## Commutative algebra – dimension of a given finite generated quotient module over a local ring.

I have dealt with the following question from the theory of dimension in commutative algebra.

To let $$(A, m)$$ be a local ring and $$M$$ a finally generated $$A$$-Module.

given $$x_1, …, x_r in M ​​$$, Prove that $$dim ( frac {M} {(x_1, …, x_r) M}) geq dim (M) – r$$with equality, if and only if {$$x_1, …, x_r$$} is part of a parameter system for $$M$$,

Now I can show that, though $$A$$ is a $$mathbf {regular}$$ local ring so $$frac {A} {(x_1, …, x_r) A}$$ is a regular local ring with dimension $$dim (A) – r$$ then and only if {$$x_1, …, x_r$$} is part of a parameter system for $$A$$, But I do not know how to show that for the given case. I also can not show the inequality. I could not find any proof, so I would be grateful for any help!

## Homological Algebra – Bimodul Resolutions

Let A be a finite-dimensional algebra. Let M be a left A-module and
Let N be a real A-module. Choose an injective resolution $$E_1 ^ *$$ from $$M$$ in the $$A$$-mod and an injective resolution $$E_2 ^ *$$ from $$N$$ in mod$$A$$, Then we have $$E_1 ^ * otimes E_2 ^ *$$ is an injective resolution of $$M otimes N$$ in the category of $$A$$$$A$$-Bimodule? I could show that $$E_1 ^ * otimes E_2 ^ *$$ are injecting objects in the category $$A$$$$A$$-Bimodule. But how can you show that there is a solution?

Does that also apply to the projective solution?

## Linear Algebra – Blockwise inversion of the matrix with rectangular blocks

Given a matrix in block form:

$$W = left ( begin {array} {cc} A & B \ C & D end {array} right)$$

from where $$A, D$$ Are square matrices, we can write the inverse in the same block form:

$$W ^ {- 1} = left ( begin {array} {cc} A ^ {- 1} + A ^ {- 1} B (D – CA ^ {- 1} B) ^ {- 1} CA ^ {- 1} & – A ^ {- 1} B (DC A ^ {- 1} B) ^ {- 1} \ – (D – C A ^ {- 1} B) ^ {- 1} C A ^ {- 1} & (D – C A ^ {- 1} B) ^ {- 1} end {array} right)$$

provided that $$A$$ and $$D-CA ^ {- 1} B$$ are invertible.

Is there a more general version of the block inversion that allows it? $$A$$ or $$D$$ be rectangular?

## linear algebra – Prove that \$ text {(null} ~ T ^ *) ^ perp subseteq text {range} ~ T \$

To let $$T in L (V, W)$$ from where $$L (V, W)$$ denotes the set of linear maps of $$V$$ to $$W$$, Prove that $$text {(null) ~ T ^ *) ^ perp subseteq text {range} ~ T$$ from where $$T ^ *$$ is the adjoint operator ( Not related to the adjoint matrix) and $$A ^ perp$$ refers to the orthogonal complement of $$A$$,

Attempt: To let $$w_2 in text {(null} ~ T ^ *) ^ perp$$,

Our goal is to show that $$exists v in V$$ so that $$Tv = w_2$$,

We can express $$W = text {null} ~ T ~ oplus ~ text {(null) ~ T ^ *) ^ perp$$, So, leave $$w = w_1 + w_2$$ from where $$w_1 in text {null} ~ T$$,

$$T ^ * (w) = T ^ * (w_2) = v$$ for some $$v in V$$

If we prove it somehow $$<(Tv-w_2),~(Tv-w_2)>= 0$$

But the extension of the above just seems to make it more complicated.

Any ideas on how to move forward?

Many thanks

## linear algebra – limiting the spectral gap of a simple symmetric matrix

I have a seemingly innocent linear algebra problem that I can not solve, and I hope you kindly give me an insight. Here is the description: Let $$mathbf {a} = (a_1, a_2, dots, a_d) ^ {T}$$ be a positive probability vector, $$d. H.$$ $$Vert mathbf {a} vert_1 = 1$$ and $$a_i> 0$$ for all $$i$$, Leave matrix $$A$$ be defined as follows: $$A = textrm {diag} ( mathbf {a}) – mathbf {a} mathbf {a} ^ {T}$$ from where $$textrm {diag} ( mathbf {a})$$ means the diagonal matrix with the $$i$$the diagonal entry is $$a_i$$, It's easy to show that $$mathbf {1} _d$$, the all-one vector of dimension $$d$$is an eigenvector of $$A$$ of eigenvalue $$0$$, And the Gershgorin circle set shows that too $$A$$The eigenvalues ​​of are greater or equal $$0$$, My question is:

What is the smallest eigenvalue of $$A$$ that is not null

I did the calculation when $$d = 3$$ and realized that there may not be a simple analytical formula for it and therefore a nice lower bound is much appreciated.

Thank you very much!

## Modules – DG algebra and its zeroth cohomology are derived equivalently

This is somewhat related to this question: Can an algebra be equivalent to the dg extension? ,

Suppose we have a DG algebra $$A$$, so that $$H ^ 0 (A)$$ is Noetherian (left and right) and $$H ^ bullet (A)$$ is gradually limited. Suppose further that the limited derived category is finally generated $$H ^ 0 (A)$$ The module corresponds to the category of the (finally generated) limited DG $$A$$Modules. Can we conclude something about it? $$H ^ i (A)$$ to the $$i neq 0$$ (for example, they must be zero)?

## Linear Algebra – Find a quadratic function to fit this data

Match a quadratic function of the form f (t) = c0 + c1t + c2t ^ 2 to the data points (0, 0), (1, -9), (2, -2), (3, 19) on the smallest squares.

I would like to know how to use least squares to do this. Many thanks.

## Abstract Algebra – Equivalent Order Definitions

To let $$K$$ be an algebraic number field and $$L$$ the algebraic integers in $$K$$, Acceptance of an order $$O$$ is defined as a subring of $$L$$ whose field of breaks is $$K$$I want to prove that it is a subgroup of the finite index additive group $$L$$?

The opposite is clearly the case, but I have trouble proving that this definition is equivalent. Thanks a lot!

## linear algebra (It is very important) linear transformation

TL (V). The T-composition operator with itself is represented by T2; i.e. T2 = T∘T and for I the operator identity I (v) = v for all v∈V.

If T2 = I and different —> ± I

prove that v∈V ∖ {0}, so that T (v) = v

I did …

Since T2 (v) = v, we see that (T – I) (T + I) (v) = 0 for each v 0V.

Let T (v) ≠ v for every v∈V ∖ {0}. We want to show that this is impossible.

Note that (T – I) (T + I) (v) = 0, ∀v∀V, so T + IV sends into the subspace of V consisting of elements x of V such that (T – I ) (x) = 0, ie
T + I: V → Ker (T – I): = {x – V – (T – I) x = 0}.
Under the assumption, the RHS subspace {x {V∈ (T – I) x = 0} = {0}.

Consequently, (T + I) (v) = 0, ∀v∀V. This means that T = – I, is a contradiction. Therefore T (v) = v for some v∈V ∖ {0}.

But in the last steps I have to say .. Why can I confirm this statement …

Under the assumption, the RHS subspace {x {V∈ (T – I) x = 0} = {0}.
Why is {0} the x value?

## Abstract Algebra – If X is a non-empty subset of R, show that C (X) = \$ {c in R | cx = xc \$ for all \$ x in X } \$ is a subring of R.

We have to check in 4 steps, if this is a subring.

$$0 in C (X)$$ there $$x.0 = 0.x$$,

$$1 in C (X)$$ there $$x.1 = 1.x$$,

Now we have to prove the conclusion with two operations. To let $$c_1 in C (X)$$ and $$c_2 in C (X)$$, Now we know that

begin {align} c_1 & = xc_1x ^ {- 1} \ c_2 & = xc_2x ^ {- 1} end
So we have to show that $$c_1 + c_2 in C (X)$$, Now,

begin {align *} c_1 + c_2 & = xc_1x ^ {- 1} + xc_2x ^ {- 1} \ & = x (c_1x ^ {- 1} + c_2x ^ {- 1}) \ & = x (c_1 + c_2) x ^ {- 1} end {align *}
That's why, $$c_1 + c_2$$ is in $$C (X)$$, We have to show that too $$c_1c_2$$ is in $$C (X)$$, Now,
begin {align *} c_1c_2 & = xc_1x ^ {- 1} xc_2x ^ {- 1} \ & = xc_1c_2x ^ {- 1} end {align *}
Therefore, $$c_1c_2$$ is in $$C (X)$$,
Since all four conditions for the subring are met, the ring is $$C (X)$$ is a subring of R.

This was my solution, but apparently it's wrong. How do we solve this problem?