New answers tagged

0

Hint: One way (perhaps not the most general, though) is using the fact that an exact sequence of modules $$0\longrightarrow M\stackrel{f}{\longrightarrow} P\stackrel{g}{\longrightarrow} N\longrightarrow 0$$ splits if and only if there exists a map $g':N\to P$ such that $g\circ g'=\operatorname{id}_N$.


1

In fact, what happens if we only define a "subset" of $M$ as $S=\{x\in M|l_R(x)\neq 0\}$? You can certainly define this set, and it could be considered a set of "torsion elements" inside $M$, but the set does not have as many nice properties in general. As in the example of $\mathbb Z/6\mathbb Z$ given above, $2$ and $3$ are torsion but $3-2=1$ is not, so ...


1

I'm on a cell phone so let the kernel be $A$. Using the axiom of choice, construct a set $N'$ consisting of one element from each coset of $A$. There is an obvious bijection of $N'$ with $N$. Then every element of $M$ can be represented uniquely as the sum of an element of $N'$ and an element of $A$, hence there is a bijection between $N'\times A$ and $M$ ...


0

If you had instead chosen $$ A= \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} $$ when defining $K^2$ as a $K[x]$-module, you would have obtained a decomposable one. So you need to say more than the fact that the above matrix has two eigenvalues $0$. What is true is that a module gotten in this way will always decompose into eigenspaces, but ...


3

Recall that if $M$ is a module over a ring $R$, then: $$\mathrm{ann}(M)=\{r\in R\mid \forall_{m\in M}rm=0\}$$ This is two sided ideal of $R$ called anihilator of $M$. Now if $M_1\cong M_2$ as $R$-modules, then: $$\mathrm{ann}(M_1)=\mathrm{ann}(M_2)$$ Annihilator of $\mathbb{C}[x,y]/(x,y)$ is $(x,y)$ and annihilator of $\mathbb{C}[x,y]/(x-1,y-1)$ is $(x-1,y-1)...


2

Finite dimensional algebras have integrals, and using one you can show selfinjectivity at once —in fact, they are Frobenius. See for example the Lectures on Hopf algebras by Schneider, which Google will find for you, or Susan Montgomery's book in Hopf algebras.


0

Consider the short exact sequence $0\xrightarrow{}M_1\xrightarrow{}M_1\oplus M_2\xrightarrow{\pi}(M_1\oplus M_2)/M_1\xrightarrow{}0$, then $M_1$ and $(M_1\oplus M_2)/M_1$ are both Artinian. Let $N_1\supset N_2\supset\cdots$ be the descending chain, let $N_i' = N_i\cap M_1$, let $N_i'' = (N_i+M_2)/M_2$, then it is easy to see that $\{N_i'\}_{i=1}^\infty$ and ...


1

For a vector space, having finite dimension is the same as being finitely generated. One direction is obvious: if the vector space has a finite basis it clearly is finitely generated. Suppose $V$ is finitely generated. From any finite spanning set $S$, one can extract a minimal spanning set $B$, in the sense that $B\subseteq S$ is a spanning set and no ...


0

What is more general, the theory of modules over algebras or over arbitrary rings? Yes, every ring is a $\Bbb Z$-algebra, and its modules are automatically $\Bbb Z$-algebra modules. And of course every $\Bbb Z$-algebra is a priori a ring. With this in mind, then neither theory is more general than the other: they are the same. The class of algebras which ...


1

$M$ is similar to the matrix $$ J = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2 \end{pmatrix} $$ hence, it has minimal polynomial $(x-1)(x-2)^3$ and characteristic polynomial $(x-1)^2(x-2)^3$. So, as a $\...


1

Yes, of course, but it's not that useful. Suppose you have two ring homomorphisms $f\colon A\to E(M)$ and $g\colon A\to E(N)$. Suppose you also have a group homomorphism $h\colon M\to N$ is a module. For each $a\in A$, you have $f(a)\colon M\to M$ and $g(a)\colon N\to N$; so you can form the square $$\require{AMScd} \begin{CD} M @>f(a)>> M \\ @VhVV ...


3

This is a very general question and can not be answered in a few words. So, may be let me describe one aspect. If $R$ is a commutative ring and $M$ an $R$-module, giving an $R$-algebra homomorphism $A\to \operatorname{End}_R M$, where $A$ is an $R$-algebra makes $M$ into an $A$-module. In the example you write above, $R=\mathbb{Z}$. Now, let me look at a ...


3

Ok Slup, here goes. Let $R$ be any commutative ring and let $A$ be a polynomial ring over $R$. Let $P$ be any projective module over $R$. Then Quillen (and Suslin a bit later in this generality) proved that if for every maximal ideal $\mathfrak{m}$ of $R$, $P_{\mathfrak{m}}$ is of the form $Q\otimes_{R_{\mathfrak{m}}} A_{\mathfrak{m}}$ for some projective $...


1

By the Serre-Swan theorem, at least if $M$ is closed, taking smooth sections defines an equivalence of categories between smooth vector bundles over $M$ and finitely generated projective modules over $C^{\infty}(M)$. This is more or less equivalent to the claim that every smooth vector bundle is a subbundle of a trivial vector bundle.


1

It should be true for any PID. See the book by Lam, Serre's Conjecture


2

Let us say that a ring $A$ satisfies condition $(S)$ if every finite type projective $A$-module is free. Clearly a necessary condition for $(S)$ to hold is $K_0(A)=\mathbf{Z}$. Example. (i) If $A$ is local Noetherian then $(S)$ holds. (ii) If $A$ is a Dedekind domain then $K_0(A)=\mathbf{Z}\oplus\mathrm{Pic}(A)$, and thus a necessary condition for $(S)$ is $...


-1

How about another proof? If $M$ is a free module over a commutative ring with identity $A$, then any two bases of $M$ have the same cardinality. This is a general fact about commutative rings ("invariant basis number"). That cardinality is called the rank of $M$. If $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ is an exact sequence of ...


1

HINT: Do you see how to generate $1$ using $2$ and $3$? That is, do you see how to write $1$ as a $\mathbb{Z}$-linear combination of $2$ and $3$? If so, do you see why this means that $\{2, 3\}$ spans?


2

I will prove that given a family $\{M_j\}_{j \in J}$ of $R$-modules then $Hom(\oplus M_{j},N) \cong \prod Hom(M_{j},N)$ for any given $R$-module. Then your question follows by setting $M_{j}=R$ for all $\thinspace$ $j \in J$. Assume that $i_{j} : M_{j} \rightarrow \oplus M_{j}$ is the canonical inclusion. Now define the above homomorphism $\phi: Hom(\oplus ...


1

You can use an injective resolution for $N$: let $E$ be injective and $0\to N\to E\to E/N\to 0$ be exact. Then the long exact sequence $$\DeclareMathOperator{\E}{Ext}\DeclareMathOperator{\H}{Hom} 0\to \H_R(\bigoplus_{i\in I}M_i,N)\to \H_R(\bigoplus_{i\in I}M_i,E)\to \H_R(\bigoplus_{i\in I}M_i,E/N)\to\\ \E_R^1(\bigoplus_{i\in I}M_i,N)\to \E_R^1(\bigoplus_{i\...


1

The proof is correct. If $(C_i)_{i\in I}$ is any family of cochain complexes, then by writing out the definitions you immediately see $H^n(\prod C_i)=\prod H^n(C_i)$. For if $D$ denotes the differential on $\prod C_i$, $d_i$ the differential on $C_i$, then $H^n(\prod C_i)=Ker D/Im D=\prod Ker d_i/\prod Im d_I=\prod Ker d_i/ Im d_i=\prod H^n(C_i).$


0

To complete the hints in the comments, note that $B = A[b_1, ..., b_n]$ for certain $b_i \in B$ since $A\rightarrow B$ is of finite type. Now the statement follows via induction on $n$.


1

The annihilator of $0\in M$ is $R$, which is always going to be essential. The singular submodule always has at least that.


1

You distribute the $n$ and the $x$ by using your very own definition: $$(x+y)(n+1) = (x+y)n + (x+y)$$ for non-negative $n$, and in the second case $$x(n+m+1) = x(n+m) + x$$ for $n+m$ non-negative. Use induction. For the negative case, use induction there too, together with your definition of what $xn$ means for negative $n$.


0

The tensor $\sum_{i=1}^n b_i \otimes m_i$ is zero exactly when, whenever $G$ is an abelian group and $\phi : B \times M \to G$ is a bilinear map such that $$\forall a \in A, \forall b \in B, \forall m \in M, \phi(ba,m) = \phi(b,am), \tag{*}$$ then $\sum_{i=1}^n \phi(b_i, m_i) = 0$. Now fix some $b_0$ and some bilinear map $\phi$ as above, and define $\psi : ...


1

Let $M$ be a nonzero finitely generated submodule of $R^+$; you can assume a set of generators is $$ \left\{\frac{a_1}{d},\frac{a_2}{d},\dots,\frac{a_n}{d}\right\} $$ by using a common denominator. The $R$-homomorphism $M\to R$ defined by $x\mapsto dx$ is injective, so $M$ is isomorphic to a nonzero ideal of $R$. (The assumption $M\ne\{0\}$ is of course ...


0

I really feel like I have seen this called "the annihilator of $\mathfrak{a}$ in $M$" and denoted $Ann_M(\mathfrak{a})$, but I have been unable to find a source quickly. That would be a reasonable name in any case.


2

Edit As @Batominovski and @TobiasKildetoft remarked, one should assume characteristic zero here. In positive characteristic $p$ and non-trivial $G$, the statement is indeed not true, at least if one takes the naive definition of characters: For example, the $p$-fold sum of the trivial representation satisfies the assumption but is not of the form $KG^{\oplus ...


4

Slup has given you the answer where it is mostly used, though it is not in general true that Trace maps $B$ to $A$, unless you assume something more (typically, one assumes that $A$ is integrally closed). One has standard counterexamples for general cases. For example take a ring $A$ which has a non-free projective module $P$ such that $A\oplus P$ is free. (...


0

This does not cover the full range of your question but at least gives an affirmative answer to an interesting special case. I am denoting the structural morphism by $j:A\rightarrow B$. Suppose first that $A$ is an integral domain. If $B$ is a free $A$-module, then for every nonzero element $b\in B$ we have $ab=0$ for some $a\in A$ if and only if $a=0$. So $...


1

Here's the way I determine the Smith normal form. First find the gcd of all entries of the matrix - its easy to see this is 1. So $f_1(x) = d_1(A)/d_0(A) = 1/1 = 1$ Now you need to find the gcd of determinants of ALL $2 \times 2$ submatrices. There are 7 such submatrices. The determinant of the upper left matrix is 2. The determinant of the upper right ...


3

Newman says that Smith form can be accomplished with elementary row and column operations as long as the coefficient ring is Euclidean, as here. Since we are not going to change the determinant, this means diagonal $(1,1,6).$ Newman's assurance means that we can accomplish this for the two by two square with entries $(2,3).$ Take distinct positive integers $...


0

We can always find a minimal projective presentation $U_{0}\overset{u}{\to}U_{1}\overset{w}{\to}\text{Tr}(M)\to 0$ of $\text{Tr}(M)$ (take a projective cover of $\text{Tr}(M)$, then a projective cover of the kernel of that epimorphism). So then we have the following setup: $$ \require{AMScd} \begin{CD} @. P_{1}^{t}\\ @. @VV \pi_{M} V\\ U_{1} @&...


2

Note that the image of your diagonal matrix is $\Bbb Z\oplus 2\Bbb Z\oplus 2\Bbb Z$. Hence the cokernel is $$ \frac{\Bbb Z\oplus\Bbb Z\oplus\Bbb Z}{\Bbb Z\oplus 2\Bbb Z\oplus 2\Bbb Z}\simeq(\Bbb Z/\Bbb Z)\oplus(\Bbb Z/2\Bbb Z)\oplus(\Bbb Z/2\Bbb Z)\simeq\Bbb Z/2\Bbb Z\oplus\Bbb Z/2\Bbb Z $$


0

$E$ can be thought of as a complex (in this case a very special complex). For example, $E$ could stand for any complex of $A$-modules, $\{M_n\}, n\in\mathbb{Z}$ with module maps, say $\partial_n:M_n\to M_{n-1}$ for all $n$. It is a complex just means $\partial_{n-1}\circ\partial_n=0$ for all $n$. For any other module $P$, one has $\mathrm{Hom}(E,P)$ which is ...


3

A structure of (unitary left) $R$-module over an abelian group $G$ (written additively) is determined by a (unitary) ring homomorphism $R\to\operatorname{End}(G)$, where $\operatorname{End}(G)$ consists of the endomorphisms of $G$ under the standard sum operation and map composition. To see why, suppose $G$ is an $R$-module. For $r\in R$, define $\lambda_r\...


1

In the particular example you gave, in fact you can make $\mathbb{Z} / 5 \mathbb{Z}$ into a $\mathbb{Z}[i]$-algebra by "mapping $i$ to 2" -- the specific formula for module multiplication would be: $$ (a+bi) \cdot c := (a+2b)c $$ for $a+bi \in \mathbb{Z}[i]$, $c \in \mathbb{Z} / 5 \mathbb{Z}$. (Here, the essential property of 2 which will make $\alpha \cdot ...


0

Have I misunderstood something fundamental here or don't we use the OR operation in the initial boolean algebra? Consider the $\vee$-$\wedge$ definition of a Boolean algebra. It turns out that you can turn this into a real boolean ring by defining $a\cdot b=a\wedge b$ and $a+b=(a\vee b)\wedge\neg(a\wedge b)$. As a result, one of the boolean algebra laws ...


0

This is true for the canonical surjective homomorphism $f:M\to M/M'$ simply because $y$ is not in the kernel.


2

Hint: Take $f:M\rightarrow M/M'$ be the canonical projection.


4

No. For example, suppose $N=S$ is simple, so that $\operatorname{soc}N=S$, $M'$ is a non-split extension of $S$ by another simple module $T$, so that $\operatorname{soc}M'=T$ and there is an epimorphism $\alpha:M'\to N$ which is zero on $\operatorname{soc}M'$, and that $M=M'\oplus S$. Then the map $\begin{pmatrix}\alpha&\operatorname{id}_S\end{pmatrix}:...


1

In the presentation $\mathbb{Z}_2=\{-1, 1\}$ you cannot write your (group) homomorphism as a matrix! In fact in the presentation $\mathbb{Z}_2=\{0,1\}$ you cannot do so either if you are considering $\mathbb{Z}_2$ as a group not a ring! To make this clear consider a homomorphism $\phi: \mathbb{Z}_2^2\to \mathbb{Z}_2^2$ sending $(1,0)\mapsto (a,b)$ and $(0,1)...


0

You should complete the following steps: $x+z$ is regular on $R$. $R/(x+z)$ has depth $0$. Deduce that $R$ has depth $1$. Since you have computed the depth of $M$ to be $2$, Auslander-Buchsbaum cannot hold, i.e. the projective dimension of $M$ cannot be finite.


0

Indeed one has to be careful with the notion of dual module: In the context of representation theory of groups (more generally, when studying modules over Hopf algebras), the dual of a $G$-module $V$ is usually defined as having the vector space dual $V^{\ast}$ as its underlying vector space, and with the $G$-action defined by $(g.\varphi)(v) := \varphi(g^{...


1

Eisenbud's statement is morally correct, but not strictly true with the usual set-theoretic representations of functions (or any representation that I know of). Writing $X \to Y$ for the set of set-theoretic functions from $X$ to $Y$, then there is a 1-1 correspondence between $X \times Y \to Z$ and $X \to (Y \to Z)$ that associates $f:X \times Y \to Z$ with ...


2

Assume $\varphi: M\times N \to P$ is a bilinear map, then the corresponding homomorphism is $\phi: M\to \hom_R(N,P)$, with $\phi(m)(n)=\varphi(m,n)$. We can check $\phi(m)$ is a homomorphism from $N$ to $P$ using the fact that $\varphi$ is bilinear. Convesely, if $\phi: M\to \hom_R(N,P)$ is a homomorphism, then the corresponding bilinear map is $\varphi: M\...


1

Given a map $f : M\times N \to P$ which is linear in the second argument, there is an associated map $F : M \to \operatorname{Hom}_R(N, P)$ defined by $F(m)(n) = f(m, n)$. If $f$ is also linear in the first argument (i.e. $f$ is bilinear), then $F$ is a homomorphism. Conversely, given a map $G : M \to \operatorname{Hom}_R(N, P)$, there is an associated map $...


1

Hint: To a bilinear map $f:M\times N\rightarrow P$ associated $H_f:M\rightarrow Hom_R(N,P)$ defined by $H_f(m)(n)=f(m,n)$. And to $H:M\rightarrow Hom_R(N,P)$ associate $H_b:M\times N\rightarrow P$ defined by $H_b(m,n)=H(m)(n)$ show that the correspondence are inverse each other


1

This is partially an answer for your question: If we consider the additional condition that $A$ is noetherian, we'll be able to prove that $\tau$ is injective (in fact under this condition it is possible to show that $\tau$ is $S^{-1}A$-isomorphism): Since under new hypothesis, $A$ is noetherian and $M$ is a finitely generated $A$-module we know that the ...


2

The following is primarily first principles. It is not the most elegant way to answer the question if you have knowledge of short exact sequences, etc., but it seems to be more what you are looking for. Choose any $m\in f^{-1}(1)$. Then Let $L=Rm$. Let's verify that $M=\ker f \oplus L$. 1) If $f(rm)=0$, then $0=rf(m)=r1=r$, so $rm=0$, i.e. $\ker f \cap L=...



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