R be a ring and M_i,N_i for i = 1,2,3 be R-modules. Consider the diagram with exact rows of R-mod homomorphisms:
0\rightarrow M_1 \xrightarrow{f_1} M_2 \xrightarrow{f_2} M_3 \rightarrow 0

0\rightarrow N_1 \xrightarrow{g_1} N_2 \xrightarrow{g_2} N_3 \rightarrow 0
Suppose that there exist \phi_1 \in Hom_R(M_1,N_1), \phi_2 \in Hom_R(M_2,N_2) such that g_1\phi_1 = \phi_2f_1. Prove that there exists \phi_3 \in Hom_R(M_3,N_3) that makes the diagram commute. Which conditions should \phi_1 and/or \phi_2 satisfy to ensure that \phi_3 is surjective.
First we define such a function. Let \phi_3 : M_3 \rightarrow N_3 by \phi (m_3) = g_2\phi_2 (m_2) where m_2 \in \{m | f_2(m) = m_3\}. This can be done because f_2 is assumed to be surjective. We must show that this function is well defined. If m_3 = m_3' then there exist m_2, m_2' \in \{m \in M_2 | f_2(m) = m_3\} which define \phi_3 (m_3) and \phi_3 (m_3'). So to show well defined we must show that g_2\phi_2 (m_2) = g_2\phi_2 (m_2'). We have m_3 = m_3'f_2 (m_2) = f_2(m_2') \Rightarrow m_2-m_2'\in Kerf_2 =Imf_1. So then there exists m_1 \in M_1 such that m_2 -m_2' =f_1(m_1). Then \phi_2(m_2-m_2') = \phi_2f_1(m_1)  = g_1 \phi_1 (m_1) by the commutativity of the first square. Then we have g_2\phi_2(m_2-m_2') = g_2g_1\phi_1(m_1) = 0 because the bottom row is exact and so g_2g_1 = 0 always. Therefore g_2\phi_2(m_2-m_2') = 0 \Rightarrow g_2\phi_2(m_2) = g_2\phi_2(m_2'). So the map is well defined. It is a composition of R-mod homomorphisms so it is also one. The diagram commutes by construction.
Now, claim that \phi_2 surjective gives that \phi_3 surjective. Let n_3 \in N_3. Since g_2 is surjective we have there exists a n_2 \in N_2 such that g_2(n_2) = n_3. Since \phi_2 is surjective there exists a m_2 \in M_2 such that \phi (m_2) = n_2. Then n_3 = g_2\phi_2 (m_2) and hence \phi_3f_2 (m_2) = n_3 which gives the map to be surjective.

Qual 2009 #1 (Final #10): Let R be a ring, M an R-mod, M_1,M_2 \subset M. Define a natural homomorphism M_1 \bigoplus M_2 \rightarrow M where this is the external direct sum. Describe it’s image and kernel and use it to prove the formula dim_D(V_1+V_2) = dim_D(V_1\cap V_2) = dim_DV_1 + dim_DV_2 where D is a division ring and V_1,V_2 are vector subspaces.
Let \phi : M_1\bigoplus M_2 \rightarrow M be the natural homomorphism defined by (m_1,m_2) \mapsto m_1 + m_2. Then clearly Im(\phi ) = M_1 + M_2, and \phi ((m_1,m_2)) = m_1 + m_2 = 0 \Leftrightarrow m_1 = -m_2 \Leftrightarrow m_1 \in M_1 \cap M_2 = \{0\}. Therefore the Kernel is trivial. Now let D be a division ring, then any vector space is free and we have dim_D(V\bigoplus W) = dim_DV + dim_DW and dim_D(V/W) = dim_DV - dim_DW. Apply \phi to the vector subspaces V_1/(V_1 \cap V_2) \bigoplus V_2/(V_1 \cap V_2) \rightarrow V/(V_1\cap V_2) then by the rank nullity theorem we have dim_D ( V_1/(V_1\cap V_2) \bigoplus V_2/(V_1 \cap V_2)) = dim_D (Im(\phi )) +dim_D (Ker(\phi )) Then using the above facts this gives that:
dim_DV_1 + dim_DV_2 - 2 dim_D (V_1 \cap V_2 ) = dim_D (V_1+V_2) - dim_D ( V_1 \cap V_2) which implies dim_DV_1 + dim_DV_2 = dim_D(V_1+V_2) + dim_D( V_1 \cap V_2).

2011 Final #2(i): Is \mathbb{Q} free as a \mathbb{Z} \text{-module}?
Suppose that the statement were true. If it were a free \mathbb{Z} \text{-module} then it would have a basis. Let X = \{\frac{p}{q}\} be a basis for \mathbb{Q} as a \mathbb{Z} \text{-module}. Let \frac{p}{q+1} \in \mathbb{Q}, and suppose there exists an element m\in \mathbb{Z}, such that \frac{p}{q+1} = m \frac{p}{q} \Rightarrow m = \frac{q}{q+1} but this is not an integer so contradiction. Thus if the rationals have a basis it cannot have cardinality 1. Suppose then that X = \{\frac{p}{q}, \frac{r}{s}\} is a basis. We can write (rq)\frac{p}{q} - (ps)\frac{r}{s} = rp-rp = 0 so that the set is not linearly independent. Thus if |X| \geq 2 X will not be linearly independent. Therefore the rationals do not have a basis and so cannot be free as a \mathbb{Z} \text{-module}.

Qual 2013 #4 (Hungerford 4.2.14, Final #12): Let R be commutative and let E, F be free R-modules of the same finite rank. Prove that if f\in Hom_R(E,F) is surjective then it is an isomorphism. Why do we need to assume that R is commutative?
Let rankE = rankF =n. Let \{x_1,\ldots ,x_k\} be a basis for Ker(f). Then because the kernel is a submodule, we can extend this to a basis for E, X = \{x_1,\ldots , x_k, x_{k+1}, \ldots , x_n\}. Next, claim that \{f(x_{k+1}), \ldots , f(x_n)\} is a basis for Im(f) = F. Let r_{k+1}f(x_{k+1}) + \ldots + r_nf(x_n) = 0. Then because f is an R-module homomorphism we have f( r_{k+1}x_{k+1} + \ldots + r_nx_n) = 0. This tells us that r_{k+1}x_{k+1} + \ldots + r_nx_n \in Ker(f). Then we can use the basis of the kernel to write: r_{k+1}x_{k+1} + \ldots + r_nx_n = r_1x_1 + \ldots r_kx_k. Then subtracting we get r_1x_1 + \ldots r_kx_k - r_{k+1}x_{k+1} - \ldots - r_nx_n = 0. Because X is a basis for E we get that r_i = 0 \forall i. Thus \{f(x_{k+1}), \ldots , f(x_n)\} is linearly independent. Next, suppose b\in F, then there exists an element a\in E such that f(a) = b. Since X is a basis of E we can write f(r_1x_1 + \ldots + r_nx_n) = b. Then by the homomorphism we have r_1f(x_1) + \ldots + r_kf(x_k) + r_{k+1}f(x_{k+1}) + \ldots + r_nf(x_n)= b. Then because \{x_1,\ldots ,x_k\} \subset Ker(f), we have that the first k elements in the sum are 0. Thus r_{k+1}f(x_{k+1}) + \ldots + r_nf(x_n)= b so that b is written as a linear combination of elements in the set \{f(x_{k+1}), \ldots , f(x_n)\}. Thus we have that this is a basis of F. However we know rank(F) = n \Rightarrow k = 0. Thus Ker(f) = \{0\} \Rightarrow f is one to one. Thus f is an isomorphism.
We need commutativity in order to have the invariant rank property for R, which allows us to have basis’ for E and F with n elements.

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