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12

Nakayama Lemma. Let $N$ be a finitely generated $R$-module, and $J\subseteq R$. Suppose that $J$ is closed under addition and multiplication and $JN=N$. Then there is $a\in J$ such that $(1+a)N=0$. (Here by $JN$ we denote the subset of linear combinations of $N$ with coefficients in $J$.) Cayley-Hamilton Theorem. Let $A$ be a commutative ring, $I$ an ...

10

No: For a local ring $(R,{\mathfrak m})$ with ${\mathfrak m}^2=0$ you have $\text{Hom}_R(R/{\mathfrak m},R)\cong\{x\in R\ |\ {\mathfrak m}x=0\}={\mathfrak m}$, so it suffices to choose $R$ such that ${\mathfrak m}$ is not finitely generated, e.g. $R := {\mathbb k}[x_1,x_2,\ldots]/(x_i^2, x_i x_j)$.

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 ...

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

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 $M = ... 8 Sometimes it is better to look at a more general situation. In fact, this makes it easier to see what is really going on. 1) Let$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$. 2) If ... 7 It is enough to prove it preserves short exact sequences:$\;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$. ... 7 Proposition:$\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}$is the filtered colimit of the free modules ... 7 The only other examples of non IBN rings that I am aware of are built using Leavitt path algebras, and they do exactly what you want. It is possible to specify positive integers$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)$pair is called the 'module type' in the ... 7 This is not true in general. For example,$\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 $is one such submodule, as it is a ... 6 For the second question, the finite rings are precisely those for which every finitely generated module has finitely many submodules. For a ring$R$and$r\in R$, let$M_r$be the submodule of$R\oplus R$generated by$(1,r)$. Then$(1,r)$is the only element of$M_r$whose first coordinate is$1$, and so$M_r\neq M_s$for$r\neq s$, and so if$R$is ... 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 ...

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 $R = ... 5 Hint: If one can find a pair of cardinals$\kappa$and$\lambda$such that$\kappa\neq\lambda$but$2^\kappa=2^\lambda$, then$\mathbb Z_2^{(\kappa)}$and$\mathbb Z_2^{(\lambda)}$give a counterexample. Since ZFC theory doesn't violate the existence of such pair, I think your conjecture is not true, but I'm not sure if it's false... By the way, this ... 5 Since you are considering not necessarily commutative ring and thus is forced to taking hom-set and tensor product of abelian group, it's not reasonable to expect that$\hom_R(M,M')\otimes\hom_R(N,N')$and$\hom(M\otimes_RN,M'\otimes_RN')$are comparable. For example, taking$M:=R_R,N:=_RR$, then the two become$R\otimes_{\mathbb Z}\hom_R(M',N')$and ... 5 It is still true because: Any$A$-module is the direct limit of its finitely generated submodules (for any ring$A$). Tensor products commute with direct limits. Direct limit is an exact functor. Btw, a submodule of an$A$-module with this property (the morphism$M'\hookrightarrow M$is universally exact) is called a pure submodule of$M$, and the ... 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 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$\text{Ext}^1(-, -)$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 ...

5

Consider $\mathbb{Z}/2\mathbb{Z}$ over $\mathbb{Z}$. Why does it not have a basis?

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