Math Studies Algebra II
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1 Math Studies Algebra II Prof. Clinton Conley Spring 2017 Contents 1 January 18, Logistics Modules January 20, Submodules Module Homomorphisms January 23, Generation of Modules Free Modules January 25, Free Modules cont January 27, Free Modules cont Dimension January 30, Finitely Generated Modules over Principle Ideal Domains February 1, February 3, February 6, Finite Dimensional Vector Spaces February 8, February 10, Feburary 13, Tensor Products February 15, February 17, Exact Sequences Splitting February 20, Topological Vector Spaces February 22,
2 17 February 24, February 27, Representation Theory Decomposition of Representations March 1, Maschke s Theorem March 3, Artin-Wedderburn Theorem March 6, March 8, Examples Using C[G] March 20, March 22, Extensions of Fields March 24, Extensions of Fields (cont.) Straight Edge and Compass Constructions March 27, Algebraic Extensions Splitting Fields March 29, Algebraic Closures March 31, Linear Orders Algebraic Closures (cont.) April 3, Separable Polynomials April 5, April 7, Galois Extensions April 10, Galois Extensions (cont.) April 12, Fun with Galois Theory Normal Subgroups Fundamental Theorem of Algebra April 14, Radical Extensions Solvable Groups April 17,
3 37 April 19, The Converse of Galois April 24, April 26, Ultrafilters (cont.) Some Combinatorics April 28, The Space of Ultrafilters Crash Course in Topology Topology on Space of Ultrafilters May 1, Semigroups May 3, Minimal Subsemigroups May 5, Hindman s Theorem
4 1 January 18, Logistics Office: Wean 7121 Office Hours: Tuesday 1:30-3:00 Andrew ID: clintonc Grading: 20% homework (graded for completeness), 20% 2 midterms, 40% final 1.2 Modules Definition. A module is something for a ring to act on. Informally, modules are the equivalent of group actions for rings. Definition. In this course, a ring is not commutative and they don t necessarily have an identity. Definition. Let R = (R, + R, R ) be a ring with identity. A (left, unital) R-module is an abelian group M = (M, + M ) with an action : R M M such that: (a) (r + R s) m = r m + M s m (b) (r R s) m = r (s m) (c) r (m + M n) = r m + M r n (d) 1 m = m. Sometimes, item (d) is dropped. Example For any ring R, M = {0} with operation 0 + M 0 = 0 and R-action r 0 = 0 is an R-module. 2. For any ring R, M = R with the same operation + and action r m = rm is an R-module. Example 1.2. Let R = Z. If M is a unital Z-module, we have that More generally, for z Z, 1 m = m = 2 m = (1 + 1) m = m + m. z m = m + + m. }{{} z times Thus, the Z-action is determined completely by the group (M, + M ). In summary, abelian groups and Z-modules are the same thing. Example 1.3. Let R = R and consider the Euclidean plane R 2 = {(x, y) : x, y R}. Consider the R-action via r (x, y) = (rx, ry). This makes R 2 into an R-module. More generally, R-modules are exactly real vector spaces. In even more generality, if R is a field F, then F-modules are simply vector spaces over F. Example 1.4. Let R = F[x], the polynomial ring in one variable over a field F. Recall that F[x] is a Euclidean domain, and thus a PID, UFD, etc. Suppose we have an F[x]-module. This module is also an F-module, so we may as well view the module as an F-vector space V. We investigate x v for v V. We know that for α, β, w, v V. If we write x v as T (v), we have that x (αw + βv) = x (α w) + x (β v) T (αw + βv) = αt (w) + βt (v), which is exactly what it means for T to be linear. So, the map v x v is a linear transformation of the vector space V. Conversely, if V is an F-vector space and T : V V is a linear transformation, we get an F[x]-module structure on V via x v = T (v). Thus, F[x]-modules and F-vector spaces with a single designated linear transformation are equivalent notions. 4
5 Definition. Let R be a commutative ring with 1. An R-algebra is a (not necessarily commutative) ring A with 1 equipped with a unital (f(1) = 1) ring homomorphism f : R A wose image is in the center of A. Remark 1.5. On the homework, we will see that R-algebras are really special R-modules such that A is a ring R acts on A (which makes A a module) r (ab) = (r a)b = a(r b) in A. 2 January 20, 2017 Today, we discuss submodules and module homomorphisms. 2.1 Submodules Let R be a ring, M a R-module, and consider N M. We say that N is an R-submodule if N is an R-module with the same operations. Proposition 2.1 (Closure properties). N is an R-submodule if N is an abelian subgroup and R acts on N, i.e. for all r R and for all n N, r n N. Example 2.2. Let R = R, M = R 2 with the action begin scalar multiplication. 1. N 0 = {(x, x) : x R} is an R-submodule; it is indeed a subgroup and closed under the action. 2. N 1 = Q 2 is not an R-submodule; it is not closed under the action. 3. N 2 = {(x, y) : x = 0 y = 0} is not an R-submodule; it does not form a subgroup. Proposition 2.3 (Submodule criterion). Let M be a unital R-module and let N M. Then N is an R-submodule if and only if N for all r R and for all x, y N, x + r y N. 2.2 Module Homomorphisms Definition. Let R be a ring and let M, N be R-modules. A map ϕ : M N is an R-module homomorphism if it preserves the R-module structure. That is, ϕ is a group homomorphism from M to N ϕ preserves the R-action. Definition. An R-module isomorphism is a bijective R-module homomorphism. If there is an isomorphism between two R-modules M and N, we write M = N. Definition. The kernel of a function ϕ : M N for M, N R-modules is ker(ϕ) = {m M : ϕ(m) = 0 N }. Proposition 2.4. The kernel of an R-homomorphism is a submodule. Proposition 2.5. Let M be an R-module and let N be a R-submodule. Then, the quotient group M/N can be given an R-module structure by r (m + N) = r m + N. Upon doing so, the map ϕ : M M/N via ϕ(m) = m + N becomes an R-module homomorphism with ker(ϕ) = N. Definition. Suppose ϕ : M N is a function. Then, im(ϕ) = ϕ(m) = {ϕ(m) : m M}. Proposition 2.6. If ϕ : M N is an R-module homomorphism, then im(ϕ) is an R-submodule of N. 5
6 Theorem 2.7 (First Isomorphism Theorem for R-modules). If ϕ : M N is an R-module homomorphism, then M/ ker(ϕ) = im(ϕ). Definition. Let M, N be R-modules. Then, hom R (M, N) denotes the set of all R-module homomorphisms from M to N. Proposition 2.8. The set hom R (M, N) forms an abelian group under the operation (ϕ + ψ)(m) = ϕ(m) + ψ(m). Proposition 2.9. If R is commutative, the action (r ϕ)(m) = r [ϕ(m)] makes hom R (M, N) an R-module. 3 January 23, Generation of Modules For this section, assume that R is a ring with identity. Definition. Let M be a unital R-module and let A M. Then, RA is the R-submodule generated by A, the smallest R-submodule of M containing A. Formally, RA = N N S where S is the set of R-submodules of M containing A. Equivalently, Remark 3.1. Note that A RA because 1 R. RA = {r 0 a r k a k : k N, r i R, a i A}. Remark 3.2. By convention, we set the empty sum to be 0. Definition. An R-submodule N M is finitely generated if there exists a finite A M such that N = RA. Definition. An R-submodule is cyclic if there exists an a M such that N = R{a} = {r a : r R} (we usually write Ra := R{a}). Remark 3.3. Cyclicity of Z-submodules coincides with the cyclicity of subgroups (abelian subgroups). Example 3.4. Consider the ring R = R and the R-module M = R 2. Then M is generated by A 0 = {(1, 0), (0, 1)}. M is also generated by A 1 = {(1, 1), (1, 1)}. 3.2 Free Modules Definition. Let M be an R-module. Then A M is said to freely generate M if for all x M there is a unique k N, unique and distinct a 0,..., a k A, and unique and nonzero r 0,..., r k R such that x = k i=0 r a. Proposition 3.5. Given a A, there is a canonical way to formally construct an R-module freely generated by A. Proof. Define the set F R (A) = {f : A R : {a A : f(a) 0 R } < }. Note that F R (A) is an abelian group via the operation (f + g)(a) = f(a) + g(a). Let R act on F R (A) by (r f)(a) = r(f(a)). It is easy to see that this makes F R (A) into an R-module. We identify a copy of A inside F R (A) via associating a A with f a F R (A) where { 1 R b = a f a (b) = 0 R b a. (1) Then, f F R (A) has the form f(a 0 ) f a0 + + f(a k ) f ak where a 0,..., a k list the elements of f(a i ) such that f(a i ) 0. 6
7 4 January 25, Free Modules cont. Remark 4.1. If A is finite, say A = n, then F R (A) = {f : A R} = R n with coordinatewise operations. To check whether the same generating set A freely generates M, it is enough to check that whenever a 0,..., a k A are distinct and r 0,..., r k such that k r i a i = 0, then r 0 = = r k = 0. i=0 Whether some set A generates M depends critically upon the acting ring. For instance, let M = R 2. Then, 1. As an R-module, M is 2-generated. 2. As a Q-module, M is not finitely (not even countably) generated. Theorem 4.2 (Universal Property of Free R-modules). Suppose A is some set, R is a ring with 1, M is some R-module, and g : A M is some function. Then there exists a unique R-module homomorphism γ : F R (A) M extending g, i.e. g(a) = γ(a) for all a A. Moreover, where B = im(g). Proof. We first prove existence and then uniqueness. Existence im(γ) = RB For f F R (A), define γ(f) = a A f(a) g(a) A, which is a finite sum of nonzero terms. Note that γ is indeed an R-module homomorphism since and γ(r f) = a A Then, γ indeed extends g since Uniqueness γ(f 0 + f 1 ) = a A(f 0 (a) + f 1 (a)) g(a) (rf(a)) g(a) = a A = f 0 (a) g(a) + f 1 (a) g(a) a A a A = γ(f 0 ) + γ(f 1 ) r (f(a) g(a)) = r f(a) g(a) = r γ(a). γ(f a ) = 1 g(a). Let γ 0, γ 1 : F R (A) M be R-module homomorphisms extending g and consider the new function δ = γ 0 γ 1, which is also an R-module homomorphism δ : F R (A) M. Note that for all a A, δ(a) = γ 0 (a) γ 1 (a) = g(a) g(a) = 0, i.e. ker(δ) is an R-submodule of F R (A) which contains A. Thus, ker(γ) = F R (A) and thus δ = 0 and we conclude as desired. a A 7
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