HW8

1. Prove: a UFD is integral closed.

\begin{proof} Let $A$ be a UFD, and let $K = Frac (A)$ be its fraction field. To show $A$ is UFD, it suffices to show for any $r \in K$ which is integral over $A$ , $r \in A$ . Assume that $r = r_{1} / r_{2}$ with $r_{1}, r_{2} \in A$ and $g cd (r_{1}, r_{2}) = 1$ . If $r$ is integral over $A$ , there exist $a_{0}, \dots, a_{n - 1} \in A$ such that

\frac{r _{1}^{n}}{r _{2}^{n}} + a_{n - 1} \frac{r _{1}^{n - 1}}{r _{2}^{n - 1}} + \dots + a_{0} = 0 ⟹ - a_{0} r_{2}^{n} = r_{1} (r_{1}^{n - 1} + a_{n - 1} r_{2} r_{1}^{n - 2} + \dots + a_{1} r_{2}^{n - 1}) \in A .

Since $A$ is a UFD, we know $r_{2} ∣ r_{1} (r_{1}^{n - 1} + a_{n - 1} r_{2} r_{1}^{n - 2} + \dots + a_{1} r_{2}^{n - 1})$ and so $r_{2} ∣ r_{1}^{n}$ . It deduces that $r_{2} = 1$ by $g cd (r_{1}, r_{2}) = 1$ . Then $r = r_{1} \in R$ and we have done. \end{proof}

2. Give an example of an $R$ -module $M$ such that the subset

tor (M) := {m \in M ∣ Ann_{R} (m) \neq = 0}

is not a submodule.

\begin{proof} Let $R = Z_{6}$ and $M = Z_{6}$ be an $R$ -module. Then $2, 3 \in tor (M)$ as $3 \cdot 2 = 0, 2 \cdot 3 = 0$ . If $tor (M)$ is a submodule, then $2 + 3 = 5 \in tor (M)$ , which is a contradiction. Hence $tor (M)$ is not a submodule. \end{proof}

3. Prove: Assume $R$ is an integral domain. For any $R$ -module $M$ we have:

(a) $tor (tor (M)) = tor (M)$ .
(b) $tor (M / tor (M)) = {0}$ .

\begin{proof} (a) Since $R$ is an integral domain, $tor (M)$ is an $R$ -submodule of $M$ . We have $tor (tor (M)) \subseteq tor (M)$ , as $tor (tor (M))$ is an $R$ -submodule of $tor (M)$ . On the other hand, for any $m \in tor (M)$ , there exists non-zero $r \in R$ such that $r \cdot m = 0$ . So $m \in tor (tor (M))$ and so $tor (tor (M)) = tor (M)$ . Now we finish the proof.

(b) If there exists $m + tor (M) \in M / tor (M)$ such that $m + tor (M) \in tor (M / tor (M))$ , then there exists nonzero $r \in R$ such that $r (m + tor M) \in tor (M)$ , that is, $r m \in tor (M)$ . So there exists nonzero $r_{1} \in R$ such that $r_{1} r m = 0$ . Since $R$ is an integral domain, $r_{1} r \neq = 0$ and so $m \in tor (M)$ . Hence, $m + tor (M) = 0 \in M / tor (M)$ for all $m + tor (M) \in tor (M / tor (M))$ . Therefore, $tor (M / tor (M)) = {0}$ . \end{proof}

4. Show: Let $M$ be a finitely generated torsion free $R$ -module, let $V =$ $FM$ (so $M$ is identified with an $R$ -submodule of $FM$ ).

(a) If $λ M \subset ⨁_{i = 1}^{r} R e_{i}$ for some basis $(e_{i})_{1 \leq i \leq r}$ of $V$ and some $λ \in R$ , $λ \neq = 0$ , then whenever $(f_{i})_{1 \leq i \leq r}$ is a basis of $V$ , there exists $μ \in R$ , $μ \neq = 0$ , such that

μ M \subset i = 1 ⨁ r R f_{i}

(b) If $M_{1}$ and $M_{2}$ are fractional modules in $V$ , then $M_{1} + M_{2}$ and $M_{1} \cap M_{2}$ are also fractional modules.

\begin{proof} (a) Since both $(e_{i})$ and $(f_{i})$ are bases of $V$ , there exists an invertible matrix $A = (a_{ij}) \in GL_{r} (F)$ such that $f_{i} = \sum_{j = 1}^{r} a_{ij} e_{j}$ for each $i \in {1, \dots, r}$ . Let $A^{- 1} = (b_{ij})$ be the inverse matrix, where $b_{ij} \in F$ can be written as $b_{ij} = c_{ij} / d_{ij}$ for some $c_{ij}, d_{ij} \in R$ with $d_{ij} \neq = 0$ . Define $d = \prod_{i, j = 1}^{r} d_{ij} \neq = 0$ . Then $d \cdot b_{ij} \in R$ for all $i, j$ and so

d e_{j} = k = 1 \sum r (d b_{jk}) f_{k} \in i = 1 ⨁ r R f_{i} .

For any $m \in M$ , we have $λm \in \oplus_{i = 1}^{r} R e_{i}$ and so $λm = \sum_{i = 1}^{j} r_{j} e_{j}$ with $r_{j} \in R$ . It deduces that

d λm = j = 1 \sum r r_{j} (d e_{j}) = j = 1 \sum r r_{j} (k = 1 \sum r (d b_{jk}) f_{k}) = k = 1 \sum r (j = 1 \sum r r_{j} (d b_{jk})) f_{k}

Since $d b_{jk} \in R$ and $r_{j} \in R$ , the coefficients $\sum_{j} r_{j} (d b_{jk})$ belong to $R$ . Therefore $d λm \in ⨁_{i = 1}^{r} R f_{i}$ and $d λ$ is what we desired.

b) Assume that $M_{1}$ and $M_{2}$ are fraction modules in $V_{1}, V_{2}$ , respectively. Then there exist bases $(e_{i})_{i = 1}^{r}$ and $(f_{j})_{j = 1}^{s}$ such that $M_{i}$ generates $V_{i}$ , $λ_{1} M_{1} \subseteq \oplus_{i = 1}^{r} R e_{i}$ and $λ_{2} M_{s} \subseteq \oplus_{j = 1}^{s} R f_{j}$ . Then $M_{1} + M_{2}$ generate $V_{1} \oplus V_{2}$ and $λ_{1} λ_{2} (M_{1} + M_{2}) \subseteq (\oplus_{i = 1}^{r} R e_{i}) \oplus (\oplus_{j = 1}^{s} R f_{j})$ , where $(e_{i})_{i = 1}^{r} \cup (f_{j})_{j = 1}^{s}$ is a basis of $V_{1} \oplus V_{2}$ . Hence $M_{1} + M_{2}$ is a fractional $R$ -module in $V_{1} \oplus V_{2}$ .

Similarly, $M_{1} \cap M_{2}$ generates $V_{1} \cap V_{2}$ and assume $(v_{i})_{i = 1}^{t}$ is a set of basis of $V_{1} \cap V_{2}$ . Then

λ_{1}^{'} λ_{2}^{'} (M_{1} \cap M_{2}) \subseteq (\oplus_{i = 1}^{r} R e_{i}^{'}) \cap (\oplus_{j = 1}^{s} R f_{j}^{'}) = \oplus_{i = 1}^{t} R v_{i}

where $(e_{i}^{'})$ and $(f_{j}^{'})$ are bases of $V_{1}$ and $V_{2}$ , respectively, which are extensions of the basis $(v_{i})$ of $V_{1} \cap V_{2}$ . Furthermore, $λ_{1}^{'}, λ_{2}^{'} \in R$ satisfies $λ_{1}^{'} M_{1} \subseteq \oplus_{i = 1}^{r} R e_{i}^{'}$ and $λ_{2}^{'} M_{s} \subseteq \oplus_{j = 1}^{s} R f_{j}^{'}$ . Thus $M_{1} \cap M_{2}$ is a fractional $R$ -module in $V_{1} \cap V_{2}$ . \end{proof}

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HW8

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