# 线性代数和几何代写|LINEAR ALGEBRA AND GEOMETRY MATH244 University of Liverpool Assignment

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In abstract algebra, a vector space is defined as a collection of objects, called vectors, which can be added together and multiplied by scalars (usually real numbers or complex numbers). These operations satisfy certain axioms, such as closure, associativity, commutativity, distributivity, and existence of identity and inverses.

The key idea behind abstract vector spaces is that the specific nature of the vectors themselves is not important, only the operations that can be performed on them. In other words, abstract vector spaces are defined purely in terms of their algebraic properties, without any reference to their physical or geometric interpretation.

By working in this abstract setting, mathematicians can develop powerful tools for studying linear phenomena in a wide variety of contexts. For example, abstract vector spaces are used to study linear transformations, eigenvalues and eigenvectors, inner products, and many other topics in linear algebra and beyond.

For an algebraic variety $X$ over a field $k$ of characteristic $p$ the Frobenius twist $X^{\prime}$ of $X$ is defined as follows. $X^{\prime}=X$ as a topological space. A function $f$ on $U^{\prime} \subset X^{\prime}$ is regular iff $f(x)=\phi(x)^p$ where $\phi$ is a regular function on $U=U^{\prime} \subset X$. The identity map $X \rightarrow X^{\prime}$ defines a morphism $F r: X \rightarrow X^{\prime}$ called the Frobenius morphism. [Notice though that it does not define a morphism from $X^{\prime}$ to $X$.] (a) Check that if $X$ is a closed subvariety in $\mathbb{A}^n$ or $\mathbb{P}^n$ whose ideal is generated by polynomials with coefficients in $\mathbb{F}_p$, then $X^{\prime} \cong X$. Moreover, we have an isomorphism such that that composition $X \stackrel{F r}{\longrightarrow} X^{\prime} \cong X$ is given by $\left(x_i\right) \mapsto\left(x_i^p\right)$.

Let $X$ be a closed subvariety in $\mathbb{A}^n$ or $\mathbb{P}^n$ whose ideal is generated by polynomials with coefficients in $\mathbb{F}_p$. Then $X$ is defined by polynomials of the form $f(x_1,\ldots,x_n)^p – f(x_1,\ldots,x_n)$ where $f$ has coefficients in $\mathbb{F}_p$. Let $U \subseteq X$ be an open set and $\phi \in \mathcal{O}_X(U)$ be a regular function. Then $\phi = g^p$ for some $g \in \mathcal{O}_X(U)$, which implies that $\phi(x) = g(x)^p$ for all $x \in U^{\prime} = U$. Thus $\phi$ is a regular function on $U^{\prime}$ as well, and so $U^{\prime}$ and $U$ have the same regular functions. Hence $X^{\prime} \cong X$ as varieties.

Moreover, if we take the composition $X \xrightarrow{Fr} X^{\prime} \cong X$ and restrict it to an affine chart $U = \operatorname{Spec} A$, where $A = k[x_1,\ldots,x_n]/I$ and $I$ is generated by polynomials with coefficients in $\mathbb{F}_p$, then $Fr$ sends $U$ to $\operatorname{Spec} A^{\prime}$, where $A^{\prime} = k[x_1^{\prime},\ldots,x_n^{\prime}]/I$, with $x_i^{\prime p} = x_i^p$ for all $i$. It follows that the map $X \to X^{\prime}$ sends $\left(x_i\right) \mapsto\left(x_i^p\right)$.

(b) Let $X$ be a normal irreducible variety of dimension $n$. Prove that $F r: X \rightarrow X^{\prime}$ is finite, find its degree and prove that every point is its ramification point.

(b) By the Noether normalization theorem, there exists a finite surjective morphism $\pi: X \rightarrow \mathbb{A}^n$. Since $\mathbb{A}^n$ is smooth and $X$ is normal, we have that $\pi$ is a finite morphism. Therefore, it suffices to prove that $F r: \mathbb{A}^n \rightarrow (\mathbb{A}^n)^{\prime}$ is finite, where $(\mathbb{A}^n)^{\prime}$ is the Frobenius twist of $\mathbb{A}^n$.

Let $x_1, \ldots, x_n$ be the coordinates on $\mathbb{A}^n$. Then the coordinate ring of $(\mathbb{A}^n)^{\prime}$ is given by $k[x_1^p, \ldots, x_n^p]$, and the Frobenius morphism is given by $F r(x_i) = x_i^p$. Thus, the map $F r$ is induced by the $p$-th power map on the coordinate ring, which is a finite morphism of rings. Therefore, $F r$ is a finite morphism.

To compute the degree of $F r$, note that $F r$ is a finite morphism between irreducible varieties of the same dimension, so its degree is given by the cardinality of the fiber over a generic point. Let $k’$ be the algebraic closure of $k$, and let $\overline{x}$ be a generic point of $\mathbb{A}^n_{k’}$. Then the fiber of $F r$ over $\overline{x}$ is given by the set of points $y$ in $(\mathbb{A}^n_{k’})^{\prime}$ such that $F r(y) = \overline{x}$. But $F r(y) = (\phi_1(y)^p, \ldots, \phi_n(y)^p)$, where $\phi_1, \ldots, \phi_n$ are regular functions on $y$. Therefore, $F r(y) = \overline{x}$ if and only if $\phi_1(y) = \cdots = \phi_n(y) = \overline{x}^{1/p}$. This is a system of $n$ equations in $n$ variables, and its solutions are the $p^n$ points $y$ in $(\mathbb{A}^n_{k’})^{\prime}$ with coordinates of the form $\alpha_1^{1/p}, \ldots, \alpha_n^{1/p}$, where $\alpha_1, \ldots, \alpha_n$ are the $p$-th power of elements in $k’$. Therefore, the degree of $F r$ is $p^n$.

(c) Describe the intersection points of the graph of Frobenius $F r: \mathbb{A}^1 \rightarrow$ $\mathbb{A}^1$ with the diagonal and check that each one has multiplicity one.

The Frobenius morphism for $\mathbb{A}^1$ is the map $F r: \mathbb{A}^1 \rightarrow \mathbb{A}^1$ given by $F r(x) = x^p$. The graph of $F r$ is the subset of $\mathbb{A}^1 \times \mathbb{A}^1$ defined by the equation $y = x^p$. The diagonal of $\mathbb{A}^1 \times \mathbb{A}^1$ is the subset defined by the equation $y = x$.

To find the intersection points of the graph of $F r$ with the diagonal, we need to solve the system of equations $y = x^p$ and $y = x$. Substituting $y=x$ into the first equation gives $x = x^p$, which implies that $x^{p-1} = 1$. Since we are working over a field of characteristic $p$, we have $p-1 \neq 0$, so $x^{p-1} = 1$ has exactly $p-1$ solutions in $\mathbb{A}^1$, namely $x = 0, 1, \omega, \omega^2, \dots, \omega^{p-2}$, where $\omega$ is a primitive $p$-th root of unity.

For each of these solutions $x$, we have $y=x^p$, so the intersection point is $(x, y) = (x, x^p)$. To check that each intersection point has multiplicity one, we need to compute the Jacobian matrix of the system of equations $y = x^p$ and $y = x$ at each intersection point. This matrix is $\begin{pmatrix} px^{p-1} & -1 \end{pmatrix}$, which has determinant $-p x^{p-1}$. Since $p$ is nonzero in the field we are working over, and $x^{p-1}$ is nonzero for each intersection point, the determinant is nonzero at each intersection point. Therefore, each intersection point has multiplicity one.