# Tag Info

1

Not necessarily. The kernel of the map $f$ is exactly the annihilator of $x\in M$. This can be non trivial. An example of this is if you take $\mathbb{Z}/3\mathbb{Z}$ as a $\mathbb{Z}$-module. Then the annihilator of $1$ is the the ideal $3\mathbb{Z}$ and so the map $\mathbb{Z}\to \mathbb{Z}/3\mathbb{Z}$ given by $a \mapsto a\cdot 1$ is not injective.

1

I will work in a category $\mathsf{Alg}(\tau)$ of algebraic structures of a given type $\tau$ (in the sense of universal algebra). Let $A$ be a finitely presented algebra. Choose some surjective homomorphism $\phi : \langle x_1,\dotsc,x_n \rangle \to A$ such that the kernel is a finitely generated congruence. Let $y_1,\dotsc,y_m$ be another generating set ...

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 ...

0

Hint: A more general statement for any commutative ring $R$ is that $\operatorname{Hom}_R(R, R) \cong R$. The key idea here is that any module homomorphism $\phi:R \rightarrow R$ is uniquely determined by $\phi(1)$. Think about how you can use this information to define an isomorphism $\Psi: \operatorname{Hom}_R(R, R) \rightarrow R$, or in your case, an ...

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 ...

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.

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} ... 0 If M is Artinian, then any nonempty collection S of submodules of M has a minimal element. For suppose S has no minimal element; let N_0\in S be any element. Since N_0 is not minimal, there is some N_1\subset N_0 in S. Since N_1 is not minimal, there is some N_2\subset N_1 in S. Continuing by induction, you get an infinite strictly ... 2 A homomorphism R^{(S)}\to R is determined as soon as an arbitrary function S\to R is selected. The set of functions S\to R is precisely R^S. The isomorphism can be described explicitly: if f\colon R^{(S)}\to R is a homomorphism, consider \varphi(f)=(f(e_s))_{s\in S}, where e_s is the basis element corresponding to s\in S. This defines a map ... 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. 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 ... 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. 1 For any V, we have |V^*|=|k|^{\dim V}\geq 2^{\dim V}>\dim V. The point of the assumption \dim V\geq |k| is that it implies (via |V|=\max(|k|,\dim V)) that \dim V=|V|, so we get |V^*|>|V|, which then clearly implies \dim V^*>\dim V. The case of arbitrary commutative rings actually follows more or less immediately from the field case. ... 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 ... 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. 1 Bad news It is impossible to prove it with the strategy you chose: in fact R is a perfect ring, and it is known that perfect rings satisfy the DCC on principal ideals. Your descending chain of ideals will therefore have to be more sophisticated. Triangular rings The ideal structure of this type of construction is explained in both of the following ... 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)} ...

1

To do this by Fitting, $P_S= \operatorname{im} (qp)^r \oplus \ker (qp)^r$ for some $r$, and one of these is zero as $P_S$ is indecomposable. Since $\lambda (qp)^r=\lambda$ we can't have $(qp)^r=0$ so $\operatorname{im}(qp)^r=P_S$ and $q$ is surjective. Now as Matthias says $$0 \to \ker q \to P_n \to P_S \to 0$$ is exact and $P_S$ projective, so $P_S | P_n$. ...

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 $f'$ ...

1

If $M$ is finitely generated, then clearly $M/A_+M$ is finitely generated, since it is a quotient of $M$. Conversely suppose that $M/A_+M$ is finitely generated, let $B=\{b_1,\dots,b_n\}$ be a finite set of homogeneous generators of cardinal $n$, and let $\phi:A^n\to M$ be the unique $A$-linear map such that $\phi(e_i)=b_i$ for each $i\in\{1,\dots,n\}$. The ...

0

Hint: Consider the subset of matrices of your matrix ring which are zero outside of the first column. It's easy to see this is a left ideal of the matrix ring and hence the left module over the matrix ring. Wouldn't you agree this suggests a natural action on $R^n$?

1

In fact the conclusion of Maschke's theorem never holds in the modular case, that is, suppose that ${\rm char}\, k =p\mid G$. Consider the element $\eta=\sum_{g\in G}g\in kG$. Then $g\eta=\eta$ for every $g\in G$ and then the ideal generated by $\eta$ is the same as the $k$-subspace generated by $\eta$, i.e. $\langle \eta\rangle =(\eta)$. Moreover ...

1

if the group is $\{1,a\}$ with $a^2=1$ then over $F_2$ $$(1+a)^2= 1 + 2a+a^2 \equiv_2 0$$ since the element $1+a$ is nilpotent the group algebra is not semisimple

1

Both $2\mathbb{Z}$ and $\mathbb{Z}$ are free rank $1$ $\mathbb{Z}$-modules, with $2\mathbb{Z}\subsetneq \mathbb{Z}$.

1

The rank of a submodule $N$ of $M$ will be equal to the rank of $M$ if and only if the quotient module $M/N$ is a torsion module, by the fundamental theorem of finitely generated modules over P. I. D.s. Hence its rank will be strictly less than the rank of $M$ if and only if $M/N$ contains non-torsion elements.

0

take $Z+Z+Z=M$ and $2Z+3Z+4Z=N$. Then $N$ is proper in the free $M$ but has same rank $3$ of $M$.

1

Here's another answer, motivated by a more recent question that asked for a more "concrete" example. Let $R$ be the polynomial ring $\mathbb Z[x]$. One $R$-module is just $R$ itself, and the other is the ideal $(2, x)$ in $R$. (This ideal consists of all integral polynomials with even constant coefficient.) These two $R$-modules are not isomorphic, as ...

0

First, you want to check that $\phi: \text{Hom}_R(R,M) \to M$ given by $\phi(f) = f(1_R)$ is indeed $R$-linear. So, first, we have to show that $\phi(f+g) = \phi(f) + \phi(g)$. Recall that we define $f+g$ "point-wise" that is: $(f+g)(r) = f(r) + g(r)$ (the right-hand sum takes place in $M$). So $\phi(f+g) = (f+g)(1_R) = f(1_R) + g(1_R) = \phi(f) + ... 1$\phi: Hom_R(R,M)\rightarrow M$is given by$\phi(f)=f(1)$for all$f\in Hom_R(R,M)$To check it is an$R-module$map (i)$\phi(f+g)=(f+g)(1)=f(1)+g(1)=\phi(f)+\phi(g)$(ii)$\phi(rf)=rf(1)=r.f(1)=r.\phi(f)$To check it is bijective: (i) injective: If$\phi(f)=0 \implies f(1)=0$Now,$f$is R-module homomorphism, i.e.,$f(r)=f(r.1)=rf(1)=r.0=0$... 2 It is good to always know different equivalent definitions for a single concept An$R-$module$M$is an abelian group$M$with a map$f:R\times M \rightarrow M$satisfying$f(a+b)=f(a)+f(b)$and$f(rm)=rf(M)\forall a,b,m\in M $and$r\in R$Equivalently,An$R-$module$M$is an abelian group$M$with a ring homomorphism$\phi:R \rightarrow End(M).$As you ... 0 Let$R=KG$with$K$a field (with char$K \ne 2$) and$G$cyclic of order two, and let$M_1$and$M_2$be the trivial and nontrivial$1$-dimensional (over$K$)$KG$-modules. So, in$M_2$,$G$acts by multiplying by$-1$. 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: ... 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 ... 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 ...

0

I will assume that $R$ is a ring with unit and that homorphisms preserve the unit. Let us try and prove that $R$ is free $R$-module on one generator. In particular let us show that the map $i:\{*\} \to R$ sending $*$ to $1$ has the following universal property: For each $R$-module $M$ and each map $f:\{*\}\to M$ there exists a unique homomorphism $\bar f: R ... 1 I doubt this is possible because the generators need not be independent. I don't think$m_i^*$is even well-defined for this reason. Given a free module$M$with basis$e_1, \ldots, e_n$, recall that we define the dual basis$e_1^*, \ldots, e_n^*$by $$e_i^*(e_j) = \begin{cases} 1 & i = j\\ 0 & i \neq j \end{cases}$$ and then extending ... 1 If$H$is the path algebra of the quiver$\bullet\to\bullet$over a field$k$, then there is a triangle-equivalence$F:D^b(H)\to D^b(H)$with$F(k\to k)=(k\to0)$and$F(0\to k)=(k\to k)$, so$Fdoesn't preserve sincerity. 2 We have \begin{align*}\phi:A\times\operatorname{Mat}_n(K)&\to\operatorname{Mat}_n(A)\\ (a,M)&\mapsto aM\end{align*} a well-defined bilinear homomorphism. By definition of the tensor product this means there is a unique linear map\varphi$, such that commutes. This map is given by ... 2$G$-modules, defined as abelian groups with an associative unital distributive action of$G$, are the same thing as$\mathbb{Z}G$-modules. For a$\mathbb{Z}G$-module restricts to a$G$-module via the inclusion of$G$into the group ring, a$G$-module induces a$\mathbb{Z}G$-module by linearity, and these operations are mutually inverse. 1 Let$K\subset P_N$be the kernel of the map$f:P_N\to M$. Since$p:P_N\to N$is a projective cover,$p(K)$must be a proper submodule of$N$, and hence$0$since$N$is simple. It follows that$p$factors through$P_N/K=M$to give a nonzero map$M\to N$. 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 ... 0 So, a couple days later, a little wiser. Here it goes. Indeed, as Gerry suggested, inverting$(2)$makes$I$equal to$(\tfrac{1}{2}(1+\sqrt{-5}))$and inverting$(3)$makes$I$equal to the whole ring since$I$contains a unit, so in both cases$I$is principal and thus free as a module. Theorem: Let$R$be a ring and$M$a finitely generated$R$-module. ... 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

A basic fact about category theory states that a morphism $f \colon A \to B$ is an epimorphism in a category $\mathcal C$ if and only if the pushout of it with itself (i.e., the cobase change of $f$ along itself) exists and $\require{AMScd}$ \begin{CD} A @>f>> B\\ @V f V V @VV \text{id}_B V\\ B @>>\text{id}_B> B \end{CD} is ...

1

You can use the universal property applied to $A \hookrightarrow F_1$. The function $\Phi: F_1 \rightarrow F_1$ is the identity on $A$, so it satisfies the universal property for the map $A \hookrightarrow F_1$. In other words $\Phi$ extends $A \hookrightarrow F_1$ to a map $F_1 \rightarrow F_1$. But there is already such a map, and that is the identity $F_1 ... 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 This boils down to understanding the bimodule structure on$\hom_k(M,N)$. Given such$k$-linear map$\varphi:M\to N$, we define$(g\varphi )(x)=g \varphi(x)$and$(\varphi g )(x)=\varphi(gx)$. The$G$-invariants are thus the$k$-linear maps that satisfy$\varphi(gx)=g\varphi(x)$for every$x\in M$and$g\in G$, i.e. the$kG$-linear maps. More generally, ... 4$\mathbb Q \otimes_{\mathbb Z} \mathbb Q = \mathbb Q$gives a negative answer to both questions. 2 Let$f\colon R\otimes M\to M$be the map. The analogues are:$f(s\otimes f(r\otimes m)) = f(sr\otimes m)f(1\otimes m) = m$1 There's nothing really mysterious. For every module$M$there exists an epimorphism$g\colon A^{(I)}\to M$, for some set$I$. Consider the kernel$K$of this epimorphism and apply the same fact, to get a homomorphism$f\colon A^{(J)}\to A^{(I)}$having$K\$ as its image. Then the sequence $$A^{(J)}\xrightarrow{f}A^{(I)}\xrightarrow{g}M\to 0$$ is exact. ...

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