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7

Zorn's lemma tells you that if you consider the set of all submodules, then this set has a maximal element. It does not tell you whether every nonempty collection contains this maximal element, so you cannot conclude that the collection has one. Now, the collection indeed does have an upper bound (as you've remarked, $ M $ is one such submodule, as it is a ...


5

For a very simple counterexample to both statements, take $R=F=\mathbb{Z}$. Then $\{2,3\}$ is a generating set which contains no basis, and $\{2\}$ is a linearly independent set which cannot be extended to a basis.


5

$\prod_{i=1}^\infty \mathbb{Z}$ is a counterexample. It is torsion free, not free (Why isn't an infinite direct product of copies of $\Bbb Z$ a free module?), and every nonzero element is only divisible by finitely many integers, so your extra hypotheses hold.


4

The fact that $\mathbb{Z}[i]\otimes_{\mathbb{Z}}\mathbb{R}\cong\mathbb{C}$ only as $\mathbb{Z}$-modules follows as $\mathbb{Z}[i]\cong\mathbb{Z}^{\oplus 2}$ and $\mathbb{Z}^{\oplus 2}\otimes_{\mathbb{Z}}\mathbb{R}\cong\mathbb{R}^{\oplus 2}\cong\mathbb{C}$. If you need to use the universal property of tensor products somewhere, it can be used in showing that ...


4

It only holds for the trivial group. As it was mentioned in the comments, we consider $\mathbb{Z}$ as a trivial $\mathbb{Z}G$-module and in this way the augmentation map $\epsilon \colon \mathbb{Z}G \to \mathbb{Z}$ is a morphism of $\mathbb{Z}G$-modules. Let $ f \colon \mathbb{Z} \to \mathbb{Z}G$ be a morphism of $\mathbb{Z}G$-modules and say $ f(1) = ...


4

The map you wrote is not an isomorphism. It is instead a map that arguably parametrizes (though not uniquely) all multiples of the diagonal map $M\rightarrow M^{\oplus n}$ (where the multiple, depending on an element $(r_1,\ldots,r_n)$ of $R^{\oplus n}$, is $\prod_ir_i$). This is why, as you observe, you could make a similar argument for $R^{\oplus n}\times ...


4

Schur's lemma tells you that an endomorphism of $M$ is either zero or invertible. That means precisely that $\text{End}_R(M)$ is a division ring. If $R$ is noncommutative this is the most that you can say. But in fact, because $R$ is commutative, the simple $R$-modules have the form $R/m$ where $m$ is a maximal ideal, and their endomorphism rings are again ...


4

$\mathbb{R}[x]$-submodules of $\mathbb{R}^2$ are vector subspaces $V$ such that $T(V)\subset V$. In particular, $V$ other than $0$ or $\mathbb{R}^2$ would be a line. Now can you find a line fixed by a rotation?


3

The condition is equivalent to being a torsion $\mathbb{Z}$-module. If $X$ is a torsion module, then any finitely generated submodule is a finitely generated torsion $\mathbb{Z}$-module : by the structure theorem for finitely generated modules over PID, it's finite. If $X$ is not a torsion module, then any non-torsion element generates an infinite ...


3

Let $R$ be a ring such that every left $R$-module is free, and let $I \subset R$ be a maximal left ideal. Then $R/I$ is a simple nonzero $R$-module, and is free by hypothesis, so $R/I$ has a basis. Take any basis element $x$, and let $\varphi \colon R \to R/I$ be the $R$-module homomorphism given by $\varphi(r) = rx$. Since $x$ is nonzero and $R/I$ is ...


3

Since $a \text{ mod } p = -1$, you already know that $p| a+1$. Since $p|-p$, you can infer $p| ( a+1)+(-p)$ and $p| a-(p-1)$, that is to say $a \text{ mod } p = p-1$.


3

First, note that $ S $ must contain a non-unit element of $ \mathbb{C}[x] $ by its definition, let this element be $ p $. (We may take $ p $ to be monic by multiplying by a unit if necessary.) Then, $p $ can be viewed as a polynomial with coefficients in $ \mathbb{C} $, and $ x $ is a root of the polynomial $ p(y) - p = 0 $. (Here, $ p(y) $ is $ p $ ...


3

The set of $R$-bilinear maps $f: R\times N\to P$ is in natural bijection with the set of $R$-linear maps $g:N\to P$. The correspondence is obtained by setting $f(r,n) = g(rn)$.


3

Zorn's lemma says that if $X$ is a poset and every chain has an upper bound then $X$ has a maximal element. It does not say that every chain has a maximal element, which is what's relevant here.


3

Zorn's lemma If every chain in a partially ordered set has an upper bound, then the partially ordered set has a maximal element. The condition you are mentioning is about any (non empty) collection of submodules having a maximal element. Such collections need not satisfy the condition that every chain has an upper bound, so Zorn's lemma cannot apply. ...


3

Since $I^n=0$ the ideal $I^{n-1}$ is a finitely generated $R/I^{n-1}$-module: it is finitely generated (any power of a finitely generated ideal is finitely generated) and $I^{n-1}\cdot I^{n-1}=0$; see also here. Now use the exact sequence of $R$-modules $$0\to I^{n-1}\to R\to R/I^{n-1}\to 0.$$


3

One reason why one needs the bilinear map (or multilinear in general), instead of just going ahead to define a map on the tensor product, is that one needs to show that the latter is well-defined. So it may be easy to define some $A \otimes B \rightarrow C$, but it actually may be quite hard to prove that it is well-defined. On the other hand, once a ...


3

How do you use the splitness of your exact sequence? Once done, it is easy. So, let me call the middle term with basis $e_1,e_2$ and the map on the left by $i$. Then $i(1)=ae_1+be_2$. This splits says we have a map $j:Ae_1\oplus Ae_2\to A$ with $j\circ i$ identity. If $j(e_i)=c_i$, we see that $ac_1+bc_2=1$. Let $v_1=i(1), v_2=-c_2e_1+c_1e_2$. Then using the ...


2

Since $x^p$ and $y^p$ are in the center, they act as scalars on your simple module $V$. This means that in fact $V$ is a module over the algebra $k\langle x,y\mid yx-xy-1,x^p-\alpha,y^q-\beta\rangle$ forsome scalars $\alpha$, $\beta$ in the field. Show that this algebra is central and simple (imitating the proofs for the Weyl algbra in characteristc zero, ...


2

Note that since $A$ is not only semi-simple but also simple, there is a unique simple $A$-module. Call it $I$. Then any finitely generated $A$-module is isomorphic to $I^n$ for some $n$ (in particular they are all projective). Then $\operatorname{End}_A(P) \simeq \operatorname{End}_A(I^n) = M_n(\operatorname{End}_A(I))$ (the last equality is valid for any ...


2

Since $A\simeq B/(\{0\}\oplus M)$ we have an exact sequence of $B$-modules (hence of $A$-modules, too): $$0\to M\to B\to A\to 0.$$ "$\Rightarrow$" $B$ is a noetherian ring, and $A$ is q quotient of $B$, so $A$ is also a noetherian ring. Moreover, the ideal $\{0\}\oplus M$ of $B$ is finitely generated hence $M$ is a finitely generated $A$-module. ...


2

If you look at the dimensions : $\dim(C) = |S|$, $\dim(D) = |T|$, and $\dim(A\otimes B) = |S|+|T|$. But then $\dim(C\otimes D) = |S|\cdot |T|\neq \dim(A\otimes B)$ (if the dimesnions are finite).


2

In general this is not true: for instance that $\mathbb{Z}_p \otimes_\mathbb{Z} \mathbb{Z}_q \cong \{ 0\}$ if $\gcd(p,q)=1$.


2

With conclusion $\,a_1\equiv a_2,\,$ it is true iff $\,b\,$ is invertible mod $n\,$ (iff $\,\gcd(b,n)=1)$ Else $b,n$ share a divisor $c>1$ so $\,\color{#c00}{(n/c)}b = n(b/c)\equiv 0\equiv \color{#c00} 0\cdot b,\,$ but $\,\color{#c00}{n/c\not\equiv 0}\pmod n$ But your hypothesis does not imply that $b$ is invertible. Indeed it is true for all ...


2

This is definitely not true in general unless $\gcd(n, b) = 1$. For instance $2 \cdot 3 = 4 \cdot 3 \pmod 6$ but $2 \not \equiv 4 \pmod 6$. The problem is that here $3$ is not invertible since it is not relatively prime to $6$.


2

No. To show it's injective, you have to show that, if $\;\sum r_i\otimes m_i\mapsto\sum r_im_i=0$, then $\;\sum r_i\otimes m_i=0$. But that is because $\;\sum r_i\otimes m_i=\sum 1\otimes r_im_i=1\otimes0=0$.


2

Consider $G=\Bbb Z[x], H= \Bbb R$ treated as $\Bbb Z$-modules. Then for everything to be an elementary tensor would mean that every polynomial in $\Bbb R[x]\cong\Bbb Z[x]\otimes_{\Bbb Z}\Bbb R$ is of the form $r\cdot p(x)$ for some $r\in\Bbb R$ and $p(x)\in\Bbb Z[x]$.


2

Check that for any $$\;m\Bbb Z\le\Bbb Z\;\;,\;\;2\Bbb Z\cap m\Bbb Z\neq0$$ and thus $\;2\Bbb Z\;$ , or for that matter any non-trivial subgroup of the integers, cannot be a non-trivial direct summand.


2

It isn't true that if every module of the form $R/aR$ is projective, then $R$ is semisimple. For instance, let $R=\mathbb{F}_2^X$ for some infinite set $X$. Then for any $a\in R$, the ideal $aR$ is a direct summand of $R$, with complement $(1-a)R$ (this follows from the fact that $a^2=a$). Thus $R/aR\cong (1-a)R$ is projective. But not every ideal in $R$ ...


2

By Schur's Lemma, the two endomorphism algebras on the right are isomorphic to $\mathbb{C}$ (so long as $\dim M$ and $\dim N$ are finite, which you assumed in the question). So in that case you only need that $M\otimes N$ is a simple $R\otimes S$-module. Let $\sum_i m_i \otimes n_i$ be a non-zero element of a submodule $U$ of $M\otimes N$ with the $n_i$ ...



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