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## Hot answers tagged modules

6

For question 2, $F\otimes_A -$ has a left adjoint iff $F$ is finitely generated, and the left adjoint is always exact. For if $(f_i)\in F^I$ is an element of an infinite product of copies of $F$, then it is easy to see $(f_i)$ is in the image of the canonical map $F\otimes_A A^I\to F^I$ iff $\{f_i\}$ is contained in a finitely generated submodule of $F$. ...

6

The argument for modules generalizes trivially, once you phrase it properly. For any object $A$, there is an epimorphism $p:F\to A$ from a free object $F$. If $A$ is projective, then applying the definition of projectivity to $p$ and the identity map $1:A\to A$, there exists a map $i:A\to F$ such that $pi=1$. That is, $A$ is a retract of $F$. (In the ...

5

Not necessarily. If there is a nontrivial ring honomorphism $\psi: A\to A$, then composing an evaluation homomorphism with $\psi$ gives you something that is not an $A$-algebra homomorphism. For example, in the case $A=\mathbb C$, $n=1$ and $\psi$ being complex conjugation we could have $$\phi(a_0 + a_1X + a_2X^2 + \cdots + a_kX^k) \mapsto \overline{a_0} ... 4 If R is noncommutative and P is a right R-module then \text{Hom}_R(P, R) naturally has the structure of a left R-module (and vice versa if P is a left R-module), so you can't even ask for this isomorphism because the two objects belong to different categories. But this is still false if R is commutative. Take R = \mathcal{O}_K to be the ... 4 If A is a finitely generated abelian group, every subgroup of A is finitely generated (because A is abelian, see proof here). So A is Noetherian as a \Bbb Z-module. For Noetherian modules, every epimorphism is an isomorphism, see proof here. We're using that a module is Noetherian if and only if every submodule is finitely generated. The analogous ... 4 K[X,Y,Z]=A[Z] is free over A, hence flat. A[Z]/(Z^2) is also free over A. Since flat modules are torsion-free, the last module is not flat. 4 Since \Bbb Q-modules are in fact vector spaces and are therefore torsion free, and \Bbb{Q/Z} is not a vector space over \Bbb Q as it is all torsion. The answer is that the question does not make sense. 4 These categories are not just equivalent but isomorphic. You have defined an operation on objects Mod(A)\to Mod(\tilde{A}), and it is easy to see that it also preserves maps and so gives a functor; the inverse is simply given by taking an \tilde{A}-module and restricting its module structure to A\subset\tilde{A}. More generally, this works with ... 4 What is interesting is the pair consisting of this category and its forgetful functor to rings; this exhibits the functor R \mapsto \text{Mod}(R) as a fibered category or Grothendieck fibration. 4 \mathbb Q \otimes_{\mathbb Z} \mathbb Q = \mathbb Q gives a negative answer to both questions. 3 What you have described is basically the "pullback" of N to the category of R-modules via the homomorphism \varphi|_R : R\rightarrow S. Also, I don't like your \sqcup notation. Really you're specifying the data of two things: A homomorphism \varphi : R\rightarrow S, and An R-linear map M\rightarrow \varphi^*N, where \varphi^*N is just N ... 3 The term "duality" in this context means "contravariant equivalence", i.e. a duality from \mathcal{C} to \mathcal{D} is an equivalence \mathcal{C}\to\mathcal{D}^{op}. Your "dual" functor on modules is just a contravariant functor, not an equivalence. 3 If Q is a projective module and f : Q \to S is a homomorphism, then there exists a map f' : Q \to P_S with f = \pi \circ f' with \pi : P_S \to S the canonical epimorphism. The important thing to note is that if f is surjective, then \text{im}(f') + \ker(\pi) = P_S. As \ker(\pi) is superfluous (it is the radical of P_S) we conclude that f' ... 3 Suppose P is a finitely generated projective \mathbb{T}-algebra. Then there is a finitely generated free \mathbb{T}-algebra F and a surjective homomorphism r : F \to P, and since P is projective, there is also a homomorphism s : P \to F such that r \circ s = \mathrm{id}_P. Then, r : F \to P is the coequaliser of \mathrm{id}_F and s \circ ... 3 Let me remove most of your assumptions and work with an arbitrary module M over an arbitrary commutative ring R. We'd like to know when the functor M \otimes_R (-) has a left adjoint. The answer is iff M is finitely presented projective, in which case the left adjoint is M^{\ast} \otimes_R (-) where M^{\ast} = \text{Hom}_R(M, R). You can ... 3 Suppose that equation holds. Choose a homogeneous basis for the graded vector space M/A_{>0}M and lift it to a set S of homogeneous elements of M. There is then a canonical map \varphi:F(S)\to M of graded A-modules, where F(S) is the free module on S (with the obvious grading). That equation says exactly that F(S)_i and M_i have the ... 3 Let f:A \to B be a bijective group homomorphism (or ring homomorphism, the argument is the same for every algebraic structure) and g:B \to A be an inverse of f as a function of sets. Then$$g(b_1b_2)=g(f(a_1)f(a_2))=g(f(a_1a_2))=a_1a_2=g(b_1)g(b_2)$$where f(a_1)=b_1 and f(a_2)=b_2 exist and are unique because f is bijective. Thus g is ... 2 First of all, the result you're trying to prove isn't true as stated. It is only true if you restrict \mathcal{C} to be the category of k-linear colimit-preserving functors. Second, you don't need a natural map M\otimes_k G(R)\to M; you need a natural map M\otimes_k G(R)\to G(M). To construct such a map, first consider the case M=R. In that ... 2 Hint: The fact that the injective \Bbb Z-modules are precisely the divisible \Bbb Z-modules comes in very handy at times like this. This, along with the elementary fact that injective submodules split out of the modules that contain them, will bring you to your solution. 2 This is not generally true over any ring: it is if the ring is hereditary, which means that submodules of projective modules are projective or, equivalently, that quotients of injective modules are injective. Over non commutative rings the notion is not symmetric: a ring can be right hereditary but not left hereditary. Let Q be an injective right module ... 2 You need to set up the correct hom–tensor adjunction. Recall that, for rings R, S, T, an (R, S)-bimodule M, (S, T)-bimodule N, and (R, T)-bimodule P, we have:$$\mathrm{Hom}_{(R, T)} (M \otimes_S N, P) \cong \mathrm{Hom}_{(R, S)} (M, \mathrm{Hom}_{(\mathbb{Z}, T)} (N, P))\mathrm{Hom}_{(R, T)} (M \otimes_S N, P) \cong \mathrm{Hom}_{(S, T)} ...

2

If $C=B/f(A)$ is not torsion, let $T$ be its torsion subgroup and use the quotient map $B\to C/T$ to show $f$ is not an epimorphism. Conversely, if $g_0,g_1:B\to D$ are such that $g_0f=g_1f$ and $D$ is torsion-free, $(g_0-g_1)f=0$, so $g_0-g_1$ factors through $B/f(A)$. Now use the fact that any map from a torsion group to a torsion-free group is $0$.

2

Set $A=\Bbb Z[x,y]/(xy)$ with $M=A$ and $M_0=(x,y)$. For an infinite example, instead use $A= \Bbb Z[x_1,x_2,\ldots]/I$ where $I$ is the ideal generated by all cross-products $x_ix_j, i\neq j$. Similarly use $M=A$ and $M_0=(x_1,x_2,\ldots)$.

2

Yes, Hopf algebras generally act on the field via the counit which in the case of group algebras can be defined as the linear extension of the map $$\epsilon:\mathcal{G}(kG)\longrightarrow k$$ $$x \mapsto 1_k$$ where $\mathcal{G}(kG)$ denotes the set of all group-like elements, that is, elements $x\in kG$ such that $\Delta (x)= x\otimes x$, where $\Delta$ ...

2

Very generally, any model of a finitary algebraic theory is a filtered colimit of finitely presented objects. For any such object $X$ has a presentation $\langle G \mid R\rangle$, where $G$ is some set of generators and $R$ is some set of relations (and each relation only involves finitely many generators). Now consider the directed poset $P$ consisting of ...

2

The expression $S \otimes_A Hom_A(S,S)$ doesn't make sense, because if $S$ is an irreducible right $A$-module, there is usually not any natural left $A$-module structure on $Hom_A(S,S)$. For instance, if $A=M_n(\mathbb{C})$ for some $n>1$ and $S=\mathbb{C}^n$, then $Hom_A(S,S)=\mathbb{C}$ cannot be made into an $A$-module (in any way compatible with the ...

2

This problem hinges on interpreting the Smith normal form of a matrix As you have said, we need to study the homomorphism $\varphi: \mathbb{Z}^3 \to \mathbb{Z}^4$ mapping the standard basis vectors to generators of the subgroup $H$. With respect to the standard bases $\mathcal{E}$ and $\mathcal{F}$ for $\mathbb{Z}^3$ and $\mathbb{Z}^4$, $\varphi$ has ...

2

While I can't think of a sensible way to generally define a morphism $M_R \to {}_R N$, here is a hint to help you prove the claim you were originally interested in. Note that module homomorphisms $R_R^n \to R_R^m$ correspond to $m \times n$ matrices with entries in $R$, if we view the modules as column vectors. Similarly, module homomorphisms ${}_R R^n \to ... 2 Let$R$be a ring. Given a right$R$-module$M$and a left$R$-module$N$, we can form their tensor product denoted$M\otimes_R N$. If$R$is a field, then a module over$R$is simply a vector space over that field. As$\mathbb{Z}$is not a vector space over$\mathbb{R}$, it is not an$\mathbb{R}\$-module, so the expression ...

2

I think the problem here is understanding Macaulay2's notation. Let's take an example: i1 : R = QQ[x,y,z] o1 = R o1 : PolynomialRing i7 : I = ideal(random(2,R), random(3,R), random(2,R)); o7 : Ideal of R i8 : betti res I 0 1 2 3 o8 = total: 1 3 3 1 0: 1 . . . 1: . 2 . . 2: . 1 1 . 3: . . 2 . 4: ...

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