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Math 463/563 - Homework 7
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\textbf{Artin \S 14, 6.1} Let $V \subseteq \C^n$ be the locus of common zeros of an infinite set of polynomials $f_1, f_2, \ldots$ and consider the ideal $I = (f_1, f_2, \ldots) \subseteq \C[x_1, \ldots, x_n]$. Since $\C[x_1, \ldots, x_n]$ is Noetherian, $I$ is finitely generated, and in particular we can pick the generators from the set of $f_i$ (consider the containments $(f_1) \subset (f_1, f_2) \subset \cdots$ and apply the ACC). After relabeling, say $I = (f_1, \ldots, f_m)$. Then every element of $V$ is a common zero of the elements of $I$ by assumption, and since each each $f_i$ can be written as a linear combination of the $f_1, \ldots, f_m$, if $\alpha \in \C^n$ is a common zero of $f_1, \ldots, f_m$, it is also a zero of all of the $f_i$. So $V = \operatorname{Zeros}(f_1, \ldots, f_m)$. \bigskip
\textbf{Additional Problems.} Let $M$ be a module over an integral domain $R$.\bigskip
\textbf{1.} Let $a \in \Tor(M)$. Then there exists a nonzero $r \in R$ such that $ra = 0$, hence $\{a\}$ is not linearly independent for any $a \in R$, so $\Tor(M)$ has rank 0.\bigskip
\textbf{2.} It is clear that $\rk (M/\Tor(M)) \le \rk(M)$, since the representatives of any set of independent cosets of $M/\Tor(M)$ must be linearly independent in $M$. So it remains to show that $\rk(M/\Tor(M)) \ge \rk(M)$.\bigskip
Let $\{x_1, \ldots, x_n\}$ be a set of linearly independent elements of $M$. It suffices to show that their images in $M/\Tor(M)$ are linearly independent, so suppose $r_1 \overline{x_1} + \cdots + r_n \overline{x_n} = 0$ for some $r_i \in R$. This is equivalent to saying that $r_1 x_1 + \cdots + r_n x_n \in \Tor(M)$. Hence there exists $r \in R$ nonzero such that $r r_1 x_1 + \cdots + r r_n x_n = 0$. But since the $x_i$ are linearly independent and $r \ne 0$, we must have $r_i = 0$ for all $i$. Hence the $\overline{x_i}$ are linearly independent.\bigskip
\textbf{3.} Suppose that $M$ has rank $n$ and that $\{x_1, \ldots, x_n\}$ is a a maximal set of linearly independent elements of $M$. Let $N = Rx_1 + \cdots + Rx_n$. Then $N$ is generated by the $x_i$, which are linearly independent by assumption since $N\subseteq M$, so $N$ is free of rank $n$. Thus $N \simeq R^n$. \bigskip
Let $\overline{a} \in M/N$. Then $r\overline{a} = 0$ in $M/N$ if and only if $ra \in N$. So we need to show that for every element of $M$ there exists a nonzero $r \in R$ such that $ra \in N$. Suppose for the sake of contradiction that there exists $a \in M$ such that $ra \notin N$ for any nonzero $r \in R$. Then the set $\{x_1, \ldots, x_n, a\}$ is linearly independent, since every element of $N$ is a linear combination of the $x_i$, which contradicts the maximality of $\{x_1, \ldots, x_n\}$.\bigskip
\textbf{4.} Suppose $N \subseteq M$ with $N \simeq R^n$ and that $M/N$ is torsion as an $R$-module. Suppose for the sake of contradiction that $\{x_1, \ldots, x_n, x_{n+1}\}$ is a set of linearly independent elements of $M$. Consider their images $\{\overline{x_1}, \ldots, \overline{x_{n+1}}\}$ in $M/N$. Since $M/N$ is torsion, there exists $r \in R$ nonzero such that $r x_1 + \cdots + r x_{n+1} \in N$, but this contradicts the fact that $N$ is free of rank $n$, so $M$ has rank at most $n$. Finally, since $N \subset M$ contains $n$ linearly independent elements, it must have rank exactly $n$.
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