# Tag Info

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 ith 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$\RightarrowK$noetherian... (ii)$Z$noetherian$\RightarrowK\$ ...

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