# Tag Info

10

No: For a local ring $(R,{\mathfrak m})$ with ${\mathfrak m}^2=0$ you have $\text{Hom}_R(R/{\mathfrak m},R)\cong\{x\in R\ |\ {\mathfrak m}x=0\}={\mathfrak m}$, so it suffices to choose $R$ such that ${\mathfrak m}$ is not finitely generated, e.g. $R := {\mathbb k}[x_1,x_2,\ldots]/(x_i^2, x_i x_j)$.

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 $a ... 9 Nakayama Lemma. Let$N$be a finitely generated$R$-module, and$J\subseteq R$. Suppose that$J$is closed under addition and multiplication and$JN=N$. Then there is$a\in J$such that$(1+a)N=0$. (Here by$JN$we denote the subset of linear combinations of$N$with coefficients in$J$.) Cayley-Hamilton Theorem. Let$A$be a commutative ring,$I$an ... 8 If$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]$. We can also see this isomorphism more directly by ... 7$\mathbb{Z}$-modules are precisely abelian groups. As every ring is an abelian group, it is a$\mathbb{Z}$-module. It is entirely possible to be a module over more than one ring. For example, if$M$is an$R$-module then it is also an$S$-module for any subring$S$of$R$(you seem to be interested in the case where$M = R$). Another example is given by ... 7 No, your argument isn't correct: if$1\otimes x=1\otimes y$then not necessarily$x=y$. I'd do this as follows: if$M$is a$\mathbb Z$-module, then$\mathbb Q\otimes_{\mathbb Z}M\simeq S^{-1}M$, where$S=\mathbb Z-\{0\}$. The isomorphism is given by$\dfrac ab\otimes x\mapsto\dfrac{ax}{b}$. This way$1\otimes(1,1,\dots)$corresponds to the fraction ... 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 ... 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 Clearly, the only candidate for a basis is$\{1\}$. However, is$\{1\}$a linearly independent set? That is, is it true that for all$n \in \Bbb Z$,$n\cdot 1 = 0 \iff n = 0$? 6 For every nonzero$x, y\in\mathbb{Q}$, there are nonzero integers$m, n$such that $$mx=ny$$ (where$mx, ny$are interpreted in the obvious way). Now, any homomorphism$f$between$(\mathbb{Q}, +)$and another structure$(G, *)$must preserve multiplication by integers: $$f(mx)=mf(x).$$ So if$(\mathbb{Q}, +)\cong(\mathbb{Q}^n, +)$, then$(\mathbb{Q}^n, +)$... 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

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

$M$ 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

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

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

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

For the second question, the finite rings are precisely those for which every finitely generated module has finitely many submodules. For a ring $R$ and $r\in R$, let $M_r$ be the submodule of $R\oplus R$ generated by $(1,r)$. Then $(1,r)$ is the only element of $M_r$ whose first coordinate is $1$, and so $M_r\neq M_s$ for $r\neq s$, and so if $R$ is ...

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 ...

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 ...

5

It is enough to prove it preserves short exact sequences: $\;0\to M\to N\to P\to 0$. As the tensor product is right-exact, and $S^{-1}M\simeq M\otimes_A S^{-1}A$, it is even enough to prove it preserves injectivity. So consider an injective morphism $\varphi\colon M\to N$ and suppose $\;S^{-1}\varphi\Bigl(\dfrac ms\Bigr)=0$ in $S^{-1}N$. This means there ...

5

If $m = n$, then $0 \to \mathbb{Z}/n\mathbb{Z} \xrightarrow{\operatorname{id}} \mathbb{Z}/n\mathbb{Z} \to 0$ is a free resolution. If $m < n$, it will be useful to write $n = km$, where $k > 1$. The first step of the resolution looks like $F_0 \xrightarrow{\epsilon} \mathbb{Z}/m\mathbb{Z} \to 0$ for some free $\mathbb{Z}/n\mathbb{Z}$-module $F_0$. ...

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

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

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 ...

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