Math 654: Graduate Algebra II Some questions and answers about the material and homeworks -------------------- 8.3.21 ----------------------------------------------------------- Q: I am having trouble even figuring out how to work with the quotient field F[t]/(f) A: Every polynomial has a unique remainder when you divide it by f, which is a polynomial of degree < degree f. Thus the monomials 1, t, ..., t^r where r+1=deg f form a natural basis for F[t]/(f). To do arithmetic, add/multiply these normal forms, and reduce mod f. Q: How do I consrtuct the matrix R ? A: If you multiply these monomials by t, all except one is another monomial in the list (it is the next one). Thus the first r columns have exactly one 1 in them, and it is just below the diagonal. The last column is more interesting, t*t^r = t^(r+1), and we need to reduce this modulo f, which amounts to subtracting f (the monic leading term vanishes). You may need to recall how a linear transformation and an ordered basis gives rise to a matrix representing that transformation in the given basis. -------------------- 8.3.21 ----------------------------------------------------------- Q: As Hom_R(R,A) \simeq A as an R-module only when R is commutative, what is the meaning of the answer you gave to the third question? A: I fixed this. The isomorphism in general is only as abelian groups. However, a sequence of R-modiules is exact if and only if it is exact as abelian groups. -------------------- 28.2.21 ----------------------------------------------------------- Q: Homework 6 question 3. Is this correctly stated? Consider the hilbert space \ell^2 of square summable series. In Functional analysis, we learn that \ell^2 is isomorphic to its dual? A: You are correct about \ell^2, but this duality is as Banach spaces (only continuous linear functionals are allowed). In the category of vector sapces, \ell^2 has a basis that is the cardinality of the real numbers, C. Consequently, its linear dual has a significantly larger independent set. The question that I posed was correctly written. ---------------------- 4.2.21 ----------------------------------------------------------- Q: Problem 2 says that ring homomorphism induced by $x\toT$, what is $x\to T$, I mean, it maps $x$ to which $T$? Or we need to define this map by ourselves? A: As F[x] is a ring that is freely generated by x, maps from F[x] are determined entirely by the image of x (it is important here that scalar multiplication by F is a vetor space endomorphism) That is a+bx + c x^2 |-> a + b T + c T^2, and etc. T is an arbitrary linear transformation; the answer about submodules will necessarily be in terms of T. Q: In Hungerford’s book p184, he says that the free module over a commutative ring with identity has invariant dimension property but this contradicts Problem 8. I don’t know if I misunderstand Hungerford’s book or Problem 8. A: My mistake, the ring R is definitely not commutative for problem 8; as yo unote, it necessarily must be noncommutative. ---------------------- 23.1.21 ----------------------------------------------------------- Q: This field of fractions construction has the scent of the adjoint of a forgetful functor, but I am not sure exactly what an adjoint functor is. A: What you are noticing is that these constructions satisfy a universal property. We could rephrase the main theorem about the construction of R[S^-1] thussly: Let C be the category whose objects are ring maps R -> T in which the image of S lies in the units (invertible elements) of T, and the morphisms are maps T -> W which make the obvious diagram commute. Theorem: C has an initial object, which is R -> R[S^-1]. Adjoint functors are a different beast, but this all has a categorical flavor. Functors f : A -> B and g : B -> A are adjoint if Hom_A( X, g(Y) ) is naturally identified with Hom_B(f(X), Y) When A = sets and g is the forgetful functor, f is the free functor; f(X) is free on X. ---------------------- 22.1.21 ----------------------------------------------------------- Q: In some treatments of the quotient construction, S is a monoid (contains 1) and in others (e.g. yours) it is only required to be a multipicative subsemigroup (does not necessarily contain 1). Is there a difference? What is the multiplicative identity in R[S^{-1}] ? A: It is worth verifying that if S is a subsemigroup, and we let U be S \cup \{1\} (or in fact the sub semigroup generated by S and any subset of units of R), then we may canonically identify R[S^{-1}] with R[U^{-1}]. (I did not check this, but the outlines are clear; I expect that the inclusion R\times S -> R\times U induces a bijection of equivalence classes.) The multiplicative identity in R[S^{-1}] is \iota(1_R) = s/s for any s\in S (the equivalence class of (s,s).) You can verify this, using that multiplicative identities are unique in a monoid. -------------------------- 21.1.21 ------------------------------------------------------- Q: I am a little bit confused about a notation in problem 6 in the homework, that is, the subring $\FF[[x]][x^{-1}]$ of the quotient field of $\FF[[x]]$. What is the multiplicative subset regarding $\FF[[x]][x^{-1}]$? I guess it might be the set $\{x^k: k\in \NN\}$, is that correct? A: Yes you are right. The notation I used for localization is intended to fit well with the other common notation that this problem uses. F[[x]] is a domain, so it has a quotient field. In that field, the element x has an inverse, x^{-1}. Then F[[x]][x^{-1}] is the subring of the quotient field containing F[[x]] and generated by x^{-1}. Equivalently, consider the polynomial ring over a domain: F[[x]][y] with one indeterminant, y. This has a homomorphism to the quotient field in which elements of F[[x]] are sent to themselves, but y is sent to x^{-1}. Then F[[x]][x^{-1}] is its image. The kernel is generated by xy-1. ----------------------------------------------------------------------------------- Q: You mentioned that all rings have an identity in your course and I was wondering how that would affect how we look at ideals. Since rings must have identity, then subrings must have an identity as well. Would this imply then that all ideals from here on out are trivial ideals? A: Ideals in rings with identities are not subrings, under this definition of rings. We do not lose anything, as that ideals are closed under multiplication is a consequence of their more important property---that they are closed under multiplication by the ambient ring.