## abstract algebra – Completed tensor product discrete topology

Hi I have a question it should be easy to answer but I dont see how to do it Let us suppose that $$k$$ is a field and $$A$$ is pre-adic $$k$$-algebra, that is a $$k$$-algebra linearly topologized such that there exist an ideal of definition I such that the collection $${I^n}_{ngeq 1}$$ is a neighborhood basis for $$0$$. And let us suppose that we have another $$k$$-algebra $$B$$ without any specified topology on it. We can attach to $$B$$ the discrete topology. My question is if in this situation it holds that
$$Awidehat{otimes}_{k}Bsimeq varprojlim (A/I^notimes_k B)$$

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## abstract algebra – How to prove that \$varphi\$ is an injective homomorphism and why it is surjective?

Here is the question I want to solve:

Prove that the subgroup of $$SL_2(mathbb F_3)$$ generated by $$begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix}$$ and $$begin{pmatrix} 1 & 1 \ 1 & -1 end{pmatrix}$$ is isomorphic to the quaternion group of order $$8.$$(use a presentation for $$mathcal{Q}_8$$)

Here is a solution I found online:

A presentation for $$mathcal{Q}_8,$$ is $$langle i,j | i^2 = j^2, i^4 = 1, ij = -ji rangle$$
Which means that $$i,j$$ generates $$mathcal{Q}_8$$ and they satisfy the relations $$i^2 = j^2, i^4 = 1, ji = -ij = i^3j.$$

Let $$A = begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix}$$ and $$B = begin{pmatrix} 1 & 1 \ 1 & -1 end{pmatrix}.$$ Define $$varphi: {i,j} rightarrow langle A, B rangle$$
as $$varphi(i) = A$$ and $$varphi(j) = B.$$

Now, note that by direct calculation , the subgroup generated by $$A$$ is $$langle A rangle = { begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix}, begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}, begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix}, begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix} }$$
And also, by direct calculation , the subgroup generated by $$B$$ is $$langle B rangle = { begin{pmatrix} 1 & 1 \ 1 & -1 end{pmatrix}, begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}, begin{pmatrix} -1 & -1 \ -1 & 1 end{pmatrix}, begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix} }$$
And it is clear that $$|A|=|B| = 4$$ and $$A^2 = B^2.$$

Also, by direct calculation, we have that $$BA = begin{pmatrix} 1 & -1 \ -1 & -1 end{pmatrix} = A^3B.$$

Here by the lemma $$varphi$$ extends to an injective homomorphism $$bar{varphi} : mathcal{Q}_8 rightarrow langle A, B rangle$$
Also, by construction $$bar{varphi}$$ is surjective. Hence the required is proved.

My questions are:

1-I do not know what lemma says that $$varphi$$ is an injective homomorphism, tthis problem is #10 in section 2.4 of Dummit and Foote (3rd edition) and I could not find any lemma that said this. Could anyone help me in proving that $$varphi$$ is an injective homomorphism please?

2- Why by construction is $$bar{varphi}$$ is surjective? what confuses me is that $$langle A, B rangle$$ is the subgroup generated by $$A$$ and $$B$$ and not the set that contains $$A$$ and $$B,$$ Could anyone explain to me this please?

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## abstract algebra – How do I find the isomorphisms between the quotient groups in D12?

I am given $$𝐷_{12}={1,𝑥,𝑥^2,𝑥^3,𝑥^4,𝑥^5,𝑦,𝑥𝑦,𝑥^2𝑦,𝑥^3𝑦,𝑥^4𝑦,𝑥^5𝑦}$$

I know that the normal subgroups are as follows: $$⟨𝑥⟩,⟨𝑥^2⟩,⟨𝑥^3⟩,⟨𝑥^2,𝑦⟩,⟨𝑥^2,𝑥𝑦⟩$$.
First, I just need some help understanding why these are the normal subgroups. After this, I would appreciate help identifying how to find the elements within these normal subgroups. I know that $$langle x^3rangle={1,x^3}$$, but some help identifying the others, especially the last two would help a bunch.

After I have this information, I am to identify the quotient groups within \$D_{12} and then identify which groups the quotient groups are isomorphic to. How would I go about doing this? It can be as simple as using Cayley tables. I am just lost on how to start.

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## abstract algebra – Second Isomorphism Theorem and Jordan-Holder

In another posting, there was a question about the following:

Let $$G$$ be a finite non-trivial group with the following two composition series:

$${e} = M_0 triangleleft M_1 triangleleft M_2 = G$$

$${e} = N_0 triangleleft N_1 triangleleft cdots triangleleft N_r = G.$$

Prove that $$r=2$$ and that $$G/M_1 cong G/N_1$$ and $$N_1/N_0 cong M_1/M_0$$.

The person posting the question went on to state that “By the second isomorphism theorem I know that $$M_1N_2/N_2 cong M_1/(N_2cap M_1)$$

Here is my question. In order to use the second isomorphism theorem with $$M_1$$ and $$N_2$$ don’t we need to know that $$M_1 leq N_G(N_2)$$? And if so, then how do we know that $$M_1$$ actually is in the normalizer of $$N_2$$ in $$G$$?

If I can get over this hurdle, I understand the remainder of the original posting. Perhaps this is something obvious, but please help me see whatever it is that I am missing.

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## c# – Refactoring a class to an abstract base class – keep default implementation in place or move it to an an other class?

I have a class which aggregates some temporal data from a database and provides a bunch of methods to query said data. It looks like this:

``````public class InfoProvider
{
public IEnumerable<Contract> GetContracts() => GetContracts(DateTime.Today);
public IEnumerable<Contract> GetContracts(DateTime at) => GetContracts(at, at);
public IEnumerable<Contract> GetContracts(DateTime from, DateTime to) { /* query contracts */ }

public IEnumerable<Job> GetJobs() => GetJobs(DateTime.Today);
public IEnumerable<Job> GetJobs(DateTime at) => GetJobs(at, at);
public IEnumerable<Job> GetJobs(DateTime from, DateTime to) { /* query jobs */ }

/* more methods like above ... */
}
``````

Now I need to extend my application to support different representations of the data in the database. Those representations are identical for the most part but have some subtle differences. For example `GetContracts` in representation 1 should return contracts of type A, B or C while in representation 2 it should return contracts only of type A or B. On the contrary, `GetJobs` would be the same in both representations.

I want to do this extension by refactoring `InfoProvider` into an abstract class and derive concrete classes, `InfoProviderRepresentation1` and `InfoProviderRepresentation2` for each representation. But I waver where to put the existing code. I see those two options:

1. Leave it in the original class
``````public abstract class InfoProvider
{
public IEnumerable<Contract> GetContracts() => GetContracts(DateTime.Today);
public IEnumerable<Contract> GetContracts(DateTime at) => GetContracts(at, at);
public virtual IEnumerable<Contract> GetContracts(DateTime from, DateTime to) { /* query contracts */ }

public IEnumerable<Job> GetJobs() => GetJobs(DateTime.Today);
public IEnumerable<Job> GetJobs(DateTime at) => GetJobs(at, at);
public virtual IEnumerable<Job> GetJobs(DateTime from, DateTime to) { /* query jobs */ }
}

public class InfoProviderRepresentation1 : InfoProvider
{
}

public class InfoProviderRepresentation2 : InfoProvider
{
public override IEnumerable<Contract> GetContracts(DateTime from, DateTime to) { /* ... */ }
}
``````
1. Move the code to an other abstract class and derive my concrete classes from there:
``````public abstract class InfoProvider
{
public IEnumerable<Contract> GetContracts() => GetContracts(DateTime.Today);
public IEnumerable<Contract> GetContracts(DateTime at) => GetContracts(at, at);
public abstract IEnumerable<Contract> GetContracts(DateTime from, DateTime to);

public IEnumerable<Job> GetJobs() => GetJobs(DateTime.Today);
public IEnumerable<Job> GetJobs(DateTime at) => GetJobs(at, at);
public abstract IEnumerable<Job> GetJobs(DateTime from, DateTime to);
}

public abstract class DefaultInfoProvider
{
public virtual IEnumerable<Contract> GetContracts(DateTime from, DateTime to) { /* query contracts */ }
public virtual IEnumerable<Job> GetJobs(DateTime from, DateTime to) { /* query jobs */ }
}

public class InfoProviderRepresentation1 : DefaultInfoProvider
{
}

public class InfoProviderRepresentation2 : DefaultInfoProvider
{
public override IEnumerable<Contract> GetContracts(DateTime from, DateTime to) { /* ... */ }
}
``````

I find option 2 a bit more readable. On the other hand it needs more boilerplate code than option 1. Does option 2 have any more advantages over option 1?

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## abstract algebra – Centralizer of subalgebras in tensor product

Let $$k$$ be a field, $$E,F$$ two non-zero $$k$$-algebras and $$C,D$$ their respective sub-$$k$$-algebras. Write $$C’$$ for the centralizer of $$C$$ in $$E$$ and $$D’$$ for the centralizer of $$D$$ in $$F$$. I want to show that
$$C’otimes_kD’=(C’otimes_kF)cap(Eotimes_k D’).$$
The author asserts that this is an application of the following result:

Let $$k$$ be a field, $$E$$ a right $$k$$-vector space, $$F$$ a left
$$k$$-vector space, $$(M_alpha)_{alphain A}$$ a family of subspaces of
$$E$$ and $$(N_beta)_{betain B}$$ a family of subspaces of $$F$$. Then
$$(bigcap_{alphain A}M_alpha)otimes_k(bigcap_{betain B}N_beta)=bigcap_{(alpha,beta)in Atimes B}(M_alphaotimes_k N_beta).$$

Let $$M_0:=E$$, $$M_1:=C’$$, $$N_0:=F$$ and $$N_1:=D’$$. Then
$$C’otimes_kD’=(C’cap E)otimes_k(D’cap F)=(M_0cap M_1)otimes_k(N_0cap N_1).$$
The last part is equal to $$(Eotimes_kF)cap(Eotimes_k D’)cap(C’otimes_kF)cap(C’otimes_kD’)$$. This is not what I needed. Am I applying the quoted result incorrectly?

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## Relationship among different abstract algebra constructs

I have rudimentary understanding of rings. I wish to know how they are related to other abstract algebra constructs. To that end, I tried to use a Hasse diagram. I welcome corrections and comments. What is the most hazy to me is that how a vector space is equivalent to a module. Posted on Categories Articles

## Logic of defining a function on an abstract set

Obviously in practice it is common to define functions given a set $$S$$, however I have never thought about the concept behind this and why “I am allowed to do that”. Please keep in mind that I have never read any logic lecture notes or visited any lecture on those topics, besides the standard material. If one is in the setting of an algebraic structure, e.g. a group $$G$$, and one wants to prove a statement, e.g. $$varphi:G to G, g mapsto 2g$$ is a function, what is the logic behind this being true for an arbitrary group? It is standard to prove this by assuming a group $$G$$ is given and then showing the statement.

Q$$1$$: Is this to be interpreted to be given an explicit but arbitrary/unknown group? What I mean by that is that this is to be interpreted as $$G$$ being any explicit group, such as a set consisting of a multiplication map satisfying the usual properties of the definition?

In this setting one does not know what set one is working with, however one would know that an “explicit” set $$G$$ is underlying, which means that it is known which elements belong to $$G$$ and which do not. Therefore it is possible to speak of “$$g in G$$“, is this correct? As far as I can tell, this is necessary in order to do this. This allows to even make sense of a statement for all $$g in G : P(g)$$. Since one knows which elements belong to $$G$$ and which do not, one can then define an assignment for those elemnts, an example given above.

One could sum up that I am confused whether it needs justification to talk about “for all elements of the set the property $$P$$ holds” without knowing the set and why one is able to do so. Since defining a function is sort of a statement of that kind, I found this to be fitting. This is then solved since one knows that $$G$$ is some explicit but unknown/arbitrary set. If anything of what I said is wrong I would appreciate corrections, since those are only my thoughts. Thank you in advance for any comments!

## abstract algebra – Co-isosimple modules are co-cyclic.

First I want to give the definitions of co-isosimple and co-cyclic modules.

If a non-zero module $$M$$ is isomorphic to all its non-zero quotients,then it is called $$textbf{co-isosimple}$$.

A module $$M$$ is called $$textbf{co-cyclic}$$ if there is a simple module which is essential in $$M$$.

In the paper which this lemma stated, they just said that since a co-isosimple module is both Artinian and uniserial, then it is co-cyclic. If I can prove this last sentence, then I complete the proof.

I know that every Artinian module has a simple submodule and every submodule of a uniserial module is essential. Therefore if a module is both Artinian and uniserial, then it has a simple essential submodule, that is, it is co-cyclic. However, I cannot prove a co-isosimple module is artinian and uniserial. I need some hints.

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## abstract algebra – why we take \$p ge n+m-1\$ ? why not \$p ge n+m\$?

I have some confusion in this statement

Theorem :The subset of nilpotent elements of $$R$$ form an ideal of $$R.$$

It is written that

Let $$N$$ be the subset of nilpotent element
let $$x,y in N$$ .let $$x^n=0$$ and $$y^m=0$$
By the Binomial theorem $$(x-y)^p=0$$ for $$p ge n+m-1$$
Thus $$x-y in N$$

My question : Im not getting why we take $$p ge n+m-1$$ ? why not $$p ge n+m$$ ?

My thinking : Here $$x^n=y^m=0 implies x=y$$.Now take $$p = n+m-1$$ then $$(x-y)^{n+m-1}= frac{(x-y)^{n+m}}{(x-y)}=frac{(x-y)^{n+m}}{0}=infty$$

This leads to a contradiction to $$(x-y)^p=0$$