# Tag Info

4

Defining a structure of module over $F[x_1,\dots,x_n]$ on $V$ (assuming the action of $F$ is preserved) is the same as specifying an $F$-algebra homomorphism $$\varphi\colon F[x_1,\dots,x_n]\to \operatorname{End}_F(V)$$ (the codomain is the ring of $F$-endomorphisms of $V$). This is equivalent to specifying endomorphisms $T_1,\dots, T_n$ that are pairwise ...

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

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

$\mathbb{Z}\subset\mathbb{R}$ are both $\mathbb{Z}$-modules, and every homomorphism $\mathbb{R}\to\mathbb{Z}$ has an image of zero. A sufficient condition is for $\hom_R(-,A')$ to be an exact functor, since then $\hom_R(A,A')\to\hom_R(A',A')$ would be surjective, and $\hom_R(A',A')$ contains the identity map, which must therefore be the zero map.

3

First notice that $\phi$ must be supposed injective, else it is impossible to extend $\phi$ to the fraction fields. If this is the case then, yes, the extended morphism $\text{Frac }\phi:\text{Frac } A\to \text{Frac }B$ is a finite extension. The trick is to consider the multiplicative set $S=A\setminus \{0\}$ and the morphism $S^{-1}\phi: ... 3 For$A$-modules and homomorphisms$0\to M'\stackrel{u}{\to}M\stackrel{v}{\to}M''\to 0$is exact, if$M'$and$M''$are fintely generated then$M$is finitely generated. There is exact sequence:$0 \to M_1 \cap M_2 \to M_1 \oplus M_2 \to M_1 + M_2 \to 0 $. 3 It suffices to show that if$X$is affine and$F_0$is free, then$F$is free. Since$F_0$is free over$X_0$, we can find an isomorphism$\mathscr{O}_{X_0}^r\rightarrow F_0$. Arbitrarily lift this to a map$\mathscr{O}_X^r\rightarrow F$. We claim that this lift is an isomorphism. Let$K$be the kernel and$M$be the cokernel. First, we see that, since ... 3 Let$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$is ... 2 Let$\phi_i(x_i) = x = \phi_j(x_j)$. Since$I$is a direct system, there's$k \in I$such that$i,j \le k$. Then$\phi_k \circ \phi_{k,i}(x_i) = \phi_i(x_i) = \phi_j(x_j) =\phi_k \circ \phi_{k,j}(x_j) $. But$\psi \circ \psi_i = \phi_i$on$N_i$and$\rho \circ \rho_i = \phi_i$on$P_i$. Thus,$\psi \circ \psi_i (y_i) = \psi \circ \psi_k \circ \psi_{k,i} ...

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

Let's write them differently, so it seems like we're "doing something". So we will write an element of $M_1 + M_2$ as $m_1+m_2$ with $m_1 \in M_1,m_2 \in M_2$, and an element of $M_1 \oplus M_2$ as $(m_1,m_2)$ (with the same convention). So our map $\phi: M_1 \oplus M_2 \to M_1+M_2$, will be, surprisingly enough: $\phi(m_1,m_2) = m_1 + m_2$. I will take ...

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

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

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

This community wiki solution is intended to clear the question from the unanswered queue. You have used the Axiom of Choice to show that there exists a function $f_1 : N \to M$. You have not shown it is a module homomorphism, which may not be the case unless you add more hypothesis.

2

Hints. Let $p_N:F\to N$ be the projection on $N$. Use a basis of $F$ and find $g:F\to M$ such that $f\circ g=\phi\circ p_N$. Now set $\psi=g_{|N}$, the restriction of $g$ to $N$.

2

A $\mathbb Z[i]$-module structure on $F$ is the same as a ring map $\mathbb Z[i] \to \operatorname{End}(F)$. We have $Z[i] = \mathbb Z[X]/(X^2+1)$, i.e. a ring map $\mathbb Z[i] \to \operatorname{End}(F)$ is the same as a ring map $\mathbb Z[X] \to \operatorname{End}(F)=F$, which maps $X$ to an element, which satisfies $x^2+1=0$. Hence $F=\mathbb ... 2 Consider$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]$. 2 Assuming$R= I \oplus J$, it follows that there exist unique$e \in I$,$f \in J$such that$1=e+f$. Recall that for all$x \in I, y \in J$you have $$xy \in IJ \subseteq I \cap J = 0$$ so that$xy =0$. Now, $$e=e(e+f)=e^2+ef=e^2$$ so that$e$is an idempotent. Moreover, for all$x \in I$$$x= x(e+f) = xe + xf = xe \in eR$$ so that$x \in eR$. By ... 2 This is true, because your "third condition" always holds for sets$C$satisfying your first two conditions. Indeed, if$C$is nonempty (say$c\in C$) and satisfies your first two conditions, then it contains$c^0=1$. Now suppose$g^k\in C$for some$k>0$. Note that $$g^k=(g^k)^1\cdot 1^{k-1},$$ so by the second condition on$C$(with$m=2$,$k_1=1$, ... 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 You can easily determine the cokernel if you first find the Smith normal form of the map's matrix, which in your case is \begin{equation*} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T ~. \end{equation*} This means that the two columns$u_1$and$u_2$of your matrix can ... 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 The main point is to realise that linear combinations can by definition only have finitely many nonzero coefficients. They must be defined this way, because in pure linear algebra there is no way to take the sum of infinitely many nonzero vectors (this cannot be defined by repeated addition: one never reaches the goal). In analysis some (convergent) infinite ... 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 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 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). 1 [...] a module$N$is flat if and only if the functor$U(N)\colon \mathbb R \to \mathbf{Set}$is flat. I think this is false. I change a bit the notations. Let$\mathbb A = (A, +_A, \cdot, 1_A)$be a commutative unitary ring and$\mathbb M = (M, +, 0_M)$a$\mathbb A$-module. I denote by$\mathbf A$the category (monoid) with a single object$\ast$... 1 You want to find two vectors which, together with$(m,n,k)$make a matrix with determinant$\pm 1$. Then the three vectors will form a free basis of${\mathbb Z}^3$. One way of doing that is to perform some elementary unimodular column operations on$(m,n,k)$to transform it to$(1,0,0)$. This you can easily extend to a free basis of${\mathbb Z}^3$. Then, ... 1 Any two free abelian groups$A$and$B$of rank$n$, with$A$a subgroup of$B$, are isomorphic as abelian groups: both are isomorphic to$\mathbb{Z}^n$. But you can still have$A \subsetneq B$; in other words,$A$and$B\$ can be abstractly isomorphic (that is, there is some function between them which is a group isomorphism), but the inclusion homomorphism ...

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