Hot answers tagged modules
7
If $I$ is an ideal of $R$, the dual of $R/I$ is isomorphic to $\mathrm{Ann}(I) = \{r \in R : rI = 0\}$, and this doesn't have to be finitely generated. Take for instance $R = k[y,x_1,x_2,\dotsc]/(y x_i : i \geq 1)$ and $I=(y)$.
A more natural question would be: How can we characterize commutative rings with the property that duals of f.g. modules over that ...
6
The key thing to remember about the operation "mod" is that it behaves "well" with respect to product (hence powers), and of course addition. This means that if you can simplify your life a lot distributing the calculation into many steps and taking "mod" at each stage.
To compute $20^{15}$, you can first notice that $20 \pmod{17} = 3$. Then $20^{15} ...
6
$\require{AMScd}$The factorization you ask about is an instance of the slogan "distinguished triangles are just like exact sequences".
To wit, the axiom on existence of morphisms yields that in a distinguished triangle
\begin{CD}
X @>u>> Y @>v>> Z @>w>> X[1]
\end{CD}
$v$ is a weak cokernel of $u$. For if $f \colon Y \to W$ is a ...
5
First check $\bf C$-linearity (this should be obvious). The only other step needed to check that $\Gamma$ is a homomorphism of ${\bf C}[G]$-modules is to check that it is $G$-equivariant. We write out
$$\quad \Gamma(h m)=\sum_{g\in G}\phi(g^{-1}h m)g $$
Consider replacing $g$ with $hg$ (this is merely a substitution) so that the sum looks different. What ...
5
The exact sequence $0 \to I/I^n \to R/I^n \to R/I \to 0$ shows that it suffices to prove that $I/I^n$ and $R/I$ are noetherian $R/I^n$-modules, as you have already observed.
Since $R/I$ is noetherian over $R/I$, it is also noetherian over $R$ and over $R/I^n$ (the submodules with respect to all these rings are just the same (as sets)).
To show that $I/I^n$ ...
4
To understand $\Hom$, think about generators. Since any generator of $\mathbb{Z}_{p^\infty}$ is torsion, it has to map to $0$ in $\mathbb{Z}$. So,
$$
\Hom(\mathbb{Z}_{p^\infty}, \mathbb{Z}) = 0.
$$
In order to understand $\Hom(\mathbb{Z}_{p^\infty}, \mathbb{Z}_{p^\infty})$, consider the definition of $\mathbb{Z}_{p^\infty}$ as a direct limit.
$$
...
4
Take $f: A \rightarrow B$ to be a ring homomorphism such that the corresponding morphism of affine schemes $Spec B \rightarrow Spec A$ is not surjective, but only flat. There is an easy way to do this: Remember that localizing a ring R in a multiplicative subset $S$ gives a flat ring homomorphism $R \rightarrow S^{-1}R$. However, this ring homomorphism is ...
4
Here are counterexamples showing what happens when we drop any of the hypotheses listed by David Speyer in the comments.
$G$ finite: The key point here is that characters are only defined for $V$ finite-dimensional, and it is possible to write down infinite groups $G$ which have no nontrivial finite-dimensional representations whatsoever. See, for example, ...
4
I guess you really mean $n>1$, and that is rather a lot of questions. I'll try to hit them all.
Well the nicest thing about $\Bbb Z/n\Bbb Z$ is that it's a quasi-Frobnius ring meaning that it's Artinian and self-injective. This is in fact equivalent to the projective modules being precisely the injective modules. Actually, all of these are equivalent:
...
4
As I already mentioned in the comment, the projective objects get isomorphic to $0$ since the identity map $P\to P$ obviously factors through a projective module and therefore the identity map is the zero map, which is only true for the zero module.
As you can see from that since every projective module gets "killed" when going to the stable module ...
3
Just for the fun of it, let me put this in a broader context:
Let $A$ be a ring (with $1$), and let $B$ be a ring (with $1$) equipped with a morphism $A \to B$ (respecting $1$). If $N$ is a left $A$-module, then Hom_A(B,N) (Hom of left $A$-modules) is naturally a left $B$_module. (We define $b \cdot \phi$, for $b \in B$ and $\phi$ in the Hom set,
by $(b ...
3
Your first part is correct.
As for the second part, think bigger: show that any two elements of an ideal $I$ in a commutative ring are linearly dependent. (Hint: this is just as easy as the first part!)
As for the third part: in view of the second part, in any commutative ring the ideals which are free as $R$-modules are necessarily principal ideals. (In ...
3
Let me expand on the comment I made (the other answer is great, by the way. But, I did have fun working it all out by hand).
First, since $M$ is cyclic it is of the form $\mathbb{Z}/m\mathbb{Z}$ for some $m \mid n$ and $m > 1$. I claim that $M$ is projective (resp. flat, injective) if and only if $(m,n/m) = 1$.
Suppose that $m\mid n$ and $m'\mid n$. The ...
3
Here is an abstract way to think about this (which might not be what you're looking for):
The condition that $f\circ v=g\circ v\Rightarrow f=g$ for all $f,g\in\operatorname{Hom}(M'',N)$ says that $v$ is right-cancellative, that is, $v$ is an epimorphism in the category of $A$-modules. Epimorphisms in the category of $A$-modules are surjective on underlying ...
3
I think I have found the solution using Zach L's hint.
Let $N=\operatorname{coker}(v)=M''/\operatorname{Im}(v)$, and let $p\in\operatorname{Hom}(M'', N)$ be the canonical map $p: M''\to M''/\operatorname{Im(v)}=N$. We observe for every $x\in M$, we have
$$p(v(x))=v(x)+\operatorname{Im}(v)=0+\operatorname{Im}(v)=0_{M''/\operatorname{Im(v)}}$$
So $p\circ ...
3
A possibly instructive, concrete example is $\Bbb{Q}$ as an abelian group (i.e. as a $\Bbb{Z}$-module). Of course, this is just a special case of Dedalus's answer (and YACP's comment), but let's still prove the result directly.
Clearly $\Bbb{Q}$ is not a faithfully flat abelian group, e.g. the zero morphism
$$
\Bbb{Z}/2\Bbb{Z} \rightarrow 0
$$
is not ...
2
If the extension $S\subset R$ is finite, then every Noetherian, respectively Artinian and finitely generated $R$-module is Noetherian, respectively Artinian and finitely generated as an $S$-module.
If the extension is not finite, then there is no much hope to get this. For instance, consider $R=\mathbb Z\subset \mathbb Q=S$ and $M=\mathbb Q$. Then $M$ is ...
2
Well, since in this case $R$ is commutative, it satisfies the invariant basis number property, so that it is meaningful to speak of the rank of $M$. This is probably what is meant by the "dimension" of $M$ over $R$, i.e., the number $n$ such that $M \cong R^n$. In this case a basis of $M$ means a set of $n$ linearly independent elements that generate $M$.
2
Hints:
$$20=3\pmod{17}\;,\;\;16=-1\pmod{17}\implies$$
$$20^{15}+16^{18}=3^{15}+(-1)^{18}=(**)$$
But for any integer $\,a\;,\;\;(a,17)=1\,$ , we have that $\,a^{16}=1\,$ , so
$$(**)=3^{16}\cdot 3^{-1}+1=1\cdot 6+1=7\ldots$$
and so the claim is false: the remainder is $\,7\,$ .
2
Let $R$ be a PID $M$ a free $R$-module of finite rank $n$. Let $\{x_1,\dots,x_k\}$ be $k\leq n$ elements in $M$, and denote $N$ their $R$-span in $M$. Then the claim is that the $k$-ple $\{x_1,\dots,x_k\}$ can be completed to a basis of $M$ over $R$ if and only if
$$
rk(N)=k\qquad\text{and}\qquad\text{$M/N$ is torsion-free.}
$$
Indeed, if ...
2
As far as I know, field should not matter, unless your algebra has "weird relations", and in most case, you are pretty safe as long as characteristic of $k$ is not 2. I hope somebody can give some supplement answer here...
Back to your specific example, (say if we work over characteristic 0..) the (minimal) projective resolution of $S(1)$ is actually:
$$
...
2
(1) No, $R$ does not have to be a subset of $M$. The map $R \times M \rightarrow M$
does need to satisfy some properties, namely:
$\bullet$ For all $a \in R$ and all $x,y \in M$, $a(x+y) = ax + ay$.
$\bullet$ For all $a,b \in R$ and all $x \in M$, $a(bx) = (ab)x$.
$\bullet$ For all $x \in M$, $1 x = x$.
When you talk about "the map $\mathbb{Q} \times ...
2
To say that $M$ contains $k$ linearly independent elements $\{x_1, \cdots, x_k\}$ is to say that the homomorphism $R^k \hookrightarrow M$, defined by sending the $i$th basis element to $x_i$, is injective. As you mention above, if $M$ is generated by $n$ elements and contains $n+1$ linearly independent elements, it follows that we may "lift" to $R^n$ so ...
2
I think it could be easier to work directly with the basic definition of injectivity, i.e. that if $A \hookrightarrow B$ is an embedding, then any $A \to J$ extends to $B$.
Now you are supposed to reduce to the case when $A \to J$ is surjective,
and $B$ is projective. You can always just add a copy of $J$ to $A$ and $B$ to
get surjectivity. And you can ...
2
Injectivity. Let $\frac{r_1}{s_1},\frac{r_2}{s_2}\in S^{-1}B$, and suppose
that
\begin{align*}
\phi\left(\frac{r_1}{s_1}\right)&=\phi\left(\frac{r_2}{s_2}\right) \\
\frac{r_1}{f(s_1)}&=\frac{r_2}{f(s_2)}
\end{align*}
So $\exists$ $t\in T$ such that
\begin{equation*}
t\left(f(s_2)r_1-f(s_1)r_2\right)=0
\end{equation*}
Since $t\in T=f(S)$, $\exists$ ...
2
If you do a search in Amazon for "consequences of the axiom of choice",
the first responses are:
Consequences of the Axiom of Choice (Mathematical Surveys and Monographs) by Paul Howard ($108.00)
The Axiom of Choice (Dover Books on Mathematics) by Thomas J. Jech ($11.05, also Kindle)
Zermelo's Axiom of Choice: Its Origins, Development, and Influence ...
2
If you ask the question
if $M$ is an $R$-module of projective dimension $n$, is $\operatorname{Ext}^n(M,R)\neq0$?
then the answer is surpring: it depends. When $R=\mathbb Z$, this is essentially known as Whitehead's problem and Shelah proved that its answer depends on the specific set theory that you choose. Indeed, he showed that depending on the ...
2
First note that that the condition $A \otimes_R K \to A \otimes_R F$ is injective for all free $R$-modules $F$ and submodules $K$ is equivalent to $A$ being flat. So you will definitely need to assume that $A \otimes_R I \to A \otimes_R R$ is injective for all ideals $I \subseteq R$.
We will prove that $\operatorname{Tor}_1^R(A, F/K) = 0$ for any such $F$ ...
2
When you do the operation $4 R_2 - 5 R_1$, you are altering the determinant by a factor of $4$, since you are multiplying your matrix on the left by
$$
\begin{bmatrix}1 & 0 & 0\\-5&4&0\\0&0&1\end{bmatrix}.
$$
Ditto for $3 R_1 + R_2$, which alters the determinant by a factor of $3$. These two operations have introduced the spurious ...
2
The "adjoint" here means "classical adjoint". For a matrix $A$, the classical adjoint is the matrix whose $i,j$ entry is $(-1)^{i+j} A_{ji}$, where $A_{ji}$ means the $j,i$ cofactor---the determinant of the matrix obtained by omitting row $j$ and column $i$. The point is that the product of the classical adjoint and $A$ is a diagonal matrix with ...
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