# Tag Info

6

From the definitions, there's no reason to expect projective/injective objects of a subcategory to be projective/injective in the ambient category. If $P, B \in \mathcal{A}$, then any map $P \to B$ factors through any epi $A \twoheadrightarrow B$ with $A \in \mathcal{A}$, but if $A, B$ are not in $\mathcal{A}$, there is no reason to expect a lift. For a ...

3

An object $P$ is projective when, given any morphism $f: P \to A$, and any epimorphism $g: B \to A$, there is a unique morphism $h: P \to B$ such that $g \circ h = f$. Even if an object $P$ is projective in a full subcategory of R-mod, then, it might not be projective in R-mod itself, because in that case the objects $A$ and $B$ might be outside of the ...

3

$4^3$ ("Qubic") is a win for the first player. According to this link, it was first proved by Oren Patashnik in 1980. The proof is complicated. It took 12 years for this proof to be converted into a practical computer algorithm; I was present at the 1992 Computer Olympiad where the program of Victor Allis and Patrick Schoo romped to victory.

3

How about using the Cayley Hamilton theorem?

3

Let the dimension of $V$ be $n$. Then the dimension of $\End(V)$ is $n^2$. Hence the endomorphisms $I, A, A^2, \dots A^{n^2}$ must be linearly dependent. The coefficients of the dependence relation are the coefficients of a polynomial $f$ for which $f(A) = 0$. As pointed out in Voldemort's answer, by the Cayley-Hamilton theorem, you can actually stop at $n$ ...

3

As I said in the comments your claim is false. Here is an explicit counterexample: Let $G = \{ \pm 1 \}$, and let $B = \mathbb{Z}/4$, $C = \mathbb{Z}/2$ be $G$-modules in the obvious way. Then $h : B \to C$ with $h(1) = 1$ is an epimorphism of $G$-modules, but $h^G : B^G \to C^G$ is the zero map.

2

Let $R$ be a ring with $1$. If $M$ is a group, written multiplicatively, equipped with an $R$-action, then, for any $x,y\in M$, $(xy)^2 = (1+1)\cdot (xy) = [(1+1)\cdot x][(1+1)\cdot y] = x^2 y^2$. So $yx=xy$, and $M$ is abelian. If $R$ doesn't have a $1$, then we conclude only that $rM$ is abelian for all $r\in R$. In this case, we might be able to ...

2

Aside from the observation that has already been made in other comments that the other module axioms already force a module to be abelian, let me make the following philosophical remarks. Rings naturally want to act on abelian groups, in the following sense: you can make sense of the notion of monoid $M$ in any monoidal category $(V, \otimes)$, and once you ...

2

Since $I$ and $J$ are maximal, the modules $R/I$ and $R/J$ are simple. A module $M$ is simple when its only submodules are $\{0\}$ and $M$; simplicity of $R/I$ is exactly the same as maximality of $I$, by the homomorphism theorems. If $f\colon R/I\to R/J$, then you have just two possibilities: $\ker f=\{0\}$; in this case $f$ must be an isomorphism (prove ...

2

This is not true. For example, $\{1\}$ and $\{2,3\}$ are minimal generating sets of the $\mathbb{Z}$-module $\mathbb{Z}$.

2

Based on what I remember from playing such games twenty-five years ago, The $3^3$ version is a guaranteed win for the first player, by going in the middle square. There are so many lines through it that the first player can always force moves. After the 2nd player places his irrelevant O, the first player chooses a plane through the middle X that doesn't ...

2

Let's prove that $\operatorname{Hom}_{\mathbb Z}(R,Q)$ is an injective $R$-module provided $Q$ is an injective $\mathbb Z$-module. In order to do this we shall use Baer's criterion (see Proposition 36(1) from the same book). Let $I\subset R$ be a left ideal and $f:I\to\operatorname{Hom}_{\mathbb Z}(R,Q)$ an $R$-linear mapping. It suffices to find ...

2

The direct limit of $\Bbb Z$-modules: $$\Bbb Z/p\to \Bbb Z/p^2\to \Bbb Z/p^3\to\cdots$$ is not finitely generated as a $\Bbb Z$ module but every proper submodule is isomorphic to $\Bbb Z/p^k$ for some $k$. Edit: For more information, please see this wiki page.

2

What you have shows that for a fixed $m \in M$, there is $s \in S$ so that $sm = 0$. It does not show that $sM = 0$. To show this, you need to use the fact that $M$ is finitely generated. Apply what you have to a generating set of $M$.

2

Well this is certainly wrong as long as you don't assume M to be finitely generated (just take an infinite dimensional vector space). If M is finitely generated this should be true, even without the assumption that supp(M) is finite (which will rather be a consequence). First note that this is obviously true if R is artinian,since M is a quotient of some ...

1

For every $i$ we have $C_i=B_i/A$. Taking direct limit, when $B'$ denotes the limit of $(B_i)$, we get $$C=B'/A,$$ and $B'$ is a submodule of $B$. But we know that $C=B/A$, hence $B'=B$.

1

Given any ascending sequence of sets $X_1\subset X_2\subset X_3\cdots\subset X$ with $\bigcup_{n\ge1}X_n=X$ and any collection of functions $g_i:X_i\to Y$ such that $g_i|_{X_{i-1}}=g_{i-1}$, we can form the map $g:X\to Y$ defined by the relation $g(x)=g_i(x)$ whenever $x\in X_i$. Each function $g_i$ is a bigger and bigger "glimpse" of $g$. One can check ...

1

Let $N_n=\sum_{i=1}^{k}c_iR$; then $D$ is the union of the increasing family of submodules $N_n$ and as such it is its direct limit with inclusions as transition maps. If you consider the restriction $h_n$ of $g_n$ to $N_n$, the given condition translates into the fact that $h_n\colon N_n\to B$ is a family of morphisms compatible with the inclusion maps, so ...

1

When I google "absolutely irreducible module" the second hit I get is a passage in Lam's Exercises in classical ring theory which explains that an irreducible module $M$ over a $k$ algebra $R$ ($k$ a field) is called absolutely irreducible if $M\otimes_k K$ is irreducible over $R\otimes_k K$ for every extension field K of k. I would guess the FL part is a ...

1

Let $R = \mathbb Z_6$ and $M = \mathbb Z_2=\{0,3\}$. Then $2m = 0$ $\ \forall m \in M$, but $2 \ne 0$ in $R$. More generally, if $R$ is not an integral domain - say $xy = 0$ where $x \ne 0\ne y$, then if $M$ is the ideal generated by $y$ viewed as an $R$- module, then we will have $x\cdot m = 0$ $\ \forall m \in M$. $R$ can even be an integral domain: ...

1

There is the following thing you can do. Given a category $C$, the free abelian group $\mathbb{Z}[C]$ on the morphisms of $C$ can be equipped with the following multiplication: two morphisms have product their composition if it exists, and $0$ otherwise. This gives you a not-necessarily-unital ring, the "category ring" of $C$, which has a unit iff $C$ has ...

1

What you're asking is if $S^{-1}R$ is a faithfully flat $R$-module. It isn't, take $R=\mathbb{Z}$, $A=B=C=\mathbb{Z}/2\mathbb{Z}$, and the maps $A\to B$ and $B\to C$ being the zero maps, and consider $S=\mathbb{Z}-\{0\}$.

1

The hint they give isn't aimed at showing that $M$ isn't free, rather that it isn't free of rank one. The point is that if $m \in M$ then the submodule $Rm$ cannot be all of $M$: there is some $x \in [0,1]$ with $m(x)=0$ by the intermediate value theorem, thus every element of $Rm$ vanishes at $x$, and it follows $Rm \neq M$ as $M$ contains functions not ...

1

I assume (as rschwieb points out) that the question concerns the category of $\mathbb{Z}$-modules. 1) $\mathbb{Z}_{p^\infty}$ is the injective hull in the case when $n=p^k$ as well. Basically because we have $\mathbb{Z}_p\subseteq \mathbb{Z}_{p^k} \subseteq \mathbb{Z}_{p^\infty}$ and the extension $\mathbb{Z}_{p} \subseteq \mathbb{Z}_{p^\infty}$ is ...

1

The same idea for $\mathbb{Z}$ extends to a polynomial ring over a field: for the ideal $(x) \subseteq k[x]$, $\{x\}$ and $\{x^2, x + x^2\}$ are both minimal generating sets. In general, given a generating set for an ideal $I$, say $I = (a_1, \ldots, a_n)$, one cannot conclude that $I$ can be generated by a proper subset of the $a_i$, even if $I$ is known ...

1

$C'$ is not a direct summand in $M$: if there is $C''$ such that $C'+C''=M$ and $C'\cap C''=0$, then $|C''|=4$, so $C''$ is either cyclic or Klein. Since the elements of order four are $(2,0)$, $(2,1)$, $(6,0)$, respectively $(6,1)$ it's clear that $C''=\langle(2,0)\rangle=\langle(6,0)\rangle$. But $|C'+C''|=8$, so $C'+C''\ne M$. When $C''$ is of Klein type, ...

1

Another useful characterization of uniform dimension is this: $u.dim(M)$ is the supremum of the set $\{k\mid M \text{ contains a direct sum of$k$nonzero submodules}\}$ (This can be found on page 214 in Lam's Lectures on modules and rings.) This being the case, take a direct sum of $n-1$ submodules of $I_0$ and call it $U_1$. If $U_R:=ann(a_0)\cap I$ ...

1

I think the general statement is that $\sqrt{\operatorname{Ann}_R M}$ is the intersection of the supporting primes (EDIT: never mind, this is false!), and, when $M$ is finitely generated, the minimal supporting primes are associated primes. To see what can go wrong when $M$ is not finitely generated, take $R=\mathbb{Z}$, $M = \bigoplus_{k\geq ... 1 Pick subspaces$S$and$T$of$V$which are both of the same dimension as$V$and such that$V=S\oplus T$. Decompose$End(V)$with respect to that direct sum decomposition. 1 Apologies in advance for this answer. The problem with it is that it is too advanced, and also relies on a lemma that is very similar to your question. I will continue to seek a more elementary answer. Lemma: Every left$R$module over a left Artinian ring$R\$ has a projective cover. Lemma: Every nonzero projective module has a maximal submodule. Lemma: ...

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