Galois Connection between ann( - ) and M( - )

Let $M$$M$ be a module over a commutative ring $R$$R$, let $(\mathrm{sub}(M),\subseteq )$$(\mathrm{sub}(M),\subseteq)$ be the lattice of submodule of $M$$M$, $(I(R),\supseteq )$$(I(R),\supseteq)$ be the lattice of sub module(i.e. Ideals) of $R$$R$.

The following definition comes from this two module homomorphism:

$$R\to M,r⟼rm,M\to IM,m⟼im$$

Define

$$\mathrm{ann}(M):=\{r\in R:rM=0\}=\bigcap _{m\in M}\ker (r⟼rm)$$

If $N_1\subseteq N_2$$N_1\subseteq N_2$, then $\mathrm{ann}(N_1)\supseteq \mathrm{ann}(N_2)$$\mathrm{ann}(N_1)\supseteq\mathrm{ann}(N_2)$.

Define

$$M(I):=\{m\in M:Im=0\}=\bigcap _{i\in I}\ker (m⟼im)$$

If $I\supseteq J$$I\supseteq J$, then $M(I)\subseteq M(J)$$M(I)\subseteq M(J)$.

Then we cliam that $\mathrm{ann}(-)$$\mathrm{ann}(-)$ is the left adjoint of $M(-)$$M(-)$.

$$\mathrm{ann}(N)\supseteq J⟺N\subseteq M(J)$$

Proof.

If $N\subseteq M(J)$$N\subseteq M(J)$, then $JN=0$$JN=0$, hence $\mathrm{ann}(N)\supseteq J$$\mathrm{ann}(N)\supseteq J$. If $\mathrm{ann}(N)\supseteq J$$\mathrm{ann}(N)\supseteq J$, then $M(\mathrm{ann}(N))\subseteq M(J)$$M(\mathrm{ann}(N))\subseteq M(J)$ and $N\subseteq M(\mathrm{ann}(N))$$N\subseteq M(\mathrm{ann}(N))$. $◻$$\square$