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## Hot answers tagged modules

9

Let $M, N$ be $R-$modules. Then the following holds. If $M$ and $N$ is flat, then so is $M\otimes_{R}N$: see related question here. If $M$ and $N$ are projective, then so is $M\otimes_{R} N$. Indeed, writing $M\oplus M'=F,\ N\oplus N'=F'$, for free $R-$modules $F,\ F'$, one has that $$F'':=F\otimes_{R}F'$$ is free (tensor product of free modules) and ...

9

The simpler tensor product $\mathbb{Z}[i] \otimes \mathbb{Z}[i]$ is free abelian on the generators $1 \otimes 1, 1 \otimes i, i \otimes 1, i \otimes i$. As either a left or a right $\mathbb{Z}[i]$-module it is free on two generators $1 \otimes 1, i \otimes i$. This tensor product is the quotient of the above tensor product by the additional relation that $a ... 8 Consider the exact sequence$M \to N \to C \to 0$, where$C$is the cokernel of$v$. Then for any$A$-module$P$we have an exact sequence$\DeclareMathOperator{\h}{Hom} 0 \to \h(C,P) \to \h(N,P) \to \h(M,P)$. By assumption the last map is injective, so we have$\h(C,P) = 0$. Since this holds for all$P$it holds especially for$P = C$, so we conclude$C=0$... 8 If$R=\mathbb{Z}[2i]$, then$\mathbb{Z}[i] \cong R[X]/I$, where$I=(X^2+1,2X-2i)$. So we have$\mathbb{Z}[i] \otimes_R R[X]/I \cong \mathbb{Z}[i][X]/I$. Substituting$Y=X-i$, this is$\mathbb{Z}[i][Y]/(Y^2+2iY,2Y) \cong \mathbb{Z}[i][Y]/(Y^2,2Y) \cong \mathbb{Z}[i] \oplus \mathbb{Z}[i]/2\mathbb{Z}[i]$. We can also see this isomorphism more directly by ... 8 Let$0_M$and${\rm id}_M$denote the zero map and identity map on an$A$-module$M$. We have $$M=0\iff 0_M={\rm id}_M.$$ Since$F$is a functor,$F({\rm id}_M)={\rm id}_{FM}$. Since it's also additive,$F(0_M)=0_{FM}$. 7 Yes, it may even happen that$N'=0$, i.e. that$M$is isomorphic to$M/N$but$N \neq 0$. Take$M=R \oplus R \oplus \dotsc$and$N = R \oplus 0 \oplus 0 \oplus \dotsc$. 7 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 ... 6 No, if the ring is commutative, yes, if the ring is noncommutative. The easiest counterexample is the endomorphism ring$R$of$V=F^{(\omega)}$, where$F$is a field and$F^{(\omega)}$denotes a direct sum of countably many copies of$F$(as vector space). Let homomorphisms act on the left, so$V$becomes a left$R$-module. Then$R\cong R^2$as left ... 6 Given a finitely generated module$M$over a noetherian ring$A$, there exists a filtration of$M$by submodules$M=M_0\supset M_1\cdots \supset M_n=0$such that$M_i/M_{i+1}\cong A/\mathfrak p_i$for some prime ideals$\mathfrak p_i\subset A$(Bourbaki, Commutative Algebra, Chapter IV, §1, Theorem 1, page 261) Now$M$has finite length iff all the ... 6 Here is a proof that doesn't involve going through$\mathbb{Q}$(and works for any PID): The image of$f$is a submodule of a free module, so it is itself free (since$\mathbb{Z}$is a PID). Therefore the short exact sequence$0 \to \operatorname{ker} f \to G \to \operatorname{im} f \to 0$is split, and therefore (general fact about split exact sequences of ... 6 Here's one (class of) example(s). Let us restrict our attention to$\mathbb Z$-modules, i.e. abelian groups. Note that two abelian groups of order$p$are always isomorphic, thus it is sufficient to find a group$P$of order$p^i$(for any$i$) with two subgroups$A$and$B$of order$p^{i-1}$such that$A\not\cong B$. For instance, if$P = \mathbf C_4 ...

6

$M$ is module isomorphic to $R/I$, but that doesn't mean it has suddenly become a field. Now, you can treat the module $M$ as a ring by transferring the structure of $R/I$ back to $M$ through the bijection, but until you do that, $M$ does not have any binary multiplication operation, and hence it does not make any sense to call it a field (or a ring.)

6

If $A$ is any ring, then $A$ and $M_n(A)$ have equivalent categories of modules, and usually $A$ and $M_n(A)$ are not isomorphic. This is the simplest example of a Morita equivalence.

6

The set $\{1\}$ isn't independent, because $2(1)=1+1=0$. (The $2$ here lives in $\Bbb{Z}$, so the "multiplication" is the module action; all $1$s and $0$s live in $\Bbb{Z}/2\Bbb{Z}$.)

6

For the non-free part: Take any two nonzero elements $x, y ∈ ℚ$ and show they satisfy $λx + μy = 0$ for some nonzero $λ, μ ∈ ℤ$, hence any two elements are linearly dependent. Thus, since $ℚ$ is not cyclic, it cannot have a basis. For the non-finitely-generated part: If $ℚ$ was finitely generated then without loss of generality (by finding the common ...

6

Yes, there is such a finite subset. Because $X$ is a generating set for the module, given any $m \in M$ there exist $x_1,\dots, x_k \in X$ and scalars $r_1,\dots,r_k \in R$ such that $m = r_1 x_1 + \cdots + r_k x_k$. Now suppose $m_1,\dots, m_n \in M$ is a finite generating set. Then for each $m_i$ there exist $x^{(i)}_1, \dots, x^{(i)}_{k(i)} \in X$ such ...

6

In general , there is no one big theorem about monoidal closed categories that justifies or illuminates them. The internal hom does what it says on the tin: it lets you represent your homsets as objects of your category. This allows you to state questions and theorems that simply wouldn't make sense in a non-closed setting. My advice would simply be to keep ...

6

Your idea is right: If $x$ has finite order and your group homomorphism is $\phi$, then $\phi(x)$ has an order that is less than or equal to that of $x$. This is true regardless of whether $\phi$ is 1-1 or not. Thus you can't map onto elements of infinite order. When I read the original post, I misread that the indexing was over $p$! But of course, if you ...

5

A prominent counter-example is the following: Take $R := {\mathbb R}[x,y,z]/(x^2+y^2+z^2-1)$, the ring of real-valued polynomial functions on the $2$-sphere, and consider the following short exact sequence: $$(\ast)\quad\quad 0\to P\to R\frac{\partial}{\partial x}\oplus R\frac{\partial}{\partial y}\oplus R\frac{\partial}{\partial z}\xrightarrow{\alpha := ... 5 \mathbb R  over \mathbb Q is a vector space with dimension 2^{\mathbb N}. and also \mathbb R^n  over \mathbb Q is a vector space with dimension 2^{\mathbb N}. So the additive group of \mathbb R is isomorphic with additive group of \mathbb R^n. for example consider f:\mathbb R\rightarrow \mathbb R^n be such isomorphism , now define new ... 5 This is true with added generality, that is in the non commutative world. So, assume \phi\colon R\to S is a ring homomorphism between not necessarily commutative rings. If P_{R} is a projective right R-module, then, if P_{R} is a direct summand of the free right module R^{(X)} for some set X, we get that$$ P_{R}\otimes_{R} S  is a direct ...

5

In fact all submodules of free $\mathbb{Z}$-modules are projective, even free, so the experimentation you're doing won't succeed. (This is because $\mathbb{Z}$ is a principal ideal domain.) Taking $(x,y)$ a submodule of $k[x,y]$ for $k$ some field will work better. $(x,y)$ is not projective, which we can show by showing it's not flat, since projective ...

5

Freyd proved in his paper Concrenteness (1973) that an abelian category which admits a faithful exact functor to $\mathsf{Ab}$ is well-powered; actually also the converse. There are abelian categories which are not well-powered, see MO/93853. Another example is mentioned in the foreword of the tac reprint of Freyd's book Abelian categories (2003) with ...

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

It is not true that if $A$ is not finitely generated, you can find $\mathbb Z$-independent elements. For example, $\mathbb Q$ is a non finitely generated $\mathbb Z$-module, but, clearly, you cannot find two $\mathbb Z$-independent rationals $\frac ab$, $\frac cd$, for $ac = bc\times \frac ab = ad\frac cd$. When we prove that $\mathcal O_K$ is finitely ...

5

Let $R$ be ring for which $F$ is a free module. Then we have isomorphisms $R^n \rightarrow F$ and $F \rightarrow R^{n+1}$ which gives us an isomorphism $R^n \rightarrow R^{n+1}$. Thus for any $m \geq n$ we have an isomorphism $R^m \cong R^n \oplus R^{m-n} \cong R^{n+1} \oplus R^{m-n} \cong R^{m+1}$. Composing these isomorphisms we get an isomorphism $R^n ... 5 The answer is negative since$A\subset B$flat and$B$regular implies$A$regular; see Bruns and Herzog, Theorem 2.2.12. But in this case$A\simeq k[a,b,c]/(ac-b^2)$, so$A$is not regular. Edit. A simpler approach: let$I=(x^2,xy)$and$A/I\to A/I$be the multiplication by$y^2$. Since$A/I\simeq k[y^2]$this is injective, but on$A/I\otimes_AB\to ...

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

Let, for example $R = \mathbb Z/(6)$. Then, as $\def\Z{\mathbb Z}\Z/(6)$-modules, $\Z/(6) \cong \Z/(2) \oplus \Z/(3)$. But a free $\Z/(6)$-module cannot have two or three elements only.

5

More generally, for any PID $R$ and every non-zero element $e \in R$, the ring $R/(e)$ is self-injective: Baer's criterion implies that, if $S$ is a commutative ring in which every ideal is principal, an $S$-module $M$ is injective if and only if for all $a \in S$, $m \in M$ with $\mathrm{Ann}(a) \subseteq \mathrm{Ann}(m)$ we have $m \in aM$. This can be ...

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