# Tag Info

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

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.

6

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 $M = ... 5 (I'm implicitly assuming that$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$-algebra map ... 5 Let$Q$be the field of fractions of$R$, and let$F:Q^n\to Q^n$be the morphism of$Q$-modules which has the same matrix as$f$(with respect to the standard bases of$R^n$in the case of$f$and of$Q^n$in the case of$F$) Show that$F$is nilpotent, and that therefore$F^n=0$. If$i:R^n\to Q^n$is the obvious inclusion, then we have$F\circ i=i\circ ...

5

No: We have $M \otimes 0 \cong 0 \cong M' \otimes 0$ for all modules $M,M'$. Even if $N$ is a very nice $R$-module, say free of finite rank, and non-zero, then $M \otimes N \cong M' \otimes N$ does not imply $M \cong M'$: This is because there are non-isomorphic modules $M,M'$ with $M^2 \cong M'^2$, i.e. $M \otimes R^2 \cong M' \otimes R^2$ (see here, take ...

5

You are asking about vector spaces over fields and modules over rings. Since the latter are more general, let's start with those. We know that tensor products are distributive over direct sums. In other words, for every triple $A, B, C$ of $R$-modules, we have a canonical isomorphism between $A\otimes (B \oplus C)$ and $(A\otimes B) \oplus (A \otimes C)$. ...

5

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 ... 5$0$is an invariant module so there exists$x$such that$T(x)=0$, construct the sequence of invariant submodule$N_m$such that$N_1$is the module generated by$x$suppose defined$N_m$, since the map induced by$T$on$M/N_m$is not injective, if$N_m\neq M$there exists$y_m\in M$such that$y_m\neq N_m$and$T(y_m)\in N_m$. We denote by$N_{m+1}$the ... 4 Let me try to put the question in its natural frame.$M$is a finitely generated$A[X]$-module via$T$, and$T$-invariant submodules correspond to the$A[X]$-submodules of$M$. Then the multiplication by$X$on$M/N$is not injective for every proper$A[X]$-submodule$N$of$M$. You want to show that$X^nM=0$for some$n\ge1$. Let$M$be a noetherian ... 4 Sorry, my previous answer was incorrect: I attempted to be too fancy. It looks like a straightforward approach is probably best. Take a simple tensor in$R \otimes_R N$which will be of the form$r \otimes n = r(1 \otimes n) = r \iota (n)$. Arbitrary elements are finite sums of simple tensors so since$\iota$is a homomorphism, we are done. 4 Take$R$to be a field$k$and take$A = B = k^2$, with basis$e_1, e_2$. The tensor product$A \otimes_k B$is$k^4$with basis$e_1 \otimes e_1, e_1 \otimes e_2, e_2 \otimes e_1, e_2 \otimes e_2$, and most elements of it are indecomposable. For example,$e_1 \otimes e_1 + e_2 \otimes e_2$is indecomposable. There's no more reason to expect all tensors to ... 4 First of all, notice that exterior powers commute with base change (Eisenbud, Commutative Algebra with a View..., Proposition A2.2, p. 576), hence $$\Lambda^n_{F[x]} (F[x] \otimes_F V)=F[x] \otimes_F \Lambda^n_{F}V$$ You can easily check, that the following diagram (of$F$-modules) commutes ($m(\lambda)$is the map$x \mapsto \lambda$from the other ... 4 A finitely generated module over a noetherian ring is of finite length iff its support is contained into the maximal spectrum. (See here, Proposition 1.6.9.) Now let$\mathfrak p\in\operatorname{Supp}(\operatorname{Coker}f)$. If$\mathfrak p$is not maximal, then it is minimal since$\dim R\le 1$. But$(\operatorname{Coker}f)_{\mathfrak ...

4

This is not even true for vector spaces, which is just about the nicest possible case. Take $k$ a field, $M = k^2$ with basis $e_1, e_2$, and $N = \text{span}(e_1 + e_2)$.

4

The abelian group $\mathbb{Q}$ is neither artinian nor noetherian, so it contains both an infinite descending chain and an infinite ascending chain (it's not difficult to find them explicitly). If you have a vector space over a field $K$ of characteristic $0$, every nonzero subspace contains a copy of $\mathbb{Q}$ as an additive subgroup. An infinite field ...

4

Any nonzero module of finite cardinality has a simple submodule: Amongst all nonzero submodules, choose one of minimal cardinality. The statement generalizes to Artinian modules $M$. The descending chain condition is equivalent to every nonempty family of submodules containing minimal elements (with respect to set inclusion). Applied to all nonzero ...

3

Any finite partially ordered set has at least a minimal element. In this case the partially ordered set is the one of the nonzero submodules.

3

You got something wrong there. $\mathbb{Q}$ is not artinian as a $\mathbb{Z}$-module because $\mathbb{Z}\supseteq 2\mathbb{Z}\supseteq 4\mathbb{Z}\supseteq 8\mathbb{Z} \supseteq \cdots$ is a strictly descending sequence of submodules. And similarly $\mathbb{Z}_{(2)}\supseteq 2\mathbb{Z}_{(2)}\supseteq 4\mathbb{Z}_{(2)}\supseteq 8\mathbb{Z}_{(2)} \supseteq ... 3 It's not entirely clear to me what you are looking for exactly, but here is a short proof of your statement: Consider the$R$-module$N= \bigoplus_{x \in X} R_x$, where each$R_x$is just a copy of$R_R$. Denote the unit of$R_x$by$1_x$. Then the map of sets$f: X \to N, x \mapsto 1_x$gives a unique module homomorphism$M \to N$extending$f$(and which ... 3 when are ideals also rings with unity? Proposition: An ideal$I\lhd R$will be a ring with identity iff there exists a central idempotent$e$such that$eR=I$. Proof: ($\implies$) The identity of$I$, call it$e$, is an idempotent element of$R$and satisfies$I=eI$. Then$I=eI\subseteq eR\subseteq I$, so$I=eR$. Since$e\in I$, we have ... 3 Here are the basic relevant facts to get you thinking about the possibilities: A short exact sequence $$0 \to A \overset{f}{\to} M \overset{g}{\to} B \to 0$$ is the data of a submodule$f(A) \subset M$, where$f$encodes$A \simeq f(A)$, and its quotient$M/f(A) \simeq B$via the quotient map$g$. The sequence is said to split if$M$has a complementary ... 3 I want to thank Tobias Kildetoft for the useful suggestion of using the Chinese Remainder Theorem. So we have that$120 = 8 \; 3 \; 5\mathbb{Z}_{120} = \mathbb{Z}_{3} \times \mathbb{Z}_{8} \times \mathbb{Z}_{5}$And from this$\mathbb{Z}_{120}^* = \mathbb{Z}_{3}^* \times \mathbb{Z}_{8}^* \times \mathbb{Z}_{5}^*$Now, from a general result: ... 3 Consider the situation in which$R$is a field,$M$is a finite dimensional vector space,$g=0$and$f$is any non-surjective linear map ---for example,$f=0$. 3 These conditions are equivalent if$X$is a levelwise finite CW complex (finitely many cells of each dimension), since this condition ensures that each homology group is finitely generated. You can prove this using the universal coefficient theorem as described here; it ensures that the torsion subgroup of$H_k(X, \mathbb{Z})$is isomorphic to the torsion ... 2 This is true if you assume$H_n(X)$is finitely generated for all$n$(all coefficients in this post will be$\mathbb{Z}$). In particular, this holds if$X$has the homotopy type of a CW-complex with finitely many cells in each degree. To prove this, we invoke the classification of finitely generated abelian groups, which says that$H_n(X)$is a finite ... 2 The "finitely generated" isn't really important here. You have to understand what are free modules. A$R$-module$M$is free if it has a basis, i.e., there exists$\{x_i\}_I \subset M$(where$I$can be chosen the be finite in the finitely generated case) such that the$x_i$'s form a$R$-basis of$M$. This is a very special property, and every free module ... 2 Hint. Let$R=k[X,Y]$. We have$H_{R/I}(t)=1+t$, and$H_{R/J}(t)=1+2t+t^2$. On the other side, their Hilbert polynomial is$0$. Edit. It seems the OP looked for$H_I(t)$,$H_J(t)$and the corresponding Hilbert polynomial. Then he could have figured this out by himself using the exact sequence$0\to I\to R\to R/I\to 0$(and the corresponding one for$J$) ... 2 Given a prime$p$, the Prüfer$p$-group is the direct limit of the cyclic groups of order a power of$p$. If$\Omega$is an infinite set, the group of finitary permutations on$\Omega$is the direct limit of all groups of permutations on the finite subsets of$\Omega$. 2 Direct limit is generalisation of the notion of union of a family of sets. Two other examples: Lazard's theorem in commutative algebra asserts that a flat$R$-module is a direct limit of free$R$-modules. Germs of continuous function at a point$a$of a topological space is defined as the direct limit of the system of pairs$(U,f)$, where$U\$ is an open ...

Only top voted, non community-wiki answers of a minimum length are eligible