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1

$\newcommand{\tensor}{\otimes}\newcommand{\from}{\colon}$Let $M \tensor N$ be an $A$ module, and $f \from M \times N \to M \tensor N$ be a bilinear map that satisfy the universal property of the tensor product. The elementary tensor $m \tensor n$ is defined to be $f(m,n)$. We will prove using just the universal property and not the construction that: $M ...


1

There is a proof that the universal property determines the tensor product of modules up to isomorphisms. The initial construction is only used to prove the existence of the tensor product. It is not a bad thing to visualize by tensors but sometimes it is better to strictly use the universal property. $\require{AMScd}$ \begin{CD} A\times B ...


0

The local case is proved here, Lemma 10.97.2, and then extend the result to the non-local case by using that an artinian ring is isomorphic to a finite product of artinian local rings and a module $M$ over a finite product of rings $R_1\times\cdots\times R_n$ has the form $M_1\times\cdots\times M_n$ with $M_i$ an $R_i$-module.


1

$\begin{eqnarray}{\bf Hint}\quad\ x^n &\in&\! \langle x^{n-1},\ldots,x,1\rangle\, =: M \\ \Rightarrow\ x^{n+1} &\in& \langle x^n,\,\ldots,x^2,x\rangle\, \subseteq\, M\ \ {\rm by}\ \ x^n\in M\\ &\vdots&\\ \Rightarrow\ x^{n+k} &\in& M \end{eqnarray}$ Remark $\, $ Instead of using the division algorithm, we interpret the ...


1

For a monic $f$, we have division with remainder, that means for every $g\in \mathbb{Z}[X]$, there are uniquely determined $q,r\in\mathbb{Z}[X]$ with $$g = q\cdot f + r$$ and $r = 0$ or $\deg r < \deg f$ (if you use the convenient convention $\deg 0 = -\infty$, the last distinction is not necessary). That means that every element of ...


1

Everything is correct. Only suggestion I have is that once you argue that your generators can have a common denominator you might as well call them $\frac{e_i}{c}$ instead of $\frac{d_ia_i}{c}$. It just makes the equations look a bit simpler. In fact you could even argue that $\frac{1}{c}$ generates all your generators so it suffices to show that $\mathbb ...


2

It is in fact that easy. You can make it even simpler by noting you can conclude $\mathbb Q \subseteq \langle \frac{1}{c} \rangle$ from your assumption. Since $\frac{1}{2c}\in\mathbb Q$ we have $\frac{1}{2c}=\frac{k}{c}$ for some $k\in\mathbb Z$, but there is no $k\in\mathbb Z$ such that $2k=1$.


3

Your claim is NOT true. $M = \mathbb{Z}, N = \mathbb{2Z}, M' = \mathbb{8Z}, N' = \mathbb{4Z}.$ All are considered modules over $\mathbb{Z}.$ Then $\frac{M}{M'} \oplus \frac{N}{N'} \cong \mathbb{Z}_8 \oplus \mathbb{Z}_2.$ On the other hand $\frac{M}{N'} \oplus \frac{N}{M'} \cong \mathbb{Z}_4 \oplus \mathbb{Z}_4.$ So this two are not isomorphic. The mistake ...


0

Define a homomorphism $\phi: \bigoplus_{i \in I} Hom(M,N_i) \rightarrow Hom(M, \bigoplus_{i \in I} N_i)$ by sending $f = (f_k)_{k \in I^*}$ to $\phi(f) = \sum_{k \in I^*} f_k$, where $I^*$ is a finite subset of $I$. Clearly $\phi$ is injective. For surjectivity, take $f \in Hom(M, \bigoplus_{i \in I} N_i)$ and let $x_1,\dots,x_n$ be the generators of $M$. ...


3

Given a ring $R$ that's Noetherian as a $\Bbb Z$-module, it's automatically a (left- or right- as desired) Noetherian ring, as ideals are by definition closed under sum of elements; so every increasing chain of ideals is an increasing chain of $\Bbb Z$-submodules of $R$, and thus stabilizes. $\Bbb Q$ is indeed a Noetherian ring (it only has two ideals) ...


2

A ring $A$ is said to have (right) invariant basis property (IBP) if for any integers $s, t \geq 0$, $$ A^{\oplus s} \cong A^{\oplus t} \Rightarrow s = t $$ where the above isomorphism is as right $A$ modules. Any non-zero commutative ring, left Noetherian ring, semi-local ring satisfies IBP. The following link might be useful: Cohn, P. M. - Some remarks ...


5

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 ...


0

Here is a more fundamental approach to showing that $\mathbb{Q}$ is not finitely generated, based on a general method. Here's the general method. In a group $G$ with a finite generating set $\{g_1,…,g_K\}$, for any increasing sequence of subgroups $H_1 \subset H_2 \subset \cdots$ whose union $\cup_i H_i$ equals $G$, for sufficiently large $i$ the subgroup ...


1

$S^{-1}B$ and $f(S)^{-1} B$ are in fact isomorphic, simply because the action of $A$ on $B$ is defined via $f: A \rightarrow B$. To see this explicitly, note that when you gave the definition of $S^{-1}B$, you defined $(b,s)$ ~ $(b',s')$ if there exists $t \in S$ such that $t (s b' - s' b)=0$. But what does $s b'$ mean? What is the action of $s$ on $B$? This ...


3

If an abelian group $G\ne\{0\}$ is free, then it is isomorphic to $\mathbb{Z}^{(X)}$ (direct sum of copies of $\mathbb{Z}$), for some set non empty set $X$. Then $\mathbb{Z}$ is an epimorphic image of $G$. Since $\mathbb{Z}$ is not divisible, $G$ can't be divisible. But $\mathbb{Q}$ is divisible. The group $\mathbb{Q}$ is not finitely generated: if ...


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 ...


2

Show any finitely generated subgroup of $\mathbf Q$ is cyclic. Since $\mathbf Q$ is not cyclic, it cannot be finitely generated. $(1)$ It cannot be free: it is not cyclic, so any putative basis has at least two elements. But if $x,y$ are elements of the basis, we know $\mathbf Z^2\simeq \langle x,y\rangle=\langle x'\rangle\simeq\bf Z$ which is impossible. ...


0

It suffices to show that $I^{n-1}/I^n$ is finitely generated as an $R$-module, because then it is also finitely generated, hence Noetherian (since $R/I$ is Noetherian), as an $R/I$-module, and finally being Noetherian over $R/I$ or over $R$ is the same.


2

Let us assume that $g$ has the quoted property. Let $Y = N/\text{im}(g)$ and $ k : N \rightarrow Y$ the natural projection. Then $k \circ g = 0$, so $k = 0$ by assumption. But $k$ is surjective, so we have to have $Y = 0$, hence $N = \text{im}(g)$ and so $g$ is surjective.


2

First, if $\text{coker}\ f$ is Artinian, then so is $\text{coker}\ f^m$ for all $m\geq 1$: it suffices to check this for $m=2$, and in this case, note that there is an epimorphism $$\text{coker}\ f\twoheadrightarrow\text{im} f / \text{im} f^2$$ and a short exact sequence $$0\to\text{im}\ f / \text{im}\ f^2\to\text{coker}\ f^2\to \text{coker}\ f\to 0.$$It ...


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 ...


1

Let $A$ and $B$ be ideals of the ring $R$. Then $A/BA$ is a module over $R/B$ in a natural way. Indeed, for $b\in B$ and $a\in A$, the product $ba\in BA$, so $$ b(a+AB)=0+AB $$ in the module $A/AB$. Thus $\operatorname{Ann}(A/BA)\supseteq B$. If $M$ is an $R$-module and $\operatorname{Ann}(M)\supseteq B$, then $M$ is a module over $R/B$ by defining $$ ...


1

(a) Yes. (b) Yes. (c) Over a semisimple ring every non-zero module is semisimple (as a quotient of a free module which is a direct sum of copies of your semisimple ring, hence semisimple).


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 ...


-2

Thanks to the comments of Manos. I could figure out finally what was going on. Let $I$ be an ideal of $A$. The following is a condition for flatness: An $R$ module $M$ is flat iff $I\otimes_R M$ injects into $M$. Let $I = <x^2, xy>$ be the ideal of $A$. Then note that the elements $x^2 \otimes y$ and $xy \otimes x$ of $I \otimes_A B$ are mapped to the ...


2

Such a map can exist, but not for every noncommutative ring. For example, no such map can exist over a division ring. This is because the annihilator of the left tensor product module is a nonzero left ideal, hence the left tensor product is always trivial in that case. For an example where the map does exist, let $T$ be the tensor algebra over a field, and ...


3

For an explicit counterexample, consider $R = k[X,Y]$ for any field $k$, which is free over itself. Then the ideal $\mathfrak m = RX+RY$ is not free.


2

You are asking if a submodule of a free module is necessarily free. This is not true in general. It is, however, true if $R$ is a PID. A counter example is given by $R=\mathbb{Z}/\mathbb4{Z}$ as a module over itself. The submodule $2\mathbb{Z}/4\mathbb{Z}$ is not free.


2

For Dedekind domains we have $\operatorname{Pic}(R)\simeq\operatorname{Cl}(R)$, where $\operatorname{Cl}(R)$ is the ideal class group of $R$. Your case is treated here in detail.


2

There's no need to consider $J$; just assume $R$ is semisimple. Every (right) module over a semisimple ring is a direct sum of simple modules. Moreover, consider $(S_i)_{i\in I}$, a family of simple modules over $R$ such that every simple $R$-module $S$ is isomorphic to $S_i$, for some $i\in I$; if $i\ne j$, then $S_i$ is not isomorphic to $S_j$. We can ...


3

You can proceed as usual by starting with $F\stackrel{f}\to A\to 0$ and $H\stackrel{h}\to C\to 0$, where $F$ and $H$ are free of finite rank. Then show that there is an exact sequence $G=F\oplus H\stackrel{g}\to B\to 0$. Now consider $F'=\ker f$ and so on. You have a short exact sequence $0\to F'\to G'\to H'\to 0$. Now use the result for finitely generated ...


3

If I understand your question correctly, consider that $a \equiv b \pmod x$ if and only if $a-b = kx$ for some $k\in\mathbb{Z^+}$. Then $x=\frac{a-b}{k}$, hence there is precisely one $x$ for each divisor $k\in\mathbb{Z^+}$.


3

user26857's answer shows how to repair the reduction to the Noetherian case. Here is how to repair the proof of the Noetherian case: Let $M$ be a noetherian $A$-module and let $N \subseteq M$ be a submodule. Let $f : N \to M$ be a surjective linear map. Then $f$ is injective. Proof: Let $n \geq 0$. Although $f^n$ is not a well-defined homomorphism this ...


3

Let $A$ be a commutative ring. Let $M$ be a finitely generated $A$-module and $N$ be an $A$-submodule of $M$. Let $f\colon N \rightarrow M$ be a surjective homomorphism of $A$-modules Then $f$ is injective. Proof. Let $0 \neq x'_0 \in N$. It suffices to prove $f(x'_0) \neq 0$. Set $f(x'_0) = x_0$. Let $x_1, \dots, x_n$ be generators for $M$. Then ...


0

A covariant exact functor ${\bf A}\to{\bf B}$ preserves finite limits and colimits, so a contravariant exact functor does it too as a (covariant) functor on the opposite category: $F:{\bf A}^{op}\to {\bf B}$. Here ${\bf A}^{op}$ is again Abelian (as [at least one form of] the axioms are self dual). Now limits (in particular, kernels and products) in ${\bf ...


2

Show that $\hom_{R/I}(M/IM,-) \cong \hom_R(M,U(-))$, where $U$ is the forgetful functor from $R/I$-modules to $R$-modules. Hence, this is a composition of two exact functors, hence exact.


3

Define $$\phi: R\to Rx\le M\;,\;\;\phi(r):=rx$$ prove the above is a (left) $\;R$- module homomorphism, and now use the first isomorphism theorem.


2

For $i=1, \dots, n+m$ call $$c_i = \left\{ \begin{matrix} a_i & \mbox{ if } &i \leq n \\ b_{i-n} & \mbox{ if } &i \geq n+1 \end{matrix} \right. $$ and analogously $$w_i = \left\{ \begin{matrix} x_i & \mbox{ if } &i \leq n \\ y_{i-n} & \mbox{ if } &i \geq n+1 \end{matrix} \right.$$ Then $$x+y = \sum_{i=1}^{n+m} c_iw_i \in ...


1

Let $z_1=x_1,\dots,z_n=x_n,z_{n+1}=y_1,\dots,z_{n+m}=y_m$, and let $c_1,\dots,c_n,c_{n+1},\dots,c_{n+m}$ be defined similarly based on $a_i,b_j$. Then:$$x+y = \sum c_iz_i$$


0

Do you know that the Jacobson radical of a semisimple ring is zero? If so, you can easily show that $n$ would have to be a squarefree positive integer for $\Bbb Z_n$ to have Jacobson radical zero. You've already mentioned you have the converse that $\Bbb Z_n$ is a product of fields for a squarefree integer, so then you would have both directions.


0

You should just use the universal property, don't worry about exact sequences. For starters assume the $p_i$ exist. You have to show that a family of maps $N_i\colon X_i \to Z$ factors through a unique map $X \to Z$. Well using the $p_i$ you get maps $X \overset{p_i}{\rightarrow} X_i \overset{N_i}{\rightarrow} Z$ so sum this over $i$ to get $X \to Z$. ...


0

Let $A=\Bbb Z_n$. I suppose the morphisms are the inclusion and multiplication by $s$. If $n=rs$, then multiplication by $r$ induces an isomorphism, $rA\simeq \Bbb Z_s$. Similarly, multiplication by $s$ induces an isomorphism $sA\simeq \Bbb Z_r$. What you have is a SES $$0\longrightarrow rA\longrightarrow A\longrightarrow sA\longrightarrow 0$$ If we ...


0

Remember that if $0 \to A \to B \to C \to 0$ is a split exact sequence of modules, then $B \cong A \oplus C$. Suppose $(r,s)>1$. (Note that $r\Bbb Z_n \cong \Bbb Z_s$ and vice versa.) Show that $\Bbb Z_n \not\cong \Bbb Z_r \oplus \Bbb Z_s$. In the other direction, you should be able to write down an explicit section $s \Bbb Z_n \to \Bbb Z_n$.


1

We can argue as follows. Pick an $h \in ker\ g_* \subseteq Hom_R(M,M_2).$ Then we have $g\circ h = 0.$ This means $im\ h \subseteq ker\ g = im\ f.$ Since $f$ is injective, we have the $R$-homomorphism $(f|_{im\ f})^{-1}:im\ f \rightarrow M_1.$ So we can define $j := (f|_{im\ f})^{-1} \circ h: M \rightarrow M_1.$ This is an $R$-homomorphism since it's a ...


3

Here is a hint: count the elements of order $p$ in the group $$ \mathbb Z/p^{i_1} \oplus\mathbb Z/p^{i_2} \oplus \dots \oplus \mathbb Z/p^{i_n}.$$ Here's the answer to the hint: You should find that there are exactly $p^n - 1$. Namely if you write down an element as $(x_1,\dots,x_n)$ than this has order $p$ if $(px_1,\dots,px_n) = (0,\dots,0)$, i.e. if ...


0

Your idea is not quite right: the product $M\times N$ of two modules $M,N$ is the pullback of $M\to 0\leftarrow N$, but in your problem you are interested in the pullback of ${\mathbb Z}_{p^2}\to {\mathbb Z}_p\leftarrow {\mathbb Z}$, which can be realized as the submodule of ${\mathbb Z}_{p^2}\times{\mathbb Z}$ consisting of those pairs $(\overline{x},y)$ ...


4

As you say, there is no reason why this should be true. Consider on the one hand $R$ regarded as an $(R, R)$-bimodule in the usual way and on the other hand $R$ regarded as an $(R, R)$-bimodule where, say, the left $R$-module has been twisted by an automorphism $\varphi : R \to R$, which is to say that left multiplication now looks like $$L_r s = \varphi(r) ...


2

I suppose that you know that if we have a s.e.s. $0\to M'\to M\to M''\to 0$, then $M$ is noetherian iff $M'$ and $M''$ are noetherian. (If not, take a look here.) Now split your exact sequence in two s.e.s.: $0\to X\to Y\to K\to 0$ and $0\to K\to Z\to T\to 0$. (i) $Y$ noetherian $\Rightarrow$ $K$ noetherian... (ii) $Z$ noetherian $\Rightarrow$ $K$ ...


4

One way to approach the problem: If $M$ is a simple $R$-module, then $M \cong R/I$ for some maximal left ideal $I ⊂ R$. What are the maximal left ideals in $R$? Elaborating: Show that a maximal left ideal $I ⊂ R $ is already generated by any matrix of maximal rank in $I$ and find that maximal rank. Hint: Use row reduction. Then show that any two matrices ...


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 ...



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