Tag Info

New answers tagged

1

If $F$ is free of rank one over $R$ then $F=R\cdot1=R$ and $R$ was a field to begin with. Otherwise suppose it is free of rank greater than one, so it has at least two summands which are copies of $R$ which must be generated by some fractions, which means $R\frac{a}{b}\cap R\frac{c}{d}=0$ must hold for some values $a,b,c,d\in R$... is that possible?


0

Any free $R$ module is isomorphic to $F=R\oplus R\oplus R \oplus ...$ Notice that if it has more than one component it has a proper ideal $I=R\oplus O\oplus 0\oplus...$ which is not possible for a field so we must have $F=R$. Thus, $R$ also must be a field. If you mean only $(F,+)\cong (R\oplus R...R,+)$ and do not use natural multiplication on $R\oplus ...


0

I somehow find the solution myself. The example I gave is just wrong, it is not middle linear. (Thanks to @Jeremy Rickard for noticing.) And since the identity of morphism is unique (Thanks to @roman for a remind), $1_{f}$ must be the identity isomorphism on $C$. And similarly for $1_{g}$. Hence for $h$ to be an equivalence, there must exist $h'$, such ...


0

(1) More generally, if $X$ and $Y$ are $R$-modules, you have $$ \operatorname{Hom}(X\oplus Y,M)\cong \operatorname{Hom}(X,M)\oplus\operatorname{Hom}(Y,M) $$ (in module theory it's more customary $\oplus$ rather than $\times$, but they mean the same). The (group) isomorphism is obtained by considering the canonical injections $i\colon X\to X\oplus Y$ and ...


2

Hint: prove it isn't generated by a single element. Then prove that any two elements are linearly dependent. More details: You can show that $M$ is not generated by a single generator using the fact that $x$ and $y$ are irreducible in the UFD $A$. If $f$ generates $M$ then $f$ divides $x$ and also $f$ divides $y$. Since these are distinct irreducible ...


2

The problem is simple: if $A$ and $B$ are both left $R$-modules, we'd want to equate $(ra,b)$ with $(a,rb)$ in the tensor product. But then we'd have $(rsa,b)=(a,rsb)$ and $(rsa,b)=(sa,rb)=(a,srb)$! So this construction forces $R$ to act commutatively on $A\otimes B$. In general the tensor product of a right and a left $R$-module is merely an abelian group. ...


2

$\DeclareMathOperator{\h}{Hom}$ $S^{-1}M$ is characterized by the universal property $\h_{S^{-1}A}(S^{-1}M,T) \cong \h_A(M,T_{|A})$. In order to show that $L := \lim_{s \in S} M_s$ satisfies this property there are two things to do: Each $s \in S$ acts as an automorphism on $L$ (so $L$ carries a natural $S^{-1}A$-module structure). For each $A$-Module $T$ ...


1

For the second question, note that $x^4, y^4$ is a SOP. Since $R=k[x^4, x^3y,xy^3, y^4]$ is a domain, $x^4$ is a non-zero divisor on $R$. But observe that $y^4$ is a zero divisor on $R/(x^4)$ because $y^4(x^3y)^2=x^4(xy^3)^2$. Ofcourse one needs to note that $(x^3y)^2\notin (x^4)$. Hence $R$ is not CM. You already did the first part, perhaps you can now ...


1

You're being misled by the notation. The module $F$ is not $A\times B$, but the free $R$-module with the set $A\times B$ as basis. So $F$ is the set of “formal expressions“ of the form $$ r_1(a_1,b_1)+r_2(a_2,b_2)+\dots+r_k(a_k,b_k) $$ where $k$ is any natural number, $a_i\in A$, $b_i\in B$, $r_i\in R$ (for $i=1,2,\dots,k$). If $k=0$ the formal sum is empty ...


3

No, the tensor product isn't zero. Be more careful about the definition. The way tensor product is constructed is to take the free group generated by elements of $A \times B$ and what we really want is to have identities such as $(a+a',b)=(a,b)+(a',b)$ so we take the subgroup generated by those three above types of elements and then taking quotient would ...


1

Consider a field $F$ and the ring $R=F\times F$. Let $M=\{0\}\times F$ with both the ordinary right $R$-module structure $(0,m)(r,s)=(0,ms)$, and let $M'$ be the same set with another $R$-module structure given by $(0,m)(r,s)=(0,mr)$. This second structure is just given by the involution on $R$ given by $(r,s)\mapsto(s,r)$. The annihilator in $R$ of $M$ ...


0

If the original sequence was a complex, then it would be of the form $$ \dotsb\xrightarrow{g_{i-1}}A_{i-1}\xrightarrow{f_{i-1}}B_{i-1}\xrightarrow{g_i}A_i\xrightarrow{f_i}B_i\xrightarrow{g_{i+1}}A_{i+1}\xrightarrow{f_{i+1}}\dotsb $$ where $g_k\circ f_{k-1}=0$ and $f_k\circ g_k=0$. Now, consider the complex $$ ...


2

Sure, it is always true that a free module (under this definition) will be isomorphic to $A^{(I)}$ for some index set $I$, no matter if the set $I$ is finite or infinite. Suppose you are given $M=\bigoplus_{i\in I} M_i$ with isomorphisms $\phi_i:M_i\to A$. This can be composed with the map that injects $A$ into the $i$th position in $\bigoplus_{i\in I}A$. ...


2

If you interpret $\underbrace{A\oplus\cdots\oplus A}_n$ as $\bigoplus_{i\in I}A$ for infinite $n$ then you're right back at the definition.


0

I guess I found a proof. Can someone verify my steps? Let $S=A - \mathfrak{p}$. At first, suppose $(Ax)_{\mathfrak{p}}=0$. Particularly $\frac{x}{1}=0$. Hence there exists a $s \in S$ and therefore $s \notin \mathfrak{p}$ such that $sx=0$ which means $s \in \mathfrak{a}$. In conclusion $\mathfrak{p}$ does not contain $\mathfrak{a}$ since $s \in ...


1

There are ideals that cannot be generated by two elements! For example, let $I_n$ be the ideal of all linear combinations of monomials of total degree $n$ or more. e.g. $I_2 = \langle x_1^2, x_1 x_2, x_2^2 \rangle$ It's easy to see that $I_2 / I_3$ is a three-dimensional vector space over $\mathbf{C}$; its elements are in one-to-one correspondence with the ...


2

Being finitely generated says the module can be spanned by finitely many elements, but rank refers to the maximum number of elements which are linearly independent in the module. These two conditions sound similar, and indeed they coincide in linear algebra, but they can be different. See http://mathoverflow.net/a/30024/19965 and also maybe section 1c of ...


4

The set of endomorphisms of an $A$-module $M$, just like the set of linear mappings from a vector space to itself, form a ring, if you define the operations pointwise. In other words, if $\phi,\psi$ are a $A$-module endomorphism, define $$\begin{eqnarray} (\phi \mathbf{+} \psi) &\,:\,& M \to M &\,:\,& x \mapsto \phi(x) + \psi(x) \\ (\phi ...


2

I'll use $x,y,z$ for the variables of $R$. Notice that $(x)(y,z) = (xy,xz)$. If we want to form a chain of prime ideals of $R$ terminating at $I$, then they must all contain $xy$ and $xz$. But this implies that they either contain $x$, or $y$ and $z$. If they contain $x$, then so too does $I$, but this is a contradiction. So they must contain $y,z$. But then ...


0

Put more succinctly: Given $R=\Bbb Z+X\Bbb Q[X]$ and $M=X\Bbb Q[X]$, why is $M$ is an infinitely-generated $R$-module? Show the $X$-coefficients of a putative generating set for $M$ would generate $\Bbb Q$ as a $\Bbb Z$-module. (The intuition here is that "mod $X^2$" we see that $R$ acts on $M$ just like $\Bbb Z$ acts on $\Bbb Q X$.)


0

It's really a preference. If this is your first exposure to modules, and you like doing things concretely, 1. would be better. If you know about short exact sequences, then 2. will give you a more "hands off" solution.


1

It turns out to be right that $M\cong N\times M/N$, but not without justification. In fact it is somewhat specific to this case (where everything is free of finite rank); e.g. $M=\mathbb Z$ has a submodule isomorphic to $N=\mathbb Z$ such that the quotient $M/N\cong\mathbb Z/n\mathbb Z$ for any $n$, and of course $\mathbb Z\not\cong\mathbb Z\times\mathbb ...


1

Let $M,N$ be two $R$-modules such that $M \otimes N$ is free of rank $>0$. Let $p : F \to M$ be any epi with $F$ free. Then $N \otimes p : N \otimes F \to N \otimes M$ is epi, hence it splits. Then also $M \otimes N \otimes p$ splits, which implies that $M \otimes N \otimes M$ is a direct summand of $M \otimes N \otimes F$ and is therefore projective. ...


4

The map $A\rightarrow M$ is an epimorphism in your construction, since the image of $\ker(B\rightarrow C)\rightarrow N$ is $\ker(N\rightarrow K)$, which is the image of $M\rightarrow N$.


0

For the inclusion $i : A \to B$ you have an induced map $i^* : A^e \to B^e$ and you may regard every $B^e$ module as an $A^e$ module, so you have an induced map $i_0^* : M \otimes_{B^e} N \to M \otimes_{A^e} N$ for every right $B^e$ module $M$ and left $B^e$ module $N$. Here you must note that tensor product is contravariant with respecto to morphisms of ...


1

The answer to your question is that you don't view those things as $kG$-modules. To construct the Bockstein, you identify $\operatorname{Ext}_{\mathbb{F}_pG}(\mathbb{F}_p, \mathbb{F}_p)$ with $\operatorname{Ext}_{\mathbb{Z}G}(\mathbb{Z}, \mathbb{F}_p)$. You have a short exact sequence of $\mathbb{Z}G$-modules with trivial action $$ 0 \to \mathbb{F}_p \to ...


1

With regard to your first question: in that section, the assumption given at the start (that $M$ is a projective $R$-module) is in force throughout. This makes $ -\otimes_R M$ exact, since projective modules are flat. Otherwise you are right that there is no reason to think the sequence will remain exact after tensoring with $M$. For your second question, ...


3

Here is the case for abelian groups: Theorem: If $M$ is an abelian group with $M \otimes M \otimes M = 0$, then $M$ is a divisible torsion abelian group and $M \otimes M = 0$. The proof is pretty standard abelian group theory. Basic subgroups are probably not well known outside of that theory (at least I never learned about them over rings that weren't ...


1

Hint: $F/mF\cong F\otimes_R R/\mathfrak{m}$.


0

Let $J$ be any submodule containing $X$. Then show that $J$ contains span$(X)$. This is not very difficult: use the definition of span($X$) to show that since $J$ is a module containg $X$ it should contain span$(X)$.


1

Suppose $S$ spans $G$ as $\mathbb{Z}$-module. $G$ is contained in a free $\mathbb{Q}$-module $V$; $S$ spans $V$ over $\mathbb{Q}$, so $\operatorname{rank}(V) \leq |S| $ . Suppose $B$ is a $\mathbb{Z}$-basis for $G$; then $B$ is linearly independent over $\mathbb{Z} \Rightarrow $ $B$ is obviously linearly independent over $\mathbb{Q} $ $\Rightarrow |B| \leq ...


3

Update: after thinking about it for awhile, I remembered I already gave an elementary answer here. Basically, it says that the Ore condition (given here) allows you to do row reduction on matrices, generating a contradiction. Here is one elementary-ish solution. If you've studied noncommutative algebra, hopefully you've read about the problem of which ...


1

For bases of infinite cardinality it is clear. Use induction in the other cases, so that if $F$ and $F'$ are bases with cardinality $m $ resp. $n$, then induct upon $max(m,n)$. The base case is clear from part 1 of the exercise. So suppose it is true for m-1, and let us prove it for m. Then say $F=R^m$ and $F'=R^n$ and let $\phi:R^m \rightarrow R^n$ be ...


0

He is trying to introduce the zyzygys. Read about this in Rotman's book "An introduction to homological algebra", he does the details.


1

I've learned the answer to my question, which I write here for future questioners: First, we assume that $\Lambda$ is Artinian, so that any finitely generated $\Lambda$-module has a composition series, and also any submodule of a finitely generated module is itself finitely generated. We first show that any finitely generated module has a unique projective ...


2

Take $A=\mathbf Z$. Are $\mathbf Z$ and $\mathbf Q/\mathbf Z$ isomorphic?


2

There are also many differences between vector spaces and $\mathbb{Z}$-modules. Every vector space has a basis, but not every $\mathbb{Z}$-module. For example, any finite abelian group is not a free $\mathbb{Z}$-module, and the $\mathbb{Z}$-module $\mathbb{Q}$ is not free. Furthermore a free $\mathbb{Z}$-module may have a linear independent set which cannot ...


1

$\newcommand{\Z}{\mathbb{Z}}$This is a problem of computing extension groups $\mathrm{Ext}_R^1(M/N, N)$. $N$ being free doesn't imply $M \cong M'$. Consider $M = \Z \oplus \Z/2\Z$, $N = \Z \oplus 0 \subset M$ and $M' = \Z$, $N' = 2 \Z \subset M'$. Then both $N \cong N' \cong \Z$ are free, $M/N \cong M'/N' \cong \Z/2\Z$, but $M \not \cong M'$. If $M/N$ is ...


2

I like Osborne's Basic Homological Algebra very much. Check chapter 2 for a gentle introduction to proj. / inj. & flat modules. You can find some exercises at the end of it (check in particular ex. 11, aka the "flat test lemma").


1

Consider any nonzero x and the associated map $xr\mapsto r$. Since R is a domain, this is well defined. Notice 1 is in the image of the map. Apply Baer's criterion to prove x is a unit.


3

You say:"If $A[x]$ is a $A$-module,then it's the ideal of $A$." this is not true. ideals of $A$ are $A$-modules, but the converse is not always true. for the 2nd question: positive powers of $x$ are $\ge$ $n$ or are $1,2,...,n-1$. in each case they lie in the $A$-module generated by $1, x, .., x^{n-1}$ because $x^{n+r}=...$


1

Update: The previous answer was incorrect - the following is a slightly different approach. $\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{\Ass}{\operatorname{Ass}}$ $\newcommand{\Supp}{\operatorname{Supp}}$ The relevant facts to answer this question are (assuming $M$ is a finite $R$-module): 1) $\Hom$ localizes, in the sense that $\Hom_R(R/I,M)_p ...


2

If you define action of $\mathbb Z$ on $\mathbb Q$ by $q\to nq$ for $q\in \mathbb Q$ and $n \in \mathbb Z $ it is trivial to check that $\mathbb Q$ is an $\mathbb Z$ module. Now, Let $M$ be maximal submodule of $\mathbb Q$ then $\mathbb Q/M$ is an simple $\mathbb Z$ module. You can also say that it is cyclic module since by simplicity the module generated ...


0

I'm gonna go ahead and make my comment a leading answer. If we let the set of submodules $\mathcal{S}=\{B\subseteq A\mid A/B \text{ not cyclic } \}$ be partially ordered by inclusion, we should try to see if Zorn's lemma applies, so that we can extract maximal elements. First of all, the statement "$A$ is not cyclic" just means that $\mathcal{S}$ contains ...


4

Hint: $P\oplus P'=R^{(\Lambda)}$ (a free module) for some set $\Lambda$ and some module $P'$.


3

This is not possible. A ring whose left modules are free is a division ring, and so all of its right modules are also all free. This is not hard to see: let $S$ be a simple left $R$ module. Since it's free, it is a direct sum of copies of $R$. Since it's simple, it cannot be a sum of more than one copy. This means that $R$ itself is a simple left $R$ ...


0

No. Let $Ass(R)\neq Min(R)$ and $R=M$. note that $⋃_{p∈Ass R}p=zd(R)$ (the set of zerodivisors of R) example: $R=k[X,Y]/(XY,Y^2)$. then $Min(R)=(\bar Y)$ , $\bar X$ $\notin(\bar Y)$ and $\bar X$.$\bar Y = 0$ but $\bar Y \neq 0$ (here $\bar Y$ is your question's $x$ and $\bar X$ is your question's $a_{p}$ and $(\bar Y)$ is your question's $p$)


1

This is the definition for "simple module" adopted for rings which do not necessarily have identity and/or modules which are not assumed to be unital. For rings with identity, we usually assume that $1\cdot m=m$ for all $m\in M$, and in that case, the definition can omit the part about $RM\neq \{0\}$. I'm not sure where you encountered the definition, but I ...


13

$\def\id{\operatorname{id}}$Suppose $M\otimes N$ is isomorphic to $R^n$. Pick a basis $\{x_1,\dots,x_n\}$ of $M\otimes N$, with $x_i=\sum_{j=1}^{r_i}m_{i,j}\otimes n_{i,j}$ for each $i\in\{1,\dots,n\}$. Let $r=r_1+\cdots+r_n$, let $\{e_{i,j}:1\leq i\leq n, 1\leq j\leq r_i\}$ be a basis of $R^r$, and consider the map $f:R^r\to M$ which maps $e_{i,j}$ to ...


0

You don't need to assume anything about the modules neither the ring. $\varphi \otimes \varphi : M \otimes M \to N \otimes N$ is te composition of the injective maps $1_N \otimes \varphi$ and $\varphi \otimes 1_M$. It is easy to generalize this for the k-th tensor.



Top 50 recent answers are included