- #1

Math Amateur

Gold Member

MHB

- 3,998

- 48

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}##. "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:

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