83 lines
3.7 KiB
Markdown
83 lines
3.7 KiB
Markdown
# Dummit & Foote Chapter 12 - Modules over Principal Ideal Domains
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## Section 12.1 The Basic Theory
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**Definition**. The left $R$-module $M$ is said to be a *Noetherian* $R$-module if there are no infinitely increasing chains of submodules. That is, given
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$$
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M_1 \subseteq M_2 \subseteq \ldots
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$$
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there exists some $k \in \mathbb{N}$ such that given any $n \in \mathbb{N}$ with $n \geq k$, then $M_n = M_k$.
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**Definition**. A ring $R$ is *Noetherian* if it is Noetherian when viewed as a left $R$-module over itself.
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**Theorem**. Let $R$ be a ring and $M$ a left $R$-module. Then, the following are equivalent:
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1. $M$ is Noetherian
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2. Every nonempty set of submodules of $M$ contains a maximal element under inclusion
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3. Every submodule of $M$ is finitely-generated
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**Corollary**. If $R$ is a principal ideal domain (PID), then all nonempty set of ideals of $R$ has a maximal element. Additionally, $R$ is as Noetherian ring.
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**Proposition**. Let $R$ be an integral domain, and $M$ be a free $R$-module of rank $n < \infty$. Then, given $S$ is subset $M$ with $|S| > n$, the elements of $S$ are $R$-linearly dependent.
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**Definition**. Given $R$ an integral domain and $M$ an $R$-module,
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$$
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\text{Tor}(M) = \{ x \in M | rx = 0 \text{ for any } r \neq 0 \}
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$$
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This is the *torsion submodule* of $M$. If $\text{Tor}(M)$ is empty, then $M$ is *torsion-free*.
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**Definition** Let $R$ be an integral domain and $M$ be an $R$-module. Then, given a submodule $N$,
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$$
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\text{Ann}_R(N) = \{r \in R | rn = 0 \text{ for all } n \in N \}
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$$
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This ideal of $R$ is the *annihilator of $N$*. That is, $\text{Ann}(N)$ is the set of elements of $R$ such that $(r)N = \{ 0 \}$.
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Note that if $N$ is not a torsion submodule of $M$, then $\text{Ann}(N) = (0)R$. Additionally, given $N, L$ are submodules of $M$ with $N \subseteq L$, then $\text{Ann}(N) \subseteq \text{Ann}(L)$.
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Additionally, if $R$ is a PID, as $\text{Ann}_R(N)$ is an ideal, $\text{Ann}(N) = (n)R$ and $\text{Ann}(L) = (l)R$ for some $n, l \in R$ such that $n | l$.
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**Definition**. Given any integral domain $R$, the *rank* of an $R$-module $M$ is the maximum number of $R$-linearly independent elements of M.
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**Corollary**. The rank of a free module is the number of generating elements.
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**Theorem**. Let $R$ be a principal ideal domain, and $M$ be a free $R$-module of finite rank $m$, and $N$ be a submodule of $M$. Then,
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1. $N$ is a free submodule with rank $n \leq m$.
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2. There exists a basis $y_1, y_2, \ldots, y_m$ of $M$ so that $r_1 y_1, r_2 y_2, \ldots, r_m y_n$ is a basis of $N$ for some $r_i \in R$ and $r_1 | r_2 | \ldots | r_n$
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**Theorem**. Fundamental Theorem, Existence: Invariant Form. Let $R$ be a PID and $M$ be a finitely generated $R$-module. THen,
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- $M$ is isomorphic for some $r \in \mathbb{N}\cup{0}$, $a_1, \ldots, \a_m \neq 0 \in R$ such that $a_1 | a_2 | \ldots | a_m$, with
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$$
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M \cong R^{\oplus r} \oplus \frac{R}{(a_1)R} \oplus \frac{R}{(a_2)R} \oplus \ldots \oplus \frac{R}{(a_m)R}
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$$
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- $M$ is torsion-free if and only if $M$ is free
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- Note that
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$$
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\text{Tor}{M} \cong \frac{R}{(a_1)R} \oplus \frac{R}{(a_2)R} \oplus \ldots \oplus \frac{R}{(a_m)R}
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$$
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As a consequence, $M$ is a torsion module if and only if $r = 0$.
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**Definition**. In the above, $r$ is the *free rank* of $M$, and $a_1, \ldots, a_m$ are the *invariant factors* of $M$.
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**Theorem**. Fundamental Theorem, Existence: Elementary Divisor Form. The sum above can be written as
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$$
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M \cong R^{\oplus r} \oplus \frac{R}{(p_1^{\alpha_1})R} \oplus \frac{R}{(p_2^{\alpha_2})R} \oplus \ldots \oplus \frac{R}{(p_t^{\alpha_t})R}
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$$
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with $p_t$ non-unique primes and $\alpha_t$ non-unique, but with $(p_t^{\alpha_t})$ unique. These are called the *elementary divisors* of $M$.
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TODO: Incomplete for Now
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