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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 $N = R \oplus 0 \oplus 0 \oplus \dotsc$.


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 ...


5

$[R^{op}, \text{Ab}]$ is the category of right $R$-modules, which is equivalently the category of left $R^{op}$-modules. The reason to prefer taking right modules here is the same reason why presheaves are contravariant functors and not covariant functors: it's so that the Yoneda embedding, which in this case is $R \to [R^{op}, \text{Ab}]$, is covariant. In ...


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) ...


4

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 ...


4

If $R$ is a ring, then the two sided ideals in the ring $M_n(R)$ of $n\times n$ matrices are in bijection with the two sided ideals of $R$. The ring $\mathbb{Z}/2^k\mathbb{Z}$ has exactly $k$ proper ideals. If $F$ is a field, then $F\times F$ has exactly two (isomorphism classes of) simple modules (irreducible and simple are synonyms).


4

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 ...


4

Hint: Let $M_i = e_i M$, where $e_i = (0,\dotsc,1,\dotsc,0) \in R$ with $1$ in the $i$th entry. Use that $e_i$ are pairwise orthogonal idempotents with $\sum_i e_i = 1$ to show $M = \oplus_i M_i$.


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 ...


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

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.


3

You actually have a short exact sequence $0\to S\stackrel{f}\to R^2\stackrel{g}\to S\to 0$, where $g(a,b)=X^2a+X^3b$, and by tensoring this with $S$ want to prove that it is not exact, that is, $S$ is not $R$-flat. Since $R^2\otimes_RS\simeq S^2$ by $(a,b)\otimes c\mapsto(ac,bc)$, we can see $f\otimes 1: S \otimes_R S \to S^2$ sending $p\otimes_R q$ to ...


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 ...


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.


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 ...


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

If $r$ is a real number, $r$ mod $1$ is the fractional part of $r$. It can also be written $\{r\}$.


3

Let $A$ be a commutative ring and $I\subset A$ an ideal. Let us investigate whether $A/I$ is flat over $A$. Consider the injection $0\to I\to A$ and tensor it with $A/I$. Using $M\otimes_A A/I=M/IM$ (for any $A$-module $M$), we get the morphism of $A$-modules: $$I\otimes_AA/I\to A\otimes _AA/I \quad\text {identified with} \quad I/I^2\to A/I: \bar ...


3

There are two trivial answers and one more profound answer: 1) $m \otimes n = m' \otimes n'$ means that $(m,n) - (m',n')$ lies in the mentioned submodule of bilinear relations 2) $m \otimes n = m' \otimes n'$ means that $\beta(m,n)=\beta(m',n')$ for all $R$-bilinear maps $\beta : M \times N \to T$, where $T$ is any abelian group. 3) We have the following ...


3

This is true and easy to prove. Let $M\stackrel{g}\to N$, $P\stackrel{f}\to N$ be graded homomorphisms, and $P\stackrel{h}\to M$ be a homomorphism such that $f=gh$. Then there is a graded homomorphism $P\stackrel{h'}\to M$ such that $f=gh'$. For $x_n\in P_n$ we have $f(x_n)\in N_n$. From $h(x_n)=\sum y_m$ with $y_m\in M_m$ we get $f(x_n)=gh(x_n)=\sum ...


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

I would say yes, it has the empty set as a basis.


3

By definition, your ideal is isomorphic to $R^n$ but this isomorphism is quite abstract and your proof shows nothing. Counter-example : the ideal $2 \mathbb{Z}$ in $\mathbb{Z}$. Exercise : show that your property is true if and only if $R$ is a field.


3

You're doing fine so far! The next thing you have to do is to count how often a fixed isomorphism type of irreducible representations occurs on both sides. For this, use the following: If $M=M_1\oplus ...\oplus M_n$ a decomposition of a semi-simple, finitely-generated $R$-module $M$ as a sum of irreducible $R$-modules, and if $I$ is any irreducible ...


2

Yes. If $N$ were projective, then the short exact sequence $$0\to e_iJ \to S_i \to N \to 0$$ of right $R$-modules would split, so $e_iJ$ would be a right $R$-module direct summand of $R$, and therefore of the form $fR$ for some idempotent $f$. But the Jacobson radical contains no nonzero idempotents.


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.


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 ...


2

$S$ can be any commutative graded ring. Recall that for two graded $S$-modules $M,N$ the graded hom (or internal hom) $\underline{\hom}(M,N)$ is given by $$\underline{\hom}(M,N)_n := \hom(M,N[n]) \subseteq \prod_p \hom(M_p,N_{p+n}).$$If $M$ is of finite presentation, then we have $\bigoplus_n \underline{\hom}(M,N)_n = \hom(U(M),U(N))$, where $U : ...


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 ...


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 $\Rightarrow$ $K$ ...



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