## Sheaf of Kähler Differentials is Invertible in Dense Open Subset

Let $$f:S→B$$ be an elliptic fibration from an integral surface $$S$$ to integral curve $$B$$

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Here I use following definitions:

A surface (resp. curve) is a $$2$$ -dim (resp. $$1$$-dim) proper k scheme over fixed field $$k$$.

Fibration has two properties: 1. $$O_B = f_*O_S$$ 2. all fibers of f are geometrically connected

Futhermore a fibration is elliptic if the generic fiber $$S_{\eta}=f^{-1}(\eta)$$ is an elliptic curve (over $$k(\eta)$$.

Denote by $$i_S: S_{\eta} \to S$$ the canonical immersion. Here I’m ot sure to 100% but I guess that for the structure sheaf holds $$O_{S_{\eta}}= O_S \otimes_k k(\eta)$$.

Now the QUESTION:

Since $$S_{\eta}$$ is elliptic curve and therefore smooth the restriction of the Kähler differentials $$\Omega^2_{S/B} \vert _{S_{\eta}}$$ is invertible.

My question is how to see that there exist open neighbourhood $$U \subset S$$ of $$S_{\eta}$$ such that the restriction $$\Omega^2_{S/B} \vert _U$$ is still invertible?

## Divisibility of group of invertible functions

Suppose we have the group $$G$$ of invertible functions over some set $$S \subseteq \mathbb{R}$$ under composition. I’m interested in the divisibility of such group. For example, taking $$S=[-1, 1]$$, for any $$n \in \mathbb{N}$$, one might try to find some function $$f \in S$$ such that $$\underbrace{f \circ f \cdots \circ f }_{n \text{ times}}= \sin( \frac{\pi x}{2})$$ More generally, we want to satisfy $$\forall n \ \forall x \ \exists y \ y^n = x$$, the usual divisibility axiom. My guess is that this groups are far from divisible, but it is known that they can always be embedded in some divisible group $$\overline{G}$$. However, the construction of $$\overline{G}$$ I have read about requires taking a sequence of wreath products and a direct limit. This usually makes things very abstract and one might have a hard time identifying $$f$$ as an element of $$\overline{G}$$.

Is there a better way to visualize this construction? My goal is to make sense of the elements in $$\overline{G}$$ as functions over some larger (possibly not even real) $$S’ \supseteq S$$.

## Number of elements in the set of invertible lower triangular matrices over a finite field

Problem:

Let $$F_q$$ be a finite field with $$q$$ elements.

$$T_n(F_q) := \{ A = (a_{ij}) \in F^{n \times n}$$ | $$a_{ij} = 0$$ for $$i < j,$$ and $$a_{ij} \neq 0$$ $$\forall i \}$$.

Determine the number of elements in $$T_n(F_q)$$.

My solution is as follows:

Starting with the last row going upwards, there are:

$$q-1$$ possibilities for the last row;

$$(q-1)q$$ possibilities for the row before the last;

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.

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$$(q-1)q^{n-1}$$ possibilities for the first row.

Therefore, in total there are $$(q-1)^nq^{\sum_{i=1}^{n-1} i} = (q-1)^nq^{\frac{n(n-1)}{2}}$$ elements.

Could you, please, check my solution?

## Why are the Cartier divisors on a variety $X$ isomorphic to invertible subsheaves of $\mathcal{K}_{X}?$

$$\mathcal{K}_{X}$$ denotes the sheaf of total rings of fractions of $$\mathcal{O}_{x}.$$

## f(x) is invertible polynomial function of degree ‘n’ {n≥3} then f”(x) = 0 has exactly ‘n – 2’ distinct real roots if

A)f′(x)=0 has n−1/2 distinct real roots B)f′(x)=0 has n−1 distinct real roots C)all the roots of f′(x)=0 are distinct D)none of these

## The set of all invertible operators need not be dense

I want to show that the set of all invertible operators $$\mathcal{G}(\ell^2)$$ is not dense in $$\mathcal{B}(\ell^2)$$.

Consider the right shift operator $$T\in \mathcal{B}(\ell^2)$$. We also know that $$T\notin \mathcal{G}(\ell^2)$$. I want to show now that $$B=\{R\in \mathcal{B}(\ell^2):\|R-T\|<1\}$$ is disjoint with $$\mathcal{G}(\ell^2)$$.

Suppose not. Then there exists $$R\in \mathcal{G}(\ell^2)$$ such that $$\|R-T\|<1$$. How to arrive at a contradiction from here? Any help is appreciated.

## if A is an invertible matrix and B is an invertible matrix, is A+B also an invertible matrix?

the answer is obviously false. for example: $$A=\begin{pmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 1/2 \ \end{pmatrix} A^{-1}=\begin{pmatrix} 1 & 0 & 0 \ 0 & 1/2 & 0 \ 0 & 0 & 2 \ \end{pmatrix} \ B=\begin{pmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & -1/2 \ \end{pmatrix} B^{-1}=\begin{pmatrix} 1 & 0 & 0 \ 0 & 1/2 & 0 \ 0 & 0 & -2 \ \end{pmatrix} \ A+B=\begin{pmatrix} 2 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 0 \ \end{pmatrix}$$ A+B dose not have an inverse matrix(rank is not n). is there a better way to prove it other then an example?

## How can we prove that if we can change a matrix into an identity matrix by doing some elementary row exchanges, then the matrix is invertible?

I started linear algebra, and I encountered the part that using Gauss-Jordan method to compute invertible matrix. Then, how can we prove that a matrix is invertible if we can change that matrix into an identity matrix by doing some elementary row exchanges?

## Show that the functor that takes $R$ to the set of invertible elements of $R[X]/(X^2-a)$ is representable.

The following question is from the Fall 2016 UCLA algebra qualifying exam:

Let $$F$$ be a field and $$a\in F$$. Show that the functor that takes $$R$$, commutative $$F$$-algebras to the invertible elements of $$R[X]/(X^2-a)$$ is representable.

What I have so far: If $$a\in F$$, then then we have that $$R[X]/(X^2-a)\cong R\times R$$. Hence we get that $$Hom_{F-\text{alg}}(F[t,t^{-1}]\otimes_FF[t,t^{-1}],R)\cong Hom_{F-\text{alg}}(F,R\times R)\cong U(R)$$ where $$U$$ is the functor that take $$R$$ to units of $$R[X]/(X^2-a)$$. Hence in this case, the functor is representable.

I’m unsure how to extend this to the general case.

## Null spaces and invertible matrix

The question is: “If A and B are n x n matrices, show that they have the same null space if and only if A = UB for some invertible matrix U.

I started the question by saying Ax = 0 for some vector x in null(A). Now I’m lost. Could someone please help me out with this question? Thank you very much.