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这是悉尼大学交换代数课程的代写成功案例。
MATH483课程简介
This course will focus on the theory of rings and modules, with an eye to applications to algebraic number theory.
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MATH483 Commutative Algebra HELP（EXAM HELP， ONLINE TUTOR）
Let $I_1, I_2 \subset K[x]{>}$be as in Section 1.8.7. We consider the quotient of $I_1$ by powers of $I_2$ $$ I_1=I_1: I_2^0 \subset I_1: I_2^1 \subset I_1: I_2^2 \subset I_1: I_2^3 \subset \ldots \subset K[x]{>} .
$$
Since $K[x]{>}$is Noetherian, there exists an $s$ such that $I_1: I_2^s=I_1: I_2^{s+i}$ for all $i \geq 0$. Such an $s$ satisfies $$ I_1: I_2^{\infty}:=\bigcup{i \geq 0} I_1: I_2^i=I_1: I_2^s,
$$
and $I_1: I_2^s$ is called the saturation of $I_1$ with respect to $I_2$.
The minimal such $s$ is called the saturation exponent. If $I_1$ is radical, then the saturation exponent is 1 .
Problem: Given ideals $I_1, I_2 \subset K[x]_{>}$, we want to compute generators for $I_1: I_2^{\infty}$ and the saturation exponent.
Solution: Set $I^{(0)}=I_1$ and compute successively $I^{(j+1)}=I^{(j)}: I_2, j \geq 0$, by any of the methods of Section 1.8.8. In each step check whether $I^{(j+1)} \subset I^{(j)}$, by using Section 1.8.1. If $s$ is the first $j$ when this happens, then $I^{(s)}=I_1: I_2^{\infty}$ and $s$ is the saturation exponent.
Recall that a sequence of polynomials $f_1, \ldots, f_k \in K\left[x_1, \ldots, x_n\right]$ is called algebraically dependent if there exists a polynomial $g \in K\left[y_1, \ldots, y_k\right] \backslash{0}$ satisfying $g\left(f_1, \ldots, f_k\right)=0$. This is equivalent to $\operatorname{Ker}(\varphi) \neq 0$, where $\varphi: K\left[y_1, \ldots, y_k\right] \rightarrow K\left[x_1, \ldots, x_n\right]$ is defined by $\varphi\left(y_i\right)=f_i \cdot \operatorname{Ker}(\varphi)$ can be computed according to Section 1.8.10, and any $g \in \operatorname{Ker}(\varphi) \backslash{0}$ defines an algebraic relation between the $f_1, \ldots, f_k$. In particular, $f_1, \ldots, f_k$ are algebraically independent if and only if $\operatorname{Ker}(\varphi)=0$ and this problem was solved in Section 1.8.10.
Related, but slightly different is the subalgebra-membership problem.
Problem: Given $f \in K\left[x_1, \ldots, x_n\right]$, we may ask whether $f$ is an element of the subalgebra $K\left[f_1, \ldots, f_k\right] \subset K\left[x_1, \ldots, x_n\right]=K[x]$.
Solution 1: Define $\psi: K\left[y_0, \ldots, y_k\right] \rightarrow K[x], y_0 \mapsto f, y_i \mapsto f_i$, compute $\operatorname{Ker}(\psi)$ according to Section 1.8 .10 and check whether $\operatorname{Ker}(\psi)$ contains an element of the form $y_0-g\left(y_1, \ldots, y_k\right)$. That is, we define an elimination ordering for $x_1, \ldots, x_n$ on $\operatorname{Mon}\left(x_1, \ldots, x_n, y_0, \ldots, y_k\right)$ with $y_0$ greater than $y_1, \ldots, y_k$ (for example, $\left.(\mathrm{dp}(\mathrm{n}), \mathrm{dp}(1), \mathrm{dp}(\mathrm{k}))\right)$ and compute a standard basis $G$ of $\left\langle y_0-f, y_1-f_1, \ldots y_k-f_k\right\rangle$. Then $G$ contains an element with leading monomial $y_0$ if and only if $f \in K\left[f_1, \ldots, f_k\right]$.
Solution 1: Define $\psi: K\left[y_0, \ldots, y_k\right] \rightarrow K[x], y_0 \mapsto f, y_i \mapsto f_i$, compute $\operatorname{Ker}(\psi)$ according to Section 1.8 .10 and check whether $\operatorname{Ker}(\psi)$ contains an element of the form $y_0-g\left(y_1, \ldots, y_k\right)$. That is, we define an elimination ordering for $x_1, \ldots, x_n$ on $\operatorname{Mon}\left(x_1, \ldots, x_n, y_0, \ldots, y_k\right)$ with $y_0$ greater than $y_1, \ldots, y_k$ (for example, $\left.(\mathrm{dp}(\mathrm{n}), \mathrm{dp}(1), \mathrm{dp}(\mathrm{k}))\right)$ and compute a standard basis $G$ of $\left\langle y_0-f, y_1-f_1, \ldots y_k-f_k\right\rangle$. Then $G$ contains an element with leading monomial $y_0$ if and only if $f \in K\left[f_1, \ldots, f_k\right]$.
Solution 2: Compute a standard basis of $\left\langle y_1-f_1, \ldots, y_k-f_k\right\rangle$ for an elimination ordering for $x_1, \ldots, x_n$ on $\operatorname{Mon}\left(x_1, \ldots, x_n, y_1, \ldots, y_k\right)$ and check whether the normal form of $f$ with respect to this standard basis does not involve any $x_i$. This is the case if and only if $f \in K\left[f_1, \ldots, f_k\right]$ and the normal form expresses $f$ as a polynomial in $f_1, \ldots, f_k$.
Problem: Let $I_1$ and $I_2 \subset R^r$ be as in Section 2.8.3. Find a (polynomial) system of generators for the quotient
$$
I_1:_R I_2=\left{g \in R \mid g I_2 \subset I_1\right} .
$$
Note that $I_1:_R I_2=\operatorname{Ann}_R\left(\left(I_1+I_2\right) / I_1\right)$, in particular, if $I_1 \subset I_2$, then we have $I_1:_R I_2=\operatorname{Ann}_R\left(I_2 / I_1\right)$.
Solution 1: Compute generating sets $G_i$ of $I_1 \cap\left\langle h_i\right\rangle, i=1, \ldots, s$, according to Section 2.8.3, “divide” the generators by the vector $h_i$, getting the generating set $G_i^{\prime}=\left{g \in R \mid g h_i \in G_i\right}$ for $I_1:_R\left\langle h_i\right\rangle$. Finally, compute the intersection $\bigcap_i\left(I_1:_R\left\langle h_i\right\rangle\right)$, again according to 2.8 .3 .
Note that there is a trick which can be used to “divide” by the vector $h_i$ : let $G_i=\left{v_1, \ldots, v_{\ell}\right}$ and compute a generating system $\left{w_1, \ldots, w_m\right}$ for $\operatorname{syz}\left(v_1, \ldots, v_{\ell}, h_i\right)$. Then the last $(=(\ell+1)-$ th $)$ components of $w_1, \ldots, w_m$ generate the $R$-module $I_1:_R\left\langle h_i\right\rangle$.
Solution 2: Define $h:=h_1+t_1 h_2+\ldots+t_{s-1} h_s \in K\left[t_1, \ldots, t_{s-1}, x_1, \ldots, x_n\right]^r$, and compute a generating system for $\left(I_1 \cdot R[t]\right):{R[t]}\langle h\rangle{R[t]}$, as before. Finally, eliminate $t_1, \ldots, t_{s-1}$ from $\left(I_1 \cdot R[t]\right):{R[t]}\langle h\rangle{R[t]}$ (cf. Section 1.8.2).
Let $A=R / I$ for some ideal $I \subset R$ and let $M, N$ be two $A$-modules with $N \subset M$. Define the radical of $N$ in $M$ as the ideal
$$
\operatorname{rad}_M(N):=\sqrt[M]{N}:=\left{g \in A \mid g^q M \subset N \text { for some } q>0\right} .
$$
Problem: Solve the radical membership problem for modules, that is, decide whether $f \in A$ is contained in $\sqrt[M]{N}$, or not. ${ }^{16}$
Solution: By Exercise 2.8.6, $\sqrt[M]{N}=\sqrt{\operatorname{Ann}_A(M / N)}$. Hence, we can compute generators for $\operatorname{Ann}_A(M / N)=N:_A M \subset A$ as in Section 2.8.4, and then we are reduced to solving the radical membership problem for ideals which was solved in Section 1.8.6. ${ }^{17}$
MY-ASSIGNMENTEXPERT™可以为您提供wgtn MATH483 Commutative Algebra交换代数课程的代写代考和辅导服务！
