# Tag Info

0

The definition of rank is the highest order of a square non degenerate submatrix. Thus being determinant invariant under transposition, you'll always have $\operatorname{rk}A=\operatorname{rk}A^T$ for every $A\in M_{n,m}(R)$.

1

The two following properties are easy set-theoretic statements $Im(f\oplus g)= Im(f)\oplus Im(g)$ $Ker(f\oplus g)=Ker(f)\oplus Ker(g)$ Those two results give the equivalence you want.

4

Consider $R=\mathbb Z$, $M=\mathbb Z$ and $N=\mathbb Z/2\mathbb Z$.

1

Hint: Prove the statements below ( may assume $I \ne 0$ ): If $I$, $J$ are ideals of an integral domain $R$ with field of fractions $K$ then every morphism of $R$-modules $I \to J$ is given by the multiplication by an element in $K$. If $I$ is a projective module then there exists an imbedding of $I$ into a free $R$ module $i \colon I \hookrightarrow$ ...

2

This result is known as Schanuel's lemma. A quick proof is given by introducing the pullback of $f$ and $f'$. It is the submodule of $P\oplus P'$ given by $$X = \{(p,p')\in P\oplus P'\mid f(p) = f(p')\}.$$ Then, the following sequences are exact : $$0 \longrightarrow \ker(f')\simeq K' \longrightarrow X \longrightarrow P \longrightarrow 0$$ and $$0 ... 0 Assume a \in IQ and a \not\in Q and then work from the element in I which is not a zero divisor. 0 As it has been mentioned in the above comments, M\otimes_SN is an R-T bimodule with the following actions; let r\in R , t\in T and m\otimes n\in M\otimes_SN ( a pure tensor) we have$$r.(m\otimes n)=r.m\otimes n$$and$$(m\otimes n).t=m\otimes(n.t)$$these actions are also compatible, since r.((m\otimes ... 1 It is clear that the product is injective wrt itself and from a theorem that a product is injective iff each factor is injective the first part is settled. As for the second part we need to see that we cannot inject the product inside the direct sum since we have infinitely many factors, see Ribenbenboim : Rings and modules top of page 25. 1 The inclusion \bigoplus R_n \hookrightarrow \prod R_n is essential, hence \bigoplus R_n is not injective. 1 One should approch this more conceptually: The uniqueness is an immediate consequence of the surjectivity of f_2 (You can view this as a definition of the term surjective). The existence follows from the universal property of the cokernel, which is the same as the fundamental homomorphism theorem: The morphism g_2 \circ \beta annihilates anything, ... 2 I think you want I to be a right ideal and J a left ideal, not the other way round. You are almost done. Consider the canonical homomorphism \mathbf{m} :A\otimes_{A}A\rightarrow A of \mathbb{Z}-modules which sends every a\otimes b\in A\otimes_{A}A to ab\in A. It is well-known that \mathbf{m} is an isomorphism. Thus, \left( ... 1 You have confused k-vector space basis as R-module basis, hence this issue. ANy commutative ring is a free module overitself with basis the singleton \{1\}, consiting of the multiplicative identity. Given two elements a,b\in R, by commutativity we have a.b -b.a=0 and so a basis has to have less than 2 elements. 1 As you remarked, there are four non-trivial subgroups of \mathbb Z_{18} given by the divisors of 18 (others than 1 and 18). Now it's good to know what's the sum and the intersection of such subgroups. In general, in \mathbb Z_n we have \bar k\mathbb Z_n+\bar l\mathbb Z_n=\overline{\gcd(k,l)}\mathbb Z_n and \bar k\mathbb Z_n\cap\bar l\mathbb ... 0 This holds whenever R is noetherian; see Bourbaki, Commutative Algebra, Chapter 4, Exercise 17(g). (In fact, it's enough to know that P is minimal over \operatorname{Ann}(m) for some m\in M, m\ne 0.) But this isn't the case in your question. However, it also holds under the assumption M noetherian. In this case the ring R/\operatorname{Ann}(M) ... 1 To augment user26857's answer, note that: (a_1,a_2,\dots,a_n) + S = (a_1+a_2+\cdots+a_n,0,\dots,0) + (a_1-a_2-\cdots-a_n,a_2,\dots,a_n) + S = (a_1+a_2+\cdots+a_n,0,\dots,0) + S. If p(X) = q(X)(X - 1) + r (for r \in A), then p(X) + S = r + S. If B = \begin{bmatrix} x_{11} & x_{12} & x_{13} & \dots & x_{1n} \\ ... 1 If P is a finitely generated and projective R-module, then there is a canonical isomorphism$$\zeta:P^*\otimes_RM\simeq\operatorname{Hom}_R(P,M)$$given by \zeta(f\otimes m)(x)=f(x)m. (For a proof see Proposition 6 from these notes.) Now, coming back to the question we have \mathfrak a^*\otimes_RM\simeq\operatorname{Hom}_R(\mathfrak a,M). But ... 1 Let b_1+A,b_2+A,\dots,b_m+A be generators of B/A and let a_1,\dots,a_n be generators of A. If b\in B, then$$ b+A=\sum_{i=1}^m r_i(b_i+A)=\biggl(\sum_{i=1}^m r_ib_i\biggr)+A $$for some r_1,\dots,r_m\in R, which means$$ b-\biggl(\sum_{i=1}^m r_ib_i\biggr)=\sum_{j=1}^n s_ja_j $$for some s_1,\dots,s_n\in R, so$$ b=\sum_{i=1}^m ...

2

The problem asks to find an $A$-module $N$ (eventually you already know from class) such that $M/S\simeq N$ (eventually by using a well known isomorphism theorem). Hints. Define $f:A^n\to A$ by $f(a_1,\dots,a_n)=\sum_{i=1}^na_i$. Define $f:A[X]\to A$ by $f(p(X))=p(1)$. Define $f:M_n(A)\to A^n$ by $f((a_{ij})_{i,j})=(a_{11},\dots,a_{n1})$.

0

Another example is $A=\mathbb C$ and $B=\mathbb R$. One knows that $\mathbb R$ is a $\mathbb C$-vector space, but there is no ring homomorphism $\mathbb C\to\mathbb R$ (why?).

1

This is well-studied under the heading of "cancellability," and Lam's crash course on the topic is very nice. Are there any simple conditions on $R$-modules $M,A$ and $B$... The readiest one is that if $R$ has stable range 1 and $M$ is finitely generated and projective, then it cancels from $M\oplus A\cong M\oplus B$. You can find this, for example, in ...

3

An example of a difference: $\mathbb{Z}/3\mathbb{Z}$ is projective (and free) as a $\mathbb{Z}/3\mathbb{Z}$-module but is not projective (nor free) as a $\mathbb{Z}$-module. However things like simplicity and indecomposability will remain the same, which is easy to prove since the action of $A/\mathfrak a$ is given by the action of $A.$ The only thing ...

4

If you want $\operatorname{Hom}_R(P,R)\simeq P$ for $P$ a finitely generated projective $R$-module, then forget it. $\bullet$ If $R$ is an integral domain, and $I\subset R$ is a non-zero ideal, then $\operatorname{Hom}_R(I,R)\simeq I^{-1}$, where $I^{-1}=\{x\in Q(R):xI\subseteq R\}$. (Here $Q(R)$ stands for the field of fractions of $R$.) $\bullet$ If ...

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Hint: what exact sequence can you build up out of $M_1\oplus M_2, M_1 + M_2$ and $M_1\cap M_2$?

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Rephrased, to show that the function is well-defined, you have to show that whenever $g^m|(f_1-f_2)$, you also have $(t-\lambda)^m|(f_1-f_2)$. But this is clear since $(t-\lambda)^m|g^m$.

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$\mathbb{C}[M]$ will be Laurent polynomials in $x_1,\dots,x_n$. Algebra homomorphisms to $\mathbb{C}$ are thus determined by sending each $x_i$ to an element of $\mathbb{C}^\times$ and therefore the set of all such homomorphisms is associated with $(\mathbb{C}^\times)^n$ More generally, the group algebra functor is left adjoint to the functor taking an ...

0

Yes. If $R$ has infinite left global dimension, then for every $n$ there is a left module $M_n$ with projective dimension greater than $n$. But then $\bigoplus_{n\in\mathbb{N}}M_n$ has infinite projective dimension. The same argument works for weak dimension.

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In the general case, there is a canonical isomorphism: \begin{align*}P/N\cap P &\longrightarrow(P+N)/N\\ x+N\cap P & \longmapsto x+N\end{align*} Also, there's a bijection between submodules of $M/N$ and submodules of $M$ that contain $N$, and $P$ is not supposed to contain $N$. However, by the canonical homomorphism $\,p\colon M\longrightarrow M/N$, ...

1

From the artinian hypothesis you may deduce that the zero submodule is an intersection of finitely many maximal submodules. Therefore $M$ is isomorphic to a submodule of a finitely generated semisimple module, hence it is semisimple and finitely generated.

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Let $\mathcal{M}$ be the set of maximal submodules of $M$. Lemma: There exist a finite subset $\mathcal{N}$ of $\mathcal{M}$ such that $\bigcap \mathcal{N} = 0$. Proof: Let $\mathcal{C}$ be the collection of the intersection finite number of maximal submodules. As $M$ is artinian, $\mathcal{C}$ must have a minimal element $N = \bigcap \mathcal{N}$. If it ...

-1

I am not sure I fully understand the question. As Hoot said, "is $R$ artinian?" to which you replied as "yes". But you also have the finitely generated assumption. In which case the assumption that $M$ is semisimple is completely unnecessary. A finitely generated module over an Artinian ring is Artinian. Because, we have the exact sequence, $$0 \to K \to ... 2 Your method is not going to work unless you're careful about the choice of the projective modules P and Q through which gf-\text{id}_M and fg-\text{id}_N factor. Indeed, you could replace P by P\oplus P' for any projective module P', changing the isomorphism type of M\oplus P. Instead, use the fact that gf-\text{id}_M factor through a ... 1 Write M as the direct sum of its irreducible submodules. Each element of the generating set can be written as a sum of elements from finitely many of the irreducible submodules. Then M is a direct sum of finitely many irreducible submodules. This implies M is Artinian. 2 Without any assumptions about R, this is false. For example, consider \mathbb{Z} as a \mathbb{Z}-module. It's finitely generated (the set \{1\} is a generating set), but the collection of submodules$$\mathbb{Z}\supset 2\mathbb{Z}\supset 4\mathbb{Z}\supset 8\mathbb{Z}\supset\cdots$$has no minimal element. 1 In different categories (such as the category of rings or the category of modules) different substructures are important. Ideals are the natural substructures of rings, and submodules are the natural substructures of modules. These substructures are special because they are suitable for forming quotient objects in their categories (that is why I am not ... 1 When that definition says Let R be a ring and 1_R its multiplicative identity... they are mentioning (implicitly) that they are requiring R to be unital. However, it is certainly possible to define modules over a non-unital ring; just throw out statement 4 from that definition. 1 Some authors use the term "ring" to mean "ring with identity." This is neither standard nor nonstandard; there is just no consensus. Authors using this definition would use the term "rng" to denote a ring that possibly does not have an identity. 1 I don't have the book by Dummit and Foote, but the maps$$ R\to Ry_1,\quad r\mapsto ry_1\\ R\to Ra_1y_1,\quad r\mapsto ra_1y_1 $$are surjective homomorphisms and they're also injective because M is torsion-free. For the second one, note that ra_1=0 implies r=0, if a_1\ne0. If a_1=0, then Ra_1y_1=\{0\} is free as well. 0 For a module such as B, with submodules C and D, we of course have the submodule C + D of B, which is, by definition, C + D= \{ c + d \mid c \in C, d \in D \} \subset B; \tag{1} in the event that C \cap D = \{ 0 \}, the notation C \oplus D may also be used; C \oplus D is then referred to as the direct sum of C and D. We observe that ... 2 Embed L in an injective module E and consider the push-out diagram$$\require{AMScd} \begin{CD} {} @. {} @. 0 @. 0 \\ @. @. @VVV @VVV \\ 0 @>>> K @>>> L @>>> M @>>> 0 \\ @. @| @VVV @VVV \\ 0 @>>> K @>>> E @>>> N @>>> 0 \\ @. @. @VVV @VVV \\ {} @. {} @. E/L @= E/L \\ @. @. @VVV @VVV ...

2

For a $b\in B$ consider the element $x=f (g(b)) -b\in B$. We claim that this $x$ is in the kernel of $g$. Then it will follow that $b= x+f(g(b))$ where in the rhs first element is in $\ker g$ and the second element is in image of $f$, that is what you wanted to be proved. To prove the claim apply $f$ to this element $x$ and use the condition that $g$ ...

1

It seems like your question is: prove that a finitely generated free module has a finite basis. Let $E$ be a basis for such a module $M$ over $R$. Then $E$ spans $M$. Now, $M$ is finitely generated so you can find a finite ste of elements that span $M$. Write each element as (finite) linear combinations of elements from $E$. Collect all these elements from ...

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$\mathbb{Z}/6\mathbb{Z}$ is not a PID, because it's not a domain.

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I am surprised that it is not mentioned here- Example of a free module M which has bases having different cardinalities. Let $V$ be a vector space of countably infinite dimension over a division ring $D$. Let $R=End_D(V)$. We know that $R$ is free over $R$ with basis $\{1\}$. We claim that given a positive integer $n$, there is a $R$-basis $B_n=\{f_1,f_2, ... 2 We want to show${\rm Ext}^1( P, N ) = 0$for all$N$(as this is equivalent to$P$being projective). Accordingly, let $$0 \to N \to I^0 \mathop\to^{\delta^0} I^1 \mathop\to^{\delta^1} I^2 \to \cdots$$ be the beginning of an injective resolution for$N$. Suppose$f\colon P \to I^1$is a cocycle, i.e.,$\delta^1\circ f = 0$. Since the sequence is exact, ... 1 You can invert$r$as long as$r$is not nilpotent (otherwise the multiplicative set generated by$r$contains$0$and the localizations are also$0$). Since$r$is a non-zero divisor on$L$it follows that$r$is not nilpotent (unless$L=0$which I think it's not the case due to the rank condition). 1 Since$\Bbb Z$is a PID and projective modules over PIDs are free, it is clear that every finite abelian group is not projective. But in fact, you can just use the definition to prove that$\Bbb Z/2\Bbb Z$is not projective as$\Bbb Z$-module: consider the quotient map $$\pi:\Bbb Z/4\Bbb Z\longrightarrow\Bbb Z/2\Bbb Z.$$ It is easy to check that there is ... 0 Any projective module over a PID (e.g.,$\mathbb{Z}$) must be free. 2$\mathbb Z/2\mathbb Z{}{}{}{}{}{}{}{}$. 1 Your proof is incorrect, sorry. The module$M/S$is not a submodule of$M$and indeed need not be torsion-free. Example. Consider$R=\mathbb{Z}$; then$\mathbb{Z}=\langle 2,3\rangle$, but$\mathbb{Z}/\langle 3\rangle\$ is not torsion-free.

2

The usual proof uses the Structure theorem for finitely generated modules over a principal ideal domain. But there is an alternative proof based on the fact that the finitely generated torsion-free modules over an integral domain are isomorphic to a submodule of a free module of finite rank. (For a proof see here.) Since over a PID the submodules of free ...

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