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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$ (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?

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.