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5

Since you are considering not necessarily commutative ring and thus is forced to taking hom-set and tensor product of abelian group, it's not reasonable to expect that $\hom_R(M,M')\otimes\hom_R(N,N')$ and $\hom(M\otimes_RN,M'\otimes_RN')$ are comparable. For example, taking $M:=R_R,N:=_RR$, then the two become $R\otimes_{\mathbb Z}\hom_R(M',N')$ and ...


4

Trivial action of ring $A$ on Abelian group $M$: $am=0$ for all $a\in A$ and $m\in M$. If $A$ has an identity element (and the axiom $1x=x$ is posed), then this forces $m=1m=0$ for all $m$, hence $M=\{0\}$.


4

Sure, it's a submodule. In fact $\langle v_1 \rangle$ is closed under addition and scalar multiplication essentially by definition: if $x,y \in \langle v_1 \rangle$ and $r \in R,$ then $x = sv_1$ and $y=tv_1$ for some $s,t \in R,$ and so the linear combination $x + ry = (s+rt)v_1 \in \langle v_1 \rangle$ as required.


4

Commutativity of the operation of $M$ is forced from the other axioms. Indeed, if $x,y\in M$, then $$ 0 = 0_R(x+y) = (1_R-1_R)(x+y) = 1_R(x+y) - 1_R(x+y) = x+y-x-y. $$ Thus, adding $y$, then $x$ on the right, we get $y+x=x+y$.


4

That's because $\,m \otimes \Bigl(\sum\limits_{i=0}^n a_i t^i\Bigr) $ can be written as $\,\sum\limits_{i=0}^n (a_im)\otimes t^i$, identified with $\,\sum\limits_{i=0}^n (a_im)t^i$.


4

$\mathbb{Z}/6\mathbb{Z}$ is not a PID, because it's not a domain.


4

If you want $\operatorname{Hom}_R(P,R)\simeq P$ for $P$ a finitely generated projective $R$-module, then forget it. $\bullet$ If $R$ is an integral domain, and $I\subset R$ is a non-zero ideal, then $\operatorname{Hom}_R(I,R)\simeq I^{-1}$, where $I^{-1}=\{x\in Q(R):xI\subseteq R\}$. (Here $Q(R)$ stands for the field of fractions of $R$.) $\bullet$ If ...


3

Did you mean to ask if any quotient module can be seen as submodule? Not in general: $\mathbb{Z}$ has $\mathbb{Z}/(2)$ as a $Z$-module (abelian group) quotient, but the latter is not a subgroup of $\mathbb{Z}$.


3

For a $b\in B$ consider the element $x=f (g(b)) -b\in B$. We claim that this $x$ is in the kernel of $g$. Then it will follow that $b= x+f(g(b))$ where in the rhs first element is in $\ker g$ and the second element is in image of $f$, that is what you wanted to be proved. To prove the claim apply $f$ to this element $x$ and use the condition that $g$ ...


3

An example of a difference: $\mathbb{Z}/3\mathbb{Z}$ is projective (and free) as a $\mathbb{Z}/3\mathbb{Z}$-module but is not projective (nor free) as a $\mathbb{Z}$-module. However things like simplicity and indecomposability will remain the same, which is easy to prove since the action of $A/\mathfrak a$ is given by the action of $A.$ The only thing ...


3

Let $R=k[X,Y]_{(X,Y)}$ for a field $k$, $P=0$, $M=(X,Y)$ the maximal ideal of $R$. This admits no $R$-module epimorphism $M \to R$.


3

Another perspective is the following. If $A$ is an abelian group, then the endomorphism monoid $\text{End}(A)$ naturally has the structure of a ring: multiplication is composition, and addition is pointwise addition. One way to describe what a (left) $R$-module structure on $A$ is is that it is a ring homomorphism $R \to \text{End}(A)$. (Compare: one way to ...


3

There exist unique pair $l\in L, n\in N$ such that $l+n=1$. But, by the definition of ideal, $ln$ lie in both ideals, so it must be zero, since we have a direct sum. So, we get $l(1-l)=0$, i.e. $l^2=l$. Now, since for any element $a$ of the ring, we have $a=1a=la+(1-l)a$, and also we know that any element can be uniquely written as such a combination, we ...


2

This is not true for arbitrary modules over unital rings. An example is $\mathbb Q$ as a module over $\mathbb Z$, which does not have a maximal submodule (see Maximal $\mathbb{Z}$-submodules in $\mathbb{Q}$). In your proof you have not shown that $L_0\neq M$, which is the source of this contradiction.


2

For a given matrix $N\in M_n(K)$ the map $\phi_N:K^n\to K^n$ defined by $\phi_N(x)=Nx$ is a homomorphism of $K$-vector spaces. By Fitting's Lemma there are two subspaces $V,W\subset K^n$ such that $K^n=V\oplus W$, $\phi_N^k(V)=0$ for $k\ge n$, and $\phi_N$ is an automorphism of $W$. We apply the Fitting's Lemma to the map $\phi_{BA}$. From $AB=I$ we get ...


2

It's easier to find the right zero divisors, transposition will give the left zero divisors. If $A=\left[\begin{smallmatrix}a&\smash{b}\\c&d\end{smallmatrix}\right]$ is a left zero divisor, then there is a vector $v=\left[\begin{smallmatrix}x\\y\end{smallmatrix}\right]\ne0$ such that $Av=0$ and conversely, because we can use the matrix having $v$ on ...


2

Part of the motivation comes from topology, and is known as the Serre-Swan Theorem. What the theorem says is that, for any compact Hausdorff space $X$, there is a one-to-one correspondence between vector bundles over $X$ and projective modules over the ring $C(X)$ of real-valued continuous functions on $X$. Specifically, the set of all continuous sections ...


2

The problem asks to find an $A$-module $N$ (eventually you already know from class) such that $M/S\simeq N$ (eventually by using a well known isomorphism theorem). Hints. Define $f:A^n\to A$ by $f(a_1,\dots,a_n)=\sum_{i=1}^na_i$. Define $f:A[X]\to A$ by $f(p(X))=p(1)$. Define $f:M_n(A)\to A^n$ by $f((a_{ij})_{i,j})=(a_{11},\dots,a_{n1})$.


2

This will only work in general if the $e_j R$ are assumed to be pairwise nonisomorphic. As you said yourself you know that each $e_j R$ is a direct summand of $M$. So we have $M = e_1 R \oplus M_1$ and by the Exchange Lemma you get for every $j \neq 1$ that $e_j R$ appears as a direct summand of either $e_1 R$ or $M_1$ and the first case, which would imply ...


2

$K[[X]]$ is not a noetherian $K[X]$-module since it's not finitely generated. $K[[X]]$ is not an artinian $K[X]$-module: $XK[X]\supset X^2k[X]\supset\cdots$ is a strictly descending chain of $K[X]$-submodules. (You can also argue as follows: every submodule of an artinian module is artinian, so if $K[[X]]$ is an artinian $K[X]$-module, then $K[X]$ is an ...


2

I don't know the book by Curtis and Reiner, but a well-established convention in ring and module theory is to write morphisms on the opposite side to the scalars and probably the book uses it. This avoids awkward references to the opposite ring that pop out when morphisms are written on the same side. So, if left module are considered, morphisms are ...


2

The first part is correct, the second is not. Do you know when $x/1 = 0$ in $S^{-1}M$? Hint: Try to choose a non-zero element $x$ of $M$ and observe that the annihilator $\text{Ann}(x)$ of $x$ is contained in some maximal ideal $\mathfrak{m}$. Then let $S = R - \mathfrak{m}$. What can you tell about $x/1$ in $S^{-1}M$? First let us recall that an ...


2

Notice that if $v\in V_0$, we have $$\phi(v)=\sum_{g}{vg}=\sum_{g}{v}=|G|v$$ This is because we are summing $v$ with itself $|G|$ times. If we divide by $|G|$ we get a vector that hits $v$, so the map is surjective. (This only works if the characteristic of the field doesn't divide $|G|$; if it does then $\phi(v)=0$.)


2

Let $k$ be a field, and take $R=k(t)$, $R'=k[t]$, $R''=k$, $M=k(t)$, and $M''=k$. Then every non-zero element of $M$ generates an $R'$-submodule of $M$ isomorphic to $R'$, which is not simple. So there is no simple $R'$-module $M'$ with $\operatorname{Hom}_{R'}M',M)\neq0$.


2

Hint: what exact sequence can you build up out of $M_1\oplus M_2, M_1 + M_2$ and $M_1\cap M_2$?


2

There is no mathematical reason to prefer right or left modules. The symmetry is broken by the conventions of written English (or more generally, written European language) and mathematical exposition, in which a function applied to an element of a set is typically written $f(x)$ and function composition is $f(g(x))$; most of us would find the notation ...


2

You seem to be just confusing different notions of module. $G$-module would be an abelian group $M$ with the action of a group $G$ compatible with addition. (This can well be the trivial action.) $R$-module, or just module, where $M$ is an abelian group and one has scalar multiplication with the elements from the ring $R$ similarly to case of ...


2

$xR\cap yR=\{0\}$, so your ideal is the direct sum of $xR$ and $yR$, both of which are isomorphic to $R$ as right ideals, so its endomorphism algebra is isomorphic to $M_2(R)$.


2

$A$ is a module of finite length over itself. Prove, by induction on the length, that a finitely generated module over $A$ is semisimple. In particular $A$ is semisimple.


2

The usual proof uses the Structure theorem for finitely generated modules over a principal ideal domain. But there is an alternative proof based on the fact that the finitely generated torsion-free modules over an integral domain are isomorphic to a submodule of a free module of finite rank. (For a proof see here.) Since over a PID the submodules of free ...



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