# Tag Info

5

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.

4

Let $R$ be the ring of smooth functions on $S^2$. Then $R^3 \cong \operatorname{Vect}(S^2)\oplus R$ where $\operatorname{Vect}(S^2)$ is the $R$-module of smooth vector fields on $S^2$. $\operatorname{Vect}(S^2)$ is not isomorphic to $R^2$ by the hairy-ball theorem, and the standard smooth embedding $S^2\rightarrow \Bbb R^3$ gives the first isomorphism. ...

4

I agree that working with the object on the left is a bit uncomfortable, that is why we are so happy that it is isomorphic to the object on the right! However, before we 'know' (i.e. are allowed to use) that the situation is not completely hopeless either. I think what confuses is you is how to write down homomorphisms whose domain is defined as a quotient ...

3

Without using any topology, you can take $R$ the ring of endomorphisms of an infinite-dimensional vector space $V$, say over $\mathbb{R}$. Then as a (left) module over itself, $R=R\oplus 0$ is isomorphic to $R\oplus R$.

3

It might be ambiguous, but I would usually parse it as a set of generators as an $R$-algebra, i.e. that every element of $A$ can be written as a polynomial in the elements of $X$ with coefficients in $R$. For instance, $\{x\}$ generates $\Bbb Z[x]$ as a $\Bbb Z$-algebra, whereas you would need something like $\{1, x, x^2, \ldots \}$ to get a set of ...

3

Certainly not, since it has a non-zero annihilator: $\{0\}\times R_2$.

3

This is precisely the extension problem for modules; if $A/B \cong C$, the standard lingo is that $A$ is an "extension of $C$ by $B$." For fixed $B$ and $C$, the possible extensions are controlled by a group called the Ext group $\text{Ext}^1(C, B)$. The zero element of this group corresponds to the trivial extension $B \oplus C$ and the others correspond ...

3

To prove that an $R$-module $M$ is flat, it suffices to show that for every ideal $I \subset R$, the canonical map $I \otimes_R M \rightarrow M$ is injective. When $R$ is a domain and $M$ is the field of fractions of $R$, we have that every element of $I \otimes_R M$ is expressible as a simple tensor, that is, $i \otimes m$ for some $i \in I$ and $m \in M$. ...

3

Let's recall that $\operatorname{Ann}_{S^{-1}R}(S^{-1}M)=S^{-1}\operatorname{Ann}_R(M)$ for any finitely generated $R$-module $M$ and every multiplicative set $S\subset R$. Now let $S$ be the set of all non-zerodivisors in $R$, and consider the $S^{-1}R$-module $S^{-1}M$. This is finitely generated and faithful. It is well known that $S^{-1}R$, the total ...

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 $\mathbb{C}^2$.

2

Let $\sigma:R^s\to M$ be an injective homomorphism, and $\pi:R^s\to M$ a surjective homomorphism. Set $N=\sigma(R^s)$ and notice that $\sigma:R^s\to N$ is an isomorphism. Now consider $\pi\sigma^{-1}:N\to M$. This is a surjective homomorphism, hence by Orzech Theorem an isomorphism.

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 ...

2

Consider the $A$-module (or $R$-module) epimorphism $$I\oplus J \rightarrow I+J=A, \;\; (i, j)\mapsto i+j \in A,$$ compute its kernel and use the first isomorphism theorem. Now, if you know that $A$ is a projective $A$-module and know what properties projective modules have, this should give you the isomorphism you are looking for. If not, the isomorphism ...

2

Over a local commutative ring, projective modules coincide with free modules, so the question is whether $\mathfrak m$ is free. If it is free it must be of rank one, because two elements of $a,b\in A$ are necessarily $A$-linearly dependent: $a\cdot b-b\cdot a=0$ (duh!) Freeness of dimension one means that for some $m\in \mathfrak m$ the $A$-linear map ...

2

It depends on the ring. For instance, if $R=\mathbb{Z}_{(p)}$ is the localization of $\mathbb{Z}$ at the prime ideal $p$, then the maximal ideal is principal, so isomorphic as a module to the ring itself, hence projective. In case the ring is $\mathbb{Z}/4\mathbb{Z}$, the maximal ideal is not projective. Added from comment. If your aim is to discuss ...

2

Notice that $Q=S^{-1}A$ and $M\otimes_AQ\simeq S^{-1}M$, where $S=A-\{0\}$. Choose a basis of $S^{-1}M$ over $S^{-1}A$, say $\{\frac{x_1}{s_1},\dots,\frac{x_n}{s_n}\}$. Let's show that $x_1,\dots,x_n$ are linearly independent over $A$: if $\sum_{i=1}^na_ix_i=0$ (in $M$), then $\sum_{i=1}^n\frac{a_is_i}{1}\cdot\frac{x_i}{s_i}=\frac01$ (in $S^{-1}M$), so ...

2

It seems you are mixing exponents and base numbers and what is to be calculated modulo $49$, and what is to be calculated modulo $42$. This is, in my opinion, the chief difficulty working with problems like this, and it's important to use the utmost care that you actually calculate the correct power in the correct modulus. We want to know $3^{2014^{2014}} ... 2 Since the submodule is torsion free, the mapping from$R$to$\langle h \rangle$by$r \mapsto rh$is an isomorphism. 2 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$\mathbb{Q}$. 2 Just do it with two simultaneous inequalities: $$||x-2| - 3| > 1 \Leftrightarrow |x-2|-3 > 1 \text{ or } |x-2|-3 < -1$$ $$\Leftrightarrow |x-2| > 4 \text{ or } |x-2| < 2$$ $$\Leftrightarrow x-2 > 4 \text{ or } x-2 < -4 \text{ or } -2 < x-2 < 2$$ $$\Leftrightarrow x>6 \text{ or } x < -2 \text{ or } 0 < x < 4$$ So ... 2 A counterexample is$\mathbb{Q}$considered as a$\mathbb{Z}$-module. Definitely torsion-free but not free. 2 The existence of that projection is the point of Maschke's theorem, and it doesn't exist in general; it exists here becaue the group is finite and we're considering representations over a field of characteristic$0$. (Those conditions can be relaxed; Maschke's theorem is usually presented in the context of finite groups and representations over some field ... 2 Take$R$a noetherian ring (e.g.$R=\mathbb Z[X]$) and$I$a non-principal ideal (e.g.$I=(2,X)$). 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$M/N_2\cong\mathbb{Z}_2$. 2 Let$x_1,\dots,x_s\in M$be linearly independent over$R$, and$y_1,\dots,y_t\in M$a system of generators. Now define an injective homomorphism$\sigma:R^s\to M$by$\sigma(e_i)=x_i$(here$(e_i)_{i=1,\dots,s}$is the canonical basis of$R^s$), and a surjective homomorphism$\pi:R^t\to M$by$\pi(f_j)=y_j$(here$(f_j)_{j=1,\dots,t}$is the canonical basis ... 2 You have to prove that if$r\equiv s \mod I$, then$rm=sm$for any$m\in $M$. That is equivalent to $(r-s)m=0$, which is by definition since $r\equiv s\mod I\iff r-s\in I\subseteq\operatorname{Ann}_AM$.

2

Let's call $\phi$ the map $N \to M$. Since $\bar\phi:N/\mathfrak aN\to M/\mathfrak aM$ is surjective we have $\phi(N)+\mathfrak aM=M$, that is, $\mathfrak a(M/\phi(N))=M/\phi(N)$. Nakayama Lemma (Statement 1) tells you that there is $a\in\mathfrak a$ such that $(1+a)(M/\phi(N))=0$, that is, $(1+a)M\subseteq\phi(N)$. Can you conclude from here?

2

Consider the following commutative diagram of $\Omega$-modules with exact rows. \require{AMScd} \begin{CD} 0 @>>> I_k \otimes_k \Omega @>>> k[x_1,\ldots,x_n] \otimes_k \Omega @>>> (k[x_1,\ldots,x_n]/I_k) \otimes_k \Omega @>>> 0 \\ \ @VVV @VV\sim V @VVV \ \\ 0 @>>> I @>>> \Omega[x_1,\ldots,x_n] ...

2

I'll leave well-defined and linear to you since I don't like to show those two. To show $f$ is injective, note if $x\in M/(U\cap V)$ with $f(x)=(0,0)$ then $x\in U$ and $x\in V$. So $x\in U\cap V$, i.e. $x=0\in M/(U\cap V)$. To show $g$ is surjective, note for $x\in M/(U+V)$ we can take $(x+U,0+V)$ so that $f(x+U,0+V)=x+U+V$. Take an element $m+(U\cap V)$. ...

2

Well, the first is clearly well-defined, for if $m_1 + (U\cap V)$ and $m_2 + (U \cap V)$ are representatives of the same class, then $m_1 - m_2 \in (U \cap V) \subset U$, so $m_1 + U$ and $m_2 + U$ represent the same class in $M/U$, and a similar argument works for $V$. Can you run with that to show that the second is well-defined?

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