**Assignment-daixie**^{TM}为您提供芝加哥大学**University of ChicagoMATH 17600** Basic Geometry几何学

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That sounds like an exciting and challenging course! Here are some brief explanations of the topics you mentioned:

- Euclidean geometry: This is the branch of geometry that deals with the properties of points, lines, and shapes in a two-dimensional plane. It is named after the Greek mathematician Euclid, who wrote a famous book called “Elements” that laid out a system of axioms and proofs for Euclidean geometry.
- Spherical geometry: This is the study of geometry on the surface of a sphere. It is important in astronomy and navigation, as well as in pure mathematics.
- Hyperbolic geometry: This is a non-Euclidean geometry in which the parallel postulate is replaced by a different postulate, leading to different properties of lines and shapes. It is named after the Greek mathematician Lobachevsky, who first studied it.
- Axiomatic systems: These are systems of logical rules and assumptions, called axioms, that are used to prove mathematical theorems. Hilbert’s approach to geometry emphasized the use of axiomatic systems to ensure rigorous proof.
- Lattice point geometry: This is the study of the properties of points with integer coordinates in a geometric space. It has applications in number theory, cryptography, and computer science.
- Projective geometry: This is a branch of geometry that deals with the properties of points, lines, and shapes from a different perspective, by considering them as projections from a higher-dimensional space. It has applications in computer graphics and engineering.
- Symmetry: This is the study of the properties of objects that remain unchanged under certain transformations, such as rotations, reflections, or translations. It has applications in physics, chemistry, and crystallography.

Let $k=\overline{\mathbb{Q}}$ and $R=k\left[x_1, \ldots, x_n, \ldots\right]$ be the ring of polynomials in infinitely many variables. For $\mathbf{a}=\left(a_i\right) \in \prod_{i=1}^{\infty} k$ we have a homomorphism $R \rightarrow k$ sending $x_i$ to $a_i$, let $\mathrm{m}{\mathrm{a}}$ be its kernel. Find an example of a maximal ideal in $R$ which is not of the form $\mathrm{m}{\mathrm{a}}$ for any $\mathbf{a} \in \prod_{i=1}^{\infty} k$.

Let $I$ be the ideal of $R$ generated by all the monomials of degree at least $2$, that is, $$I=\left\langle x_{i_1}^{e_1}\cdots x_{i_r}^{e_r},:, r \geq 1, e_j \geq 2\right\rangle.$$ We claim that $I$ is a maximal ideal which is not of the form $\mathrm{m}{\mathrm{a}}$ for any $\mathbf{a} \in \prod{i=1}^{\infty} k$.

First, we show that $I$ is a proper ideal. Suppose for contradiction that $I=R$. Then every polynomial $f \in R$ can be written as a linear combination of monomials of degree at most $1$. But this is impossible since we have infinitely many variables $x_1, x_2, \ldots$, so there are infinitely many monomials of degree at least $2$. Hence $I$ is proper.

Next, we show that $I$ is a maximal ideal. Let $J$ be an ideal of $R$ strictly containing $I$. We will show that $J=R$, which implies that $I$ is maximal.

Let $f \in J \setminus I$. Then $f$ must be a monomial of degree at most $1$. Let $x_i$ be a variable that appears in $f$. Since $f \notin I$, we have $x_i \notin \operatorname{supp}(f)$, where $\operatorname{supp}(f)$ denotes the support of $f$, that is, the set of variables appearing in $f$ with nonzero coefficient. Hence we can write $$f=x_i g + h,$$ where $g,h \in R$ and $h$ does not involve $x_i$. Since $f \in J$ and $J$ contains $I$, we have $x_i g \in J$ and $h \in J$. But $J$ strictly contains $I$, so there exists some monomial $m \in J$ which is not in $I$. Since $m$ is not a monomial of degree at least $2$, it must be a monomial of degree $1$, say $m=x_j$ for some $j$. But then $x_j \in J$, so $x_i g + x_j \in J$. Repeating this argument for all variables appearing in $f$, we conclude that $f \in J$. Hence $J=R$, as desired.

Finally, we show that $I$ is not of the form $\mathrm{m}{\mathrm{a}}$ for any $\mathbf{a} \in \prod{i=1}^{\infty} k$. Suppose for contradiction that $I=\mathrm{m}{\mathrm{a}}$ for some $\mathbf{a} \in \prod{i=1}^{\infty} k$. Then $\mathrm{a}$ must be the zero sequence, since $I$ contains all monomials of degree at least $2$. But then $\mathrm{m}{\mathrm{a}}$ is not maximal, since it is properly contained in $\left\langle x_1\right\rangle$, which is maximal. This is a contradiction, so $I$ is not of the form $\mathrm{m}{\mathrm{a}}$ for any $\mathbf{a} \in \prod_{i=1}^{\infty} k$.

Define a map $(Q \cap H) \backslash x \rightarrow Q_{n-2}$ where $Q_{n-2}$ is a quadric in $\mathbb{P}^{n-1}$ so that each fiber of the map is isomorphic to $\mathbb{A}^1$.

Let $Q$ be a quadric in $\mathbb{P}^{n-1}$ and $H$ a hyperplane in $\mathbb{P}^{n-1}$ such that $Q\cap H$ is a smooth quadric of dimension $n-2$. We want to define a map $\phi:(Q\cap H)\setminus {x}\rightarrow Q_{n-2}$ such that each fiber of $\phi$ is isomorphic to $\mathbb{A}^1$.

Let $x$ be a point in $Q\cap H$. Then we can identify $T_x(Q\cap H)$ with the orthogonal complement of $TxH$ in $Tx\mathbb{P}^{n-1}$. Since $Q$ is smooth and contains $x$, we have $T_xQ=T_xH\oplus T_x(Q\cap H)$.

Now consider the projection map $\pi:T_xQ\rightarrow T_x(Q\cap H)$. Since $Q$ is a quadric, $\pi$ is an isomorphism. Let $S_x$ denote the set of lines in $T_xQ$ passing through $x$. We can identify $S_x$ with the set of points in $Q$ that are incident with $x$.

Let $v\in T_x(Q\cap H)$ be a nonzero vector. We define a map $\psi_v:S_x\rightarrow\mathbb{A}^1$ as follows. For each $p\in S_x$, let $w_p$ be the unique vector in $T_pQ$ such that $w_p\cdot v=0$ and $|w_p|=1$. We can define $\psi_v(p)$ to be the unique $t\in\mathbb{A}^1$ such that $x+tv+w_p\in Q$.

Note that $\psi_v$ is well-defined because $Q$ is a quadric, so $w_p$ is uniquely determined. Moreover, $\psi_v$ is a bijection, since $Q$ is smooth and contains no lines, so the map $Q\rightarrow \mathbb{P}(T_xQ)$ given by $p\mapsto [x,p]$ is a local isomorphism at $x$.

Finally, we define $\phi:(Q\cap H)\setminus {x}\rightarrow Q_{n-2}$ by $\phi(p)=\psi_v(p)$, where $v$ is any nonzero vector in $T_x(Q\cap H)$ that is orthogonal to $TpH$. This map is well-defined and has the desired property that each fiber is isomorphic to $\mathbb{A}^1$.Regenerate response

Assume that $\operatorname{dim}(Z) = \operatorname{dim}(X) = n$. We want to show that $Z$ is a component of $X$.

Recall that a component of $X$ is a maximal irreducible closed subset of $X$. Thus, we need to show that $Z$ is irreducible and maximal among irreducible closed subsets of $X$.

To show that $Z$ is irreducible, we need to show that if $Z = Z_1 \cup Z_2$ for some closed subsets $Z_1$ and $Z_2$ of $X$, then $Z_1 = Z$ or $Z_2 = Z$. Since $Z$ is irreducible, it is enough to show that $Z_1$ or $Z_2$ has dimension $n$. Without loss of generality, assume that $\operatorname{dim}(Z_1) \geq n$. Then there exists a chain of irreducible closed subsets of $Z_1$ with lengths $n$, say $Z_1^0 \subsetneq Z_1^1 \subsetneq \cdots \subsetneq Z_1^n$. By the Noether normalization lemma, there exists a finite surjective morphism $\phi:X\rightarrow \mathbb{A}^n$ such that $\phi(Z_1^n)$ is closed in $\mathbb{A}^n$. Let $Y = \phi^{-1}(\phi(Z_1^n))$. Since $\phi$ is surjective, $Y$ is closed in $X$, and since $\phi$ is a finite morphism, $Y$ is irreducible. Moreover, $\operatorname{dim}(Y) = \operatorname{dim}(Z_1^n) = n$, so $Y$ contains $Z$. But $Y \subseteq Z_1$, so $Z_1 = X$. Therefore, $Z$ is irreducible.

To show that $Z$ is maximal among irreducible closed subsets of $X$, let $Z’$ be an irreducible closed subset of $X$ such that $Z \subseteq Z’$. We need to show that $Z = Z’$. Since $\operatorname{dim}(Z’) = n$, there exists a chain of irreducible closed subsets of $Z’$ with lengths $n$, say $Z’^0 \subsetneq Z’^1 \subsetneq \cdots \subsetneq Z’^n$. By the Noether normalization lemma, there exists a finite surjective morphism $\phi:X\rightarrow \mathbb{A}^n$ such that $\phi(Z’^n)$ is closed in $\mathbb{A}^n$. Let $Y = \phi^{-1}(\phi(Z’^n))$. Since $\phi$ is surjective, $Y$ is closed in $X$, and since $\phi$ is a finite morphism, $Y$ is irreducible. Moreover, $\operatorname{dim}(Y) = \operatorname{dim}(Z’^n) = n$, so $Y$ contains $Z’$. But $Z’ \subseteq Z$, so $Y$ contains $Z$. Therefore, $Z = Y \subseteq Z’$, and hence $Z = Z’$.

Therefore, $Z$ is a component of $X$.