An example of Galois Connection in module theory

Let $M$$M$ be a nonzero, finite generated torsion $R-$$R-$module, where $R$$R$ is a $\mathrm{PID}$$\mathrm{PID}$.

Definition. Let $S$$S$ be a subset of $M$$M$, the annihilator of $S$$S$, denote as $\mathrm{ann}(S)$$\mathrm{ann}(S)$, defined as:

$$\begin{array}{}\text{(1)}& \mathrm{ann}(S):=\{a\in R|ax=0\mathrm{\forall }x\in S\}=\bigcap _{x\in S}\mathrm{ann}(x)\end{array}$$

Proposition. If $S_1\subseteq S_2$$S_1\subseteq S_2$, then $\mathrm{ann}(S_1)\supseteq \mathrm{ann}(S_2)$$\mathrm{ann}(S_1)\supseteq\mathrm{ann}(S_2)$.

Proof. Obviously.

Definition. Let $(a)$$(a)$ be an ideal in $R$$R$, the module annihilated by $a$$a$, denote as $M(a)$$M(a)$, defined as:

$$\begin{array}{}\text{(2)}& M(a):=\{m\in M|am=0\}\end{array}$$

Proposition. If $a|b$$a|b$, i.e. $(a)\supseteq (b)$$(a)\supseteq (b)$, then $M(a)\subseteq M(b)$$M(a)\subseteq M(b)$.

Proof. Obviously.

Recall the definition of Galois Connection.

Proposition. $M(-):(I,\supseteq )\to (\mathrm{sub}(M),\subseteq )$$M(-):(I,\supseteq)\to(\mathrm{sub}(M),\subseteq)$ and $\mathrm{ann}(-):(\mathrm{sub}(M),\subseteq )\to (I,\supseteq )$$\mathrm{ann}(-):(\mathrm{sub}(M),\subseteq)\to(I,\supseteq)$​ form a pair of Galois Connection.

i.e.

$$\begin{array}{}\text{(3)}& \mathrm{ann}(F)\supseteq (a)⟺F\subseteq M(a)\end{array}$$

Proof.

$⟹$$\Longrightarrow$ Observe that $M(\mathrm{ann}(F))=F$$M(\mathrm{ann}(F))=F$.

$⟸$$\Longleftarrow$ $F\subseteq M(a)$$F\subseteq M(a)$ implies that $F$$F$ is annihilated by $(a)$$(a)$ as well. $\mathrm{\forall }x\in F,ax=0⟹\mathrm{\forall }x,a\in \mathrm{ann}(x)$$\forall x\in F,ax=0\implies \forall x,a\in\mathrm{ann}(x)$.

Hnece ${\displaystyle a\in \bigcap _{x\in F}\mathrm{ann}(x)=\mathrm{ann}(F)}$$\displaystyle a\in\bigcap_{x\in F}\mathrm{ann}(x)=\mathrm{ann}(F)$, $\mathrm{ann}(F)\supseteq (a).$$\mathrm{ann}(F)\supseteq(a).$

Corollary. Left adjoint preserve colimit and right adjoint preserve limit, hence

$$\begin{array}{}\text{(5)}& \mathrm{ann}(F+F^{\prime })=\mathrm{ann}(F)\cap \mathrm{ann}(F^{\prime }),M((a)+(b))=M(d)=M(a)\cap M(b)\end{array}$$

Here $d=gcd(a,b)$$d=\gcd(a,b)$​.

Proposition. If $gcd(a,b)=1$$\gcd(a,b)=1$, then $M(ab)=M(a)\oplus M(b).$$M(ab)=M(a)\oplus M(b).$

Proof. Observe that $M(a),M(b)\subseteq M(ab)$$M(a),M(b)\subseteq M(ab)$, hence $M(a)+M(b)\subseteq M(ab)$$M(a)+M(b)\subseteq M(ab)$.

To see the otherside, oberve that $m\in M(ab)⟹ab(m)=0⟹bm\in M(a)\vee am\in M(b)$$m\in M(ab)\implies ab(m)=0\implies bm\in M(a)\or am\in M(b)$.

Since $gcd(a,b)=1,m=(sa+tb)m=sam+tbm\in M(a)+M(b)$$\gcd(a,b)=1,m=(sa+tb)m=sa m+tb m\in M(a)+M(b)$. By $(4)$$(4)$ we see that $M(a)\cap M(b)=M(1)=0$$M(a)\cap M(b)=M(1)=0$.

Hence we prove that $M(ab)=M(a)\oplus M(b)◻.$$M(ab)=M(a)\oplus M(b)\quad\square.$

Corollary. Let $M$$M$ be a torsion module over a $\mathrm{P}\mathrm{I}\mathrm{D}$$\rm PID$, and $\mathrm{ann}(M)=(a)=(p_1^{e_1}...p_n^{e_n})$$\mathrm{ann}(M)=(a)=(p_1^{e_1}...p_n^{e_n})$

Then

$$\begin{array}{}\text{(6)}& M=M(p_1^{e_1})\oplus ...\oplus M(p_n^{e_n})\end{array}$$

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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