# Tag Info

10

No, your argument isn't correct: if $1\otimes x=1\otimes y$ then not necessarily $x=y$. I'd do this as follows: if $M$ is a $\mathbb Z$-module, then $\mathbb Q\otimes_{\mathbb Z}M\simeq S^{-1}M$, where $S=\mathbb Z\setminus\{0\}$. The isomorphism is given by $\dfrac ab\otimes x\mapsto\dfrac{ax}{b}$. This way $1\otimes(1,1,\dots)$ corresponds to the fraction ...

10

In general there is no reason for $f$ to be an $R$-module homomorphism just because it is an abelian group homomorphism. Consider complex conjugation $\bar{\cdot} : \mathbb{C} \to \mathbb{C}$. It is clearly an abelian group morphism since $\overline{z+z'} = \bar{z} + \bar{z}'$. But if you take $R = \mathbb{C}$, it's clearly not an $R$-module homomorphism, ...

9

No, for two reasons. First, $\frac{1}{q} m$ must be a $q^{th}$ root of $m$, but such a $q^{th}$ root need not exist: for example take $M = \mathbb{Z}$. Second, if $m = qn$ then $\frac{1}{q} m = \frac{1}{q} qn = n$, so $q^{th}$ roots also need to be unique, and even when they exist in some abelian group they need not be unique: for example take $M = \mathbb{Q}... 9 A simple counterexample is the ring$\mathbb{Z}$, and the$\mathbb{Z}$-modules$M=\mathbb{Z}/2\mathbb{Z}$and$N=\mathbb{Z}/3\mathbb{Z}$. There is no non-trivial$\mathbb{Z}$-module homomorphism from$M$to$N$, because there is no element of$N$that has order$2$. (One could replace$2$and$3$with any relatively prime integers.) Another counterexample ... 8 Proposition:$\mathbb{Q}$is flat as a$\mathbb{Z}$-module. (Which is to say, the functor$ \_ \otimes_{\mathbb{Z}} \mathbb{Q}$is exact.) Pf: The easiest is to observe that tensoring with$\mathbb Q$is the same as localizing at$\mathbb{Z} \setminus \{0\}$, and localization is exact. Fancier proof:$\mathbb{Q}$is the filtered colimit of the free modules ... 8 The only other examples of non IBN rings that I am aware of are built using Leavitt path algebras, and they do exactly what you want. It is possible to specify positive integers$n<m$such that$R^i\ncong R^j$for$(i,j)$less than$(n,m)$in the lexicographic order, but$R^n\cong R^m$as modules. (The$(m,n)$pair is called the 'module type' in the ... 8 Sometimes it is better to look at a more general situation. In fact, this makes it easier to see what is really going on. 1) Let$V$be a$K$-vector space and let$L/K$be a field extension. Then$L \otimes_K V$carries the structure of a vector space over$L$via (linear extension of)$\alpha(\beta \otimes x) = \alpha\beta \otimes x$. 2) If$\{b_1,\...

8

If I understand your question correctly, it seems to me that $X$ is the unique smooth projective curve with function field $K$. Your question then boils down to the following. Lemma. Let $X$ be a smooth projective curve, and let $x_1, \ldots, x_r \in X$ be distinct points. Set $U = X \setminus \{x_1, \ldots, x_r\}$. Then $\Gamma(U, \mathcal{O}_U)^\times/... 7 It is enough to prove it preserves short exact sequences:$\;0\to M\to N\to P\to 0$. As the tensor product is right-exact, and$S^{-1}M\simeq M\otimes_A S^{-1}A$, it is even enough to prove it preserves injectivity. So consider an injective morphism$\varphi\colon M\to N$and suppose$\;S^{-1}\varphi\Bigl(\dfrac ms\Bigr)=0$in$S^{-1}N$. This means there ... 7$\mathbb{Z}$-modules are precisely abelian groups. As every ring is an abelian group, it is a$\mathbb{Z}$-module. It is entirely possible to be a module over more than one ring. For example, if$M$is an$R$-module then it is also an$S$-module for any subring$S$of$R$(you seem to be interested in the case where$M = R$). Another example is given by$\...

7

For question 2, $F\otimes_A -$ has a left adjoint iff $F$ is finitely generated, and the left adjoint is always exact. For if $(f_i)\in F^I$ is an element of an infinite product of copies of $F$, then it is easy to see $(f_i)$ is in the image of the canonical map $F\otimes_A A^I\to F^I$ iff $\{f_i\}$ is contained in a finitely generated submodule of $F$. ...

7

This is not true in general. For example, $\mathbb{Q}\otimes_{\mathbb{Z}}(\mathbb{Z}/p\mathbb{Z}) \cong 0 \cong 0\otimes_{\mathbb{Z}}(\mathbb{Z}/p\mathbb{Z})$ but $\mathbb{Q}$ and $0$ are not isomorphic as $\mathbb{Z}$-modules.

7

Zorn's lemma tells you that if you consider the set of all submodules, then this set has a maximal element. It does not tell you whether every nonempty collection contains this maximal element, so you cannot conclude that the collection has one. Now, the collection indeed does have an upper bound (as you've remarked, $M$ is one such submodule, as it is a ...

7

I don't know if you consider this a special name, but the ring $\mathbb{Z}[\zeta]$ is the ring of integers for the number field $\mathbb{Q}(\zeta)$. This is (IMO, anyway) a nontrivial fact, but you can find its proof in Neukirch (see below), or (for the case $n$ prime) in Samuel's ''Algebraic Theory of Numbers''—this link might also be helpful to you. ...

6

For every nonzero $x, y\in\mathbb{Q}$, there are nonzero integers $m, n$ such that $$mx=ny$$ (where $mx, ny$ are interpreted in the obvious way). Now, any homomorphism $f$ between $(\mathbb{Q}, +)$ and another structure $(G, *)$ must preserve multiplication by integers: $$f(mx)=mf(x).$$ So if $(\mathbb{Q}, +)\cong(\mathbb{Q}^n, +)$, then $(\mathbb{Q}^n, +)$ ...

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 ...

6

No. Let $G$ be any non-zero abelian group. Then the obvious morphism $(0,G)\to (G,G)$ is a monomorphism and epimorphism, but not an isomorphism, which can't happen in an abelian category.

6

This is never true unless $\mathfrak{m}=0$, i.e. unless $A$ is a field. Let $x\in\mathfrak{m}$, and consider the short exact sequence $0\to (x)\to A\to A/(x)\to 0$. If $A/\mathfrak{m}$ is flat, we can tensor it with the short exact sequence to get a short exact sequence $0\to (x)/\mathfrak{m}(x)\to A/\mathfrak{m}\to A/\mathfrak{m}\to 0$. Exactness of this ...

6

(I'm implicitly assuming that $R$ is a commutative ring.) An $R[x_1,\ldots,x_n]$-module is the same thing as an $R$-module equipped with $n$ mutually commuting endomorphisms. One way of putting it is that a left $S$-module $M$, where $S$ is a (not necessarily commutative) $R$-algebra, is the same thing as an $R$-module $M$ together with an $R$-algebra map ...

6

Let $\mathbb{F}$ be a field and let $V$ be a module over $\mathbb{F}$ (i.e. a vector space over $\mathbb{F}$), then $V^*$ is also a module over $\mathbb{F}$. The tensor product $V^*\otimes_{\mathbb{F}}V$ is canonically isomorphic to $\operatorname{End}_{\mathbb{F}}V$ via the map induced by the bilinear map $V^*\times V \to \operatorname{End}_{\mathbb{F}}(V)$,...

6

No. Free abelian groups don't have any nonzero divisible elements, but every element of $\mathbb{R}$ is divisible. $\mathbb{R}$ is free as a $\mathbb{Q}$-module (this requires the axiom of choice), since $\mathbb{Q}$ is a field.

6

For commutative rings there is a geometric way to think about these things. Every module $M$ over a commutative ring $R$ has a support, which is the set of prime ideals $P$ such that the localization $M_P$ at $P$ is nonzero. This use of "support" is analogous to the notion of support of a function: it's "where the module is nonzero." For example, when $R = ... 6$\prod_{i=1}^\infty \mathbb{Z}$is a counterexample. It is torsion free, not free (Why isn't an infinite direct product of copies of$\Bbb Z$a free module?), and every nonzero element is only divisible by finitely many integers, so your extra hypotheses hold. 6 This is a solution to (d), and partially for (b). We use Cauchy Matrix, and its evaluation of inverse given by Schechter: If$T$is a$n\times n$Cauchy matrix on the sequences$\{x_i\}$,$\{y_j\}$, then$S=T^{-1}=[s_{ij}]$is given by: $$s_{ij} = (x_j - y_i) A_j(y_i) B_i(x_j)$$ where $$A_i(t) = \frac{A(t)}{A^\prime(x_i)(t-x_i)} \quad\text{and}\quad ... 6 Sure, the keyword is "multivariable Chinese remainder theorem." A nice exposition can be found here. The general theorem states that if all mods m_i are relatively prime and each row i has an a_{ij} that is relatively prime to m_i, then there's going to be at least one solution. The mechanical way of solving the system can be done by row reduction, ... 6 Please see the following paper by Bhargava and Satriano here, which contains the details of this computation and other related results. M. Bhargava and M. Satriano, On a notion of “Galois closure” for extensions of rings, Journal of the European Mathematical Society 16, 1881-1913 (2014). The aim of the authors is to define for commutative ring ... 5 Since M=\mathbb Z/n as a group is cyclic generated by 1, an endomorphism \psi:M\rightarrow M is determined by \psi(1). If \psi^2=-I_M, then a^2=-1 in M. Thus, you're looking for all n such that there is a\in \mathbb Z such that a^2 \equiv -1 \bmod n. For instance, when n is prime, this means that n \equiv 1 \bmod 4. But there are ... 5 If m = n, then 0 \to \mathbb{Z}/n\mathbb{Z} \xrightarrow{\operatorname{id}} \mathbb{Z}/n\mathbb{Z} \to 0 is a free resolution. If m < n, it will be useful to write n = km, where k > 1. The first step of the resolution looks like F_0 \xrightarrow{\epsilon} \mathbb{Z}/m\mathbb{Z} \to 0 for some free \mathbb{Z}/n\mathbb{Z}-module F_0. ... 5 \text{Ext}^1(-, -) sends direct sums in the first variable to direct products, and as mentioned in the comments, as an abstract abelian group$$S^1 \cong \left( \bigoplus_X \mathbb{Q} \right) \oplus \mathbb{Q}/\mathbb{Z}$$where$X$is uncountable. So it suffices to compute$\text{Ext}^1(\mathbb{Q}, \mathbb{Z})$and$\text{Ext}^1(\mathbb{Q}/\mathbb{Z}, \...

5

A ring $R$ for which every finitely generated right $R$-module has finite projective dimension is sometimes called right regular. Such rings are discussed in McConnell, Robson, Noncommutative Noetherian Rings, Chapter 7.7. In particular, Example 7.7.2 (6.4.9) gives an example of a ring of infinite global dimension which is left and right regular. The ring is ...

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