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4

Your answer is basically correct, but in a couple of places the terminology and notation are a little off. It’s not really correct to say that $\Bbb Q^+$ is a closed set: in this context closure is a joint property of set and an operation, so you should say that $\Bbb Q^+$ is closed under $*$. The statement it is associative under [the] operation ...


3

Because if there were some operation you could perform on the two quantities to get different results, they would not be equal. If two things are equal, the results of multiplying them each by $b^{-1}$, or of performing any other operation on them, must be equal, because that is what it means for two things to be equal.


3

The idea for the proof is that for $2$ sets $A,B$ $$ A=B\iff A\subset B \land B\subset A $$ To prove $A\subset B$, use $$ A\subset B\iff \forall x\in A\implies x\in B $$ So to prove $G=H$, just prove $G\subset H$ and $H\subset G$. To prove $G\subset H$, just prove that if $x\in G$ then $x\in H$.


3

The group of real numbers $(\Bbb R,+,0,-)$ with its usual topology is a topological group: The addition $+:\Bbb R\times\Bbb R \to \Bbb R$ (where $\Bbb R\times \Bbb R$ has the product topology) and the negation $-:\Bbb R \to \Bbb R$ are continuous maps. These properties of $\Bbb R$ are exactly what leads to formulas like $$\lim(x_n+y_n) = x+y\quad \text{ and ...


3

The idea is that computing ring operations in $A$, then applying $f$, is the same as first applying $f$, then computing ring operations in $B$.


2

http://www.amazon.com/gp/offer-listing/0936428066/?tag=wwwcampusboocom667-20&condition=used http://product.half.ebay.com/Between-Nilpotent-and-Solvable-by-Henry-G-Bray-John-F-Humphreys-David-Johnson-Paul-Venzke-and-W-E-Deskins-1982-Hardcover/717850&item=345068498941&tg=videtails These are some sites I found after searching, hope this helps.


2

First, since $Q$ is finite and acyclic, it has a maximal path length, so $R_Q^m=0$ for sufficiently large $m$. Thus $R_Q$ is a nilpotent ideal, so is contained in $J$. To see the reverse containment, note that $k(Q)/R_Q\simeq k^n$ for some $n$. This follows since $R_Q$ contains all nontrivial paths, so after quotienting we're just left with copies of $k$ ...


2

Module $2$ syllabus can be found in almost all books on Algebra but I recommend Dummit and Foote. For Module $1$, I would recommend "Linear Algebra done right" by Axler for abstract approach avoiding Matrices, and "Linear algebra done wrong " by Sergie Treil (Google it for e-copy). These two books covers all topics espically Sergie's book, but Axler is ...


2

I don't really know what "easiest" means: this heavily depends on what you are taking as granted. Anyway, here is a direct elementary proof. In what follows, $\Phi$ is an automorphism of $M_N(\mathbb C)$. Denote by $(E_{i,j})_{1\leq i,j\leq N}$ the "canonical basis" of $M_N(\mathbb C)$. If $(e_1,\dots ,e_N)$ is the canonical basis of $\mathbb C^N$, then ...


2

One of my teachers I think seemed to dislike "multiplying" both sides of an equation by the same element. If you share the same aversion you can almost always use something like this to avoid it. $$c = c \cdot e = c \cdot(b \cdot b^{-1}) = (c \cdot b) \cdot b^{-1} = (a \cdot b) \cdot b^{-1} = a \cdot(b \cdot b^{-1}) = a \cdot e = a$$ All we have used here ...


2

$$f(r) = f(r.1) = f(r).f(1).$$ $f(1) = 0 \Rightarrow f(r) = 0 \forall r \in R$


2

Not necessarily. Let $R = \mathbb Z$. A non-empty direct sum of copies of $\mathbb Z$ is infinite. Since $\mathbb Z / 2 \mathbb Z$ is finite, it cannot be free.


1

Here is a simple direct proof. Suppose $f:M_n(\mathbb{C})\to M_n(\mathbb{C})$ is an automorphism. For $1\leq i,j\leq n$, let $e_{ij}$ denote the matrix with $ij$ entry $1$ and all other entries $0$, and write $f_{ij}=f(e_{ij})$. Note that $e_{ij}e_{k\ell}=\delta_{jk}e_{i\ell}$, and hence by applying $f$ to both sides we have ...


1

Take a peek at some no-cost alternatives around the 'net, like Treil's "Linear Algebra done Wrong" (very nice, but might not be all the abstract you'd want). Often lecture notes are available, and add a different explanation that helps you over some rough spot.


1

We wish to find all homomorphisms $\phi: \Bbb Z_4 \to \Bbb Z_9^*$. Now first of all it suffices to know what $\phi(1)$ is since for any other element $m \in \Bbb Z_4$ we have $\phi(m) = \phi(1 + 1 + .. + 1) = \phi(1) + ... + \phi(1) = m \phi(1)$. Hence $\phi(1)$ completely determines $\phi$. Now, $[\phi(1)]^4 = \phi(1) \cdot \phi(1) \cdot \phi(1) \cdot ...


1

This is because a finitely generated projective module is finitely presented, so that for any prime ideal $\mathfrak p\in\operatorname{Spec} A$, $$\bigl(\operatorname{Hom}_A(P,A)\bigl)_{\mathfrak p}\simeq\operatorname{Hom}_{A_\mathfrak p}(P_{\mathfrak p},A_{\mathfrak p}) $$ and also $\;(E\otimes_AF)_{\mathfrak p}\simeq E_{\mathfrak p}\otimes_{A_\mathfrak ...


1

If $G$ is a finite group and $F$ is a field of characteristic zero, Then the group algebra $F[G]$ is semisimple by maschke's theorem (this is a fact that is demonstrated in any book of representation theory, i.e Fulton Representation Theory) with one matrix ring factor for each conjugacy class of $G$, so $K_0(F[G])\cong \Bbb Z^n$ by the Artin-Wedderburn ...


1

Let $F$ be the arrow ideal. The hypotheses (finite, acyclic) mean that $F$ is nilpotent and $Q/F$ is semisimple. This means that $F={\rm rad}\,Q$, since Proposition Let $A$ be an Artinian ring, and let $R={\rm rad}\,A$. $(1)$ $R$ is a nilpotent ideal. $(2)$ Any nilpotent ideal is contained in $R$. $(3)$ $A/R$ is semisimple, and if $A/I$ is semisimple, ...


1

Indeed the algebra of limits holds (for sums and scalar multiples) in any normed space. The proof is identical. The remaining algebra of limits hold in any normed algebra. Note here that the norm on an algebra imust satisfy $\lVert xy\rVert\leq\lVert x\rVert\lVert y\rVert$ for all $x,y$ in the algebra. Also note that non-zero must be replaced by invertible ...


1

The map induced by a matrix $M \in \mathrm M _{m \times n}(\mathbb R)$ (or any field, for that matter) is defined by matrix multiplication. Given a vector $v \in \mathbb R ^n$ (i.e. an $n \times 1$ vector), its image is $Mv$ (an $m \times 1$ vector). In other notation: $$ \begin{align} \phi_M : \mathbb R ^n &\to \mathbb R ^m \\ v &\mapsto Mv ...


1

If $H$ is a proper subgroup of the finite group $G$ then there is a maximal subgroup of $G$ containing $H$. Note that a subgroup M of $G$ is called maximal if $M \neq G$ and the only subgroups of $G$ that contain $M$ are $M$ and $G$) Let's prove the statement. Proof. Consider the set $S=\{K\leq G: K\neq G, H\leq K\}$. This is a subset of ...


1

Let $\rm P$ be a subgroup property (e.g. proper, nontrivial, normal, central, characteristic, finite-index, and so on). I picked up this cool term from GroupProps. Then: a "$\rm P$ maximal subgroup" is a maximal subgroup which is $\rm P$, whereas a "maximal $\rm P$ subgroup" is a subgroup which is maximal among $\rm P$ subgroups. The set of all subgroups ...


1

Since $ Q \in Syl_{2}(G^{\prime}) $, then $ Q \unlhd G^{\prime} $. So $ Q $ chr $ G^{\prime} $, since $ Q $ is normal sylow subgroup of finite group $ G^{\prime} $. Now we have $ Q $ chr $ G^{\prime} $ and $ G^{\prime} \unlhd G $. So $ Q \unlhd G $.



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