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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 6394
View attachment 6395
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 $$a \in \mathfrak{m}$$. "Can someone please explain exactly why $$a \in \text{ann}_r( R / \mathfrak{m} )$$ implies that $$a + \mathfrak{m} = ( 1 + \mathfrak{m} ) a = 0 $$ ... ... ?
Can someone also explain how this then implies that $$a \in \mathfrak{m}$$ ... ?Hope someone can help ...
Peter
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 6394
View attachment 6395
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 $$a \in \mathfrak{m}$$. "Can someone please explain exactly why $$a \in \text{ann}_r( R / \mathfrak{m} )$$ implies that $$a + \mathfrak{m} = ( 1 + \mathfrak{m} ) a = 0 $$ ... ... ?
Can someone also explain how this then implies that $$a \in \mathfrak{m}$$ ... ?Hope someone can help ...
Peter