MHB Exploring Proposition 6.1.7 and its Proof in Bland's "Rings and Their Modules"

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 6.1 The Jacobson Radical ... ...

I need help with the proof of Proposition 6.1.7 ... Proposition 6.1.7 and its proof read as follows:View attachment 6396
View attachment 6397In the above text from Bland, in the proof of (1), we read the following: " ... ... we see that $$\text{ann}_r( R / \mathfrak{m} ) = \text{ann}_r(S)$$. But $$a \in \text{ann}_r( R / \mathfrak{m} )$$ implies that $$a + \mathfrak{m} = ( 1 + \mathfrak{m} ) a = 0$$ , so $$a \in \mathfrak{m}$$.

So, it follows that $$\bigcap_\mathscr{S} \text{ann}_r(S) \subseteq J(R)$$ ... ... "
Could someone please explain why it follows that $$\bigcap_\mathscr{S} \text{ann}_r(S) \subseteq J(R)$$ ... ... ? Hope someone can help ...

Peter

 

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 6.1 The Jacobson Radical ... ...

I need help with the proof of Proposition 6.1.7 ... Proposition 6.1.7 and its proof read as follows:
In the above text from Bland, in the proof of (1), we read the following: " ... ... we see that $$\text{ann}_r( R / \mathfrak{m} ) = \text{ann}_r(S)$$. But $$a \in \text{ann}_r( R / \mathfrak{m} )$$ implies that $$a + \mathfrak{m} = ( 1 + \mathfrak{m} ) a = 0$$ , so $$a \in \mathfrak{m}$$.

So, it follows that $$\bigcap_\mathscr{S} \text{ann}_r(S) \subseteq J(R)$$ ... ... "
Could someone please explain why it follows that $$\bigcap_\mathscr{S} \text{ann}_r(S) \subseteq J(R)$$ ... ... ? Hope someone can help ...

Peter

Just some thoughts ...

Since $$\text{ann}_r( R / \mathfrak{m} ) = \text{ann}_r(S)$$

we have $$a \in \text{ann}_r( R / \mathfrak{m} )$$ means $$a \in \text{ann}_r(S)$$ ...

thus $$a \in \bigcap_\mathscr{S} \text{ann}_r(S)$$ ... ...

But ... we also have that $$a \in \text{ann}_r( R / \mathfrak{m} )$$ implies that $$a \in \mathfrak{m}$$ ... ...

But this means that $$a \in J(R)$$ ...

Thus $$a \in \bigcap_\mathscr{S} \text{ann}_r(S) \Longrightarrow a \in J(R)$$ ... ...

So $$\bigcap_\mathscr{S} \text{ann}_r(S) \subseteq J(R)$$ ...Is that correct?


Peter