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9

Let $M, N$ be $R-$modules. Then the following holds. If $M$ and $N$ is flat, then so is $M\otimes_{R}N$: see related question here. If $M$ and $N$ are projective, then so is $M\otimes_{R} N$. Indeed, writing $M\oplus M'=F,\ N\oplus N'=F'$, for free $R-$modules $F,\ F'$, one has that $$ F'':=F\otimes_{R}F' $$ is free (tensor product of free modules) and ...


8

Let $0_M$ and ${\rm id}_M$ denote the zero map and identity map on an $A$-module $M$. We have $$M=0\iff 0_M={\rm id}_M.$$ Since $F$ is a functor, $F({\rm id}_M)={\rm id}_{FM}$. Since it's also additive, $F(0_M)=0_{FM}$.


8

Consider the exact sequence $M \to N \to C \to 0$, where $C$ is the cokernel of $v$. Then for any $A$-module $P$ we have an exact sequence $\DeclareMathOperator{\h}{Hom} 0 \to \h(C,P) \to \h(N,P) \to \h(M,P)$. By assumption the last map is injective, so we have $\h(C,P) = 0$. Since this holds for all $P$ it holds especially for $P = C$, so we conclude $C=0$ ...


8

Let $Q$ be a finitely generated injective $R$-module. Suppose that $R$ is not a field and let $\mathfrak m$ be a maximal ideal of $R$. Since $\mathfrak m\ne 0$ there is $a\in\mathfrak m$, $a\ne 0$. Then we have $aQ=Q$ (injective modules are divisible), and therefore $\mathfrak mQ=Q$. Localizing we get $\mathfrak mQ_{\mathfrak m}=Q_{\mathfrak m}$, and by ...


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

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

Algebra objects or monoid objects can be defined in any monoidal category. When $R$ is a commutative ring, then the category of left $R$-modules has a monoidal structure given by $\otimes_R$, and algebras in that category coincide with $R$-algebras. But the category of left $R$-modules has no "natural" monoidal structure when $R$ is not commutative - this is ...


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

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'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

$M$ 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

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

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 ...


6

The set $\{1\}$ isn't independent, because $2(1)=1+1=0$. (The $2$ here lives in $\Bbb{Z}$, so the "multiplication" is the module action; all $1$s and $0$s live in $\Bbb{Z}/2\Bbb{Z}$.)


6

To expand on Zhen Lin's comment: Let $R$ be an integral domain, $A$ a finitely presented $R$-algebra, $Q$ the fraction field of $R$. Then: to say that $A$ is smooth over $R$ can be reinterpreted geometrically as saying that the morphism of affine schemes $\operatorname{Spec} A \rightarrow \operatorname{Spec }R$ has nonsingular schemes as fibres over ...


6

Hint: An endomorphism of $\mathbb Q$ is determined by the image of $1$. Here are some details:


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 ...


5

$\mathbb R $ over $\mathbb Q$ is a vector space with dimension $2^{\mathbb N}$. and also $\mathbb R^n $ over $\mathbb Q$ is a vector space with dimension $2^{\mathbb N}$. So the additive group of $\mathbb R$ is isomorphic with additive group of $\mathbb R^n$. for example consider $f:\mathbb R\rightarrow \mathbb R^n$ be such isomorphism , now define new ...


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

This is true with added generality, that is in the non commutative world. So, assume $\phi\colon R\to S$ is a ring homomorphism between not necessarily commutative rings. If $P_{R}$ is a projective right $R-$module, then, if $P_{R}$ is a direct summand of the free right module $R^{(X)}$ for some set $X$, we get that $$ P_{R}\otimes_{R} S $$ is a direct ...


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 ...


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

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, for example $R = \mathbb Z/(6)$. Then, as $\def\Z{\mathbb Z}\Z/(6)$-modules, $\Z/(6) \cong \Z/(2) \oplus \Z/(3)$. But a free $\Z/(6)$-module cannot have two or three elements only.


5

Yes, it is. If $a\in R$, $a\ne 0$, then $aR$ is free (as a submodule of the free $R$-module $R$), in particular torsion-free, so $a$ is a non-zero divisor. This shows that $R$ is necessarily an integral domain. Now let $I$ be a non-zero ideal of $R$. Since $I$ is free (as a submodule of the free $R$-module $R$) it must have a basis with a single element ...


5

Hint $1$: If $R$ is a domain, then $K=\mathrm{Frac}(R)$ is a flat $R$-module. Hint $2$: If $N_r=\{m\in M\mid r\cdot m=0\}$, then $N_r\otimes_{R}K=0$ for $r\ne 0$. Now for any $r\ne 0$, consider the exact sequence of $R$-modules: $$0\to N_r\to M\to rM\to 0$$ Can you finish the argument from here?



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