# Tag Info

## New answers tagged modules

0

Linear combinations are themselves by definition so that almost all summands are zero. For an algebraist the sum of infinitely many nonzero summands is not really a sum. Also, the infinite variant $R^X$ is certainly not free over the given set. As such it would have to be true that for any other module $A$ and map $f\colon X\to A$ there exists a unique ...

2

If $R$ is any ring and $n \geq 1$, then the evident map $M_n(R) \to \bigoplus_{i=1}^{n} R^n$ which decomposes a matrix into its columns is an isomorphism of left $M_n(R)$-modules. Hence, $R^n$ is a projective left $M_n(R)$-module. It is usually not free. If $R$ is commutative and non-zero (more generally, when $R$ has IBN), then free $M_n(R)$-modules of rank ...

2

$M_2(\mathbb{C})$ is a simple ring, so every module over this ring is projective and injective. Moreover, $\mathbb C^2$ is a simple $M_2(\mathbb{C})$-module, and if it is free it must be isomorphic to $M_2(\mathbb{C})$. In fact, $M_2(\mathbb{C})\simeq \mathbb C^2\oplus\mathbb C^2$, and therefore by taking the ranks we get a contradiction.

0

Assuming your definition of perfect pairing, the answer to your first question is no. You are essentially asking if $N \cong\text{Hom}(\text{Hom}(N,R), R))$ and this is not true when $N = \mathbb{Z}/2$, $M = 0$ and $R = \mathbb{Z}$. The same example with $P = \mathbb{Z}$ tells you that the answer to your second question must also be no. I don't ...

1

Since $I$ is nonzero, choose $0 \neq y \in I$. Then $y \cdot \bar x$ is a nontrivial linear combination of $\bar x$, so should be nonzero because $\{\bar x\} \subset R/I$ is independent. But this is absurd, because $yx \in I$ so $y \cdot \bar x = \overline{yx} = 0 \in R/I$.

1

As $M$ is cyclic, there is an epimorphism $R \rightarrow M$. Because $M$ is assumed to be projective, it is thus isomorphic to a direct summand of $R$. Note that every direct summand of $R$ is of the form $Re$ where $e \in R$ is an idempotent. Now we have $(1-e)e = 0$ and as $R$ was assumed to be an integral domain, it follows that $e = 1$ or $e = 0$, that ...

1

This is true because a projective module over an integral domain is torsion-free (direct summand of a free module), hence its annihilator is $0$.

0

I think yes, a projective module is a direct summand of a free module, and $A \oplus B \simeq A \times B$ as $A \times B$ modules. The isomorphism sends $(a, 0) + (0, b) \mapsto (a, b)$.

1

It's not an isomorphism, it's equality! The module $\mathfrak{a}(M/N)$ is generated by the elements of the form $a(x+N)$, for $a\in\mathfrak{a}$ and $x\in M$. Clearly $a(x+N)=ax+N\in (\mathfrak{a}M+N)/N$, so one inclusion is settled up. Conversely, an element in $(\mathfrak{a}M+N)/N$ is of the form $$y+z+N$$ for some $y\in\mathfrak{a}M$ and $z\in N$. ...

1

Let us try this : $$\mathfrak{a}M+N=M\Rightarrow(\mathfrak{a}M+N)/N=M/N$$ Now, we always have $\mathfrak{a}(M/N)\subseteq M/N$ so the only thing to show to get the equality is to show the reverse inclusion, take $m_0\in M$ then there exists $a\in\mathfrak{a}$, $m\in M$ and $n\in N$ such that : $$m_0=am+n$$ Now modulo $N$ we have $[m_0]=[am]$ hence ...

1

It is because $A\simeq\operatorname{Im}f$ and tensoring an isomorphism always gives an isomorphism. Hence it's equivalent to prove that tensoring the inclusion morphism or tensoring the original injective morphism results in an injective morphism. Btw, any rings of fractions is a flat $A$-module.

4

This is false. Taking these two facts: tensoring $- ⊗_R M$ is exact iff M is a flat module a module over a PID is flat iff it is torsion free we see that $- ⊗ \mathbb Q : \mathrm{Ab} → \mathrm{Ab}$ is an exact functor, but it doesn't preserve free objects: $\mathbb Z ⊗ \mathbb Q = \mathbb Q$, and $\mathbb Q$ is not free over $\mathbb Z$ (any two elements ...

1

I'm afraid not. The invariant basis number or IBN property is not satisfied by all rings. It is trueu for the so-called stably free rings, for which, given two $n\times n$ matrices, $AB=I_n\Rightarrow BA=I_n$. Among stably free rings are commutative rings, noetherian rings, artinian rings and semi-local rings. An example of a ring that does not satisfy ...

1

$aL=0$ and $bN=0$ implies $(ab)x=0$ for all $x\in M$:

1

Consider $R = \mathbb{Z}$ and $A=C= \mathbb{Z}/(p)$ and $B=\mathbb{Z}/(p^2)$ and the short exact sequence $$0 \to \mathbb{Z}/(p) \to \mathbb{Z}/(p^2) \to \mathbb{Z}/(p) \to 0$$ where the first morphism is given by $1\mapsto p$, the second is the projection mod$(p)$. Clearly for $p$ prime, this does not split, since $\mathbb{Z}/(p^2)$ is not isomorphic to ...

1

Hint: look at $A=F[x]/(x^2)$ with $A=M$, and $L=I=(x)/(x^2)$

1

Note that if either side is $\infty$ then the other is as well. So assume at least $\lambda(M)$ is finite. Let $$N=N_0\supset N_1 \supset \cdots \supset N_t=(0)$$ be a normal series series for $N$. By the correspondence theorem, any submodule of $M/N$ is of the form $L'=L/N$ for some submodule $L\subset M$ containing $N$. Then we have a normal series for ...

1

No, Let $A,B = \mathbb{Z}$ as $\mathbb{Z}$-modules. Let $M$ be the submodule of $A\oplus B$ generated by $(2,1)$ and let $N$ be the submodule generated by $(0,1).$ Then $M\oplus N \subset A\oplus B$ but $M$ is not a subset of $A$ or $B.$

1

No, in either cases really. Consider $V=\bigoplus_{p\in\Bbb P}\Bbb Z/p\Bbb Z$ (where $\Bbb P$ is all the prime integers). Now partition $\Bbb P$ into $A,B$ and $A',B'$ such that neither $A\subseteq A'$ nor $B\subseteq B'$, and consider the decompositions defined by $A,B$ and $A',B'$ as counterexamples.

0

So the question has an answer: Suppose $M$ is a matrix whose first column is somewhere nonzero. Then if $v=(1,0)^t$ then $Mv\neq 0$ because it is equal to the first column of $M$. A similar argument works if the second column is nonzero with the vector $(0,1)^t$. Thus no nonzero matrix annihilates the module.

3

If $R$ is a commutative ring and $M$ is some $R$-module, then $R[x]$-module structures on $M$ extending the given $R$-module structure correspond 1:1 to $R$-linear endomorphisms of $M$. So, any $\mathbb{C}$-linear map $\mathbb{C}^2 \to \mathbb{C}^2$ (and surely, there are plenty of them!) corresponds to a $\mathbb{C}[x]$-module structure on $\mathbb{C}^2$.

1

To make an abelian group $M$ into a module over $R$ we need a ring homomorphism from $R$ into the (non-commutative) ring of group endomorphisms of $R$. To make $\mathbf{C^2}$, a module we can construct a homomorphism that goes into the subring of vector space endomorphisms. Here we can easily specify a ring homomorphism $\mathbf{C}[X]\to ... 2 There exist a lot of ways to show this. I think that the simplest is to say that we can define a$\Bbb{C}[x]$-module structure on$M$by composing $$\Bbb{C}[x] \longrightarrow \Bbb{C} \longrightarrow \operatorname{End}(M)$$ where the map$\Bbb{C}[x] \longrightarrow \Bbb{C}$is any ring morphism (for example$f \mapsto f(0)$). This can be generalized, saying ... 0 Suppose$\dim_{\Bbb{C}} M = n$is finite. Then we can identify$M$with$\Bbb{C}^n$. Since$M$is a$\Bbb{C}[x]$-module,$x$represents an$n \times n$complex matrix$A$(an endomorphism of$\Bbb{C}^n$). Let$p_A(x)$be the characteristic polynomial of$A$. By Cayley-Hamilton theorem,$p_A(A)=0$. So$M$cannot be a faithful$\Bbb{C}[x]$-module, because ... 1 Suppose$M$were a finite-dimensional vector space over$\mathbb{C}$. Let$f : M \to M$be the action of$x$, ie$f(m) = x \cdot m$. Then by the Cayley–Hamilton theorem, there is a polynomial$p \in \mathbb{C}[x]$,$p \neq 0$(eg. the characteristic polynomial of$f$) such that$p(f) = 0$; in other words, for all$m \in M$,$(p(f))(m) = 0 = p(x) \cdot ...

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.

1

$${}_pM=\bigoplus_{i=1}^t(p^{e_i-1})/(p^{e_i})\simeq\bigoplus_{i=1}^tR/(p).$$ Indeed, if $\bar x=x+(p^{e_i})$ is an element of $R/(p^{e_i})$, we have: $$p \bar x=\bar 0\iff px\in(p^{e_i})\iff x\in (p^{e_i-1})$$ since $p$ is a prime element.

0

No. Consider the Prüfer group $G=\mathbb{Z}(p^{\infty})$; a quotient of a proper subgroup is cyclic, no quotient of $G$ (except $G/G$, of course) is cyclic.

1

No. Let $A = 0$. Then your question is whether, for every submodule $B \subseteq C$, we can find a submodule $L \subseteq C$ such that $C/L \cong B$. In other words, the question is whether every submodule of $C$ is isomorphic to a quotient module of $C$. This is true if $B$ is a direct summand, in which case we can take $L$ to be its complement, but not ...

0

If $$0\to A\to B\to C\to 0$$ is an exact sequence of $S$-modules, then $$0\to i^\ast(A)\to i^\ast(B)\to i^\ast(C)\to 0$$ is an exact sequences of $R$-modules, because it has the same underlying sets and the same functions.

1

If I understood well your question, the answer is negative: in $\mathbb Z[i]$ the principal ideal $I=(2+i)$ is a free $\mathbb Z$-module of rank two. On the other side, a two-generated ideal of a quadratic number field is necessarily a free $\mathbb Z$-module of rank two (why?).

1

An ideal is a $\mathbb{Z}$-submodule. It is possible for two elements in $\mathbb{Z}^2$ to generate a proper submodule, but it is not possible for three elements to be linearly independent over $\mathbb{Z}$ (we can always find a nontrivial linear combination that is $0$). Furthermore, any submodule is free and therefore has a linearly independent generating ...

1

I assume what they mean is that this homomorphism makes $M$ into an $R$-module and the kernel is the annihilator of $M$ over $R$.

0

$\ker f$ is a two-sided ideal of $M_n(K)$, so $\ker f=(0)$. (For the form of two-sided ideals in matrix rings you can look here.) This tells us that $f$ is injective. Then $B=f(M_n(K))$ is a simple subalgebra of the central simple algebra $A=M_m(K)$, and $[A:K]=[B:K][C_A(B):K]$ where $C_A(B)=\{a\in A:ab=ba,\forall b\in B\}$ is the commutant subalgebra of ...

0

If $A$ is a coherent ring, any ring of power series $A[[X_1,\dots,X_n]]$ is faithfully flat over $A$. Also, the direct sum of a flat $A$-module and a faithfully flat $A$-module is faithfully flat. The direct sum $\bigoplus\limits_{\mathfrak m\in\operatorname{Max}A} A_{\mathfrak m}$ is faithfully flat over $A$.

0

Let $(b_1, \dots ,b_k)$ be a generating set of $B$ as an $A$ module. As $L= K(\theta)$ and the extension is finite, $L=K[\theta]$, and recall $K$ is the quotient field of $A$. So $b_i= \sum_{j=1}^{l_i} d_{ij} \theta^l$ where $d_{ij} \in K$ yet then $d_{ij}= \frac{a_{ij}}{c_{ij}}$ with ${a_{ij}}, {c_{ij}} \in A$. So with $c_i = \prod_{j=1}^{l_i} ... 1 See Lemma$1$and Lemma$2$here, in Osserman's lecture notes. 2 Note that as$\mathbb{Z}$-modules,$M\cong\mathbb{Z}\oplus\mathbb{Z}$,$N_1\cong\{(a, b)\in\mathbb{Z}\oplus\mathbb{Z}| a\text{ and }b\text{ are both even}\}$and$N_2\cong\{(a, b)\in\mathbb{Z}\oplus\mathbb{Z}|a+b\text{ is even}\}$. So$M/N_1\cong\mathbb{Z}_2\oplus\mathbb{Z}_2$and$M/N_2\cong\mathbb{Z}_2$. 1 Hint 1:$2=(1+i)(1-i)$and$1\pm i$is irreducible. Use the Chinese Remainder Theorem. Hint 2:$M/N_1$has order$4$and exponent$2$. What abelian group that has that? 1 A ring satisfying 1) is said to have the "rank condition," and a ring satisfying 2) is said to satisfy the "strong rank condition." (Actually there is a right-left sidedness to the SRC which I gloss over for now.) You can read about both of these conditions in the first chapter of Lam's Lectures on modules and rings. In general, the strong rank condition ... 1 I can't comment otherwise I would. But I think it must be a typo, in the sense that (2) is supposed to be:$\mathrm{Ann}_A(M/N)$is a primary ideal of$A$. Then it seems rather tautological to prove that (1) implies (2) supposed to be that$R$is noetherian and$M$is finitely generated. Indeed, we may assume that$N=0$. Let$\mathfrak{p}$be the unique ... 2 This property does not hold in general. In fact, it holds if and only if$A$is semisimple. Using your notation, the image of the composition map is the set of morphisms$f:X\to Y$that factor through a free module. Indeed, the image of a sum$\sum_{i=1}^n g_i\otimes h_i$is exactly the composition of the morphism$$\pmatrix{g_1 \\ \vdots \\ g_n}: X\to ... 1 If$M$is isomorphic to a direct sum of$n$copies of$\mathbb{Z}$and some finite abelian group, then the Z-rank is then$n$.$\mathbb{Z}$-modules are simply abelian groups. If it is finitely generated then it must be isomorphic to$\mathbb{Z}^{\bigoplus n}\oplus\mathbb{Z}_{m_1}\oplus\cdots\oplus\mathbb{Z}_{m_k}$. After localizing to the field of fraction ... 0 Since you already obtained very good answers, let me add a short remark. In some cases, you can indeed quite easily derive a concrete construction from an universal property. For instance, let's assume you already know the universal property of the limit of a functor$F$(as a terminal$F$-cone) and you want to derive an explicit construction of the limit ... 3 Hint: for the first one, tensor product is distributive with respect to a finite sum. As for the second, write$1=\frac{n}{n}$in$\mathbb{Q}$. 0 I'm going to assume that$M,N$are right modules and$M',N'$are left modules.$(M\otimes_R N) \otimes_{A\otimes_R B} (M'\otimes_R N')$is governed$R$-multilinear maps$f:(M\times N)\times (M'\times N')\to Z$for$R$-modules$Z$that are balanced$A\otimes_R B$-linear maps with respect to the factors$M\times M'$and$N\times N'$.$(M\otimes_A M')\otimes_R ...

0

For $M,M',N,N'$ free there is an obvious, well defined isomorphism. In fact, if $M=A^{(I)}$, $M'=A^{(I')}$, $N=B^{(J)}$, $N'=B^{(J')}$, they are both naturally isomorphic to $A\otimes_R B^{(I\times I'\times J\times J')}$. In the general case, take free resolutions of $M,M',N,N'$, they induce free resolutions of both $(M \otimes_R N) \otimes_{A \otimes_R B} ... 0 You are correct about what the kernel is. When you say "Give the basis$C$..." I assume you mean "Give a basis$C$...". I.e. you have to find a basis.$T/ \text{ker} f$is one-dimensional since$\dim(T/S) = \dim(T) - \dim(S)$. Any nonzero vector of a one-dimensional space constitutes a basis. (I.e. pick any vector$v$not in$\ker f$and$v + \ker f$is ... 2 Not at all:$R[x]$is generated by$x$as an$R$-algebra, but it's a free$R$-module with an infinite basis. Actually a finitely generated$R$-algebra is finite if and only if its generators (as an algebra) are integral over$R\$.

1

It looks correct to me! Not sure what else to say as an answer.

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