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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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